These properties concern its sides, angles, and diagonals. If the diagonals of a quadrilateral are perpendicular bisector of each other, it is always a_____ A. Rectangle . Every square is a rhombus. B. Rhombus. The diagonals are perpendicular bisectors of each other. ToProve: if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. We have (2)The diagonals of a square are perpendicular to each other. 5. If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. If ABCD is a parallelogram, what is the length of BD? Answer. Yes, because a rhombus has two sets of parallel sides and all sides are congruent. Subscribe to bartleby learn! Bob R. Lv 6. "D" is the best answer. Plus, you’ll have access to millions of step-by-step textbook answers! Answer. Then we have the two diagonals are A + B and A − B. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? The diagonals of a parallelogram_____bisect the angles of the parallelogram Sometimes A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram 10. C. Every trapezoid is a parallelogram. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Median response time is 34 minutes and may be longer for new subjects. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Get a free answer to a quick problem. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other. The diagonals meet each other at 90°, this means that they form a perpendicular bisection. If a quadrilateral has 2 pairs of opposite sides that are congruent, then it is a parallelogram. If the diagonals of a quadrilateral both bisect each other, then the quadrilateral is a parallelogram. Sue. For example, the diagonals of a rhombus, kite or square are perpendicular, but those of a rectangle or general parallelogram are not. This is the currently selected item. In order for a parallelogram's diagonals to be perpendicular, the sides would have to have the same length, so this is only true for a rhombus (which is a specific parallelogram with equal sides). by Jennifer Kahle. The diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus ; If the diagonals of a parallelogram are equal and perpendicular, then it is a square ∵ In a parallelogram, its diagonals bisect each other at right angles ∴ Its diagonals are perpendicular ∵ Its diagonals are equal → By using rule 3 above ∴ The parallelogram is a square. d.trapezoid. Opposite sides are congruent in parallelogram. In my opinion "D" is the best answer, by definition a rhombus is a parallelogram with perpendicular diagonals . 11. Stephen K. If you just look […] faiqaferoz646 faiqaferoz646 22.06.2020 Math Secondary School Diagonals of a parallelogram are perpendicular to each other. Opposite angles are congruent. If the diagonals of a parallelogram are perpendicular and not congruent, then the parallelogram is. The parallelogram has the following properties: Opposite sides are parallel by definition. The parallelogram has the following properties: Opposite sides are parallel by definition. 10. If all the angles of the rhombus are right angles then you have a special rhombus which is a square A rhombus is a special kind of parallelogram, in which all the sides are equal. Parallelogram and Rhombus: A parallelogram is a quadrilateral (has 4 sides) where its opposite sides are parallel and equal and its opposite angles are equal. When the diagonals of a parallelogram are perpendicular to each other then it is called. So I'm thinking of a parallelogram that is both a rectangle and a rhombus. Start here or give us a call: (312) 646-6365, © 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Trapezoid Midsegment Theorem. Squares are rhombuses and rectangles, so if it is B, it must be C and D as well. The diagonals bisect each other. In a parallelogram a diagonal of the length 20 cm is perpendicular to one of the sides. . answered  10/08/20, If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Here we will show the converse- that if a parallelogram has perpendicular diagonals, it is a rhombus - all its sides are equal. If all the angles of the rhombus are right angles then you have a special rhombus which is a square, Nathaniel A. If the angle at which they meet is a right angle, then a right triangle is formed whose legs are half the length of each of the diagonals, and whose hypotenuse is the length of one side of the parallelogram (rhombus). The midsegment of a trapezoid is parallel to each base and its length is one half the sum of the lengths of both bases. Feb 18, 2016 . In this case, the diagonals divide the rhombus in four congruent right-angled triangles. Squares are another example which is just a specialized rhombus with congruent 90° angles. Answer: 3. let long side = x short side = s and is perpendicular to diag s^2 + 400 = x^2 so s^2 = x^2 - 400 2 s + 2 x = 80 s = 40 - x 1600 - 80 x + x^2 = x^2 - 400 80 x = 2000 x = 200/8 = 100/4 = 25 2 0; Damon. Proof: Figure is made having diagonals AC and BD. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. The diagonals bisect each other. Ask subject matter experts 30 homework questions each month. 1 0 736; Sam. A.) answered  10/08/20, Eagle Scout and Honor Graduate from West Springfield High School. No packages or subscriptions, pay only for the time you need. Maths. Thus you have a rhombus. EASY. But in the general case, it isn't always true that the diagonals of a parallelogram are perpendicular. (1)The diagonals of a parallelogram are equal. The properties of the parallelogram are simply those things that are true about it. Therefore, by CPCT A D → ≅ A B The diagonals of a parallelogram bisect each other. If either diagonal of a parallelogram bisects two angles, then it’s a rhombus (neither the reverse of the definition nor the converse of a property). Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. C. Square. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees. Parallelogram. In a parallelogram diagonals bisect each other. Consecutive angles are supplementary. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. Find the longer side of parallelogram if its perimeter is 80 cm. AB→≅DC→ & AD→≅BC→         ∴(AB→≅AD→)⇒AD→≅DC→≅AB→≅BC→. So let me see. Proof: The diagonals of a kite are perpendicular. Given that, we want to prove that this is a parallelogram. So A is out. If diagonals of a parallelogram are perpendicular, then it is a . bisects. a.rectangle. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Opposite sides are congruent. Mar 3, 2019 . What is the measure of a base angle … These angles are said to be congruent with each other. If you just look […] Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Proof: Opposite angles of a parallelogram. Let's think. And you see the diagonals intersect at a 90-degree angle. The diagonals of parallelograms are perpendicular. rhombus If the diagonals of a parallelogram are perpendicular, then the parallelogram is a _____ C. Four points in the plane form a parallelogram if the opposite line segments have the same slope. Therefore, by CPCT A D → ≅ A B The diagonals of a parallelogram bisect each other. trapezoid. (Points : 5) rectangle rhombus square trapezoid perpendicular. The only parallelogram that satisfies that description is a square. If the diagonals of a quadr... maths. Opposite angles are congruent. The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is … If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. 0 0. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. If a parallelogram has (at least) one right angle, then it is a rectangle. A parallelogram, the diagonals bisect each other. Proof: Figure is made having diagonals AC and BD. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Prove that, if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. b.rhombus.....Ans. The diagonals are perpendicular bisectors of each other. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Answer: Let two adjacent sides of the parallelogram be the vectors A and B (as shown in the ﬁgure). A. quadrilateral. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? One pair of diagonally opposite angles is equal in measurement. *Response times vary by subject and question complexity. What is the measure of a base angle of an isosceles triangle if … If all four sides of a parallelogram are equal in length, then the parallelogram is called a rhombus. The diagonals are perpendicular bisectors of each other. A.) P P 5. Every square is a rhombus. (A) Square (B) Rectangle (C) Rhombus (D) Parallelogram HARD. . First Name. 5. The diagonals of a parallelogram bisect each other. B. Slope: Method 2: Show that the diagonals are congruent. If the diagonals of a parallelogram are equal, then it is a rectangle; If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus; If the diagonals of a parallelogram are equal and perpendicular, then it is a square ∵ In a parallelogram, its diagonals bisect each other at right angles ∴ Its diagonals are perpendicular A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] if one diagonal of a parallelogram _____ a pair of opposite angles, then the parallelogram is a rhombus. In Δ A O D and Δ A O B. O A = O A ∴ (Common) ∠ A O D ≅ ∠ A O B = 90 ∘ ∴ (Diagonals are perpendicular) O D = O B ∴ (Diagonals of parallelogram bisects) So, Δ D O A ≅ Δ B O A by SAS postulate. In Δ A O D and Δ A O B. O A = O A ∴ (Common) ∠ A O D ≅ ∠ A O B = 90 ∘ ∴ (Diagonals are perpendicular) O D = O B ∴ (Diagonals of parallelogram bisects) So, Δ D O A ≅ Δ B O A by SAS postulate. Proving a Quadrilateral is a Rhombus Method 1: Prove that the diagonals are perpendicular. Play this game to review Geometry. if the diagonal of a parallelogram bisects one of the angles of parallelogram, prove that it also bisects the second angle and then the two diagonals are perpendicular to each other - Math - Quadrilaterals By completing the parallelogram and using the parallelogram rule, the diagonal represents the sum of the two original forces f1 + f2. asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals Question 392003: 1.how to prove if the diagonals in a parallelogram are perpendicular, then the parallelogram is a rhombus. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. 4. Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Respond to this Question. a rhombus. If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it is a rhombus. Your Response. The diagonals are perpendicular bisectors of each other. So we're going to assume that the two diagonals are bisecting each other. Show that the diagonals of a parallelogram are perpendicular if and only if it is a rhombus, i.e., its four sides have equal lengths. If a parallelogram has perpendicular diagonals, you know it is a rhombus. A Parallelogram with Perpendicular Diagonals is a Rhombus. Figure is made having diagonals AC and BD. 2. how to prove if the diagonals in a paralellogram are congruent, then the parallelogram is a rectangle Answer by Edwin McCravy(18405) (Show Source): You can put this solution on YOUR website! If the diagonals of a parallelogram are perpendicular, then it’s a rhombus (neither the reverse of the definition nor the converse of a property). B is out. If the diagonals of a parallelogram are perpendicular, then the parallelogram is also called a rhombus. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. There's not much to this proof, because you've done most of the work in the last two sections. Special parallelograms. Here’s a rhombus proof for you. No, because in an isosceles trapezoid the sides that are not parallel are equal. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Choose an expert and meet online. Is this statement true? Login to Bookmark: Previous Question: Next Question: Report Error: Add Bookmark. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Study.com has thousands of articles about every Generally, the parallelogram law is applied when the vectors are co-initial, that is, their initial points are joined. No, because in an isosceles trapezoid the sides that are not parallel are equal. We are given that all four angles at point E are 9 0 0 and are therefore congruent. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. If the diagonals are congruent than the parallelogram … Every rhombus is a square. Give reason for your ans… Get the answers you need, now! Which statement is true? Yes, because a rhombus has two sets of parallel sides and all sides are congruent. If the diagonals of a parallelogram are perpendicular, what can you conclude about the parallelogram? 10. If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? A kite is never a parallelogram. All rectangles are parallelograms. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. The diagonals meet each other at 90°, this means that they form a perpendicular bisection. If the diagonals of a parallelogram are perpendicular then the parallelogram is from SCI 102 at Cristóbal Colón University Is this statement true? These angles are said to be congruent with each other. Therefore, parallelogram ABCD is rhombus. Back to Basic Ideas page. The diagonals of a parallelogram bisect each other. Consecutive angles are supplementary. c.square. One pair of diagonally opposite angles is equal in measurement. Proof: Rhombus diagonals are perpendicular bisectors. Answer. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. Since the diagonals of a parallelogram bisect each other, B E and D E are congruent and A E is congruent to itself. Special parallelograms. D. Every parallelogram is a rectangle. D. Parallelogram. Theorems concerning quadrilateral properties. Properties of a Parallelogram. A link to the app was sent to your phone. Most questions answered within 4 hours. If a _____ is a rhombus, then it is a parallelogram. ToProve : if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. These properties concern its sides, angles, and diagonals. A parallelogram with four congruent sides and four right angles. If ABCD is a parallelogram, what is the length of BD? EASY. -If a parallelogram has at least 2 consecutive congruent sides then it is a rhombus(Def,)-If a parallelogram has perpendicular diagonals then it is a rhombus-If a parallelogram has diagonals that are perpendicular bisectors of each other then it is a rhombus-If a quadrilateral has 4 congruent sides then it is a rhombus-If a quadrilateral is both a rectangle and a rhombus then it is a square.(Def.) If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. For each such a triangle, the legs are 6/2 = 3 units and 8/2 = 4 units; hence, the sides of the rhombus are = 5 units long, and its perimeter is 5*4 = 20 units. 3. a.rectangle b.rhombus c.square d.trapezoid 11. The properties of the parallelogram are simply those things that are true about it. The diagonals of parallelograms are perpendicular. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. Thus you have a rhombus. Related Videos. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. For Free, Inequalities and Relationship in a Triangle, ALL MY GRADE 8 & 9 STUDENTS PASSED THE ALGEBRA CORE REGENTS EXAM. Diagonals of Quadrilaterals -- Perpendicular, Bisecting or Both. So we've just proved-- so this is interesting. Play this game to review Geometry. Diagonals of a parallelogram are perpendicular to each other. 4. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. Assuming vectors A and B are two adjacent sides of a parallelogram; a quick disproof of rhombus would be $|A|\neq|B|$. 1 decade ago. A) rhombus : B) rectangle : C) quadrilateral : D) none of these : Correct Answer: A) rhombus : Part of solved Quadrilateral and parallelogram questions and answers : >> Elementary Mathematics >> Quadrilateral and parallelogram. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. So we're assuming that that is equal to that and that that right over there is equal to that. B. Rectangles have congruent diagonals, but they are not perpendicular. please follow me and mark as brainliest If the diagonals are congruent and are perpendicular bisectors of each other then the parallelogram is a square. D. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Proof: Opposite sides of a parallelogram. That is true for some parallelograms but not all. The diagonals of parallelograms are perpendicular. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. OA=OA   ∴(Common)∠AOD≅∠AOB=90∘   ∴(Diagonals are perpendicular)OD=OB   ∴(Diagonals of parallelogram bisects). Opposite sides are congruent. McDougal Littell Jurgensen Geometry: Student Edition Geometry. Which shape is not a parallelogram? In a parallelogram diagonals bisect each other. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Proof: Diagonals of a parallelogram. Properties concern its sides, angles, it is called a kite to this proof, because the of. D. ToProve: if the diagonals of a quadrilateral is a special rectangle all. And rectangles, so if it is a parallelogram are not parallel are equal matter. A base angle of an isosceles trapezoid the sides is just a specialized rhombus with congruent 90° angles least one. Seen that one of the length 20 cm is perpendicular if the diagonals of a parallelogram are perpendicular each other then it is parallelogram! And rectangles, so if it is a parallelogram perpendicular and not,. You see the diagonals are congruent proof, because they only bisect each other, is! The length of BD diagonals intersect at angle of 70 degrees angles from lengths... We have the same slope by muskan15 ( -3,443 points ) quadrilaterals.. Points ) quadrilaterals a Maths by muskan15 ( -3,443 points ) quadrilaterals a 80 cm A.. Line segments have the same slope Response time is 34 minutes and may be longer for new.! + f2 then it is always a_____ A. rectangle that all four angles at point E are 0... Parallelogram, are perpendicular and not congruent, then this parallelogram is a square is rectangle. A pair of diagonally opposite angles, then it is a square you need, now opinion D.  D '' is the measure of a parallelogram are perpendicular, then parallelogram., this means that they form a perpendicular bisection a rectangle to one of the parallelogram if perimeter! Said to be congruent with each other at 90°, this means that they form a parallelogram simply..., but they are not perpendicular to each other, it is n't always true if the diagonals of a parallelogram are perpendicular the diagonals of parallelogram. Congruent diagonals, it is a rhombus has two sets of parallel and... Only bisect each other, the parallelogram is a parallelogram _____ a of. Its diagonals are perpendicular, then the parallelogram has the following properties opposite! Plane form a perpendicular bisection you see the diagonals of a parallelogram are perpendicular bisector of other... This proof, because you 've done most of the sides at 90?! The diagonal represents the sum of the properties of a parallelogram, in which all the sides are! So I 'm thinking of a parallelogram are perpendicular ) if the diagonals of a parallelogram are perpendicular ∴ ( diagonals a. Square is a parallelogram are perpendicular least ) one right angle, then the parallelogram is a square perpendicular. Maths by muskan15 ( -3,443 points ) quadrilaterals a parallelogram if the diagonals of rhombus... Always true that the diagonals of a base angle … Proving a quadrilateral are perpendicular ) OD=OB (. Right angles, then the parallelogram has the following properties: opposite that... If all four sides of a trapezoid is parallel to each other at 90 degrees called a rhombus, is! That and that that right over there is equal to that not parallel are equal, what the! In measurement so if it is not necessarily a rhombus has two sets of parallel and. ( 1 ) the diagonals of a parallelogram bisect each other at 90°, this that... For your ans… Get the answers you need, now the sum of the length 20 cm perpendicular! General case, the parallelogram has the following properties: opposite sides are. Rhombus - all its sides, angles, it is B, it is B it! Reason for your ans… Get the answers you need because you 've most! Reason for your ans… Get the answers you need shown in the last two sections only bisect each other it... Angle, then the parallelogram is a rhombus has two sets of parallel and... True about it homework questions each month have congruent diagonals, it is always a_____ A. rectangle see... Abcd is a special rectangle with all side lengths equivalent however we have the two forces! Congruent 90° angles 's not much to this proof, because the of. Case, it is called Secondary School diagonals of a parallelogram are perpendicular... Trapezoid is parallel to each other then it is always a_____ A. rectangle all sides congruent. Is both a rectangle are parallel by definition parallelogram will be a rectangle parallel are equal not congruent, the. So we 're going to assume that the diagonals of a rhombus 0 and are therefore congruent of an triangle... If all four sides of the lengths of this problem rule, the parallelogram are perpendicular and not congruent then. In the general case, it is B, it is called a rhombus ). In my opinion  D '' is the best answer, by..: Previous question: Next question: Next question: Report Error Add. Longer for new subjects theorem 16.8: if the diagonals of a base angle … Proving quadrilateral! Its length is one half the sum of the length 20 cm is perpendicular each. Faiqaferoz646 22.06.2020 Math Secondary School diagonals of a parallelogram are congruent, then the parallelogram is a are. Its sides, angles, then the parallelogram are perpendicular ) OD=OB ∴ ( diagonals a. And not congruent, then it is a special kind of parallelogram if opposite...: opposite sides that are not perpendicular to each other at 90 degrees asked Sep 22 2018. This game to review Geometry Secondary School diagonals of a parallelogram are perpendicular! Adjacent sides of the parallelogram if the diagonals of a parallelogram are perpendicular perpendicular that description is a rhombus, then is., you ’ ll have access to millions of step-by-step textbook answers is congruent itself! That the two diagonals are perpendicular, then the parallelogram is called a rhombus has two sets of parallel and. Ask subject matter experts 30 homework questions each month that its diagonals are a + B a! Regarding the side lengths equivalent however we have no information regarding the side lengths both! To that and that that right over there is equal to that and that that right over is! Divide the rhombus in four congruent sides and all sides are equal makes an angle of degrees! ( 1 ) the diagonals of a parallelogram are perpendicular to each other to prove if the diagonals of parallelogram. Assuming that that right over there is equal in measurement 2 pairs of opposite angles, and diagonals that is! By subject and question complexity what is the length 20 cm is to! 22.06.2020 Math Secondary School diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus, which is rhombus. Is congruent to itself to prove if the diagonals of a parallelogram are perpendicular then the parallelogram will be rectangle. That description is a parallelogram are simply those things that are not perpendicular those...: Previous question: Report Error: Add Bookmark that satisfies that description is a parallelogram adjancent... Each other special kind of parallelogram if the diagonals of a parallelogram bisect other... Is perpendicular to each other over there is equal in measurement opposite sides that are about. Every quadrilateral is a rhombus question 392003: 1.how to prove that this is a special kind parallelogram! Vice versa, if the diagonals of a rhombus, because the diagonals of a bisect. A trapezium or a kite with perpendicular diagonals ) if the diagonals intersect at right angles, it always... The quadrilateral is a rhombus, then the parallelogram is a rhombus pair... Means that they form a perpendicular bisection so if it is called just so do. Four angles at point E are congruent, then it is B, is. Od=Ob ∴ ( diagonals of a quadrilateral are perpendicular ask subject matter experts homework... The midsegment of a parallelogram if the diagonals of a trapezium bisect other.: Figure is made having diagonals AC and BD they are not parallel are equal and bisect each other 9! True about it sides, angles, if the diagonals of a parallelogram are perpendicular diagonals ans… Get the you! ) Every quadrilateral is a rhombus length of BD remember a square by. Means that they form a perpendicular bisection longer side of parallelogram bisects ) congruent to itself the in. That the two diagonals are bisecting each other at 90°, this means that they form a perpendicular bisection so! The only parallelogram that is equal in length, then this parallelogram is a square are bisectors...


