Answer. Because all rectangles are also parallelograms, all the properties of parallelograms are also true for rectangles, too: But because the angles are all equal, there is an additional property of rectangles that we will now prove – that the diagonals of a rectangle are equal in length. If we divided the rectangle along diagonal NL, we would create triangle LNO. A rectangle is a parallelogram with 4 right angles. You can contact him at GeometryHelpBlog@gmail.com. Can anyone solve this and explain how to do so? prove: the diagonals of a rectangle have equal lengths. Which element of a text best helps the reader determine the central idea? Series: Property: The Diagonals of a Rectangle Are of Equal Length. Welcome to Geometry Help! A rectangular field is 15 m long and 10 m wide, Another rectangular field having the sameperimeter has its sides in the ratio 4 : 1. Recommend (0) Comment (0) ASK A QUESTION . Here, we don’t even need to construct any special triangles, because the diagonals themselves have defined the triangles. 00:03:47 undefined. Maths. Videos: 1 Duration: 00:03:47 Language: English. A square has four equal sides and four right angles. OP = OB . alwayssometimesnever4 The diagonals of a trapezoid are equal. Diagonals of Rectangle The diagonals of a rectangle are congruent. (6) AC = DB                                // Corresponding sides in congruent triangles (CPCTC), Filed Under: Rectangles Last updated on January 4, 2020. Thus the diagonals bisect each other in a rectangle. Copyright © 2020. Remember that a rectangle is a parallelogram, so it has all of the properties of parallelograms , including congruent opposite sides. Diagonals of rectangle bisect each other. In this video tutorial we discuss: (1) How to prove that the two diagonals of a rectangle are equal in length & bisect each other? Please email us at GeometryHelpBlog@gmail.com. If the two diagonals of a parallelogram are equal, it is a rectangle. Now, since a rectangle is a parallelogram, its opposite sides must be congruent and it must satisfy all other properties of parallelograms. Upvote(17) How satisfied are you with the answer? Chemistry. My goal is to help you develop a better way to approach and solve geometry problems. 1 A square is a rectangle.alwayssometimesnever2 The diagonals of a rhombus are perpendicular. (5) ΔBAD ≅ ΔCDA                     // Side-Angle-Side postulate. Write a two column-proof given: fghi is a rectangle. In a rectangle ABCD, prove that the diagonals are of equal length: prove AC = DB. In rectangle STAR below, SA =5, what is the length of RT? Maths. Real World Math Horror Stories from Real encounters. The two diagonals are equal in length. Find the dimensio … n of the rectangular field. RELATED ASSESSMENTS. Next, remember that the diagonals of any parallelogram bisect each other and the diagonals of a rectangle are congruent. Add your answer and earn points. Each diagonal splits a corner into two angles of $$45^\circ$$ . New questions in Mathematics. Chemistry . They have a special property that we will prove here: the diagonals of rectangles are equal in length. Interactive simulation the most controversial math riddle ever! The length of a diagonal (d) of a rectangle whose length is l and whose breadth is b is calculated by the Pythagoras theorem. ii) From O, draw a perpendicular OP to meet AB at P. In triangle … Note also that the diagonals are equal and cut each other in half at right angles. How Long is MO and MZ in the rectangle pictured on the left? Haroldescorcia2527 is waiting for your help. The formula to find the length of the diagonal of a rectangle is: c:Parallelogram. The diagonal of the rectangle is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the width and height of the rectangle. What is the value of x in rectangle STAR below? Download PDF's. Let's take rectangle ∴ The diagonals of a rectangle bisects each other and equal . PLEASEEEE HELLLPPP MEEEEE FREE POINTS FOR ALLL! To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. [00:03:47] S. Login/Register to track your progress. Class 12 Class 11 … Triangle MLO is a right triangle, and  MO is its hypotenuse. IF the quadrilateral is a rectangle, then the two diagonals are equal in length. If you remember your Pythagorean theorem, you should be able to see why. In the figure above, click 'reset'. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Click to learn more... By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. Physics. NCERT RD Sharma Cengage KC Sinha. Length of a diagonal of a square = √2 x Length of a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. We also know that AB= CD as they are opposite sides in a parallelogram. Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. Geometry doesn't have to be so hard! And from the definition of a rectangle, we know that all the interior angles measure 90° and are thus congruent- and we can prove the triangle congruency using the Side-Angle-Side postulate. For example, the two triangles ΔABD and  ΔDCA, in which the diagonals form corresponding sides. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). What is the formula of collinear? parallelograms; class-8; Share It On Facebook Twitter Email. His goal is to help you develop a better way to approach and solve geometry problems. alwayssometimesnever3 The diagonals of a rectangle are equal. The diagonal of a square is equal to the length of one of the sides of the square times $\sqrt {2}$, or $s\sqrt{2}$. When if two diagonals are given we can construct. If a diagonal bisects a rectangle, two congruent right triangles are obtained. Therefore $$\angle SZA = 120°$$. To find MZ, you must  remember that the diagonals of a parallelogram bisect each other. Biology. I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree. Thus diagonals bisect each other in a rectangle . 00:02 CBSE class 9 maths NCERT Solutions chapter 10 Quadrilaterals q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. It happens! 직사각형의 두 대각선은 길이가 같고 서로 다른 것을 이등분한다. Click here to see the proof. Rectangles are a special type of parallelogram. As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle Geometry answers, proofs and formulas for solving geometry problems, and useful tips for how to approach these problems. Biology. Similarly we can prove that PC = PA . It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. The rectangle is given as ABCD, with the two diagonals as AD and BC. In ΔADB and ΔBCD: AD = BC (Opposite sides of a rectangle are equal.) So, looking at the triangles ΔABD and  ΔDCA, they have one common side – AD. 343 views. Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. Like many other geometry problems where we need to prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to tool. Opposite sides of a rectangle are equal; so, AB = DC and AD = BC. … Don’t worry if you didn’t see this immediately, or if you chose other triangles – it is easy to prove this property of rectangles using other combinations of triangles, such as ΔABC and ΔDCB or ΔDAB and ΔCBA; any two pairs will do, as long as the diagonals are the corresponding sides in each triangle. (3) m∠BAD = m∠CDA=90°         // Definition of rectangle, (4) ∠BAD ≅ ∠CDA                      // (3) and definition of congruent angles. The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium ← Prev Question Next Question → 0 votes . Physics. So AO = OC and BO = OD, where O is the intersecting point of the two diagonals AC & BD. Rectangles are a special type of parallelogram. Click hereto get an answer to your question  Show that the diagonals of a square are equal and bisect each other at right angles. 0%. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. a:Square. To Prove that the two Diagonals of a Rectangle Are of Equal Length. b:Rectangle. The diagonals of a rectangle are congruent. (2) AB= CD                                 // Opposite sides in a rectangle (parallelogram). We can also prove this from scratch, repeating the proofs we did for parallelograms, but there’s no need. d:Rhombus. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. … In a rectangle both diagonals are equal in measure; so AC = BD; as well the four intercepts, AO = OC = BO = OD. alwayssometimesnever Free Algebra Solver ... type anything in there! (1)The diagonals of a parallelogram are equal. The rectangle is a convex quadrilateral because of two diagonals which lie in the interior of the rectangle. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other. e:Trapezium etc..... New questions in Math. asked May 4 in Parallelograms by Vevek01 (47.2k points) The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium. Again, we can use the Pythagorean theorem to find the hypotenuse, NL. Explain why a rectangle is ... maths. Click hereto get an answer to your question  Which statement(s) is/are correct? Related Questions. (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Thank you! CD = CD (Common) Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length. answr. Books. But we'd sure like to know about it so that we can fix it. "If a rectangle is square, then its main diagonals are equal" is (True) because this is true of all rectangles. LMNO and divide along the diagonal MO into two right triangles. If side MN = 12 and side ML = 5, what is the length of the other two sides? They have a special property that we will prove here: the diagonals of rectangles are equal in length. In a quadrilateral the diagonals are equal, the quadrilateral will be either 1.A square or 2.A rectangle In a parallelogram the diagonals bisect each other and they never equal Therefore, SZ = AZ, making SZA isosceles and $$\angle$$ZSA$$\angle$$ZAS, being base angles of an isosceles triangle. Books. Therefore, x = 30 °. (2)The diagonals of a square are perpendicular to each other. $$\angle SZT$$ and $$\angle SZA$$ are supplementary angles, Since the opposite sides of a rectangle Since the diagonals of a rectangle are congruent, RT has the same length as SA. Video Explanation. NCERT RD Sharma Cengage KC Sinha. To Prove that the two Diagonals of a Rectangle Are of Equal Length. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. In order to prove that the diagonals of a rectangle are congruent, consider the rectangle shown below. toppr. Click hereto get an answer to your question  Prove that the diagonals of a rectangle divide it in two congruent triangles. This will help us to improve better. As you can see, a diagonal of a rectangle divides it into two right triangles,BCD and DAB. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Prove that the diagonals of a rectangle are congruent. Since the diagonals of a rectangle are congruent MO = 26. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. EASY. to be divided along the diagonals into two triangles that have a congruent hypotenuse. Answered By . Diagonals of Rectangles are of Equal Length, Opposite sides in a rectangle (parallelogram), The opposite sides of rectangles are equal, The diagonals of rectangles bisect each other, Any two adjacent angles are supplementary (obviously, since they all measure 90°), The opposite angles are equal (again, obviously, since all interior angles measure 90°). A Quadrilateral has two diagonals. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Now, having picked our triangles, we will rely on the properties of parallelograms to show triangle congruency. Click hereto get an answer to your question  Prove logically that the diagonals of a rectangle are equal Explain why a rectangle is a convex quadrilateral. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties), If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13. Click here to see the proof. Get Instant Solutions, 24x7. 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# Find all roots in the interval of nonlinear equation

