The only constraint is that, together, they sum to 360 deg. A Polygon is any flat shape with straight sides. Drag vertices to create irregular polygons. In the figure shown above, the measure of the exterior angle at vertex C is equal to the sum of the measures of the remote interior angles at vertices A and B. A trapezium has one pair of parallel sides. The exterior angles of a rectangle are each 90°. Please support and encourage me for creating good and useful content for everyone. Among them exterior angle of a regular polygon formula is one. In contrast, an exterior angle (also called an external angle or turning angle) is an angle formed by one side of a simple polygon and a line extended from an adjacent side. After Subscription please visit your email and activate it. Includes a worksheet with answers and a load of challenge questions from the UKMT papers. The sum of exterior angles in a polygon is always equal to 360 degrees. Figure out the number of sides, measure of each exterior angle, and the measure of the interior angle of any polygon. Exterior angles are created where a transversal crosses two (usually parallel) lines. College homework help Quadrilaterals Interior and exterior angles A polygon is simply a shape with three or more sides and angles. :. The sum of exterior angles of a polygon is 360°. These are not the reflex angle (greater than 180 °) created by rotating from the exterior of one side to the next. Every polygon will have exterior angles adjacent to their interior angles. An interior angle is an angle inside a shape. Answer. You will see that the angles combine to a full 360° circle. A polygon is a flat figure that is made up of three or more line segments and is enclosed. Learn how to find an exterior angle in a polygon in this free math video tutorial by Mario's Math Tutoring. For a positive directed simple polygon, convex positive angles are blue and concave ones are orange. Learn and know what is the formula for exterior angle of regular polygon. Corbettmaths Videos, worksheets, 5-a-day and much more. Let’s look at more example problems about interior and exterior angles of polygons. This has 1,2,3,4,5,6, sides and this has 1,2,3,4,5,6 sides. Pretty easy, huh? Yes, we can say what type of polygon. We know what is mean by a polygon? If we know exterior angle then can we say what type of polygon is it? MEDIUM. Fine-tune your skills using the angles in polygons worksheets with skills to find the sum of interior angles of regular and irregular polygons, find the measure of each interior and exterior angle and much more. Each interior angle of a regular polygon = n 1 8 0 o (n − 2) where n = number of sides of polygon Each exterior angle of a regular polygon = n 3 6 0 o According to question, n 3 6 0 o … Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. If a convex polygon is regular with “n” number of sides, then each exterior angle of a convex polygon is measured as 360°/n. … The Corbettmaths video tutorial on Angles in Polygons. Our tips from experts and exam survivors will help you through. Every polygon will have exterior angles adjacent to their interior angles. Exterior angles of a polygon are formed when by one of its side and extending the other side. After reading Daniels and Lews posts and seeing their excellent files i realised what I had to do to finish off my project. with the subscription you can get all my latest post updates. A lesson covering rules for finding interior and exterior angles in polygons. View Set. The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is. You will see that the angles combine to a full 360° circle. Includes a number of exercises for which solutions are in the slides. The exterior angles are the angles formed between a side-length and an extension. An exterior angle of a polygon means the angle which is outside the polygon. ACT Review - Math Formulas. Some of the worksheets for this concept are Interior and exterior angles of polygons, Interior angles of polygons and multiple choices, 6 polygons and angles, Infinite geometry, Work 1 revised convex polygons, 15 polygons mep y8 practice book b, 4 the exterior angle theorem, Mathematics linear 1ma0 angles polygons. Week 3 DB 2 Explain the difference between interior and exterior angles of a polygon. The sum of exterior angles in a polygon is always equal to 360 degrees. Recently I have created a YouTube Channel called Murali Maths Class, check for the latest Maths Videos on All the topics. The question can be answered only if the 20-gon is regular - ie all its angles are the same. The exterior angle of a regular polygon is our fourth of its interior angle. Some additional information: The polygon has 360/72 = 5 sides, each side = s. It is a regular pentagon. Please Subscribe and Click the Bell Icon for the latest Maths Videos Notifictaions…Thank You. And also the formula for the exterior angle of a regular polygon. 4.8 44 customer reviews. Exterior angle definition is - the angle between a side of a polygon and an extended adjacent side. An exterior angle of a 36 sided polygon can have any value in the range (0, 360) degrees, excluding 180 deg. The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles of the triangle. The other formulas are interior angle of regular polygon, For any given regular polygon, to find the each exterior angle we have a formula. The sum of the exterior angles of convex polygons is 360°. As a demonstration of this, drag any vertex towards the center of the polygon. As we can see in the figure... For a triangle, angle 1, 2, 3 are exterior angles of triangle ABC. The exterior angles of a square are each 90°. For example, a six-sided polygon is a hexagon, and a three-sided one is a triangle. Read more. The exterior angle sum theorem states that the sum of the exterior angles of a convex polygon is 360°. Always. Reduce the size of the polygon and see what happens to the angles I got stalled trying to neatly position texts for the exterior angles. The interior angles of an irregular 6-sided polygon are; 80°, 130°, 102°, 36°, x° and 146°. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. : pp. The exterior angle of a polygon is defined as the angle formed bt extending the sides of a polygon. a demonstration on the sum of exterior angles of any polygon The exterior angle of the regular polygon with 24 sides is given as the \frac { { 360 }^{ 0 } }{ 24 }   = { 15 }^{ 0 } . The exterior angle of the regular octagon is given as the \frac { { 360 }^{ 0 } }{ 8 }   = { 45 }^{ 0 } . 5-a-day GCSE 9-1; 5-a-day Primary ; 5-a-day Further Maths; 5-a-day GCSE A*-G; 5-a-day Core 1; More. IF that is the case, then: The sum of the exterior angles of any polygon is 360 degrees. Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180°.They are "Supplementary Angles". The formula to find the sum of the interior angles of any polygon is sum of angles = (n - 2)180° , where n is the number of sides of the polygon.The sum of exterior angles of any polygon is 360º.. Read the lesson on angles of a polygon for more information and examples. Calculate the size of angle x in the polygon. The exterior angle of the regular pentagon is given as the \frac { { 360 }^{ 0 } }{ 5 }   = { 72 }^{ 0 } . Now that you’re an expert at finding the sum of the interior and exterior angles of a polygon, how might this concept be tested on the GMAT? The sum is always 360°. A polygon has exactly one internal angle per vertex. As a demonstration of this, drag any vertex towards the center of the polygon. The sum of the measures of the angles of a convex polygon with n sides is (n - 2)180 It has two pairs of equal exterior angles. For a regular polygon, the size of each exterior angle, #theta# can be found from: #theta = (360°)/n" "larr# where n = number of sides Using this property, if you know the size of the exterior angle, you can find the number of sides. Khan Academy is a 501(c)(3) nonprofit organization. The measure of each interior angle of an equiangular n -gon is. The sum of exterior angles in a polygon is always equal to 360 degrees. Exterior Angle : An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. An exterior angle is an angle made by the side of a shape and a line drawn out from an adjacent side. Polygons are 2-dimensional shapes with straight sides. Objective: I know how to calculate the interior and exterior angles of polygons. Rule: The sum of the exterior angles of a polygon is 360°. Polygons are classified by their number of sides. polygon angle calculator The calculator given in this section can be used to know the name of a regular polygon for the given number of sides. Generally, If we extend the any one line segment associated to interior angle we will get the exterior angle. As you can see, for regular polygons all the exterior angles are the same, and like all polygons they add to 360° (see note below). The exterior angles of a polygon always add up to #360# degrees. sheehy7math. The interior angles of a shape are the angles inside the shape. Preview. Covers all aspects of the GCSE9-1 syllabus. Sector, segment and arc - Higher only – WJEC, Circles - Intermediate & Higher tier – WJEC, Home Economics: Food and Nutrition (CCEA). By using this formula, easily we can find the exterior angle of regular polygon. So from this number of sides, easily we can say the type of the polygon. Measure of a Single Exterior Angle Formula to find 1 angle of a regular convex polygon of n … Hi, am Murali a Mathematics blogger. ACT Review - Math Formulas. The sum of all the exterior angles in a polygon is equal to 360 degrees. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. finding exterior angles of a polygon worksheet, Angles in Polygons Worksheets. The sum of the exterior angles at each vertex of a polygon measures 360 o. Divide 360 by the number of sides, to figure out the size of each exterior angle in this unit of regular polygons pdf worksheets for 8th grade and high school students. For instance, in an equilateral triangle, the exterior angle is not 360° - 60° = 300°, as if we were rotating from one side all the way around the vertex to the other side. Interior and exterior angle formulas: The sum of the measures of the interior angles of a polygon with n sides is ( n – 2)180. Given this a regular polygon, all the angles are equal and all the sides are equal. Convex case. Among them exterior angle of a regular polygon formula is one. Polygons. 360 ° / n. Note : This calculator will work only for regular polygons. If it is a Regular Polygon (all sides are equal, all angles are equal) Shape Sides Sum of Interior Angles Shape Each Angle; Triangle: 3: 180° 60° Quadrilateral: 4: 360° 90° Pentagon: 5: 540° 108° Hexagon: 6: 720° 120° Heptagon (or Septagon) 7: 900° 128.57...° Octagon: 8: 1080° 135° Nonagon: 9: 1260° 140°..... Any Polygon: n (n−2) × 180° (n−2) × 180° / n These 2 tutorials and 2 worksheets can be used to develop formulae that connect the number of sides, interior angle and exterior angle of a regular polygon the sum of interior and exterior angles in any polygon. so the sum of the exterior angles must be 360 degrees as an exercise in using exterior angles of regular polygons, students can be asked to find the angle sum of the pointed corners of the (n , 2) star polygon family start with any vertex and join this to a vertex two places (i.e. Each pair of these angles are outside the parallel lines, and on the same side of the transversal. Shapes, geometry and angles are important areas of mathematics. The Corbettmaths Practice Questions on Angles in Polygons. A polygon is simply a shape with three or more sides and angles. Exterior angle: An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. WORKSHEET: Angles of Polygons - Review PERIOD: DATE: USING THE INTERIOR & EXTERIOR ANGLE SUM THEOREMS 1) The measure of one exterior angle of a regular polygon is given. A regular 6-sided polygon has exterior angles of 60o(360o/6)If it is not regular, and one interiorangleis 140o, then the exterior angle at that vertex is 40o (180-40). exterior angles the angles outside a polygon that are adjacent to the interior angles indirect measurement uses similar figures to find a missing measure when it is difficult to find directly interior angles the angles inside a polygon +17 more terms. Interior Angle : An interior angle of a polygon is an angle inside the polygon at one of its vertices. Exterior Angles of a transversal. the sum of the exterior angles is ALWAYS 360° So you can find the size of the exterior angles of a regular polygon quite easily: If there are 18 sides (n=18), then each exterior angle is: (360°)/n = (360°)/18 = 20° The sum of the exterior and interior angles is 180° because they are adjacent angles on a straight line. (adsbygoogle = window.adsbygoogle || []).push({}); Help With Math For any given regular polygon, to find the each exterior angle we have a formula. The sum of the exterior angles of any polygon is 360 degrees. Sign in, choose your GCSE subjects and see content that's tailored for you. I was previosly trying to develop a geogebra file for demonstrating the concept regarding exterior angles of a polygon. The Corbettmaths video tutorial on Angles in Polygons. Polygon explorer to learn about the properties of regular polygons. Here is a complete lesson on calculating the interior and exterior angles of a polygon. Number of interior angles and number of exterior angles will be equal and this is equal to number of sides of a polygon. The sum of the exterior angles of a polygon is 360°. So each interior angle = 180–72 = 108 deg. Exterior angles of a polygon have several unique properties. The formula for calculating the size of an exterior angle is: exterior angle of a polygon = 360 ÷ number of sides. Calculate angles of regular and irregular polygons and create tessellations and tiling patterns. And for quadrilateral Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180°. 261–264 160° An alternative method is to use the exterior angle. Rule: Interior and exterior angles add up to 180\degree 180°. Exterior Angle The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. Solution. Notice that corresponding interior and exterior angles are supplementary (add to 180°). Exterior-angle 1. With respect to polygon, we have four important formulas. positive, angle and orange ones are vice versa. A parallelogram has two pairs of equal sides. Angle Q is an interior angle of quadrilateral QUAD. Interior angle of a polygon is that angle formed at the point of contact of any two adjacent sides of a polygon. Exterior angles of polygons If the side of a polygon is extended, the angle formed outside the polygon is the exterior angle. The diagonals bisect each other at right angles. Exterior Angle of Regular Polygons. How many sides does the polygon have? The sum of the interior angles = 5*108 = 540 deg. If you find the ratio of { 360 }^{ 0 } and exterior angle, then you will get number of sides. As you can see, for regular polygons all the exterior angles are the same, and like all polygons they add to 360° (see note below). The sum of the degree measures of the exterior angles of a convex polygon is always equal to 360°. The sum of the exterior angles for each polygon is consistent for all types of polygons whether they are regular or irregular, large or small -- no matter how many sides. The exterior angle of a regular polygon = 72 deg. So each exterior angle is 360 divided by the n, the number of sides. Sum of exterior angles of a polygon. This is concave, sorry this is a convex polygon, this is concave polygon, All you have to remember is kind of cave in words And so, what we just did is applied to any exterior angle of any convex polygon. A rectangle has two pairs of equal sides. Find the nmnbar of sides for each, a) 72° b) 40° 2) Find the measure of an interior and an exterior angle of a regular 46-gon. Menu Skip to content. The sum of the exterior angles of a polygon is 360°. The exterior angle of the regular polygon with 16 sides is given as the \frac { { 360 }^{ 0 } }{ 16 }   = { 22.5 }^{ 0 } . Example 1. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. The formula . Exterior angles of a polygon have several unique properties. tells you the sum of the interior angles of a polygon, where n represents the number of sides. An exterior angle of a polygon is an angle at a vertex of the polygon, outside the polygon, formed by one side and the extension of an adjacent side. If every internal angle of a simple polygon is less than 180°, the polygon is called convex. The ratio between the exterior angle and interior angle of a regular polygon is 2: 3. The number of sides is therefore # 360/15 = … The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360°. Thesishelpers.com. are 2D shapes with four sides and angles. Welcome; Videos and Worksheets; Primary; 5-a-day. Exterior angles of a polygon have several unique properties. 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Does the Cantor set have the cardinality of the continuum?