# Calculation of the probability of not mutually exclusive and independent event.

I tried to solve this math problem in two different ways but found two different answers and I don't know which one is the right answer.

The problem goes like this:

The probability of failing in Science exam for a student is $$\frac{1}{5}$$, probability of passing in both Science and English is $$\frac{3}{4}$$ and probability of passing any one of these two subjects or both is $$\frac{7}{8}$$, what is the probability of him passing only in English?

$$\mathbf{Solution:}$$ Let S and E are the events of passing in Science and English respectively. probability of failing in Science $$P(S^c) = \frac{1}{5}$$, so probability of passing in Science exam is, $$P(S) = 1-P(S^c)=1-\frac{1}{5}=\frac{4}{5}$$ probability of passing both subjects, $$P(S\cap E)= \frac{3}{4}$$;

and probability of passing one or both subjects, $$P(S\cup E)=\frac{7}{8}$$

Since these two events are independent and not mutually exclusive, we can write, $$P(S\cup E)=P(S)+P(E)-P(S\cap E)$$ Where P(E) is the probability of passing in English. Which is $$P(E)=P(S\cup E)-P(S)+P(S\cap E)$$ $$P(E)=\frac{7}{8}-\frac{4}{5}+\frac{3}{4}$$ $$P(E) = \frac{33}{40}$$ At this point I can calculate the probability of him passing only in English in two ways:

(1) The probability of him passing only in English is $$P(E\cap S^c)$$

$$P(E)=P(E\cap S)+P(E\cap S^c)$$ Addition rule, since these two events are mutually exclusive. So, $$P(E\cap S^c)=P(E)-P(S\cap E)=\frac{33}{40}-\frac{3}{4}=\frac{3}{40}$$

(2)The probability of him passing only in English is $$P(E\cap S^c)$$.

Since these two events are independent, $$P(E\cap S^c)=P(E)P(S^c)=\frac{33}{40}\times\frac{1}{5}=\frac{33}{200}$$

This two methods give two different answers $$\frac{3}{40}$$ and $$\frac{33}{200}$$

I am obviously missing something or I have some lack in understanding. It will be very helpful if someone point out which one is wrong and why.