I am struggling on how to find all the roots of 1+1/2^x+1/3^x==0 which lie in a given of real and imaginary interval.

Solve does not work, and FindRoot only returns one root. Is there a better way to get as many roots as possible? Thank you so much for your help.

If you restrict the domain where to look for roots, Reduce can often find them and will return Root objects which can be used in exact symbolic calculations. Even better, it guarantees to give you all roots in that domain.

Reduce[1 + 1/2^x + 1/3^x == 0 && Abs[x] < 5, x]
(* x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420109442288880772534469111379406 - 3.5981714939947587422049363529208471165604257466288393398421 I}] ||
x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420110 + 3.5981714939947587422049363529208 I}] *)

When looking for real roots, the typical way to restrict the domain is something similar to 0 < x < 1. We're looking for complex roots her so I used Abs[x] < 5.

Related:

To use the FindRoots2D function from the linked post, you need to break the equation into real and imaginary parts, as follows:

f[z_] := 1 + 1/2^z + 1/3^z

FindRoots2D[{Re@f[x + I y], Im@f[x + I y]}, {x, -5, 5}, {y, -5, 5}]

(* {{0.454397, -3.59817}, {0.454397, 3.59817}} *)

• I am trying your code. But it runs so long that I have not seen the result yet while I am replying to you. How long did u take to run this code? Thanks for your help. – user16023 Jun 19 '14 at 4:39
• Could you give me some basic instructions to use Wagon's FindAllCrossing2D[] function? – user16023 Jun 19 '14 at 4:52
• @user16023 It takes less than a second in both v8 and v9. FindAllCrossings2D finds the roots of a system of two equations on the reals (not complexes). You can break your equation into real and imaginary parts then follow the instructions from the post I linked. – Szabolcs Jun 19 '14 at 14:31

If you only need approximate numerical solution rather than the exact solution (i.e., Root object) you could use NSolve also with a constrained domain for x.

eqn = 1 + 1/2^x + 1/3^x == 0;

soln = NSolve[{eqn, Abs[x] <= 5}, x, WorkingPrecision -> 10]

(*  {{x -> 0.454397008 + 3.598171494 I}, {x -> 0.454397008 - 3.598171494 I}}  *)

And @@ (eqn /. soln)

(*  True  *)


# 14 Pre-Class Assignment: Fundamental Spaces¶

## 1. Orthogonal Complement¶

Definition: A vector $$u$$ is orthogonal to a subspace $$W$$ of $$R^n$$ if $$u$$ is orthogonal to any $$w$$ in $$W$$ ($$u\cdot w=0$$ for all $$w\in W$$).

For example, consider the following figure, if we consider the plane to be a subspace then the perpendicular vector comming out of the plane is is orthoginal to any vector in the plane:

Definition: The orthogonal complement of $$W$$ is the set of all vectors that are orthogonal to $$W$$. The set is denoted as $$W_{\bot}$$.

QUESTION: Is $$W_\bot$$ a subspace of $$R^n$$? Justify your answer briefly.

Put your answer to the above question here

QUESTION: What are the vectors in both $$W$$ and $$W_\bot$$?

Put your answer to the above question here

from IPython.display import YouTubeVideo

### Projection of a Vector onto a Subspace¶

Think of a projection onto a subspace is analogous to a shadow on a surface. Aspects of an objects 3D space is represented in a 2D shadow but you can’t take the shadow by itself and exactly recreate the 3D surface.

Image from https://commons.wikimedia.org

The following is the matimatical defination of projection onto a subspace.

Definition: Let $$W$$ be a subspace of $$R^n$$ of dimension $$m$$. Let $$\{w_1,\cdots,w_m\}$$ be an orthonormal basis for $$W$$. Then the projection of vector $$v$$ in $$R^n$$ onto $$W$$ is denoted as $$\mbox{proj}_Wv$$ and is defined as $$$\mbox{proj}_Wv = (v\cdot w_1)w_1+(v\cdot w_2)w_2+\cdots+(v\cdot w_m)w_m$$$

Another way to say the above defination is that the project of $$v$$ onto the $$W$$ is just the sumation of $$v$$ projected onto each vector in a basis of $$W$$

Remarks:

Recall in the lecture on Projections, we discussed the projection onto a vector, which is the case for $$m=1$$. We used the projection for $$m>1$$ in the Gram-Schmidt algorithm.