I saw somewhere that there are no sets between $$\mathbb Q$$ and $$\mathbb R$$ in the sense that there are no set $$S\subset \mathbb R$$ s.t. $$|\mathbb Q|<|S|$$ but $$|S|<|\mathbb R|$$, i.e. all set $$S\subset \mathbb R$$ s.t. $$|\mathbb Q|<|S|$$ should have the cardinality of the continuum. Now, what about the Cantor set ? It's a set of measure $$0$$, but it's uncountable. Since it's uncountable, there are no bijection with $$\mathbb Q$$, but on the other hand, a set of measure $$0$$ that has a bijection with $$\mathbb R$$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ?

• (1) The statement that "there are no sets between $\mathbb{Q}$ and $\mathbb{R}$" in the sense you write is called the "Continuum Hypothesis". It is an statement that is independent from regular set theory (can neither be proven nor disproven), just like the parallel postulate is independent from the remaining geometric axioms. You can work in theories where it is true that no such sets exist, and you can work in theories where it is false that no such sets exist. (2) As for the Cantor set, it definitely has the cardinality of $\mathbb{R}$.(cont) – Arturo Magidin Mar 21 at 16:56
• Elements of the Cantor set are precisely those that have a ternary (base 3) expansion that does not contain any 1s. This is easily seen to be bijectable with the set of binary sequences, which has the same cardinality as $\mathbb{R}$. Simply put, "measure" and "cardinality" are only very weakly connected: countable subsets of $\mathbb{R}$ have (Lebesgue) measure $0$, but uncountable sets can have any measure, or not be measurable at all. – Arturo Magidin Mar 21 at 16:57
• @ArturoMagidin: I thought every set with cardinality less that $\mathbb R$ had Lebesgue measure $0$? Do I misremember? – celtschk Mar 21 at 19:00
• @celtschk that is precisely what Arturo Magidin said. If a set has cardinality less than $\mathbb R$ it has Lebesue measure $0$. If the set has cardinality of $\mathbb R$ it doesn't have to have measure $0$.... but it could. That Cantor set is uncountable with measure $0$. $[0,1]$ is uncountable with measure $1$. $\mathbb R$ is uncountable with infinite measure. and so on. – fleablood Mar 21 at 19:19
• @celtschk: Assuming the Continuum Hypothesis, any set with cardinality less than $|\mathbb{R}|$ is countable, hence has Lebesgue measure zero. – Arturo Magidin Mar 21 at 19:21

Yes. Cantor set has cardinality of the reals (continuum).

As Cantor Set $$\subset \mathbb R$$ it's cardinality is at most $$|\mathbb R|$$ and as it is uncountable it's reasonable that we can't have found a contradiction to "Continuum Hypothesis" and found a cardinality between $$|\mathbb Q|$$ and $$|\mathbb R|$$ so it reasonable that Cantor set has the cardinality of the reals.

But to seal the deal we need a bijection between Cantor set and $$\mathbb R$$.

Following a comment by Arturo Magidin:

If $$x \in [0,1]$$ then $$x = \sum\limits_{i=0}^{\infty} b_i 3^{-i}$$ for some sequence of $$b_i$$ where each $$b_i=0,1,2$$. If we disallow infinite tailing $$0$$s then this sequence is unique. This is just writing $$x$$ is decimal in base $$3$$. But where all terminating decimals are replaced with tailing $$2$$s.

Likewise if $$y \in [0,1]$$ then $$y = \sum\limits_{i=0}^{\infty} c_i 2^{-1}$$ for some sequence of $$c_i = 0,1$$. And if we disallow infinite tailing $$0$$s 0 this sequence is unique. This is just the base $$2$$ decimal.

If $$x = \sum b_i 3^{-i}$$ is in the Cantor set then none of the $$b_i = 1$$. That is because we removed the middle third of all segments and $$b_k = 1$$ means $$\sum\limits_{i=0}^{k-1} b_i 3^{-i} < x < \sum\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3^{-k}$$ would mean $$x$$ is in some middle third.

So let $$f(\sum b_i 3^{-i}) = \sum c_i 2^{-i}$$ where if $$b_i = 0$$ then $$c_i = 0$$ and if $$b_i = 2$$ then $$c_i = 1$$. $$f$$ is a bijection between the Cantor set and $$[0,1]$$.

but on the other hand, a set of measure 0 that has a bijection with R looks very strange as well.

Ah.... not really. It seems counterintuitive because ... to have measure $$0$$ no two points can be connected in the set so $$\color{red}{\text{for any point there must be a measurable distance before the "next" one}}$$ and there can only be countably many such points. But that clause in $$\color{red}{\text{red}}$$ is completely erroneous and is based on a naive concept of numbers must "follow each other". Uncountable numbers don't.

And the Cantor set exists merely to be a simple counter example.

• Comments are not for extended discussion; this conversation has been moved to chat. – Pedro Tamaroff Mar 23 at 13:00


# Absolute value inequality for complex numbers

I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution.

Does the inequality $$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$ hold for all complex numbers $a,b,c$ ? For real values a case analysis will verify the inequality. What is desired is a proof using the triangle inequality or a counterexample. Thanks in advance.

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See answers and comments at math.stackexchange.com/questions/793905/…. –  Dietrich Burde May 20 '14 at 12:35
You cannot prove it using just the triangle inequality, because it fails in $\mathbb R^3$ with the $l_\infty$ norm: just take the standard basis vectors for $a,b,c$. You’ll probably need to use that $\mathbb C$ is an inner product space. –  Emil Jeřábek May 20 '14 at 13:35
Once you have it for $\ell_1^n$ for all $n$ you have it for $L_1(0,1)$ by approximation. Once you have it for $L_1(0,1)$ you have it for Hilbert spaces because $\ell_2$ embeds isometrically into $L_1(0,1)$ (as the span of IID $N(0,1)$ random variables). –  Bill Johnson May 20 '14 at 14:22
If you want to be more sophisticated, once you have it for some infinite dimensional space you have it for Hilbert spaces by Dvoretzky's theorem. –  Bill Johnson May 20 '14 at 14:23
Even more sophisticated is that every two dimensional real Banach space embeds isometrically into L$_1(0,1)$, so the inequality is true in all two dimensional Banach spaces. –  Bill Johnson May 20 '14 at 14:38

It seems that your inequality is just an incarnation of Hlawka's inequality which says that for any vectors $x, y, z$ in an inner product space $V$ we have

\begin{equation*} \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|. \end{equation*}

Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP.