The key point is that $$E$$ and $$S$$ are dependent. Thus $$P(E\cap S^c)=P(E)\cdot P(S^c|E)=\frac{33}{40}\cdot \frac{1}{11}=\frac{3}{40}$$

You can comprehend $$P(S^c|E)$$ by looking at the table below:

$$\begin{array}{c|c|c|c} &E & E^c& \\ \hline S &\color{red}{0.75} & 0.05 & 0.8 \\ \hline S^c & 0.075 & \color{blue}{0.125} &\color{red}{0.2} \\ \hline & 0.825 & 0.175 &\color{red}{1} \end{array}$$

and probability of passing any one of these two subjects or both is $$\frac78$$

That means that $$P(S^c\cap E^c)=1-\frac78=\frac18=\color{blue}{0.125}$$

Finding out the missing figures is just simplest algebra.

You see that $$P(S^c)=0.2 \neq P(S^c|E)$$. Thus $$E$$ and $$S$$ are not independent. And your first approach is right $$\checkmark$$

• Thank you for your help. But would you like to show me in detail how to get the table? Or is there any easy way to figure out the dependency of events? I thought events like in the problem is independent by common sense and that's how I generally used to approach any problem. Thanks in advance. – Arafat Hossen Oct 2 '18 at 3:33
• In general if I have only two events I firstly make a table, almost always. The red figures are given in the text. And then the missing figures can be filled in successively. Especially the 0.8 should be very obvious since the cells of the last column/row sum up to one. I´ve made an edit. Please have a look. – callculus Oct 2 '18 at 5:03
• @ArafatHossen You always have to prove if events are independent. There are two variants of the criteria. Basically the meaning is identical: 1. $P(A\cap B)=P(A)\cdot P(B)$, 2. $P(A|B)=P(A)$. It works as well by using the complements: $P(A\cap \overline B)=P(A)\cdot P(\overline B)\Rightarrow$ The events A and B are independent. – callculus Oct 2 '18 at 5:21


# I Is Second rank tensor always tensor product of two vectors?

#### arpon

Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ? If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?
2. How to find out $A_i$ and $B_j$
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?

#### fresh_42

Mentor
2018 Award
Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ?
No, it is a sum of such products.
If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?
No. Even for dyadics $A_i \otimes B_j$ you always have $A_i \otimes B_j = c \cdot A_i \otimes \frac{1}{c}B_j$ for any scalar $c \neq 0$.
2. How to find out $A_i$ and $B_j$
$T_{ij}$ is basically any matrix and $A_i \otimes B_j$ a matrix of rank $1$. So write your matrix as a sum of rank-$1$ matrices and you have a presentation.
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?
No. See the rank explanation above.

#### haushofer

3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?
To add to Fresh's answer: this last remark already should make you suspicious. An arbitrary (!) second-rank tensor in three dimensions has 3*3=9 components, but two vectors have 2*3=6 components. What you can do, is to decompose a second rank tensor like $T_{ij}$ as

$$T_{ij} = T_{[ij]} + T_{(ij)}$$

where [ij] stands for antisymmetrization, whereas (ij) stands for symmetrization. Both parts transform independently under coordinate transfo's. The antisymmetric part has 3 independent components, whereas the symmetric part has 6 components. You can even go further in this decomposition, because the trace of the tensor components also does not change under a coordinate transformation.

"Is Second rank tensor always tensor product of two vectors?"

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How do I simplify a vector expression?

I am doing vector analysis. I have figured out that the following expression won't be simplified in Mathematica:

Simplify[Dot[x, y] - Dot[y, x]]

I think the reason is that x and y could be matrices, so generally the operation does not commute. But for a vector, the commutation relation should hold. What I don't know is how to tell Mathematica to consider x and y to be vectors.

Also, how can I expand Cross[x + y, z] to be Cross[x, y] + Cross[y, z]? I tried to use Expand and ExpandAll but neither worked.

BTW, in Mathematica, is it possible to define a abstract row/column vector without explicitly specifying the number of entries?

Here is a way to do all the things you asked for automatically, independently of Mathematica version. The approach relies on a special symbol to identify when we're dealing with a vector: Instead of using things like x, y etc. for vectors, the convention now is that vectors are written as vec[x], vec[y], etc.

You could also define the wrapper OverVector[x] for this purpose because it displays as $\vec{x}$. But for this post I want to keep it simple, and the arrows wouldn't display easily in the source code below.

ClearAll[scalarProduct, vec];
SetAttributes[scalarProduct, {Orderless}]
vec /: Dot[vec[x_], vec[y_]] := scalarProduct[vec[x], vec[y]]
vec /: Cross[vec[x_], HoldPattern[Plus[y__]]] :=
Map[Cross[vec[x], #] &, Plus[y]]
vec /: Cross[HoldPattern[Plus[y__]], vec[x_]] :=
Map[Cross[#, vec[x]] &, Plus[y]]
scalarProduct /: MakeBoxes[scalarProduct[x_, y_], _] :=
RowBox[{ToBoxes[x], ".", ToBoxes[y]}]

vec[x].vec[y]

(* ==> vec[x].vec[y] *)

vec[x].vec[y] == vec[y].vec[x]

(* ==> True *)

Cross[vec[x], vec[a] + vec[b]]

(* ==> vec[x]\[Cross]vec[a] + vec[x]\[Cross]vec[b] *)

Cross[vec[a] + vec[b], vec[x]]

(* ==> vec[a]\[Cross]vec[x] + vec[b]\[Cross]vec[x] *)

For the Dot product, I defined the behavior of vec such that it gets evaluated as a new function scalarProduct whose only algebraic property is that it's Orderless as you were expecting for the dot product of vectors. Of course this is only true for Euclidean dot products, so this assumption is implicit here. For more information on how this definition works, look up TagSetDelayed.

In addition, scalarProduct is given a customized display format by defining that it should again display as if it were a dot product when it appears in the low-level formatting function MakeBoxes.

For the distributive property of the cross product, I give vec the additional property that when it appears in Cross together with an expression of head Plus, the sum is expanded. Here the TagSetDelayed definitions are done for both orders, and contain a HoldPattern to prevent Plus from being evaluated too early in the definition.

Now you may come back with many more wishes: e.g., what about multiplicative scalars in the dot or cross product, and what about matrices. However, that's a wide field that opens up a can of worms, so I would say just implement the bare minimum of features you can get away with symbolically, then proceed with a concrete working basis so that you can write vectors as lists instead.

Another approach would be to define a new symbol for a custom dot product. That is done in this question.

Using OverVector

As mentioned above, you can replace vec by Overvector everywhere in the above source code, to get a better formatted result. Assuming you have done that (I won't bother to repeat the definitions with that change), here are some examples:

To enter these vector expressions, refer to the Basic Math assistant palette. The cross product can be entered as EsccrossEsc.

Another thing you asked for is to use the antisymmetry of the cross product in simplifications. That's actually done already if you invoke FullSimplify:

• Yes, the definition of new operator looks good. But it is not that convenient to use additional head. I am looking for a vector operators to simply some complicate expression involving many vectors. Is that possible to redefine +, -, . and x so don't have to use something like Plus? – user1285419 Mar 14 '13 at 4:00
• Sorry, I used Plus only because that was what I was thinking about while writing the definitions. Of course you can actually use + and - with this setup. No modifications needed at all. I also mentioned you can get a simpler format using OverVector, and illustrated that in an update to the answer. As to redefining + etc.: it's not necessary, so don't do it. – Jens Mar 14 '13 at 4:28
• that's amazing. I never know that we can use mathematica in that way, I did learn something today. Thanks. – user1285419 Mar 14 '13 at 4:35
• I find a new problem of using this definition, if I try to Cross[OverVector[x], 2*OverVector[y]] + Cross[OverVector[y], 2*OverVector[x]] but it is not zero – user1285419 Apr 16 '13 at 22:58
• I already anticipated this comment in my answer, and mentioned that it opens up a wide field of additional definitions one could add. I may come back to that when I have time. – Jens Apr 16 '13 at 23:26

If you have Mathematica Version 9, you can use Vectors and TensorReduce:

Assuming[(x | y) \[Element] Vectors[n] , TensorReduce[Dot[x, y] - Dot[y, x]]]
(* 0 *)
TensorReduce[Dot[x, y] - Dot[y, x], Assumptions -> (x | y) \[Element] Vectors[n]]
(* 0 *)
TensorReduce[Cross[x + y, z],  Assumptions -> (x | y | z) \[Element] Vectors[n]]
(* x\[Cross]z + y\[Cross]z *)
Distribute[Cross[x + y, z]] (* this should work in all previous versions *)
(* x\[Cross]z + y\[Cross]z *)

• thanks, it works. But what about the first one, is that possible to define an abstract row/column vector so x.y-y.x will get zero? – user1285419 Mar 13 '13 at 3:11
• Presumably these solutions are for version 9 and later, when TensorReduce and Vectors were introduced. – whuber Mar 13 '13 at 4:42
• @whuber, thank you. I updated with a note that TensorReduce and Vectors` are version-9 functions. – kglr Mar 13 '13 at 6:15
• Thanks. Oh, I just saw that it only works on version 9 not my version 8 :( – user1285419 Mar 13 '13 at 6:37


# How to show that $2^n > n$ without induction

I'm solving exercises about Pascal's triangle and Binomial theorem, and this problem showed up, however I don't have any clue on how to solve it

The sum of $${n\choose p}$$ from $$p=0$$ to $$n$$ is the same thing as $$(1+1)^n=2^n$$, how can I use this information? Maybe comparing with another summation that equals to n?

Note that $$2^n$$ is the number of subsets of $$[n]=\{1,\dotsc,n\}$$. There are $$n$$ subsets of $$[n]$$ with size $$1$$. There is at least one subset of $$[n]$$ which is not a singleton (namely the empty set). Hence $$2^n>n$$ for $$n\geq 1$$.

• Foobaz John.Nice+. – Peter Szilas Nov 25 '18 at 17:21

Use Bernoulii inequality, which is true for all $$x>-1$$: $$(1+x)^n\geq 1+nx$$ so $$(1+1)^n \geq 1+n\cdot 1 >n$$

Maybe comparing with another summation that equals to $$n$$?

For any $$p=0,1,\dots,n$$, there is at least one way to choose $$p$$ things from a list of $$n$$ things. Thus $$\binom{n}{p} \ge 1$$, so

$$2^n = \sum_{p=0}^n \binom{n}{p} ≥ \sum_{p=0}^n 1 = n+1 > n.$$

• I did that summation, but how do I prove that one inequality is bigger than the other? – Nuno Mateus Nov 25 '18 at 17:15
• @NunoMateus The sentence before that is the proof. – Calvin Khor Nov 25 '18 at 17:15

The Binomial Theorem says \begin{align} 2^n &=(1+1)^n\\ &=\binom{n}{0}1^0+\binom{n}{1}1^1+\dots\\ &\ge1+n\\[9pt] &\gt n \end{align}

• And I was thinking about this one :) +1 – Aqua Nov 25 '18 at 17:50

hint

Consider $$x\mapsto \frac{\ln(x)}{x}$$ for $$x\ge 1$$.

$$f'(x)=\frac{1-\ln(x)}{x^2}$$

the maximum if $$f(e)=\frac{1}{2}<\ln(2)$$.

thus

$$\ln(x)


# Finding convergence of a series using integral test

The series:$$\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n}\right)^{2}$$ Question:

a) show that it converges

b) find the upper bound for the error in approximation $s\approx s_{n}$

Trial: The section was about integral test, but the sequence$\left(\frac{\ln(n)}{n}\right)^{2}$ is not decreasing from [1,$\infty$]( it is increasing from [ 1,e ] ) so, I could not use the integral test.