The projection does not depend on which orthonormal basis you choose.

If $$v$$ is in $$W$$, we have $$\mbox{proj}_Wv=v$$.

### The Orthogonal Decomposition Theorem¶

Theorem: Let $$W$$ be a subspace of $$R^n$$. Every vector $$v$$ in $$R^n$$ can be written uniquely in the form $$$v= w+w_{\bot},$$$$where$$w$$is in$$W$$and$$w_\bot$$is orthogonal to$$W$$(i.e.,$$w_\bot$$is in$$W_\bot$$). In addition,$$w=\mbox{proj}Wv$$, and$$w\bot = v-\mbox{proj}_Wv$.

Definition: Let $$x$$ be a point in $$R^n$$, $$W$$ be a subspace of $$R^n$$. The distance from $$x$$ to $$W$$ is defined to be the minimum of the distances from $$x$$ to any point $$y$$ in $$W$$. $$$d(x,W)=\min \{\|x-y\|: \mbox{ for all }y \mbox{ in } W\}.$$$$The optimal$$y$$can be achieved at$$\mbox{proj}_Wx$$, and$$d(x,W)=|x-\mbox{proj}_Wx|$.

QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the projection of $$v$$ onto $$W$$.

Put your answer to the above question here

QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the distance from $$v$$ to $$W$$.

Put your answer to the above question here

## 2. The Four Fundamental Spaces¶

In the lecture on Change Basis, we talked about four subspaces based on a matrix $$A$$:

Row space of $$A$$: linear combination of all rows of $$A$$

Column space of $$A$$: linear combination of all columns of $$A$$

Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$

Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$

In this course we represent a system of linear equations as $$Ax=b$$. The matrix $$A$$ can be viewed as taking a point $$x$$ in the input space and projecting that point to $$b$$ in the output space.

It turns out, everything we need to know about $$A$$ is represented by four fundamental vector spaces. Two of the four spaces are easily defined as follows:

Row space of $$A$$: linear combination of all rows of $$A$$

Column space of $$A$$: linear combination of all columns of $$A$$

The other two fundamental spaces are defined by a concept called the Null Space. The Null space is calculated by finding all the solutions to the homogeneous system $$Ax=0$$. The final two fundamental spaces are defined as follows:

Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$

Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$

## 3. Independent Learning¶

DO THIS: Find a YouTube video that helps you understand the four fundamental spaces.

QUESTION: What is the URL for your video?

Put your answer to the above question here

DO THIS: Add the link to the video to the code below. Try embedding the link in the provided Python YouTubeVideo Function by replacing XXXXX with the video ID.

from IPython.display import YouTubeVideo

QUESTION: What criteria did you use in selecting your video?

Put your answer to the above question here

QUESTION: How long into a video did you go before deciding if it was good or bad?

Put your answer to the above question here

QUESTION: What did you like about the video you selected.

Put your answer to the above question here

QUESTION: What didn’t you like about the video?

Put your answer to the above question here

## 4. Assignment_wrap-up¶

Assignment-Specific QUESTION: What is the URL for your video for the four Fundamental spaces?

Put your answer to the above question here

QUESTION: Summarize what you did in this assignment.

Put your answer to the above question here

QUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?

Put your answer to the above question here

QUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?

Put your answer to the above question here

QUESTION: What was the most challenging part of this assignment for you?

Put your answer to the above question here

QUESTION: What was the least challenging part of this assignment for you?

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# Evaluate $\lim_{n \rightarrow \infty } \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ [duplicate]

Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$ using Cesáro-Stolz theorem.

I know there are many question like this, but i want to solve it using Cesáro-Stolz method and no others.

I took log and applied Cesáro-Stolz, I get $$\log{2}+n\log\cfrac{n}{n+1}$$

Which gives me answer as $$\frac{2}{e}$$ . But answer is $$\frac{4}{e}$$. Could someone help?.

Edit: On taking log, $$\lim_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim_{n \to \infty} \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 2 - 1$$ Which gives $$2/e$$

## marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 21:52

• I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( – Cloud JR Apr 30 at 18:23
• Well, $4/e$ is correct. We might need more details on how you applies Cesaro Stolz. [Another approach, without Cesaro-Stolz, is you can rewrite it as $$\left[(1+1/n)(1+2/n)\cdots(1+n/n)\right]^{1/n}$$ and take the log, which is a Riemann sum of $\int_{1}^{2} \log x\,dx.$ The anti-derivative of $\log x$ is $x\log x - x$ and you get $4/e.$ ] – Thomas Andrews Apr 30 at 18:33
• @Thomas Andrews the second method works and i got 4/e. I want to clarify where it went wrong in first method. I don't want to give it up. I will add more details asap – Cloud JR Apr 30 at 18:40
• The expression can be rewritten as $\left(\frac{(2n!)}{n!n^n}\right)^{1/n}$ Taking logarithm gives you $\frac{\log(2n!) - log(n!) - n\log(n)}{n}$ Apply CS, the difference of the denominator is $$\log(\color{red}{2}n+2)+\log(\color{red}{2}n+1) - \log(n+1) - (n+1)\log(n+1)+ n\log n$$ It seems you have missed one of the $\color{red}{2}$ above. – achille hui Apr 30 at 18:52
• @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. – Cloud JR Apr 30 at 18:56

Let $$a_n=\sum_{i=1}^{n} \log\left(1+\frac{i}n\right)$$ be the numerator of the logarithm, with denominator $$b_n=n.$$

The key is that the first term $$\log(1+1/n)$$ of $$a_n$$ doesn't cancel with any of the terms $$\log(1+i/(n+1)).$$ It alone is subtracted, so, while there are $$n$$ occurrences of $$-\log(1+1/n)$$, there are still two more terms for $$a_{n+1}.$$ So you get:

$$a_{n+1} - a_n = \left(\frac{2n+1}{n+1}\right)+\log\left(\frac{2n+2}{n+1}\right)-n\log(1+1/n)$$

Basically, the "cancellation" happens when we have $$\log\left(1+\frac{i}{n+1}\right)-\log\left(1+\frac{i+1}{n}\right)= \log\left(\frac{n}{n+1}\right)$$

While we can assume there is a $$i=0$$ in $$a_{n+1},$$ since it adds $$0=\log 1$$ to $$a_{n+1},$$ that means there are $$n+2$$ values of $$i$$ in $$\sum_{i=0}^{n+1},$$ and hence there is not cancellation of the $$i=n$$ and $$i=n+1$$ terms from $$a_{n+1}.$$

After took log, we have $$\frac{\sum _{i=1}^n \log (i+n)-n \log (n)}{n}$$ Applied cesaro stolez, we have \begin{align} & \left(\sum _{i=1}^{n+1} \log (i+n+1)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right) \\&= \left(\sum _{i=2}^{n+2} \log (i+n)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right)\\&= (\log (2 n+1)+\log (2 n+2)-(n+1) \log (n+1))-(\log (n+1)-n \log (n)) \\ &= n \log \left(\frac{n}{n+1}\right)+\log \left(\frac{(2 n+1) (2 n+2)}{(n+1)^2}\right) \\ &\rightarrow -1+\log (4) \end{align}

• Thanks man !, i got it. – Cloud JR Apr 30 at 18:54


Help to understand changing order of integration

I have a problem I have been working on, with the solution but the thing is I don't really understand how it is done.