To add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by Zurab below). The paper linked to above explores the interesting generalization: \begin{equation*} f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z), \end{equation*} where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.

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Do you know a proof of it? –  Qiaochu Yuan May 21 '14 at 0:46
For a proof please see: books.google.com/… . That link also mentions extension to Banach spaces by Lindenstrauss and Pelcynski (under certain embedability assumptions) –  Suvrit May 21 '14 at 0:50
It looks like Lindenstrauss & Pelczynski had in mind the same observations I made in comments above. That approach for extending inequalities from the real line to $L_p$ spaces has of course been around for a long time. –  Bill Johnson May 21 '14 at 4:28

In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this" = inequality where every term is the length of some linear combination of variable vectors in the space; here the vectors are $a,b,c$.) In the case of complex numbers we have $$|z| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d\theta.$$ Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces the desired inequality to the one-dimensional case. In $d$-dimensional space we'd write $C\|z\|$ as an average of $|u \cdot z|$ over $u$ in the unit sphere (for a suitable constant $C>0$).

I learned this trick at MOP 30+ years ago, and don't know or remember who discovered it. I didn't even know that the specific inequality we were assigned was due to Hlawka (if I remember right that it was the inequality $$\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|$$ quoted by Suvrit). We were shown the averaging solution after laboring to prove it bare-handed. The reference Suvrit cites does not use the averaging method, so I do not know whether it too is due to Hlawka or to another mathematician.

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Notice that $T:\ell_2^n \to L_1(S^{n-1})$ defined by $(Tx)(y):= \langle x, y \rangle$ is another (multiple of an) isometric embedding of an $n$ dimensional Hilbert space into $L_1$. So at the appropriate conceptual level, the two proofs are basically the same. –  Bill Johnson May 21 '14 at 6:35
Very nice trick! +1 –  Malik Younsi May 21 '14 at 14:09

In fact the Hlawka's inequality first appeared (as a special case of more general result) in H. Hornich, Eine Ungleichung für Vektorlängen, Mathematische Zeitschrift 48 (1942), 268-274 http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0048&DMDID=DMDLOG_0025&LOGID=LOG_0025&PHYSID=PHYS_0256 (see p. 268. P.S. as Joni Teräväinen has remarked, Hornich credits on page 274 to Hlawka an algebraic proof of this special case and reproduces it).

Hlawka's original proof, besides the book indicated by Suvrit, can be found in "Classical and New Inequalities in Analysis" by D.S. Mitrinovic, J. Pecaric and A.M Fink, p. 521 and in "Analytic Inequalities" by D.S. Mitrinovic, p.171. Both books provide Adamovic and Djorkovic generalizations of the Hlawka's inequality. Interestingly, all these generalizations are special cases of more general result given in http://www.sciencedirect.com/science/article/pii/S0022247X96904588 (Generalizations of Dobrushin's Inequalities and Applications, by M. Radulescu and S. Radulescu).

Another proof of Hlawka's inequality can be found in http://www.sbc.org.pl/Content/34160/1995_13.pdf (On two geometric inequalities, by A. Simon, P. Volkmann), and still another one in http://www.jstor.org/discover/10.2307/2310890?uid=3738936&uid=2&uid=4&sid=21104051771107 (The Polygonal Inequalities, by D.M. Smiley and M.F. Smiley).

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On page 274 it says that Hlawka had already proved the special case which is this inequality. I'm not sure though if that was published. –  Joni Teräväinen May 21 '14 at 11:38


# Math Help - Is emptyset a function?

1. ## Is emptyset a function?

My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).

So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?

(1) Is it a function? If not, explain why.
(2) If yes, what are it's domain and range?
(3) Is the function one-to-one? If not, explain why.
(4) If yes, what is the inverse function?

f. $f=\emptyset$
(1) Yes (trivially), because there are no ordered pairs in $f$, it does not violate the definition of function.
(2) dom $f$ = im $f$ = $\emptyset$
(3) Yes (trivially), since the definition of one-to-one is not violated
(4) $f^{-1}=\emptyset$

Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.

2. Originally Posted by MSUMathStdnt
My discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \in f$ and $(a,c) \in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).

So, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?
I wouldn't put it as "not violating" but rather as "does fulfill."

When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

Originally Posted by MSUMathStdnt
Above is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.
All correct, except I would modify (1) as I mentioned above.

3. Originally Posted by MoeBlee
I wouldn't put it as "not violating" but rather as "does fulfill."

When we say "S is a set of ordered pairs" we mean "For all x, if x in S, then x is an ordered pair." Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.

All correct, except I would modify (1) as I mentioned above.
I understand what you're saying. But I still don't see how to word it (although I'll probably get full credit as long as I've got the idea right). How does this sound:

(1) Yes (trivially). There are no ordered pairs in $f$, therefore; for all ordered pairs in $f$, there are none that have the same first value and a different second value.

4. This reminds me of a thread I saw recently.
http://mymathforum.com/viewtopic.php?f=22&t=18683

5. But you missed mentioning that every member of the empty set is an ordered pair.

I'll do it in English [where '0' stands for the empty set]:

For all x, if x is in 0 then, x is an ordered pair. So 0 is a relation. And for all x, y, z, if <x y> and <x z> are in 0 then y=z. So 0 is a relation that is moreover a function.

In symbols:

Ax(x in 0 -> x is an ordered pair).
So 0 is a relation.
Axyz((<x y> in 0 & <x z> in 0>) -> y=z).
So 0 is a function.

/

If you want to get more detailed, you can mention that

Ax(x in 0 -> x is an ordered pair)

Axyz((<x y> in 0 & <x z> in 0>) -> y=z)

are true because the antecedent in each is false.


# Why null space and column space?

I am not asking this question for WHAT is null space or WHAT is column space. I have finished learning about the definitions of these two concepts for a while. However, to install these concepts in my mind forever, I really want to know what the purposes are for null space and column space of a vector.

Thanks!

• Null space. They are the solutions to the equation $Ax=0$ where A and x are matrices. It's like asking why is $x= \frac{-b +-\sqrt{b^2-4ac}}{2a}$ solutions important. – The Artist Oct 23 '14 at 8:29
• math.stackexchange.com/questions/21131/… This will be helpful. Meaning of Null space is asked here. – The Artist Oct 23 '14 at 8:42
• I think you mean "null space and column space of a matrix," Justin. – Gerry Myerson Oct 23 '14 at 9:19

Perhaps an example will clarify things.