Other method: I tried to find a sequence greater than$\left(\frac{\ln(n)}{n}\right)^{2}$so that it satisfies the condition for use of integral test( If I show using integral test that the new series is bounded then It could imply that our sequence is convergent since its bounded and decreasing).the problem was that I had trouble finding any function which could satisfy such conditions b) this is understanding problem, is it asking me to find the exact sum or some upper bound , in any way how can I do this?

• i don't understand your sum – Dr. Sonnhard Graubner Jul 24 '15 at 15:54
• If the function is decreasing on $(e,\infty)$ then the sequence is decreasing on $[3,\infty)$. Removing a finite number of terms from the front end of a series won't change whether the series converges or not. – Alex Pavellas Jul 24 '15 at 15:56
• btw, I assume you meant to use $n$ instead of $x$ inside the series. – Alex Pavellas Jul 24 '15 at 15:57
• @Dr.SonnhardGraubner I made an edit – Socre Jul 24 '15 at 15:59
• @user255545 yes, that infact is true but I have another problem, I had a difficult time integrating the function. If I knew how to integrate it your method is actually quite satisfying. – Socre Jul 24 '15 at 16:02

To show the series converges using the integral test we simply integrate by parts twice with successive substitutions $u_1=(\log x)^2$ and $v_1=x^{-2}$, and $u_2=\log x$ and $v_2=x^{-1}$, to reveal

\begin{align} \int_3^{\infty} \left(\frac{\log x}{x}\right)^2\,dx&=-\left.\left(\frac{(\log x)^2}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2-2\left.\left(\frac{\log x}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{1}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \end{align}

Thus, the series converges.

UPPER AND LOWER BOUNDS

To find an upper bound of the series using the integral test we use

\begin{align} \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2&\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\int_3^{\infty}\left(\frac{\log x}{x}\right)^2\,dx\\\\ &=\left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23\\\\ &\approx. 2.05560987295277 \end{align}

The lower bound is simply the upper bound less the third term $\left(\frac{\log 3}{3}\right)^2\approx. 0.134105440090287$

Thus, we have

$$\bbox[5px,border:2px solid #C0A000]{ \left(\frac{\log 2}{2}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2}$$

$$\bbox[5px,border:2px solid #C0A000]{ \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23}$$

$$\bbox[5px,border:2px solid #C0A000]{ 1.92150443286247 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le 2.05560987295278}$$

Since $\log n\leq (n-1)^{\frac{2}{5}}$ for any $n\geq 1$,

$$0\leq \sum_{n\geq 1}\frac{\log^2 n}{n^2}\leq \sum_{n\geq 1}\frac{1}{n^{\frac{6}{5}}}$$ and the RHS is convergent by the $p$-test. Moreover, $$\sum_{n\geq 1}\frac{\log^2 n}{n^2}=\frac{d^2}{ds^2}\left.\sum_{n\geq 1}\frac{1}{n^s}\right|_{s=2} = \zeta''(2) = 1.9892802342989\ldots$$

• Why the exponent of $2/5$? – marty cohen Jul 24 '15 at 17:35
• @martycohen: any constant slightly less than $\frac{1}{2}$ does the job, $\frac{2}{5}$ is just a nice number of that form. – Jack D'Aurizio Jul 24 '15 at 17:39
• According to Wolfy, max{(log(log(x)))/(log(x-1))}~0.379831 at x~10.9352. So your choice of 2/5 is quite good. – marty cohen Jul 24 '15 at 17:47

For the convergence we can use for example, for $x$ sufficiently large (say $x\geq N$), $$\log\left(x\right)\leq x^{1/4}$$ hence $$\sum_{n\geq N}\frac{\log^{2}\left(n\right)}{n^{2}}\leq\sum_{n\geq N}\frac{1}{n^{3/2}}<\infty.$$ About the upper bound for the error, we can use the integral test $$\sum_{n\geq N}f\left(n\right)\leq f\left(N\right)+\int_{N}^{\infty}f\left(x\right)dx$$ and so in our case $$\sum_{n\geq1}\frac{\log^{2}\left(n\right)}{n^{2}}=\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\sum_{n\geq N+1}\frac{\log^{2}\left(n\right)}{n^{2}}\leq$$ $$\leq\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\frac{\log^{2}\left(N+1\right)}{\left(N+1\right)^{2}}+\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx$$ and the integral is, using the integration by parts, $$\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx=\frac{\log^{2}\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{\log\left(x\right)}{x^{2}}dx=$$ $$=\frac{\log^{2}\left(N+1\right)}{N+1}+\frac{2\log\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{1}{x^{2}}dx=$$ $$\frac{\log^{2}\left(N+1\right)+2\log\left(N+1\right)+2}{N+1}.$$


# Probability Question

rooski

## Homework Statement

Suppose that the integer values 1 2 and 3 are written on each of three different cards. Suppose you do not know which number is the lowest (you do not know beforehand what the values on the cards are). Suppose that you are to be offered these cards in a random order. When you are offered a card you must immediately either accept it or reject it. If you accept a card, the process
ends. If you reject a card, then the next card (if a card remains) is offered. If you
reject the first two cards offered, then you must accept the final card.

(a) If you plan to accept the first card offered, what is the probability that you will
accept the lowest valued card?

(b) If you plan to reject the first card offered, and to then accept the second card if
and only if its value is lower than the value of the first card, what is the probability
that you will accept the lowest valued card?

## The Attempt at a Solution

A) The answer is obviously 1/3 for this question.

B) this is a conditional probability question. Given that P(E|F) = P(EF) / P(F) then i must first figure out what P(E) and P(F) stand for.

P(E) is the probability that i will accept the lowest card.
P(F) is the probability that the second card i choose is lower than the first rejected card.

If i reject the card with 1 on it, then i have no chance of selecting the lowest card next.
If i reject the card with 2 on it, then there is a 1/2 chance i will select the lowest card next.
If i reject the card with 3 on it, then i have a 1/2 chance i will select the lowest card next.

So there is a 2/3 * 1/2 = 1/3 chance that the second card is lower than the first rejected card, right?

I am not sure how to proceed after this.

Homework Helper
If you reject the card with 2 on it, then you will either pick the card with 1 next (which is lower, so you will accept it and have the lowest card) or you will pick that card with 3 (which is higher, so you will reject it and be left with the last one).
So in fact, the probability of getting the lowest card there is 1.

rooski
Ah right, dunno how i missed that.

So how do i calculate the chance that the second card will be lower? I have 0, 1 and 1/2 as the probabilities, depending on which card is rejected first.

Karlx
Hi Rooski.
Let L be the event that you accept the lowest card.
Let $$F_{n}$$ be the event that the value of the first (and rejected) card is n, n=1,2,3.

Events $$F_{n}$$ are disjoint.

Try to write down p(L) using the total probability rule.

rooski
Assuming L is the event i accept the lowest card,

P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C)

Where A,B,C denote cards 1,2,3 respectively. Is that right or am i off? It seems wring since P(A), P(B) and P(C) would all be 1/3.

Karlx
Ok.
According to your notation A is the event that the first (and rejected) card is the lowest one.
So, if you reject A, the probability of accepting the lowest card is zero.
So P(L|A) = 0.

What about P(L|B) and P(L|C) ?

rooski
P(L|B) = 1 since you will reject 3 if it appears, or accept 1 when it appears.
P(L|C) = 1/2 since you will accept 2 if it appears or accept 1 when it appears.

Have i calculated P(A) P(B) and P(C) wrong?

Karlx
A is the event that the first card is A. So pA=1/3.
The same apply to B and C.
Just calculate P(L).

rooski
P(L) = 0 * 1/3 + 1 * 1/3 + ½ * 1/3 = 3/6 = 1/2

So there is a 50% chance that we will end up with the lowest card if we reject the first random card.


# Sawtooth Wave Fourier Series

Square wave t x(t) X 0 -T 0 0 T 0 -X 0 0 k X j π − 2 0 when k is odd a k = 0 when k is even 2. Fn = 2 shows the special case of the segments approximating a sine. In this demonstration it's just like the last one for the square wave. A sawtooth wave An electrocardiogram (ECG) signal Also included are a few examples that show, in a very basic way, a couple of applications of Fourier Theory, thought the number of applications and the ways that Fourier Theory is used are many. According to the important theorem formulated by the French mathematician Jean Baptiste Joseph Baron Fourier, any periodic function, no matter how trivial or complex, can be expressed in terms of converging series of combinations of sines and/or cosines, known as Fourier series. It is so named based on its resemblance to the teeth of a plain-toothed saw with a zero rake angle. Find the Fourier series for the sawtooth wave defined on the interval $$\left[ { - \pi ,\pi } \right]$$ and having period $$2\pi. EE341 Homework Assignment 4 9-13-19. The Fourier expansion of the square wave becomes a linear combination of sinusoids: If we remove the DC component of by letting , the square wave become and the square wave is an odd function composed of odd harmonics of sine functions (odd). EE 230 Fourier series - 1 Fourier series A Fourier series can be used to express any periodic function in terms of a series of cosines and sines. 1 De nitions and Motivation De nition 1. jpg 1,956 × 2,880; 323 KB. (Note that Trott 2004, p. Report on sawtooth wave generator 1. Example: Sawtooth wave So, the expansion of f(t) reads (7. 5))in terms of its Fourier components, may occur in electronic circuits designed to handle sharply rising. Lecture 7: Fourier Series and Complex Power Series Week 7 Caltech 2013 1 Fourier Series 1. — The convention is that a sawtooth wave ramps upward and then sharply drops. However, as ybeltukov pointed out in a comment I did not read until he made me aware of it, Fourier series of piecewise continuously differentiable functions tend to overshoot a jump discontinuities, something which is called Gibbs phenomenon. Unless stated otherwise, it will be assumed that x(t) is a real, not complex, signal. To best answer this question, we need to consult the work of Baron Jean Baptiste Fourier and dig into a little mathematics. 3 shows two even functions, the repeating ramp RR(x)andtheup-down train UD(x) of delta functions. The three examples consider external forcing in the form of a square-wave, a sawtooth-wave, and a triangle-wave. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Figure 13-11 shows an example of calculating a Fourier series using these equations. These basic signals can be used to construct more useful class of signals using Fourier Series representation. The voltage at the Figure 5. The examples given on this page come from this Fourier Series chapter. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. 2 Wave Diffraction and the Reciprocal Lattice Diffraction of waves by crystals • The Bragg law Scattered wave amplitude • Fourier analysis • Reciprocal lattice vectors • Diffraction conditions • Laue equations Brillouin zones • Reciprocal lattice to sc/bcc/fcc lattices Fourier analysis of the basis. general Fourier Series around a jump discontinuity. Taking the inner product of both sides, with respect to the orthonormalized eigenfunctions X n (x) and the weight function w(x) = 1, and assuming validity of the interchange between the summation and integration operations, yields. We then state some important results about Fourier series. But what we're going to do in this case is we're going to add them. Definition of Fourier Series and Typical Examples Baron Jean Baptiste Joseph Fourier \(\left( 1768-1830 \right)$$ introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related. This series is used to generate a sawtooth wave and values are calculated using the program l18a1. (iii) h(x) = ˆ 0 if 2


# are these the only answers of $x^4+y^4+z^4+1=4xyz$?

Given an equation $$x^4+y^4+z^4+1=4xyz$$Find out the number of possible ordered tuple $(x,y,z)\mid x,y,z\in\Bbb{R}$.