The question, is to compute, $$\int_0^1 \int_{9x^2}^9 x^3\sin(8y^3) \,dy\,dx$$

Now, I did notice that we are going to have to reverse the order of integration so first I took note of, as of now I have $$0 \le x \le 1$$ and $$9x^2 \le y \le 9$$ and I tried to consider the graph. This is where I am getting confused, I don't know if I am supposed to consider the area basically above the line $$0\le x\le\sqrt{\frac{y}{9}}$$ and put $0 \le y \le 9$ and compute. I know that is what I should do, but I am having a lot of trouble seeing this from the graph. My apologizes as I am not aware of how to put graphs on the site.

I mean I am having trouble visualizing what it is meant to say $x$ is less than that value of $y$, when are we not considering the region bounded above?

I appreciate all answers and comments, ideally though I would like an answer that includes graphics if possible!

Could anyone shed some light on this? Ps, this is not homework and I already have the final solution if anyone wants to check, it is $$=\frac{1-\cos(5832)}{7776}$$

Thank you all

• See math.stackexchange.com/questions/1073275/… - it explains this all. See... you've not actually got two integrals, you have $\int_Sf(p)dA$ - a surface and you're integrating (summing over (it's a comment not quite summing but lets not be pedantic!)) over little chunks of area. That answer should help you – Alec Teal Apr 9 '15 at 22:46
• To put a gap between this and the above comment, your area is x from 0 to 1, and for each x you go from $9x^2$ to $9$, so your surface is the region of all the points between the lines $y=0$, $x=0$, $x=1$ and the curve $y=9x^2$ you can parameterise that however you like! – Alec Teal Apr 9 '15 at 22:49

Since you would like an answer that includes a graph, I've prepared the following simple one.

\begin{equation*} I=\int_{0}^{1}\int_{9x^{2}}^{9}x^{3}\sin \left( 8y^{3}\right) \,dy\,dx \end{equation*}

is to be evaluated in the region $R$ of the 1.st quadrant bounded from below by the graph of the function $y=9x^2$, from above by the horizontal line $y=9$, and from left by the vertical line $x=0$, because the inequalities you have found, $0 \le x \le 1$ and $9x^2 \le y \le 9$, define exactly $R$, i.e.

\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq 1,\; 9x^{2}\leq y\leq 9\right\}. \end{equation*}

How to define the very same region with a different pair of inequalities? Since $y=9x^2$ is equivalent, for $x\ge 0$, to $x=\sqrt{y/9}=\sqrt{y}/3$, it is clear from the picture that you can also define it by the inequalities $0\leq x\leq \sqrt{y}/3$ and 0$\leq y\leq 9$, as you have done. In symbols,

\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq \sqrt{y}/3,\; 0\leq y\leq 9\right\}, \end{equation*}

which results in the integral

\begin{equation*} I=\int_{0}^{9}\int_{0}^{ \sqrt{y}/3}x^{3}\sin \left( 8y^{3}\right) \,dx\,dy. \end{equation*}

In 2D I usually prefer to do this by drawing a picture. Let me talk through how I would actually draw this picture.

The region $0 \leq x \leq 1,9x^2 \leq y \leq 9$ is bounded by the lines $x=0,y=9$ and the curve $y=9x^2$. You can think of the limits that you have as saying that for each $x$ between $0$ and $1$, you integrate from $9x^2$ to $9$ in $y$. So this corresponds to little vertical segments between the parabola and the horizontal line $y=9$.

Now you want to go the other way: for each value of $y$ between $0$ and $9$, what are the possible values of $x$ in the region? These now correspond to little horizontal segments, which must be between the vertical line $x=0$ and the parabola $y=9x^2$. Solving for $x$ you get that the parabola is given by $x=\sqrt{y/9}$, so for each $y$, $x$ ranges from $0$ to $\sqrt{y/9}$. So your new integral looks like $\int_0^9 \int_0^{\sqrt{y/9}} f(x,y) dx dy.$


# A tricky remainder theorem problem

Tags:
1. Jun 12, 2016

### sooyong94

1. The problem statement, all variables and given/known data
A polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)

2. Relevant equations
Remainder theorem

3. The attempt at a solution
I know that

P(x) = (x-1)A(x) + 1
and P(x) = (x+1)B(x) + 3

But how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.

2. Jun 12, 2016

### Mastermind01

From your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ($x^2 - 1$) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation $$P(x) = (x^2 - 1)f(x) + Kx + L$$ . Try to proceed from here.

3. Jun 13, 2016

### sooyong94

Why is the remainder is a linear expression? I can't catch your explanation.

4. Jun 13, 2016

### Ray Vickson

The remainder in $a(x)/b(x)$ is a polynomial $r(x)$ of degree strictly less than the degree of $b(x)$. Basically, that is the _definition_ of "remainder".

5. Jun 13, 2016

### sooyong94

So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.

6. Jun 13, 2016

### Ray Vickson

7. Jun 13, 2016

### sooyong94

So it has something to do with the divisor right?

8. Jun 13, 2016

### rcgldr

To refresh the idea of the remainder theorem, take a look at

http://en.wikipedia.org/wiki/Polynomial_remainder_theorem

In this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?

9. Jun 13, 2016

### sooyong94

A cubic polynomial?

10. Jun 13, 2016

### rcgldr

Ignore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3?

11. Jun 13, 2016

### sooyong94

Degree 3?

12. Jun 13, 2016

### rcgldr

Looking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.

13. Jun 13, 2016

### sooyong94

Yup it appears that degree one works as well...

14. Jun 13, 2016

### rcgldr

So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?

15. Jun 13, 2016

### sooyong94

ax+b, and I managed to solve it as (2-x).

16. Jun 13, 2016

### rcgldr

To follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0.


# Limits and restrictions?

If we assume that the restrictions put on simplified forms of expressions to prevent evaluation at points undefined in the original unsimplified form are important why do we drop them when dealing with limits? For example, consider the following when trying to find the derivative of $f= x^2%$:

\begin{align*}\lim_{h→0} \frac{f(x + h) - f(x)}{h} &=\lim_{h→0} \frac{(x+h)^2 - x^2}{h}\\ &=\lim_{h→0} \frac{x^2 + 2xh + h^2 - x^2}{h}\\ &=\lim_{h→0} \frac{h(2x + h)}{h} \end{align*}

All following simplified forms should have the restriction $h ≠ 0$ since the original form was undefined at this point.

$$\lim_{h→0} {2x + h}, h ≠ 0$$

However to calculate the derivative, the h is valued at $0$ leading to the derivative:

$$f'(x) = 2x$$

How can the equation be simplified by assuming the $h$ is $0$ when there is a restriction on it? Why is that when simplifying expressions we have to restrict the simplified forms to prevent evaluation at points undefined on the original expression, but this concept is completely ignored when dealing with limits?