Let's suppose that the matrix A represents a physical system. As an example, let's assume our system is a rocket, and A is a matrix representing the directions we can go based on our thrusters. So what do the null space and the column space represent?

Well let's suppose we have a direction that we're interested in. Is it in our column space? If so, then we can move in that direction. The column space is the set of directions that we can achieve based on our thrusters. Let's suppose that we have three thrusters equally spaced around our rocket. If they're all perfectly functional then we can move in any direction. In this case our column space is the entire range. But what happens when a thruster breaks? Now we've only got two thrusters. Our linear system will have changed (the matrix A will be different), and our column space will be reduced.

What's the null space? The null space are the set of thruster intructions that completely waste fuel. They're the set of instructions where our thrusters will thrust, but the direction will not be changed at all.

Another example: Perhaps A can represent a rate of return on investments. The range are all the rates of return that are achievable. The null space are all the investments that can be made that wouldn't change the rate of return at all.

Another example: room illumination. The range of A represents the area of the room that can be illuminated. The null space of A represents the power we can apply to lamps that don't change the illumination in the room at all.

Good luck!

• best answer ever :D – Justin Chan Oct 24 '14 at 1:37
• @JustinChan Thanks man! If you'd like to know more about applications of Linear Algebra like the ones I've described, reviewing the online lectures of Linear Dynamical Systems by Stephen Boyd of Stanford may be of interest after your Linear Algebra class. – NicNic8 Oct 24 '14 at 1:44
• by the way, I was looking at some questions that are quite similar to mine, and I used your answer as a quote! Hope you don't mind :D – Justin Chan Oct 24 '14 at 1:58
• @JustinChan Of course not! It's very flattering. I'm happy to help. – NicNic8 Oct 24 '14 at 3:03
• According to your explanation, there's no way Range(A) could possibly equal to Null(A), right? I also tried thinking if it's possible for Range(A) = Null(transpose(A)) I think the answers to both questions are no. – Justin Chan Oct 24 '14 at 21:11


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T B [3] Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.[5]:p. enl. {\displaystyle \triangle ABC} Pedoe, D. Circles: The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the (Kimberling 1998, pp. The touchpoint opposite C Circumcircle of a triangle. {\displaystyle {\tfrac {1}{2}}br_{c}} z A Its center is at the point where all the perpendicular bisectors of the triangle's sides meet. ) is defined by the three touchpoints of the incircle on the three sides. and s Δ A A ( Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd the length of {\displaystyle T_{B}} and {\displaystyle \Delta } {\displaystyle x:y:z} Modern Geometry: The Straight Line and Circle. , the circumradius C , {\displaystyle \triangle ABC} B where 08, Apr 17. 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# A question on function equality

1. Jul 23, 2008

### zenctheo

Hello to every one!
I have a question that came up when I was talking with a fellow mathematician.
I used to say that two functions are equal when the have the same formula and the same domain and codomain.
We read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same X.
For example
$$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \in \{0,1\}$$because f(0)=g(0)=0 and f(1)=g(1) even though their formula is different.
I thought that this definition of equality is incomplete because by saying that f(x)=g(x) then

$$\frac{df}{dx}=\frac{dg}{dx}$$ but on point x=1 $$\frac{df}{dx}=2$$ and $$\frac{dg}{dx}=3$$.

Thus we derive two different results from to equal quantities. Therefore two functions in order to be equal should also have the same formula.
Can you please give any insight on this?
Akis

2. Jul 23, 2008

### Ben Niehoff

The derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative:

$$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$

But for nearly all $\Delta x$, $x + \Delta x$ lies outside your domain. Therefore, you can't take the limit. :)

So, you are correct: Two functions are equal if and only if they have the same domain and their values are equal at every point within the domain.

3. Jul 23, 2008

### zenctheo

Thanks a lot for the reply.
You that I am wrong because I was the one saying that the functions should also have the same formula.
In order to get things straight: You mean that the above two functions are equal.... or not?

4. Jul 23, 2008

### LukeD

The functions are in fact equal. Also, as Ben said, those functions don't have derivatives because they're not defined on an open interval of the real numbers.

As another example, would you consider these to be the same function?
Let's say f and g are functions from the real numbers to the real numbers defined as

f(x) = x
g(x) = x when x^2 >= 0 and -x when x^2 < 0

Since the functions are only defined on the real numbers, there are no points where they'd differ.

On a related note: "Having the same formula" is not a well-defined concept. Most (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas.

5. Jul 23, 2008

### zenctheo

Ok. It's nice to learn a new thing. Even if I am proven wrong

Thanks a lot.


+0

Difficult logarithmic equation without a calculator

0
682
9

The following equation needs to be done without the use of a calculator and for the life of me I can't figure out how. If anyone has an idea of how to do it, please let me know. Log bases are in square brackets here.

4^(0.5log[4](9) - 0.25log[2](25))

The answer is apparently 3/5 but I can't figure out how to do it without the use of a calculator. Thanks.

Guest Jul 17, 2014

#2
+93305
+13

I really appreciate how well you have presented your problem.

People often leave brackets out with questions like these and they become very ambiguous.

I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct.

4^(0.5log[4](9) - 0.25log[2](25))

$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\ =4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;3-0.5log_2\;5}\\\\$$

Now, I can't do this unless I can get the bases the same.

$$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^y\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$

-------------------------------

so

$$=4^{log_{4}\;3-0.5log_2\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$

calculator check - using the web2 site calculator.

$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$

The calc has a little rounding error - the answers are the same.

Melody  Jul 17, 2014
#1
+1314
0

4(0.5log49 - 0.25log225)

Is the above the right equation?

Stu  Jul 17, 2014
#2
+93305
+13

I really appreciate how well you have presented your problem.

People often leave brackets out with questions like these and they become very ambiguous.

I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct.

4^(0.5log[4](9) - 0.25log[2](25))

$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\ =4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\ =4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\ =4^{log_{4}\;3-log_2\;5^{0.5}}\\\\ =4^{log_{4}\;3-0.5log_2\;5}\\\\$$

Now, I can't do this unless I can get the bases the same.

$$\begin{array}{rll} let\;\; y&=&log_2 5\\\\ 5&=&2^y\\\\ 5&=&4^{0.5y}\\\\ log_4 5&=&log_4 4^{0.5y}\\\\ log_4 5&=&0.5ylog_4 4\\\\ log_4 5&=&0.5y\\\\ y&=&2log_4 5\\\\ log_2 5&=&2log_4 5\\\\ \end{array}$$

-------------------------------

so

$$=4^{log_{4}\;3-0.5log_2\;5}\\\\ =4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\ =4^{log_{4}\;3-log_4\;5}\\\\ =4^{log_{4}\;(3/5)}\\\\ =\frac{3}{5}$$

calculator check - using the web2 site calculator.