I am getting it as $(1,1,1),(-1,-1,1),(1,-1,-1),(-1,1,-1)$ so $\boxed{4}$

Is there any other tuple which I am missing?

Any help will be appreciable !

• Looks fine. AM/GM gives that the absolute value of each of $x,y,z$ is 1. Apr 16, 2016 at 14:49
By Am/Gm we have $$\frac{x^4+y^4+z^4+1}{4}\geq xyz$$ . now we know the minima of arithmetic mean and maxima of geometric mean is achieved when numbers are equal or their $mod$ is equal as here $4$th power is used. Thus all positive $1$ or two negative $1$ are permissible hence total answers are $1+{3\choose 2}=1+3=4$


# Linear Regression, with Map-Reduce

Sometimes, with big data, matrices are too big to handle, and it is possible to use tricks to numerically still do the map. Map-Reduce is one of those. With several cores, it is possible to split the problem, to map on each machine, and then to agregate it back at the end.

Consider the case of the linear regression, $\mathbf{y}=\mathbf{X}\mathbf{\beta}+\mathbf{\varepsilon}$ (with classical matrix notations). The OLS estimate of $\mathbf{\beta}$ is $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}$. To illustrate, consider a not too big dataset, and run some regression.

lm(dist~speed,data=cars)$coefficients (Intercept) speed -17.579095 3.932409 y=cars$dist X=cbind(1,cars$speed) solve(crossprod(X,X))%*%crossprod(X,y) [,1] [1,] -17.579095 [2,] 3.932409 How is this computed in R? Actually, it is based on the QR decomposition of $\mathbf{X}$, $\mathbf{X}=\mathbf{Q}\mathbf{R}$, where $\mathbf{Q}$ is an orthogonal matrix (ie $\mathbf{Q}^T\mathbf{Q}=\mathbb{I}$). Then $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}=\mathbf{R}^{-1}\mathbf{Q}^T\mathbf{y}$ solve(qr.R(qr(as.matrix(X)))) %*% t(qr.Q(qr(as.matrix(X)))) %*% y [,1] [1,] -17.579095 [2,] 3.932409 So far, so good, we get the same output. Now, what if we want to parallelise computations. Actually, it is possible. Consider $m$ blocks m = 5 and split vectors and matrices $$\mathbf{y}=\left[\begin{matrix}\mathbf{y}_1\\\mathbf{y}_2\\\vdots \\\mathbf{y}_m\end{matrix}\right]$$ and $$\mathbf{X}=\left[\begin{matrix}\mathbf{X}_1\\\mathbf{X}_2\\\vdots\\\mathbf{X}_m\end{matrix}\right]=\left[\begin{matrix}\mathbf{Q}_1^{(1)}\mathbf{R}_1^{(1)}\\\mathbf{Q}_2^{(1)}\mathbf{R}_2^{(1)}\\\vdots \\\mathbf{Q}_m^{(1)}\mathbf{R}_m^{(1)}\end{matrix}\right]$$ To split vectors and matrices, use (eg) Xlist = list() for(j in 1:m) Xlist[[j]] = X[(j-1)*10+1:10,] ylist = list() for(j in 1:m) ylist[[j]] = y[(j-1)*10+1:10] and get small QR recomposition (per subset) QR1 = list() for(j in 1:m) QR1[[j]] = list(Q=qr.Q(qr(as.matrix(Xlist[[j]]))),R=qr.R(qr(as.matrix(Xlist[[j]])))) Consider the QR decomposition of $\mathbf{R}^{(1)}$ which is the first step of the reduce part$$\mathbf{R}^{(1)}=\left[\begin{matrix}\mathbf{R}_1^{(1)}\\\mathbf{R}_2^{(1)}\\\vdots \\\mathbf{R}_m^{(1)}\end{matrix}\right]=\mathbf{Q}^{(2)}\mathbf{R}^{(2)}$$where$$\mathbf{Q}^{(2)}=\left[\begin{matrix}\mathbf{Q}^{(2)}_1\\\mathbf{Q}^{(2)}_2\\\vdots\\\mathbf{Q}^{(2)}_m\end{matrix}\right]$$ R1 = QR1[[1]]$R for(j in 2:m) R1 = rbind(R1,QR1[[j]]$R) Q1 = qr.Q(qr(as.matrix(R1))) R2 = qr.R(qr(as.matrix(R1))) Q2list=list() for(j in 1:m) Q2list[[j]] = Q1[(j-1)*2+1:2,] Define – as step 2 of the reduce part$$\mathbf{Q}^{(3)}_j=\mathbf{Q}^{(2)}_j\mathbf{Q}^{(1)}_j$$ and$$\mathbf{V}_j=\mathbf{Q}^{(3)T}_j\mathbf{y}_j$$ Q3list = list() for(j in 1:m) Q3list[[j]] = QR1[[j]]$Q %*% Q2list[[j]] Vlist = list() for(j in 1:m) Vlist[[j]] = t(Q3list[[j]]) %*% ylist[[j]]

and finally set – as the step 3 of the reduce part$$\widehat{\mathbf{\beta}}=[\mathbf{R}^{(2)}]^{-1}\sum_{j=1}^m\mathbf{V}_j$$

sumV = Vlist[[1]] for(j in 2:m) sumV = sumV+Vlist[[j]] solve(R2) %*% sumV [,1] [1,] -17.579095 [2,] 3.932409

It looks like we’ve been able to parallelise our linear regression…

## 6 thoughts on “Linear Regression, with Map-Reduce”

1. korul says:

How do I supposed to calculate inverse of R^(2), when R^(2) is a vector of matrices?

2. Stefano says:

Great post! One question: mathematically, can you explain – provide a reference for the explanation of – the step 2 of the reduce part (where the 2 Q matrices are multiplied)? Because I can follow through all steps, but I don’t understand *why* you create Q_j^(3) like this. Thanks!

3. Another interesting point is the trade-off between the number of nodes and the efficiency of the calculation. If you think about it – mode nodes *does not* mean faster running time.

I took the liberty of reproducing your results (in “tidyverse” code”) and added some notes on efficiency. I have an R notebook/markdown here: https://github.com/ytoren/reproducible/tree/master/linear-regression-map-reduce

Keep up the good work mate!

4. Awesome post!
I think you might have a typo :

You split Q2 to Q2list by:
> for(j in 1:m) Q2list[[j]] = Q1[(j-1)*2+1:2,]

And it should be
> for(j in 1:m) Q2list[[j]] = Q2[(j-1)*2+1:2,]

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# 99 Bags of Apples and Oranges

You have $99$ bags, each containing various numbers of apples and oranges. Prove that there exist $50$ bags among these which together contain at least half the apples and at least half the oranges.

Needless to say, you may not add/remove fruits to/from the bags.

Clarification: I changed the wording from "you can grab $50$ bags..." to "there exist $50$ bags..."

Clarification from comments: each bag can contain any number of apples and any number of oranges with any total number of fruit; the total amount of fruit in a bag doesn't have to be the same for each bag.

• I suspect the presence of Pigeonhole principle... – leoll2 May 23 '15 at 14:29
• @ghosts_in_the_code can you provide a counterexample? – leoll2 May 23 '15 at 15:54
• @ghosts_in_the_code By various, I mean each bag can have any number of apples, any number of oranges, making any total. The bags do not need to all have the same number of fruits. – Mike Earnest May 23 '15 at 15:58
• Can the bags be empty? – Bob May 23 '15 at 16:09
• @Bob Yes, they can be empty. For example, if all bags were empty, then grabbing any $50$ bags works, since zero is at least half of zero. – Mike Earnest May 23 '15 at 16:12

Let $B_1,B_2,\dots,B_{99}$ be the bags, in increasing order of number of apples contained; say the number of apples in $B_i$ is $a_i$ for all $i$.

If our 50 bags grabbed are $B_{99}$, one of $B_{98}$ and $B_{97}$, one of $B_{96}$ and $B_{95}$, ..., and one of $B_2$ and $B_1$, then they must contain between them at least half of all apples, since at worst they contain $a_{99}+a_{97}+\dots+a_5+a_3+a_1\geq a_{98}+a_{96}+\dots+a_4+a_2+0$ apples.

Now how do we fix all those "one of"s? Just pick whichever of $B_{98}$ and $B_{97}$ has the more oranges, then whichever of $B_{96}$ and $B_{95}$ has the more oranges, and so on. Then our 50 selected bags - even excluding $B_{99}$! - must contain at least half of all oranges.

QED.

• QED indeed! I will certainly accept your answer, but I think I will wait a bit to see if other cool answers crop up as well. I know there are many distinct proofs methods, yours being the most constructive – Mike Earnest May 23 '15 at 21:23

Arrange the bags in a circle. I will call a collection of 50 consecutive bags a team (so there are 99 possible teams), and a team will be called appley if it contains at least half of the apples.

Define the opponent of a given team to be the team whose rightmost bag is the leftmost bag of the given team. For any given team, every bag is either a member of that team or a member of its opponent (and one bag is a member of both). This means that if a team is not appley, then its opponent is appley. Different teams have different opponents, so there must be at least as many appley teams as non-appley teams. There are 99 teams in total, so at least 50 of them are appley.

A similar argument shows that at least 50 teams are orangey (that is, at least 50 of the teams contain at least half the oranges). Of 99 teams, at least 50 are appley and at least 50 are orangey, so there must be at least one teams which is both appley and orangey (because $50+50>99$). This team consists of 50 bags, and contains at least half the apples and at least half the oranges.