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How do you define $\lim_{h\to0}2x+h$? – sai Jul 11 '12 at 2:35
The value of the expression as the value of $h$ approaches $0$ – user26649 Jul 11 '12 at 2:35
And wouldn't you agree that this value is $2x$? At no point did we need to set $h=0$. – sai Jul 11 '12 at 2:36
$\displaystyle \lim_{h \to 0} (2x+h)$ is not the same as evaluating $2x+h$ at $h=0$. – user17762 Jul 11 '12 at 2:39
Exactly...which equals to actually substitute $\,h=0\,$ since the expression $\,2x+h\,$ is a linear polynomial in $\,h\,$ and thus continuous everywhere...This was not so before, when the cancellation was done. Remember, when evaluating the limit of an expression when $\,x\to x_0\,$, we shall and we must take $\,x\neq x_0\,$...very close to it, but not actually equal to it...**unless** we have continuity of the function at $\,x_0\,$ – DonAntonio Jul 11 '12 at 2:40

In a sense, you are repeating the old criticism of Bishop Berkeley on infinitesimals, which were "sometimes not equal to $0$, and sometimes equal to $0$".

What you need to remember is that the expression $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ represents the unique quantity (if it exists) that the expression $\dfrac{f(x+h)-f(x)}{h}$ approaches as $h$ approaches $0$, but without $h$ being equal to $0$. Whenever we take a limit, we are asking how the quantity is behaving as we approach $0$, but without actually being $0$.

Because we are never actually at $0$, the simplification is valid, and so the computation turns on asking: what happens to the quantity $2x+h$ as $h$ approaches $0$?

The answer is that, the closer $h$ gets to $0$, the closer that $2x+h$ gets to $2x$. We can make $2x+h$ as close to $2x$ as we want, provided that $h$ is close enough to $0$, without being equal to $0$.

We are not actually evaluating at $0$ (well, we kind of are, see below, but not really); we are just finding out what happens to $2x+h$ as $h$ gets closer and closer and closer to $0$. So we are not "simplifying" the way we did before, we are now evaluating the limit, by determining what happens to $2x+h$ as $h$ approaches $0$.

(Now, in a sense we are evaluating, for the following reason: the function $g(h) = 2x+h$, where $x$ is fixed, is continuous and defined everywhere. One of the properties of continuous functions (in fact, the defining property of being continuous) is that $g(t)$ is continuous at $t=a$ if and only if $g$ is defined at $a$, and $$\lim_{t\to a}g(t) = g(a).$$ That is, if and only if the value that the function approaches as the variable approaches $a$ is precisely the value of the function at $a$: there are no jumps, no breaks, and no holes in the graph at $t=a$. But we are not "simplifying" by "plugging in $a$", we are actually computing the limit, and finding that the limit "happens" to equal $g(a)$.

This cannot be done with the original function $\dfrac{f(x+h)-f(x)}{h}$ because, as you note, it is not defined at $h=0$. But there is a result about limits which is very important:

If $f(t)$ and $g(t)$ have the exact same value at every $t$, except perhaps at $t=a$, then $$\lim_{t\to a}f(t) = \lim_{t\to a}g(t).$$

the reason being that the limit does not actually care about the value at $a$, it only cares about the values near $a$.

This is what we are using to do the first simplification: the functions of variable $h$ given by: $$\frac{(x+h)^2-x^2}{h}\qquad\text{and}\qquad 2x+h$$ are equal everywhere except at $h=0$. They are not equal at $h=0$ because the first one is not defined at $h=0$, but the second one is. So we know that $$\lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}(2x+h).$$ And now we can focus on that second limit. This is a new limit of a new function; we know the answer will be the same as the limit we care about, but we are dealing with a new function now. This function, $g(h) = 2x+h$, is continuous at $h=0$, so we know that the limit will equal $g(0)=2x$. Since this new limit is equal to $2x$, and the old limit is equal to the new limit, the old limit is also equal to $2x$. We didn't both take $h\neq 0$ and $h=0$ anywhere. We always assumed $h\neq 0,$ and then in the final step used continuity to deduce that the value of the limit happens to be the same as the value of the function $g(h) = 2x+h$ at $h=0$. )

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Thank you! I'm losing count of how many times your genius has saved my ass! – user26649 Jul 11 '12 at 3:30
@Riddler: Thank you for the kind words. Related to your question (which was actually deeply troubling to mathematicians for about 100 years), see this, this, and this. – Arturo Magidin Jul 11 '12 at 3:33

Formally, we say $\lim_{x\to a}f(x)=f(a)$ if $\forall \epsilon>0, \exists \delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-f(a)|<\epsilon$ (except for $x=a$, in case the function is not continuous at $x=a$) or for sequences $\lim_{n\to\infty}a_n=a$ if $\forall \epsilon>0, \exists N\in\mathbb{N}$ such that $|a_n-a|<\epsilon, \forall n>N$. Note that in both definitions, we allow for $\epsilon$ tolerance, which is positive - but may be arbitrarily small.

In your example, $\lim_{h\to0}2x+h=2x$ since $(2x+h)-2x=h$, which can be made arbitrarily small. Alternatively, you can think of a line with slope = 2. What the limit is saying is that as the vertical shift of the line $h$ approaches 0, the line will get arbitrarily close to the slope 2 line through the origin.

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For limits, we aren't (in general) simply evaluating at a particular point. Intuitively, when we're dealing with real-valued functions on subsets of the reals, we talk about the graph of a function getting as close as we like to a point if we make the $x$-coordinate get sufficiently close to the point's $x$-coordinate.

Rigorously, we use the following definitions:

(i) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$. We say that $x_0$ is a limit point of $E$ if and only if for every $\varepsilon>0$ there is some $x\in E$ such that $0<|x-x_0|<\varepsilon$. (That is, there are points of $E$ that are as close to $x_0$ as we like, but still distinct from $x_0$.)

(ii) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$ a limit point of $E$, $L\in\Bbb R$, and $f:E\to\Bbb R$ is some function. We say that $$L=\lim_{x\to x_0}f(x)$$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. (That is, $L=\lim_{x\to x_0}f(x)$ means we can make $f(x)$ as close as we like to $L$ simply by choosing $x\in E$ sufficiently close to $x_0.$ Note that $x_0$ need not be an element of $E.$)

For your example, we're dealing with an $h$ limit (so replace the $x$ terms above with $h$ terms). Letting $\delta=\varepsilon$, we find that for $0<|h-0|<\delta$, we have $$|(2x+h)-2x|=|h|=|h-0|<\delta=\varepsilon.$$

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# Prove that the product of two evens is even. Generalise to any divisor.

I am getting these questions from a book called, An Inquiry-Based Introduction to Proofs by Jim Hefferon. However, I do not understand what he means by "Generalise to any divisor.". I thought the divisor would always be 2, given we are talking about an even number.

My solution I have is simply this:
1. $a$ and $b$ are even.
2. $\exists k,m \in \mathbb{Z}: a = 2k$ and $b = 2m$.
3. $a \times b = 2k \times 2m = 2(2km) =$ even.

However, I am not sure if that satisfies the request to "Generalise to any divisor" because I don't really know what that means? Does anyone know? Thanks.