$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$

The calc has a little rounding error - the answers are the same.

Melody  Jul 17, 2014
#3
+1314
0

I think there is easier way. I'm going to work on this for my own revision. Those logs are what I always forget how to and what to and where to, but don't wait for an answer. Good luck.

Stu  Jul 17, 2014
#4
0

@Melody

Jesus, that was amazing. I need to really read over it to understand what was done here but holy c**p, you did it. It's possible. Thanks so, so much :)

Guest Jul 17, 2014
#5
+93305
0

You are very welcome

I really appreciate your enthusiasm but even so a little less swearing would be good.

I am sure that you do not want to offend anyone.

Melody  Jul 17, 2014
#6
0

Melody - noted, and thanks again

Stu - if you have an easier way I'd really like to see it so I'll stay posted on this page if you ever get around to it.

Guest Jul 17, 2014
#7
+26971
+10

Here's a slightly different approach, though it ultimately amounts to the same as Melody's:

Alan  Jul 17, 2014
#8
+88848
+10

Here's my (belated) take on this one:

We can write:

4^(0.5log4(9) - 0.25log2(25))  ...as.....

[4^log4(3)] / [ 4^ log2(5)(.5)]

4log4(3)/ [4(log2(5)/2)]    ......   the numerator simplifies to 3

Note that log2(5) is just a number.....call it "a'  ....so we have

4(a/2) = [4^(1/2)]^a = 2(a) =

2log2 5 = 5

So our answer is just  .....  3/5

CPhill  Jul 18, 2014
#9
0

Sorry I haven’t gotten back sooner. I was so amazed by Melody’s answer I didn’t check back to see if anyone else posted. Two more great answers.

Melody’s was like a dissection to find the answer. Alan’s was like a resection.

CPhill’s answer was like he chopped it up with a machete and found the answer hidden inside. Jesus Christ, CPhill, you really are fucking amazing! You are like the guy I watched whacking the s**t out of a coconut with an axe and rock, then after a few minutes, my grandmother’s face appeared.

Grandma didn’t think it looked like her but everyone else did. After she bitched for awhile, my brother says, well Grandma, if you’d go to a plastic surgeon he might make you look as good as the coconut. My other brother says, s***w that, send her to the guy with the axe and rock, he does great work and he’s cheap. Everyone thinks this is hilarious, except Grandma, of course. She got really pissed about it.

If I, or any of my friends have another problem like this, you can bet your sweet a*s I’ll send them here.

Thank you all, so very much.

Charlotte

Guest Aug 7, 2014


Back to tutorial index

# Finite precision point arithmetic

## Introduction

We all know that when we use the value of pi in Matlab, or compute cos(3.4), we are not getting the "exact" value of $\pi$ or the cosine function, but rather an approximation. In fact, we don't expect to be able to compute any irrational function to all of its digits, if only because we know that such values are nonterminating, non-repeating decimals values. In typical practical situations, our answers will not be significantly affected by the approximations that are made. What may be suprising, however, is that even typical "rational" numbers, such as 0.1 are only approximately represented on the computer.

In this lab, we will explore the number system represented by floating point arithmetic, and discuss some of the consequences for scientific computing. The ideas presented here extend to most modern computing systems, not just Matlab.

clear all
format long e

Back to the top

## Examples : Floating point arithmetic

### Example 1

Compute the following.

$$x = 0.1 + 0.1 + 0.1 + ... + 0.1 \qquad \mbox{(10 times)}$$

x = 0;
for i = 1:10,
x = x + 0.1;
end

What is the difference between $x$ and 1?

fprintf('%g\n',abs(x-1))
1.11022e-16

### Example 2

Compute $x$.

$$x = 2 - 3\left(\frac{4}{3} - 1\right)$$

% Algebra tells us that this should be 1
x = 2 - 3*(4/3 - 1);

What is the difference between $x$ and 1?

fprintf('%g\n',abs(x-1));
2.22045e-16

### Example 3

Verify the following exact mathematical expression for the value $a = 0.3$.

$$1 + a + a^2 + a^3 + a^5 = \frac{1-a^6}{1-a}$$

a = 0.3;
S_left = sum(a.^(0:5));
S_right = (1-a^6)/(1-a);

How close are the left and right sides of this expression?

fprintf('%g\n',abs(S_left - S_right));
2.22045e-16

### Example 4

For the function $f(x) = x$, we can compute the derivative $f'(x)$ exactly using the formula $$\begin{eqnarray*} f'(x) & = & \frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x}{h} = 1 \end{eqnarray*}$$

Verify this formula for $x = 1$ and $h = 10^{-3}$.

% Compute the derivative of f(x) = x using the secant method
x = 1;
h = 1e-3;
dfdx = ((x + h) - x)/h;    % This should be exactly 1

Do we get $f'(x) = 1$?

fprintf('%g\n',abs(dfdx-1));
1.10134e-13

Back to the top

## More examples : Is floating point arithmetic commutative and associative?

You may recall from your first introduction to algebra that we can often re-arrange the order of operations like addition or multiplication. This communative property of these operations allows us to write $$a + b + c = a + c + b = b + c + a$$

and so on. Unfortunately, this is not always true for floating point arithmetic.

The associative property that we learned in algebra, i.e. that $$a + (b + c) = (a + c) + b$$

may not always hold either.

### Example 5

The order in which we add numbers can matter. Consider these two expressions $$x = 10^{16} + 1 - 10^{16}$$ $$y = 10^{16} - 10^{16} + 1$$ $$z = 10^{16} - (10^{16} - 1)$$

x = 1e16 + 1 - 1e16;
y = 1e16 - 1e16 + 1;
z = 1e16 - (1e16 - 1);  % Test the associative property

Are $x$, $y$ and $y$ equal?

fprintf('x = %g\n',x);
fprintf('y = %g\n',y);
fprintf('z = %g\n',z);
x = 0
y = 1
z = 0

### Example 6

The above can happen for much smaller values as well

x = 1 + 0.1 - 1;
y = 1 - 1 + 0.1;
z = 1 - (1 - 0.1);

Comparing the differences of these three values, we get

fprintf('x-y = %g\n',x-y);
fprintf('y-z = %g\n',y-z);
fprintf('x-z = %g\n',x-z);
x-y = 8.32667e-17
y-z = 2.77556e-17
x-z = 1.11022e-16

### Example 7

In this example, we add up 100 random numbers in different orders.

rand('seed',1110);          % Get the same random numbers each time
x = rand(100,1);
xperm = x(randperm(100));   % Permute the values in array x
sum_orig = sum(x);
sum_perm = sum(xperm);

Compare the sum of the original array and the permuted array.

fprintf('%g\n',abs(sum_orig-sum_perm));
1.42109e-14

Back to the top

## Examples (modified)

In some of the examples above, a slight change of the constants involved can fix the problems that we saw above. Below are modified versions of some of the above examples.