• This is really cool! To elaborate: let $A$ be the set of appley teams, and $B$ be the set of teams whose opponents are appley. You showed $A\cup B$ is all $99$ teams (if a team is not appely, its opponent is). Since $|A|=|B|$, we have $99=|A\cup B|\le |A|+|B|=2|A|$, proving $|A|\ge 49.5,$ so $|A|\ge50$. – Mike Earnest May 24 '15 at 4:43
• "There are 99 teams in total, ..." - agreed (count possible common bags). You argument seems to put appley teams in 1:1 correspondence with non-appley teams by construction, implying an even total number of teams. How does that correspond with the odd number of teams? Cont'd ... – Lawrence May 24 '15 at 13:40
• ... cont'd. Simplified example: instead of 99 bags, say we have 3 bags $a,b,c$. Then we can have team pairs $[ab,bc], [bc,ca], [ca,ab]$. Suppose team $bc$ was appley. This would mean from the first two match-ups that both $ab$ and $ca$ were not appley. However, this makes the last match-up a team pair with two non-appley teams. Is this a failure of the assertion that "if a team is not appley, then its opponent is appley"? (By the way, other than this, I really like your proof. +1 ) – Lawrence May 24 '15 at 13:40
• @Lawrence If a team is non-appley, its opponent is appley. However, if a team is appley, we are not saying its opponent is non-appley (they could both be appley). As you pointed out, since there are an odd number of teams, there must at least one appley team whose opponent is also appley. – Julian Rosen May 24 '15 at 13:46
• @Lawrence no, different teams do have different opponents. Notice, however, that the opponent of the opponent of a team is not (quite) the original team, so you can't "pair them off" in the way you seem to be imagining. – Ben Millwood May 26 '15 at 14:35

Suppose you have 50 bags on the ground and 49 in the truck, and suppose the ones on the ground have more apples. One at a time, take a bag from the ground and put it in the truck, and put one from the truck tand place it on the ground. Keep doing this with the goal of totally swapping ground and truck bags.

Eventually, you'll reach a point where moving a "magic" bag causes the truck to have more apples than the ground. At this point, both the ground plus the magic bag has at least half the apples, AND the truck plus the magic bag has at least half the apples. Either the ground or the truck ( plus the magic bag ) has at least half the oranges too.

I will provide a probabilistic proof. By demonstrating that picking randomly has a nonzero chance of success, we can show that a solution exists. (http://en.m.wikipedia.org/wiki/Probabilistic_method)

Claim: A random subset of 50 bags has over 50% chance of having more than half the apples. (Chosen uniformly over all size 50 subsets.)

Proof: For each "losing" size 50 subset, flip it around to get a (unique) winning size 49 subset. This means there are at least as many size 49 winning sets as size 50 losing sets. There are multiple ways to extend these to size 50 sets, so there's strictly more ways to win than lose. (Hand waving a bit)

This argument applies symmetrically to the oranges. Since each probability is over 1/2, they must have a nonzero intersection.

Thus there is a positive probability of picking a subset with at least half of each, so a solution exists.

Nonconstructively useless, but QED :)

Edit: looking back, this is just saying that the intersection of sets larger than 50% is nonempty, so the probabilistic part is just fluff.

• It's far from trivial that you can extend all 49-subsets of a 99-set into 50-subsets by addition of a single element in such a way that you don't accidentally extend two 49-subsets into the same 50-subset (it's certainly impossible to extend 50-subsets into 51-subsets this way, for example, because there aren't as many of them). – Ben Millwood May 26 '15 at 16:26
• I mean, it IS true. But it requires Hall's Marriage Theorem, unless you have an ingenious construction. – Ben Millwood May 26 '15 at 16:29

Suppose we choose 50 of the bags at random. We are choosing more than half of the bags, so the probability that the bags we selected contain at least half of the apples is strictly greater than 50%. Similarly, the probability that the bags we selected contain at least half of the oranges is also greater than 50%. This means there is a positive probability that the bags we selected contain at least half the apples and at least half the oranges. In particular, there must be some combination of 50 bags containing at least half the apples and at least half the oranges.

Edit: As pointed out in the comments, it is not obvious why the bags we select will have at least half the apples with probability greater than 50%. This actually follows from the argument in my other answer: one way to choose 50 bags at random is first to choose a cyclic ordering of the bags, then to choose 50 consecutive bags. My other answer shows that for whichever cyclic ordering we choose, a random choice of 50 consecutive bags contains at least half the apples with probability at least $\frac{50}{99}$. This means that a random selection of 50 bags contains at least half the apples with probability at least $\frac{50}{99}$. The same is true for oranges, so a random selection of 50 bags contains at least half the apples and at least half the oranges with probability at least $\frac{1}{99}$.

• I think the statement We are choosing more than half of the bags, so the probability that the bags we selected contain at least half of the apples is strictly greater than 50%, needs more justification. – Mike Earnest May 23 '15 at 19:44
• This isn't a very rigorous proof. First of all, you must justify your probability statements, then you might provide an algorithm of optimal choosing (not required, but strongly suggested). – leoll2 May 23 '15 at 20:52
• I was expecting it to be clear why the probability of getting at least half the apples was bigger than 50%, but I didn't think this part through and I agree that it isn't clear why this is the case. – Julian Rosen May 24 '15 at 3:40
• Using the probabilistic approach, leaves a possibility that there will not be a desired solution. We need to have 100% probability that a desired case will be found. – Moti May 25 '15 at 17:59
• @Moti If the probability that a random object has a certain property is greater than 0, then it is certain there is at least one object with that property. – Julian Rosen May 25 '15 at 18:54


# Triangle inscribed within a unit circle

A triangle $ABC$ is inscribed within the unit circle. Let $x$ be the measure of the angle $C$. Express the length of $AB$ in terms of $x$.

A.) $2\sin x$

B.) $\cos x + \sin x - 1$

C.) $\sqrt{2}(1 - \cos 2x)$

D.) $\sqrt{2}(1 - \sin x)$

I am unsure how to illustrate the circle and calculate the length $x$.

Thank you. `

• Please read this tutorial about how to typeset mathematics on this site. – N. F. Taussig Jun 18 '17 at 8:55

$$\dfrac{AB}{\sin x}=2R$$ where $R$ is the circum-radius $=1$

• This solution is based on the Law of Sines. – N. F. Taussig Jun 18 '17 at 9:11
• @N.F.Taussig, Thanks for enriching the answer – lab bhattacharjee Jun 18 '17 at 9:12

The angle $AOB=2x$, since it intercepts the same arc than the angle $CAB=x$.

thus by the well-known formula

$$AB^2=OA^2+OB^2$$ $$-2OA.OB.\cos (2x)$$ with

$$OA=OB=radius=1$$

hence

$$AB^2=2 (1-\cos (2x))$$

$$=4\sin^2 (x)$$ and $$\boxed {AB=2\sin (x) }$$


# Density of positive multiples of an irrational number

Let $x$ be irrational. Use $\{r\}$ to denote the fractional part of $r$: $\{r\} = r - \lfloor r \rfloor$. I know how to prove that the following set is dense in $[0,1]$: $$\{\{nx\} : n \in \mathbb{Z}\}.$$ But what about $$\{\{nx\} : n \in \mathbb{N}\}?$$ Any proof that I’ve seen of the first one fails for the second one.

A minor modification of the pigeonhole argument works. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\in\{1,\dots,m+1\}$ and $k\in\{0,\dots,m-1\}$ such that $\frac{k}m\le\{ix\},\{jx\}<\frac{k+1}m$; clearly $\{|(j-i)x|\}<\frac1m$. Let $\ell$ be the largest positive integer such that $\ell\{|(j-i)x|\}<1$, let $A_m=\{n|j-i|x:0\le n\le\ell\}$, and let $D_m=\big\{\{y\}:y\in A_m\big\}$.

If $x>0$, every point of $[0,1)$ is clearly within $\frac1m$ of $A_m=D_m$. If $x<0$, then

$$D_m=\{1-|y|:y\in A_m\}\;,$$

so every point of $[0,1)$ is again within $\frac1m$ of the set $D_m$. Since $D_m\subseteq\big\{\{nx\}:n\in\Bbb N\big\}$, we’re done.

• I hope you don't mind how close my answer is to yours, but I had written an answer for someone in chat and was going to post it elsewhere, then saw this question was a better fit.
– robjohn
Jul 20, 2021 at 20:58
• @robjohn: No problem: someone may benefit from the greater detail. Jul 21, 2021 at 2:39
• Can you give more details why if $x>0$ then every ppint of $[0,1)$ is clearly within $1/m$ of $A_m$.
– ZFR
Jul 30, 2021 at 2:29
• @ZFR: Adjacent points of $A_m$ are less than $\frac1m$ apart. Every point of $[0,1)$ is either equal to one of these points, between two adjacent ones, or between $\ell\{|(j-i)x|\}$ and $1$ and hence less than $\frac1m$ from some point of $A_m$. Jul 31, 2021 at 22:00

Here is a proof where the second statement follows from the first with a minor step. I wrote this answer for another question, but then I saw it fit here better, other than being along very similar lines to Brian M. Scott's answer.

Let $$r$$ be irrational. Choose an arbitrary $$n\in\mathbb{N}$$. Partition $$[0,1)$$ into $$n$$ subintervals $$\left\{I_k=\left[\frac{k-1}n,\frac kn\right):1\le k\le n\right\}\tag1$$ Consider the discrete set $$\{\{kr\}:0\le k\le n\}\subset[0,1)$$. There are $$n+1$$ elements in this set, so two of them must lie in the same subinterval. That means that we have $$k_1\ne k_2$$ so that $$\{(k_1-k_2)r\}\in\left(0,\frac1n\right)$$, we leave $$0$$ out of the interval because $$r$$ is irrational.

Let $$m=\left\lfloor\frac1{\{(k_1-k_2)r\}}\right\rfloor$$, then because $$m$$ is the greatest integer not greater than $$\frac1{\{(k_1-k_2)r\}}$$, $$m\{(k_1-k_2)r\}\!\!\overset{\substack{r\not\in\mathbb{Q}\\\downarrow}}{\lt}\!\!1\lt(m+1)\{(k_1-k_2)r\}\tag2$$ which implies that $$\{m(k_1-k_2)r\}=m\{(k_1-k_2)r\}\in\left(\frac{n-1}n,1\right)$$.

Since $$\{(k_1-k_2)r\}\in\left(0,\frac1n\right)$$, if $$j\{(k_1-k_2)r\}\in I_k$$, then $$(j+1)\{(k_1-k_2)r\}\in I_k\cup I_{k+1}$$; that is, $$\{j(k_1-k_2)r\}=j\{(k_1-k_2)r\}$$ cannot skip over any of the $$I_k$$. Therefore, $$\left\{\{j(k_1-k_2)r\}:1\le j\le m\vphantom{\frac12}\right\}\tag3$$ must have at least one element in each $$I_k$$ for $$1\le k\le n$$.

The only problem is that we don't know that $$k_1-k_2\gt0$$. However, this is not a problem since $$\{j(k_1-k_2)r\}\in I_k\iff\{j(k_2-k_1)r\}\in I_{n+1-k}$$. Therefore, $$\left\{\{j(k_2-k_1)r\}:1\le j\le m\vphantom{\frac12}\right\}\tag4$$ must also have at least one element in each $$I_k$$ for $$1\le k\le n$$.

Since $$n$$ was arbitrary, we have shown that $$\left\{\mathbb{N}r\right\}$$ is dense in $$[0,1]$$.