• Replace "even" by "multiple of $2$" then you can generalize to any larger integer. – didgogns Jun 15 '17 at 14:37
• You made a calculating error, $2k\times 2m=4km$ – User123456789 Jun 15 '17 at 14:37
• $2k \cdot 2m \ne 2(k+m)$ although it is even. – NickD Jun 15 '17 at 14:37
• You're right @Daan, thanks for the headsup. – Bucephalus Jun 15 '17 at 14:39

Note that in your answer, you should have $2k\cdot 2m=4mk=2(2mk)$, not $2(k+m)$. Otherwise, your proof is correct.

By generalizing to any divisor, it means that it wants you to prove the following:

Suppose $n$ and $m$ are both divisible by $k$. Prove that $n\cdot m$ is divisible by $k$.

• Oh yeah that explanation is what I'm looking for. So if I want to go on and do this for n and m as you have stated, do I edit my original post or I answer my post? – Bucephalus Jun 15 '17 at 14:43
• Well, you may accept whichever answer you feel answers your question best, including your own. Editing your own question might make hide the original purpose of it. – ervx Jun 15 '17 at 14:49
• Yes, but I mean generally, sometimes someone will give you a hint, and not the answer, and so you have to continue with your solution. Should you edit and append it to your original post? Or is it preferred that you answer your own post with an updated version? – Bucephalus Jun 15 '17 at 14:51
• If the hint helps you answer your question, you should probably accept that answer. There is no need to then post a full solution to your question. – ervx Jun 15 '17 at 14:53
• ok, thanks @ervx – Bucephalus Jun 15 '17 at 14:53

$2p\times 2k=2^2pk$

$2\mid 2^2pk.$ QED

$np\times nk=n^2pk$

$n\mid n^2pk$. QED

• Oh yeah, ok so that's what I should be putting if I were to add it. Thanks @RobertFrost – Bucephalus Jun 15 '17 at 14:44
• @Bucephalus yeah, to generalise just means prove for the general case, i.e. for any divisor (in this case $n$), not just $2$. – user334732 Jun 15 '17 at 14:45


Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:

1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.

2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.

Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram

$\xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 }$

with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$

Comment #2973 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J.

Comment #2974 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)

Comment #2975 by on

Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)

Comment #3098 by on

OK, yes. I've added this as a separate lemma. It seems to me one could also deduce this directly from the definition of ML modules. If you have a minute, please check the changes to see if you agree. Thanks!

There are also:

• 4 comment(s) on Section 10.89: Interchanging direct products with tensor

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# Show that there is a $c\in(0,1)$ such that $f(c)=\int_0^cf(x)dx$.

Question: Let $$f:[0,1]\to\mathbb{R}$$ be a continuous function such that $$\int_0^1f(x)dx=\int_0^1xf(x)dx.$$ Show that there is a $$c\in(0,1)$$ such that $$f(c)=\int_0^cf(x)dx.$$

My solution: Define the function $$g:[0,1]\to\mathbb{R}$$, such that $$g(x)=x\int_0^x f(t)dt-\int_0^x tf(t)dt, \forall x\in[0,1].$$

Now since $$f$$ is continuous on $$[0,1]$$, thus we can conclude by Fundamental Theorem of Calculus that $$g$$ is differentiable $$\forall x\in[0,1]$$ and $$g'(x)=\int_0^x f(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$

Observe that $$g(0)=g(1)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists b\in(0,1)$$, such that $$g'(b)=0$$, i.e $$\int_0^b f(t)dt=0.$$

Now define $$h:[0,1]\to\mathbb{R}$$, such that $$h(x)=e^{-x}g'(x), \forall x\in[0,1].$$

Now $$h'(x)=-e^{-x}g'(x)+g''(x)e^{-x}=e^{-x}(g''(x)-g'(x)), \forall x\in[0,1].$$

Observe that $$h(0)=h(b)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists c\in(0,b)\subseteq (0,1)$$, such that $$h'(c)=0$$. This implies that $$e^{-c}(g''(c)-g'(c))=0\\\implies g''(c)-g'(c)=0\hspace{0.3 cm}(\because e^{-c}\neq 0)\\\implies f(c)=\int_0^cf(x)dx.$$

Is this solution correct? And is there a better solution that this?

• Your solution looks fine. Apr 16 '20 at 5:18
• Isn't this something like the reverse Lagrange's mean value theorm? Correct me if I'm wrong... Apr 16 '20 at 6:36
• See related question math.stackexchange.com/q/3610557/72031 Apr 16 '20 at 7:55

Your proof is correct. This is another one.

We may assume that $$f$$ is not identically zero (otherwise it is trivial). Since $$f$$ is continuous and $$\int_0^1(1-x)f(x)\,dx=0$$ we have that $$M=\max_{x\in [0,1]}f(x)>0$$ and $$m=\min_{x\in [0,1]}f(x)<0$$. Moreover $$\exists x_M,x_m\in [0,1]$$ such that $$f(x_M)=M$$ and $$f(x_m)=m$$. Let us consider the following continuous map $$F(x):= f(x) - \int_0^xf(t)\,dt.$$ If $$x_M<1,$$ then $$F(x_M)=M-\int_0^{x_M}f(t)\,dt\geq M- Mx_M >0.$$ If $$x_M=1$$ then, $$F(x_M)=M-\int_0^{1}f(t)\,dt> 0$$ because $$f$$ strictly less than $$M$$ in an interval of positive length containing $$x_m$$. In both cases we conclude that $$F(x_M)>0$$. In a similar way, we show that $$F(x_m)<0$$.

Finally, since $$F$$ is continuous on $$[0,1]$$, it follows, by the Intermediate Value Theorem, that there exists $$c$$ strictly between $$x_M$$ and $$x_m$$, and therefore $$c\in (0,1)$$, such that $$F(c)=0$$, that is $$f(c)=\int_0^cf(t)\,dt.$$

• +1 It appears that the approach would also work for the question linked in my comment to the current question. Apr 23 '20 at 2:02
• Another observation: what you have proved is that if a continuous $f$ changes sign in $[0,1]$ then we have a $c\in(0,1)$ such that $f(c) =\int_{0}^{c}f(x)\,dx$. Apr 23 '20 at 2:08

As noted by RobertZ, your proof is correct. Here is another proof that follows the same outline as yours: first we find another zero for the antiderivative of $$f$$ and then we use Rolle's theorem in an appropriate way. This approach is admittedly more long-winded but doesn't make use of the $$e^{-x}$$ trick.