### Example 1 (modified)

Compute the following.

$$x = 0.125 + 0.125 + 0.125 + ... + 0.125 \qquad \mbox{(8 times)}$$

x = 0;
for i = 1:8,
x = x + 0.125;
end

What is the difference between $x$ and 1?

fprintf('%g\n',abs(x-1))
0

### Example 2 (modified)

Compute $x$.

$$x = 2 - 2\left(\frac{3}{2} - 1\right)$$

% Algebra tells us that this should be 1
x = 2 - 2*(3/2 - 1);

What is the difference between $x$ and 1?

fprintf('%g\n',abs(x-1));
0

### Example 3 (modified)

Verify the following exact mathematical expression for the value $a = 0.0625$.

$$1 + a + a^2 + a^3 + a^5 = \frac{1-a^6}{1-a}$$

a = 0.0625;
S_left = sum(a.^(0:5));
S_right = (1-a^6)/(1-a);

How close are the left and right sides of this expression?

fprintf('%g\n',abs(S_left - S_right));
0

### Example 4 (modified)

For the function $f(x) = x$, verify the following formula for $x = 1$ and $h = 0.015625$. $$\begin{eqnarray*} f'(x) & = & \frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x}{h} = 1 \end{eqnarray*}$$

x = 1;
h = 0.015625;
dfdx = ((x + h) - x)/h;    % This should be exactly 1

Do we get $f'(x) = 1$?

fprintf('%g\n',abs(dfdx-1));
0

Back to the top

## Machine epsilon

As as way to gauge how close two numbers are to each other, we can use the Matlab function eps(x). This measures the distance to the next representable number. For example,

### Example 7

In this example, we want to see if the following mathematical expression holds for all $\x > 0$.

Compute $x$.

$$1 + x > 1, \qquad x > 0$$

x = 1;
while (1 + x > 1)
x = x/2;
end

Is $x$ equal to 0?

fprintf('x = %8.4e \n',x);
x = 1.1102e-16

Back to the top

## Lab exercises

1. Can we modify examples 5,6 and 7 above to fix the apparent problems?
2. Using a loop similar to the one above to compute the smallest value we could add to 1 and still have something greater than 1, try to find the smallest value we can add to $10^6$ and still get something larger than a million?


# If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine.

Given $$\sec\theta=-\frac{13}{12}$$ find $$\cos{\frac{\theta}{2}}$$, where $$\frac\pi2<\theta<\pi$$.

If the $$\sec\theta$$ is $$-\frac{13}{12}$$ then, the $$\cos \theta$$ is $$-\frac{12}{13}$$, and the half angle formula tells us that $$\cos{\frac{\theta}{2}}$$ should be

$$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$

which gives me $$\sqrt{\dfrac{1}{26}}$$ which rationalizes to $$\dfrac{\sqrt{26}}{26}$$.

The worksheet off which I'm working lists the answer as $$\dfrac{5\sqrt{26}}{26}$$.

Can someone explain what I've done wrong here?

• I'm pretty sure there's a typo. I think your answer is correct. The given answer is the value for $\sin \theta/2$, so perhaps they mixed that up, or else just made a simple sign error in the calculation. Feb 18, 2019 at 15:17
• Is an intervall for $\theta$ given? Feb 18, 2019 at 15:19
• $\frac{\pi}{2} < \theta < \pi$ Feb 18, 2019 at 15:21
• @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large Feb 18, 2019 at 15:22
• My edit was to put brackets in the half-angle formula so you do not have "$+-$". An alternative would be to type \frac {-12}{13} instead of -\frac {12}{13}. Feb 18, 2019 at 21:24

The answer they gave $$\left(\frac {5 \sqrt{26}}{26}\right)$$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$
EDIT (thanks, DanielWainfleet!): For the range $$\pi/2 \lt \theta \lt \pi$$, $$\pi/4 \lt \theta/2 \lt \pi/2$$, so $$\cos \dfrac {\theta}{2}$$ will be positive. Thus, your answer will be $$\left(\frac {\sqrt{26}}{26}\right).$$
• If $\pi/2<\theta<\pi$ then $\pi/4<\theta/2<\pi/2$ so $\cos (\theta /2)>0.$ Feb 18, 2019 at 21:31
Your answer is correct and the answer given in the working list is wrong because it's the value of $$\sin \frac {\theta} {2}.$$


# Probability density problem

Suppose that the random variable $X$ is uniformly distributed on the interval $[0,1]$ (i.e. $X \sim U(0,1)$) and suppose that $$Z=min(2,2X^2+1)$$

(a) Explain why $Z$ does not have a density function.

(b) Find $\mathbf{E}Z$.

Hint: Use the fact that $\mathbf{E}Z=\int_\mathbf{R}{zdF_Z}$

Is $E(Z)=0.9428+2$? Thanks for helping.

• integrate 2x^2+1 from interval 0 to square root 1/2 , plus integrate 2 from interval square root 1/2 to 1 , as pdf of X is 1 – Wei Sheng Sep 24 '15 at 2:30
• Note that $\min(2,2X^2+1)$ cannot exceed 2 so its expectation also cannot exceed 2. – Glen_b -Reinstate Monica Sep 24 '15 at 2:55
• Is the answer 0.94+0.58? – Wei Sheng Sep 24 '15 at 2:57
• @WeiSheng It would help if you could detail out your approach. – rightskewed Sep 24 '15 at 3:07
• Z= 2X^2+1 for interval [ 0, sqr(1/2) ] , Z= 2 for interval [sqr(1/2) ,1] E(Z)=E(2X^2+1)+E(2) is that correct ? – Wei Sheng Sep 24 '15 at 4:43

Now, there are two ways to find the expectation of this random variables. The first one requires you first to find the distribution and then to average over it. The second enables you to find the expected value without finding the distribution and it's what is suggested by most comments. So let's do it both ways and verify the result.

First the "conventional" way. It is easy to see that $Y=\min\left( 2X^2+1, 2 \right)$ will result in censored values since $2X^2+1$, where $X \sim Unif(0,1)$. could very well exceed $2$. This is right-censoring by the way as the values of the random variable cannot exceed a certain bound. Thus the event $\left\{Y=2\right\}$ occurs if and only if $\left\{2X^2+1 \geq 2 \right\}$. Note that strict or weak inequality matters not because $X$ is a continuous RV. We then have

$$P\left(Y=2 \right)=P\left(2X^2+1 \geq 2 \right)=\left(X \geq \frac{1}{\sqrt{2}}\right)=1-\frac{1}{\sqrt{2}}$$

by the properties of the uniform distribution.