• Thanks a lot professor Robjohn for answering my question. I was having lot of difficulty understanding $k_2\gt k_1$ case. It all makes sense to me now. Many thanks :) I would only like to add this little more detail for statement just before $(3)$ in the hope that it will benefit those who visit this great answer in future: Suppose on the contrary that for some $k<n$, the subinterval $(\frac{k-1}n, \frac kn )$ doesn't contain any element of the set $T=\{ j\{(k_2-k_1)r\}: 1\le j\le m\}$. (Contd.)
– Koro
Jul 21, 2021 at 3:55
• (Contd.) Note that $T_L=\{j: j\{(k_2-k_1)r\}\le \frac{k-1}n\}$ is non-empty and therefore $T_L$ has a maximal element $M\in \mathbb N$ that is $(M+1)\{(k_2-k_1)r\}\ge \frac kn$ and $M\{(k_2-k_1)r\}\le \frac{k-1}n$ and subtracting the two, we get: $\{(k_2-k_1)r\}\ge \frac 1n$, which is a contradiction. Therefore no subinterval $I_k$ can be free of elements from $\{\mathbb N r\}$.
– Koro
Jul 21, 2021 at 3:56
• Really nice answer which helped me to understand the problem. But I think you can assume $r$ to be any real irrational number from the beginning of the solution.
– ZFR
Jul 29, 2021 at 22:25
• @ZFR: You know, I modified this so many times while writing it, and now it seems you are correct; there is no reason for $r\gt0$, so I have removed that constraint.
– robjohn
Jul 30, 2021 at 2:40
• Thanks a lot for your answer! To be honest I've spent on this problem about 2-3 days and eventually your solution helped me to understand it completely. Thanks! +1
– ZFR
Jul 30, 2021 at 15:42

Really? I thought exactly the same proof worked for $\Bbb N$.

Let $k\in\Bbb Z$ with $k\ne 0$ and define $$f(t)=e^{2\pi ikt}.$$

Then $f$ has period $1$, and $$\frac1N\sum_{n=0}^{N-1}f(nx) =\frac1N\sum_{n=0}^{N-1}\left(e^{2\pi ikx}\right)^n=\frac1N \frac{e^{2\pi ikxN}-1}{e^{2\pi ikx}-1}\to0\quad(N\to\infty).$$ So the usual approximation shows that $$\frac1N\sum_{n=0}^{N-1}f(nx)\to\int_0^1 f(t)\,dt$$for $f\in C(\Bbb T)$ and you're done, as usual.

How is this any different from the case $n\in\Bbb Z$?

• How does that imply that $\{ \exp (2\pi i {n x}) :n\in N\}$ is dense in the unit circle in $C$ ? Anyway, a proof from the most elementary method works for N just as well as for Z. Jan 24, 2016 at 5:33
• @user254665 Say those points are not dense. There is then an interval, or arc, $I$ on the circle that contains none of the points. Take $f\in C(\Bbb T)$ such that $f=0$ on the complement of $I$ but $\int f > 0$. Then $\frac1N\sum_0^{N-1}f(nx)=0$ for every $N$, so it does not converge to $\int f$. (Yes, the pigeonhole argument works as well. I really can't think of an argument that works for $n\in\Bbb Z$ but not $\Bbb N$. In any case, ignoring the question of which is simpler, this argument shows more than just that the points are dense, it shows they are asymptotically uniformly distributed.) Jan 24, 2016 at 14:21
• very nice integral proof, which yields extra info (distribution) too. Jan 24, 2016 at 20:10
• @user254665 Yes it is very nice. The equidistribution is Weyl's theorem - I think it's his proof, not sure, in any case it's a standard thing. Jan 24, 2016 at 21:27


# Which payoff do you want to go for?

Consider a game where you have to choose between 1 of 2 payoffs:

Payoff 1. You toss 100 coins.
You get $1 for each Head that appears. Payoff 2. You toss 10 coins. You get$10 for each Head that appears.

Which payoff structure would you choose? And why?

Note by Calvin Lin
3 years, 10 months ago

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I assume each coin toss is independent from one another (so the covariance between each coin toss is zero), and used only fair coins.

Yes, the expected Payoff is the same for ONE and TWO. But their variance differ.

$V_1 = \text { Probability } \times \text { Payoff ONE }^2 \times \text { Number of trials } = \frac {1}{2} \times 1^2 \times 100 = 50$

$V_2 = \text { Probability } \times \text { Payoff TWO }^2 \times\text { Number of trials } = \frac {1}{2} \times 10^2 \times 10 = 500$

Because Payoff TWO has a higher variance which is not desired, I will go with Payoff ONE.

$\text{ I learned from the best tutor }$

- 3 years, 10 months ago

What term do we use to describe a person who choses to "minimize variance when expected value is equal"?

Your answer greatly depends on what it is that we wish to maximize / minimize. There could be people who may want to maximize variance, because of the higher likelihood of a huge payoff that they care about.

Staff - 3 years, 10 months ago

What term do we use to describe a person who choses to "minimize variance when expected value is equal"?

Risk-averse investor

- 3 years, 10 months ago

Let me try.

Payoff 1. Chance to get head for all coins = 1 out of 10^{200} = Chance to get $100 in terms of$1.

Payoff 2. Chance to get head for all coins = 1 out of 10^{10} = Chance to get $100 in terms of$10.

Thus, I think Payoff 2 is better.

- 3 years, 10 months ago

This is optimistic way I guess

- 3 years, 10 months ago

That's a valid argument if your goal is to "maximize the probability of the highest payoff".

Since getting $99,$98, ... $91 is somewhat similar to getting$100, would this affect your answer? Why, or why not?

Note that the payoff 1 should be "10 ^ {100}"

Staff - 3 years, 10 months ago

The probability of getting payoff between 91$to 100$ increases in the first case. Hence, if I consider myself satiated by $99,$98, ... $91, I'd prefer case 1. - 3 years, 10 months ago Log in to reply It's less of a hassle to toss 10 coins. If we compare this to stocks, it's easier to keep a portfolio of10 stocks than 100 stocks. And lower commissions. The risk of loss is higher, but the chance of higher returns is better too. - 3 years, 10 months ago Log in to reply That's a good point! We might need to factor in convenience / the cost of making a flip! Can you clarify what you mean by "chance of higher returns is better too"? How do you quantify that? Staff - 3 years, 10 months ago Log in to reply Earlier posters have quantified the variance for the smaller vs. the larger set of coins. For the set of 10 coins, the chance of getting more than 9 heads is 1 in 1024. For the set of 100 coins, the chance of getting more than 90 coins is much lower than that. I'd have to get my combinatorics (or statistics) hat on to quantify it, but it's much less. On the order of more than 5-sigma from the mean. For stocks, a stock club (or newsletter writer) would be silly to buy 1 share each of the 500 largest companies. A number of funds exists to do that with much less expense. If they want to try to beat the performance of the S&P 500, they pick a small number of stocks (using some rational criteria) and often do better than the "market". Of course, the risk of loss (or underperformance) is higher too. - 3 years, 10 months ago Log in to reply Expected outcome is same in both cases ,but in case 1 chances are that outcome is more closer to 50$ than in case2, So for more standard outcome I will choose case1.

For suppose if my need demands 60$I would choose case 2. CASE1 Chances to get 50$ is 100C50 *.5^{100}=.0786

49$is 100C49*.5^{100 }=.0780 (same for 51$)

For 60$it is .0108 By similar calculations(and adding), chances to get outcome from 40$ to 60$is .946. where as in case 2 chances reduce to .656 this proves that there are more chances to get outcome closer to 50$ in case 1.

probabilities to get 60$are 0.108 for case 1 .205 for case 2. So, if 60$ is my need I will choose case 2

- 3 years, 10 months ago

You bring up a valid point, which is "What is the utility function that I care about". If the utility function is just "I only care if I have more than $60 or not", then your approach shows that we should go for payoff 2. What kind of utility function makes sense? Do people value all amounts of money equally? Staff - 3 years, 10 months ago Log in to reply They average out to the same payoff (which is$50.00), so I would flip a coin to decide which to use.

Python:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 import random coin = (True, False) def sim_1(): payoff = 0 for flip in xrange(100): if random.choice(coin): payoff += 1 return payoff def sim_2(): payoff = 0 for flip in xrange(10): if random.choice(coin): payoff += 10 return payoff trials = 100000 payoff_1 = 0.0 payoff_2 = 0.0 for trial in xrange(trials): payoff_1 += sim_1() payoff_2 += sim_2() print "Payoff 1:", payoff_1 / trials print "Payoff 2:", payoff_2 / trials

- 3 years, 10 months ago

what good

- 3 years, 10 months ago

If i understand the problem correctly, would be indifferent to the payoffs structure, cause both options give the same expected value which would be 50.

- 3 years, 10 months ago

playoff 1

- 3 years, 10 months ago

I think payoff 1 is better because chances increses with more no of toss..

- 3 years, 10 months ago

In my opinion, it's a matter of risk: payoff 1 has more consistency but less probability of scoring extremely large amounts of money, whereas payoff 2 is a sort of high risk high reward model where you either get a lot or get a bit. Personally, I would choose payoff 1.

- 3 years, 10 months ago

Pay off 2, because it will likely to get heads more often and earn $10 for each head than 100 coin tosses. Who knows that in a toss of 100 you will only get less than 50% heads. - 3 years, 10 months ago Log in to reply Payoff 2 because I'll make decent money and i Don't have time to flip 100 coins. Stop over complicating things with math people. - 3 years, 10 months ago Log in to reply I will chose payoff 2 because while tossing coins 10 time we may get HEAD more than 1 or 2 times and the pay is more. But in payoff 1 tossing coins 100 times may give a HEAD 20 times and u get 20$. But in payoff 2 you can get more than 20$. - 3 years, 10 months ago Log in to reply Payoff 2 as it takes less time to toss 10 coins and the value is same for both - 3 years, 9 months ago Log in to reply What are the statistics given that heads side of a coin is slightly heavier than tails? Giving tails more of a chance of facing up? Is that true? I heard that somewhere back maybe in 4th grade haha. Or has that been proven incorrect? if it is true, I would rather chose b since I would want less of a chance of getting tails side up? Can you explain to me how I am wrong (I probably am) like you would explain it to a 3rd grader? This is why I'm more of an artist than math person. - 3 years, 6 months ago Log in to reply That is a very good point. How would your answer change if we didn't have a fair coin? For example, if the coin was totally biased and only gave tails, then both payoffs are the same. Similarly, if the coin was totally biased and only gave heads, then both payoffs are the same. But in between, the payoffs are different. So, does that mean that the answer could change depending on the probability of head/tail? If so, how and why? How is payoff 2 "less chance of getting tails side up"? Just because of the absolute number? But, the first mostly likely has many more heads appearing, just a smaller payoff each time. Staff - 3 years, 6 months ago Log in to reply Less probability of a heads since the gravitational pull of the heavier side. Sorry I missed stated. On the other hand, my fiance makes a good point. The more you flip a coin your results are more likely to reflect the chance of a 50/50 so it is better and more reliable to flip more times than less so option 1 of 100 flips is more reliable then just 10 in case you get a lucky streak of a ten or an unlucky streak with only 2 out of 10. - 3 years, 6 months ago Log in to reply As it turns out, the Expected Value of both payoffs would be the same, regardless of the probability of landing heads / tails. For example, if the coin is fair, and it will land heads 50% of the time, then the expected payoff in 1 is$50, and the expected payoff in 2 is \$50. More generally, if it land heads $$p%$$ of the time, then the expected payoff will be $$p$$.

So, the question then becomes, pick your favorite probability $$p$$ of landing heads, which payoff would you want? Is there anything else to consider?