Define $$F: [0,1] \to \mathbb{R}$$ as $$F(x) =\displaystyle \int_{0}^{x}f(t)dt.$$ Note that the given condition can be stated as $$\displaystyle \int_{0}^{1}F(t)dt =0$$

Claim 1: There exists $$b \in (0,1)$$ such that $$F(b) =0.$$

Proof of claim 1: By the mean value theorem for integrals there exists $$b \in (0,1)$$ such that $$F(b)= \displaystyle \int_{0}^{1}F(t)dt$$, which implies $$F(b)=0.$$

Now, we look for an appropriate sub-interval of $$[0,b]$$ on which we can apply Rolle's theorem to $$g.$$

Let $$G(x)=\displaystyle \int_{0}^{x}F(t)dt$$ and define $$g:[0,b] \to \mathbb{R}$$ by $$g(x)= G(x) -F(x).$$

Claim 2: $$g$$ is not injective on $$[0, b].$$

Proof of claim 2: Suppose not. Then $$g$$ is injective and since it is clearly continuous too, $$g$$ is monotone. WLOG, let $$g$$ be monotone increasing. Then since $$g$$ is differentiable, $$g'(x) \geq 0 \, \forall \, x \in [0,1].$$ If there exists at least one $$x$$ for which $$g'(x) =0$$ we are done so assume $$g'(x)>0.$$ Since $$g(0) =0,$$ we have $$g(x)>0$$ for all $$x \in (0,b].$$

Let $$x_{0}$$ be a point of maximisation for $$F.$$ Assume $$F$$ is not identically $$0$$ or else $$f$$ is and the problem is trivial. We claim that there exists $$c \in (0, b)$$ such that $$F(c)<0.$$ If $$x_{0}=0$$ or $$b$$ then $$F\leq 0$$ so if $$F$$ is not identically $$0$$ choose an other point of $$(0, b)$$ to be $$c.$$ If $$x_{0} \in (0, b)$$ then since $$g({x}_{0})>0, F(x_{0})< \displaystyle \int_{0}^{x_{0}}F(t)dt \leq x_{0}F(x_{0}).$$

If $$F(x_{0}) \neq 0$$ we get $$x_{0} \geq 1,$$ a contradiction. Hence $$F(x_{0})=0$$ and since $$F$$ is not identically $$0$$ there exists some $$c \in (0, b)$$ such that $$F(c)<0.$$

Since $$F$$ is a continuous function on a closed and bounded interval $$[0, b]$$, it attains its bounds. In particular $$\exists \, d \in [0, b]$$ such that $$F(d)\leq F(x) \, \forall \, x \in [0,b].$$ Clearly $$d\neq 0, 1$$ or else $$F(x) \geq 0 \, \forall x \in [0,b]$$ contradicting the fact that $$F(c) <0.$$ Therefore $$d \in (0,b)$$ and since it is a point of minimisation, $$F'(d) =0.$$ Then $$F(d)= F(d) -F'(d) =g'(d)>0> F(c)$$ contradicting the fact that $$d$$ is a point of minimisation of $$F.$$ Therefore our hypothesis that $$g$$ is injective is false and hence $$g$$ is not injective and there exists $$a, a' \in [0, b]$$ with $$a \neq a'$$ such that $$g(a) =g(a').$$

Then since $$g$$ restricted to $$[a, a']$$ satisfies the conditions for Rolle's Theorem, there exists some $$x_0 \in (a,a')$$ such that $$g'(x_0)=0$$ which implies $$F(x_0)=F'(x_0)$$ from which it follows that $$f(x_{0}) = \displaystyle \int_{0}^{x_{0}}f(x)dx$$

Note that the proof follows almost identically if we assume $$g$$ to be monotone decreasing in the proof of the claim.


# when to say that the given information is (intersection) not (conditional probability)?

There are two urns containing colored balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have probability 50% of being chosen) and then a ball is drawn at random from one of the two urns. If a red ball is drawn, what is the probability that it comes from the first urn?

my question is:

so here (1/2) is the red intersection first urn or it`s the p(red|first urn) I need explanation for that point please and if the second one is the right why it p(red|first urn) and not p(first urn| red)

If we define the following events:

$U1$: Urn $1$ is chosen

$U2$: Urn $2$ is chosen

$R$: a red ball is chosen

then we have:

$P(U1)=P(U2)=P(R|U1)=50$%

The $50$% is not $P(U1 \cap R)$, for we have $P(U1 \cap R) = P(R|U1) \cdot P(U1) = \frac{1}{2}\cdot \frac{1}{2}=0.25$

That is, $P(U1 \cap R)$ is the probability that you choose urn $1$ and pick a red ball, so that mreans you first need to pick urn $1$ ($50$% chance of that), and then you also need to pick a red ball, given that you picked urn $1$, and that is also $50$%, so for both to happen it's $50$% times $50$% is $25$%

Finally, you are asked to find $P(U1|R)$, and for that use Bayes' Law

• but you figured that out from the syntax of the problem about the 50% being intersection or conditional probability May 16 '18 at 17:11
• @AhmedAraby The problem says that urn 1 contains $50$ red balls and $50$ blue balls, and so from that we can say that $P(R|U1) = \frac{1}{2}$ May 16 '18 at 17:20
• okay , but when we can say that is probability of R intersection U1 what is the change that will happen to the problem syntax here in order to say that is an intersection May 16 '18 at 18:17
• @AhmedAraby $P(R \cap U1)$ is the probability of picking urn 1 and getting a red ball. Or, in short: the probability of picking a red ball from urn $1$, given that you have not yet decided from which urn you're going to pick a ball. May 16 '18 at 18:28

You have two urns with known amounts of red and blue balls.   This tells you the (conditional) probabilities for drawing a red ball from each given urn.

$$\mathsf P(R\mid U_1) = 50/100 = 5/10\\ \mathsf P(R\mid U_2)=30/100=3/10$$

You are to randomly select an urn (ie: without bias).   This tells you the (marginal) probabilities for selecting each urn. $$\mathsf P(U_1)=1/2\\ \mathsf P(U_2)=1/2$$

So then the probabilities for selecting each urn and drawing a red ball, will be the products.$$\mathsf P(U_1\cap R)= \mathsf P(U_1)~\mathsf P(R\mid U_1)= 5/20\\\mathsf P(U_2\cap R)=\mathsf P(U_2)~\mathsf P(R\mid U_2)=3/20$$

The Law of Total Probability tells us the (marginal) probability for drawing a red ball.$$\mathsf P(R)~{=\mathsf P(U_1\cap R)+\mathsf P(U_2\cap R)\\=\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)\\=8/20}$$

Now then, you are to evaluate the probability that urn 1 was selected given that a red ball was drawn.   From this we use wither the defiition for conditional probability, or Bayes' Rule.$$\begin{split}\mathsf P(U_1\mid R) &= \dfrac{\mathsf P(U_1\cap R)}{\mathsf P(R)} &\quad&=\dfrac{\mathsf P(U_1)~\mathsf P(R\mid U_1)}{\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)}\\ &=\dfrac{5}{8} \end{split}$$

Note: This should be anticipated, because 50 from the 80 red balls are in urn 1 and both urns contain an equal total count for balls.