Then for the event $\left\{2X^2+1 \leq 2 \right\}$ and for $y \in \left(1,2\right)$ we may compute the CDF of $Y$ as follows

\begin{align} P\left( Y \leq y \ \cap \min\left( 2X^2+1, 2 \right) = 2X^2+1 \right) &= P\left(2X^2+1 \leq y\right) \\ &= P \left(X \leq \sqrt{\frac{y-1}{2}} \right) \\ &= \sqrt{\frac{y-1}{2}} \end{align}

And so the distribution of $Y$ is given by

$$f_Y(y) = \begin{cases} \frac{1}{2^{3/2} \sqrt{y-1}} & 1<y<2 \\ 1-\frac{1}{\sqrt{2}} & y=2 \end{cases}$$

This is a mixed continuous-discrete distribution as the event $\left\{Y=2\right\}$ has positive proability and this I believe answers the first question. Now, the expectation is given by

$$E(Y)=2 \left(1-\frac{1}{\sqrt{2}} \right) + \int_{1}^2 y \frac{1}{2^{3/2} \sqrt{y-1}} \mathrm{dy}$$

and integrating by parts, it is easy to see that the last expression equals $\sqrt{2}-\frac{\sqrt{2}}{3}$, hence the expectation equals $\sqrt{2}-\frac{\sqrt{2}}{3} +2 \left( 1-\frac{1}{\sqrt{2}} \right)$.

Here is what the density looks like in case you are curious (the plot was made in R). Notice the discontinuity at $2$.

   x<-runif(5000, 0, 1)
y <- ifelse(2*x^2+1<=2, 2*x^2+1,  2)
hist(y, prob = T)
curve(2^(-3/2)*(x-1)^(-1/2), add = T, col = "blue", xlim = c(1,1.9), lwd = 2)

Using the Law of unconscious statistician, one may arrive at the last result simply by writing

\begin{align} E\left(Y \right) = \int_0^{\frac{1}{\sqrt{2}}} \left(1+ 2x^2 \right)\ \mathrm{dx} + \int_{\frac{1}{\sqrt{2}}}^1 2 \ \mathrm{dx} \end{align}

where we have effectively split the expectation to account for each case. The result is the same and you get here way faster. Of course I find the long way a bit more instuctive. In an exam, however, definitely do it as fast as possible.

Hope this helps.


# GATE1997-13

2.3k views

Let $F$ be the set of one-to-one functions from the set $\{1, 2, \dots, n\}$ to the set $\{1, 2,\dots, m\}$ where $m\geq n\geq1$.

1. How many functions are members of $F$?

2. How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i, 1\leq i \leq n$?

3. How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \ \ 1\leq i \leq j \leq n$?

edited
0

In above question we have calculate the number of strictly-increasing functions. But if you also want to understand that how to calculate the total number of monotonically increasing functions (or say non-decreasing functions) then refer to below link :-

https://math.stackexchange.com/questions/1396896/number-of-non-decreasing-functions

It is difficult to understand. But after reading from above link, you will able to remember the generalized formula easily.

(a) A function from A to B must map every element in A. Being one-one, each element must map to a unique element in B. So, for $n$ elements in A, we have $m$ choices in B and so we can have $^m\mathbb{P}_n$ functions.

(b) Continuing from (a) part. Here, we are forced to fix $f(i) = 1$. So, one element from A and B gone with $n$ possibilities for the element in A and 1 possibility for that in B, and we get $n \times$ $^{m-1}\mathbb{P}_{n-1}$ such functions.

(c) $f(i) < f(j)$ means only one order for the $n$ selection permutations from B is valid. So, the answer from (a) becomes $^m\mathbb{C}_n$ here.

edited
14
For case (C). i , j  are  from set A(i.e. from n) which is domain of any one to one function , but mapped element f(i) , f(j) are  from range to specific one to one function .
I've considered an example , with n=3(1,2,3) and m=4(1,2,3,4)
for strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.}
Similary , for element 3 , I can't map with below with element 3 of set B ,  so , remaining number elements is 4 only . so , it should be (3,4).

final mapping ,
example :
A(1,2,3)         B(1,2,3,4)
for the satsfying condition f(i)<f(j)  .where i , j are from set A and f(i) , f(j) from set B
Total number of such functions are :
1.{(1,1) ,(2,2), (3,3)}
2.{(1,1),(2,2),(3,4)}
3.{(1,1),(2,3),(3,4)}
4.{(1,2),(2,3),(3,4)}

(1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element
Only 4 such possible functions .
So , the possible functions are choose n element from m element  , i.e., mCn
2
Yes. Also, we can consider all permutations of the range- and only 1 is valid.
0
yes, dividing by $n!$.
0
Nice Explanation.
0
thats indeed a nice way of thinking !!
20
part C:- They are talking about strictly increasing functions, strictly increasing functions are always One-One, therefore when i am dealing with strictly increasing then i do not need to think about One-One.

In case function is monotonically increasing ($f(i) \leq f(j)$) then total number of such functions are = $m+n-1\choose n-1$
18
Yes Sachin Sir, In case of monotonically increasing functions (f(i) <= f(j)), the total no of such functions will be Selecting N element from the set of distinct M element such that repetition is allowed.

N element in domain and M element in co-domain. This will be  $\binom{M + N - 1}{N}$. which is also same as $\binom{M + N - 1}{M - 1}$
0
Well explained .Thank u Sir.
0
Can anyone provide more clarification for c?
2

Option B) can also be written as P(m,n) - P(m-1,n) ...

0
@hemant , u r applying (m+n-1,n-1) but this is choclate problem where any person might not get any choclate , but here it has said that f(i)<f(j) so u cant apply this above formula since equality has not given
12

option C is correct, you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing order, so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n)

0

Best explained @Shubhanshu Thanks

0

Proofs of the number of strictly increasing and monotonically increasing functions. - https://gateoverflow.in/215132/isi-2014-pcb-a2

5
0
@Arjun sir, please solve option c. I am not getting doubt in option c.
1
@ayush... It is given $1\leqslant i\leqslant j\leqslant n$. Suppose a function f maps i=1 f(i=1) to x. But it says j can be equal to i. If j=1 then f(j)= y where f(i)< f(j) i.e x is less than y. But this violates the condition of function as the same value is getting mapped to two different value.
0

for all i,j  1≤ijn?

Doesn't that imply that no such function exists?Because when i=j, f(i)<f(j) cannot happen.

0
Should not (1,3) (2,2) (3,4)  be included as one of the function
1 vote

Below image contain the answers

A) mPn

B) mPn - m-1Pn

C) (m*(m-1))/2

0
Option C ans is surely incorrect !  B does  not look promising either !
0
i am not getting b and c can someone explain?

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