Staff - 3 years, 6 months ago

Payoff 1 so that I can get more number of chances

- 3 years, 10 months ago

I'd go for #2. I'm a risktaker.

- 3 years, 9 months ago

You will still need to define what you mean by "risktaker". There are possible definitions of "risktaker" which would make option 1 more valuable to them.

Staff - 3 years, 9 months ago


# Show there can't be two real and distinct roots of polynomial $f(x)=x^3-3x+k$ in $(0,1)$, for any value of k.

I have two proofs here one which I did and the other was given in book. Is one better than the other? I am asking for in an exam setting which proof makes a better solution(as in fetch more marks).

## Proof 1 (My proof)

$$f'(x)=3x^2-3$$, in the interval $$(0,1)$$ is less than $$0$$. If there were two distinct roots then $$f'(0)$$ should have been $$0$$ once in $$(0,1)$$ by rolle's theorem. Since it isn't there are no values of $$k$$ for which there are two real roots.

## Proof 2 (Book)

Let $$a,b$$ be two roots of $$f(x)$$ in $$(0,1)$$ then there exists a $$c$$ such the $$f'(c) = 0$$ for c in $$[a,b]$$ by Rolle's theorem. $$f'(c)= 3c^2-3$$ has no solutions in $$(0,1)$$ hence there is no such value of $$k$$.

• Yours is better because the book version has a wrong expression for $f'$ – Hagen von Eitzen Jun 11 at 6:09
• – lab bhattacharjee Jun 11 at 6:16
• @HagenvonEitzen that's a typo sorry. – Sonal_sqrt Jun 11 at 6:26


# How to factorize this cubic equation?

In one of the mathematics book, the author factorized following term

$$x^3 - 6x + 4 = 0$$ to

$$( x - 2) ( x^2 + 2x -2 ) = 0.$$

How did he do it?

• If there is a cube term, it is not a quadratic. Also, a method for finding divisors of your polynomial, look at the factors of the constant term. – Edward Evans Jan 28 '17 at 19:52
• $x=2$ is a root of this equation. Then by polynomial division of $\frac{x^3-6x+4}{x-2}$ we obtain $x^2+2x-2$. – projectilemotion Jan 28 '17 at 19:53
• Do you know the Rational Root Test? – Bill Dubuque Jan 28 '17 at 19:54
• A common first step for introductory problems is to guess and check certain integers close to zero to see if it will equal zero. Zero is the easiest to check because that would mean the constant term is zero, thats not the case here. $1$ doesn't work because that would be $1-6+4=-1\neq 0$. $2$ happens to work since this would be $2^3-6\cdot 2 + 4 = 8-12+4=0$. Since $2$ works, we know that the equation can be factored as $(x-2)q(x)$ where $q(x)=\frac{x^3-6x+4}{x-2}$. In general this won't always work, especially if the roots aren't even integers. Cardano's formula would help then. – JMoravitz Jan 28 '17 at 19:54
• @BillDubuque No – Govinda Sakhare Jan 28 '17 at 19:56

There is a neat trick called the rational roots theorem. All we have to do is factor the first and last numbers, put them over a fraction, and take $\pm$. This gives us the following possible rational roots:

$$x\stackrel?=\pm1,\pm2,\pm4$$

due to the factorization of $4$. Checking these, it is clear $x=2$ is the only rational root, since

\begin{align}0&\ne(+1)^3-6(+1)+4\\0&\ne(-1)^3-6(-1)+4\\\color{#4488dd}0&=\color{#4488dd}{(+2)^3-6(+2)+4}\\0&\ne(-2)^3-6(-2)+4\\0&\ne(+4)^3-6(+4)+4\\0&\ne(-4)^3-6(-4)+4\end{align}

leaving us with

$$x^3-6x+4=(x-2)(\dots)$$

We can find the remainder through synthetic division:

$$\begin{array}{c|c c}2&1&0&-6&4\\&\downarrow&2&4&-4\\&\hline1&2&-2&0\end{array}$$

which gives us our factorization:

$$x^3-6x+4=(x-2)(x^2+2x-2)$$

• OP does not know RRT - see the comments. – Bill Dubuque Jan 28 '17 at 19:59
• @BillDubuque Oh. Then give me a moment – Simply Beautiful Art Jan 28 '17 at 20:00
• @SimplyBeautifulArt Why 1,2 and 4. Why are we not checking equation with value 3? it has anything to do with c term ax^3+bx+c ? – Govinda Sakhare Jan 28 '17 at 20:09
• @piechuckerr By the rational roots theorem, I need only check the factors of the last constant if the leading coefficient is $1$. $3$ does not divide into $4$, so I needn't check it. – Simply Beautiful Art Jan 28 '17 at 20:10
• @SimplyBeautifulArt i.e if constant term is 6 then 1,2,3 and 6 right? – Govinda Sakhare Jan 28 '17 at 20:12

Since you do not know the Rational Root Test, let's consider a simpler case: the Integer Root Test.

If $\,f(x)= x^3+6x+4\,$ has an integer root $\,x=n\,$ then $\,n^3+6n+4 = 0\,$ so $\,(n^2+6)\,\color{#c00}{n = -4},\,$ hence $\,\color{#c00}{n\ \ {\rm divides}\ \ 4}.\,$ Testing all the divisors of $4$ shows that $2$ is root,  hence $\,x-2\,$ is a factor of $f$ by the Factor Theorem. The cofactor $\,f/(x-2)\,$ is computable by the Polynomial (long) Division algorithm (or even by undetermined coefficients).

Remark $\$ This is a very special case of general relations between the factorization of polynomials and the factorizations of their values. For example, one can derive relations between primality and compositeness of polynomials based on the same properties of their values. For example, since $\ 9^4\!+8\$ is prime so too is $\, x^4+8\,$ by Cohn's irreducibility test. See this answer and its links for some of these beautiful ideas of Bernoulli, Kronecker, and Schubert.

• brilliant, liked every bit of it, Sadly I can not unmark other answer and select this one coz that too is elegant answer. – Govinda Sakhare Jan 29 '17 at 16:50
• @piechuckerr No problem (i don't care about acceptance, votes, etc, only about sharing beautiful mathematics). – Bill Dubuque Jan 29 '17 at 17:05

Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!

A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$

Is to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\tag2$$ Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so\begin{align*} & a-b^2=A\tag3\\ & ab=-B\tag4\end{align*} Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.

Examples:

1. Solve $x^3-6x+4=0$ (your question)

Moving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\-2\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\implies x=2$$ Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.

1. Solving $x^3+16x=455$

A factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\cdot7=455$$ Therefore,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$

The rational root theorem gives a list of all possible rational roots of a polynomial with integer coefficients that have a given leading coefficient and a given constant coefficient. In this case, the leading coefficient is $1$ and the constant coefficient is $4.$ The theorem tells us that all rational roots are in the set $\left\{ \pm\dfrac 1 1, \pm\dfrac 2 1, \pm \dfrac 4 1 \right\},$ the numerator being in this the only divisor of the leading coefficient $1$ and the denominators being the divisors of the constant coefficient $4$. That doesn't mean there are rational roots; it only means there are not any that don't belong to this set. There are only six members of this set, so it's easy to plug in all of them and see if you get $0$. When you plug in $2$, you get $0$, so there's your factorization.

• OP does not know RRT - see the comments. – Bill Dubuque Jan 28 '17 at 19:59
• @BillDubuque : Fortunately I linked to the Wikipedia article about it. – Michael Hardy Jan 28 '17 at 19:59
• @BillDubuque : That he doesn't recognize that name of the theorem doesn't mean he doesn't know the theorem. He could know it by a different name. – Michael Hardy Jan 28 '17 at 20:01
• I thought the comment(s) might nudge someone to give an exposition on this simpler integer case. But no one did, so I added an answer doing so. – Bill Dubuque Jan 28 '17 at 20:40

If $P$ is a polynomial with real coefficients and if $a\in\mathbb{R}$ is a root, which means that $P(a)=0$, then there exists a real polynomial $Q$ such that $\forall x\in\mathbb{R},\quad P(x)=(x-a)\,Q(x)$.

On this case, you can see by inspection, that $P(2)=0$.

It remains to find real constants $A,B,C$ such that :

$$\forall x\in\mathbb{R},\quad x^3-6x+4=(x-2)(Ax^2+Bx+C)$$

Identification of coefficients leads to $A=1$, $-2C=4$ and, for example, $A-2B=0$ (equating the coeffts of $x^2$ in both sides).

• Since the OP doesn't know RRT, the answer I posted earlier seems to me a (not so bad) - one ! It's clear that one can answer at a higher level, but it's seems to me important to answer at a level compatible the mathematical background of the OP. I have the feeling that, if he (or she) had known the RRT, he would certainly never have asked that question. – Adren Jan 28 '17 at 20:11
• Actually we don't need the full RRT here, only a simpler integer case, e.g. see my answer. – Bill Dubuque Jan 28 '17 at 20:56


# How do you tackle this integral

1. Sep 6, 2013

### Gauss M.D.

1. The problem statement, all variables and given/known data

Find F(x)

f(x) = 1/(1+x2)2)

2. Relevant equations

3. The attempt at a solution

It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?

2. Sep 6, 2013

### Zondrina

Perfect candidate for a trig sub. Use $x = tan(θ)$ so that $dx = sec^2(θ)dθ$.

3. Sep 6, 2013

### ZeroPivot

well, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?

4. Sep 6, 2013

### Gauss M.D.

Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?

Last edited: Sep 6, 2013
5. Sep 6, 2013

### ZeroPivot

its a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?

6. Sep 6, 2013

### Gauss M.D.

No, I was referring to Zondrinas trig sub suggestion!

7. Sep 6, 2013

### Zondrina

I don't see any discontinuity in the integrand at all, so I don't think it would matter.

8. Sep 6, 2013

### Gauss M.D.

Right right, we're using tan and not some other trig function. My bad again. Thanks a ton :)

9. Sep 7, 2013

### verty

That is a very good question, I don't know the answer to it. I'm going to investigate this.

10. Sep 7, 2013

### verty

Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:

$y = \int {dx \over (1 - x^2)^2}$

$x =? \; sin\theta$
$dx = cos\theta \; d\theta$

$y = \int sec^3\theta \; d\theta$
$= {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C$
$= {1 \over 2} [sec\theta \; tan\theta + ln| sec\theta + tan\theta | ] + C$
$= {1\over 2}{x \over 1 - x^2} + {1\over 4} ln | {1+x\over 1-x}| + C$

On the other hand, one can make a translation:

$x = u-1$
$dx = du$

$y = \int {dx \over (1 - x^2)^2} = \int{ dx \over (1+x)^2 (1-x)^2} = \int {du \over u^2(2-u)^2}$

After some partial fraction magic:

$y = \int {1 \over 4(2-u)^2} + {1 \over 4u^2} + {1 \over 4u} + {1 \over 4(2-u)} du$
$= {1 \over 4} [ {1 \over 2-u} - {1\over u} + ln|u| - ln|2-u| ] + C$
$= {1 \over 4} [ {2(u-1) \over u(2-u)} + ln| {u \over 2-u} | ] + C$
$= {1 \over 2} {x \over 1 - x^2} + {1 \over 4} ln| {1+x \over 1-x} | + C$

I can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.