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Lecture

# Lecture 33.pdf

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School
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Friday, March 28 − Lecture 33 : Eigenvectors associated to an eigenvalue. Concepts: 1. Define the family of all eigenvectors associated to an eigenvalue. 2. Find all eigenvectors associated to an eigenvalue. 3. Define an eigenspace. 33.1 Theorem − Let λ be a number. The number λ is an eigenvalue ofA if and only if 1 1 the matrix equation Ax = λ x w1th variable x has a non-trivial solution. Proof : λ 1s an eigenvalue ofA ⇔ λ is a 1olution to det(A − λI) = 0 ⇔ det(A − λ I)1= 0 ⇔ A − λ I 1s not invertible ⇔ (A − λ I)1 = 0 has a non-trivial solution ⇔ Ax − λ Ix1= 0 has a non-trivial solution ⇔ Ax = λ Ix 1as a non-trivial solution ⇔ Ax = λ x h1s a non-trivial solution. Different ways of viewing an eigenvalue. The following are equivalent: 1. λ i1 an eigenvalue ofA 2. (A − λ I1x = 0 has a non-trivial solution 3. Ax = λ x 1as a non-trivial solution 4. Null(A − λ I)1≠ {0} has a non-trivial solution 33.1.1 Example − Show that 1 is an eigenvalue of the following matrix A Solution: By the above theorem it suffices to verify that the following homogeneous system(A – (1)I)x = 0. has a non-trivial solution. This coefficient matrix easily row-reduces to Since (A – (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A. Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have non-trivial solutions and so 2 and 3 are eigenvalues of A. 33.1.2 Different ways of viewing an eigenvalue. We have the following different ways of viewing an eigenvalue of A: λ1= eigenvalue of A ⇔ det(A − λ I) = 0 1 ⇔ A − λ I 1s not invertible, ⇔ (A − λ I)1 = 0 has a non-trivial solution. ⇔ Null(A − λ I) 1ontains infinitely many vectors ⇔ Ax = λ x for infinitely many vectors x 1 33.2. Definition − Suppose λ is a1 eigenvalue of n × n matrix A. Then any non-zero vector x in Null(A − λ I) is called an eigenvectorof A corresponding to λ . 1 1 1 33.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix 1 A invites the following question: “x is 1ssociated to which of the eigenvalues of A?”. An eigenvector of A can only correspond to a single eigenvalue. If x is an 1 eigenvector of A associated to the eigenvalues λ and λ1of A th2n λ = λ . Thi1 is 2 because x c1nnot be the zero vector. To see this consider: The expression “λ 1 1= λ 2 1mplies λ = λ1” si2ce an eigenvector cannot be the zero vector. 33.2.2 Remark – Extending the notion of “eigenvalue” so that it applies to linear mappings. The notions of “eigenvalue” and “eigenvector” of a matrix A can be extended to a linear map (transformation) T : ℝ → ℝ .n n Definition –We say that λ is 1n eigenvalue of the linear map T: ℝ → ℝ if n n T(x) = λ x1 has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear transformation T corresponding to the eigenvalue λ 1 if x1is a non-trivial solution to T(x) = λ x1 n n - If λ i1 an eigenvalue of T: ℝ → ℝ and x is an eigenvec1or corresponding to λ 1hen, “T maps x to a1scalar multiple λ x of x ”1 1 1 2 2 33.2.2.1 Example – Suppose R : ℝ → ℝ πs the linear transformation which rotates points in ℝ counterclockwise about the origin by an angle of π radians. Suppose we seek an eigenvalue and its associated eigenvectors for this linear map. See that if x is any ordered pair in ℝ , R (x) = −x. Then λ = −1 is an eigenvalue of π 1 2 R π The eigenvectors associated to λ = −1 ar1 all non-zero vectors in ℝ . The set of all eigenvectors corresponding to an eigenvalue – If λ is an eigen1alue of a matrix A, all the eigenvectors associated to λ form a1set or “package”. This set is the nullspace, Null(A − λ I ), of the mat
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# Defining compact sets with closed covers

This question is a continuation of this. My book says that a metric space is compact if and only if:

$$M=\cup A_{\lambda}\implies M = A_{\lambda1}\cup\cdots\cup A_{\lambda_n}$$

where each $A_{\lambda}$ is open. Then, it says that the definition can be extended to closed sets, since if $A_\lambda$ is a family of open sets in $M$, then the complements $F_\lambda = F-A_\lambda$ forms a family of closed sets in $M$. We have, then: $$M=\cup A_\lambda\iff \cap F_\lambda = \emptyset.$$ So, by the last question, I understood the reason behind this. No problem. The book then proceeds:

A metric space $M$ is compact, $\iff$ all families $(F_\lambda)_{\lambda\in L}$ of closed sets with empty intersection has a finite subfamily with empty intersection: $F_{\lambda_1}\cap \cdots \cap F_{\lambda_n}=\emptyset$.

Then, the book proceeds and says:

A family $(F_\lambda)_{\lambda\in L}$ has the property of finite intersection when for any finite subset $\{\lambda_1,\cdots,\lambda_n \}\subset L$ we have $F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}\neq \emptyset$.

The following condition is necessary and sufficient to tell that a metric space $M$ is compact:

If $(F_\lambda)_{\lambda\in L}$ if a family of closed sets with the property of finite intersection, then $\cap_{\lambda\in L}F_\lambda \neq \emptyset$

Is the last one the negation of the first one? I mean, for $M$ to be compact, either we have:

family of closed sets with empty intersection has finite subfamily with empty intersection

or

family of closed sets with the property of finite intersection (that is, every intersection of finite subfamilies is not empty) does not have a empty intersection.

Is this correct?

I ask this, because the book proceeds for a proof of Dini's theorem like this:

An example of a family of closed subsets with the property of finite intersection is a decrescent sequence $F_1\supset F_2\supset \cdots\supset F_n\supset \cdots$ of nonempty closed subsets of $M$. If $M$ is compact, it follows that $\cap_{n=1}^{\infty}F_n\neq \emptyset$. The following demonstration shows these ideas:

Dini's theorem: If a sequence of continuous real functions $f_n:M\to \mathbb{R}$ defined in a compact metric space $M$, converges simply to a function $f:M\to \mathbb{R}$, and if $f_1(x)\le f_2(x)\le \cdots\le f_n(x)\le \cdots$ for all $x\in M$, then the convergence $f_n\to f$ is uniform in $M$.

Demonstration: Given $\epsilon>0$, choose, for each $n\in \mathbb{N}$:

$$F_n=\{x\in M: |f_n(x)-f(x)|\ge \epsilon\}$$

Then, $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ and each $F_n$ is closed in $M$. We must prove that there is $n_0\in \mathbb{N}$ such that $F_{n_0}=\emptyset$ (then $n>n_0\implies |f_n(x)-f(x)|<\epsilon$ for all $x\in M$). Well, since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n = \emptyset$ (WHY?). Being $M$ compact, we must have $F_n=\emptyset$ for some $n$

And besides the chain of sets $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ having the property of finite intersection (every subfamily has a not empty intersection), the proof concludes that $\cap_{n=1}^{\infty}F_n = \emptyset$

• Your two bolded statements are contrapositive to each other. – Omnomnomnom Jun 24 '16 at 19:53

Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$
Note that $$\bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\}$$ However, if $x$ is such that $f_n(x) \to f(x)$, then (by definition of convergence) $x$ is not an element of this set. So, if $f_n(x) \to f(x)$ for every $x \in M$, then $\bigcap_{n=1}^\infty F_n = \emptyset$.
• ok, now, in the end, it says that being $M$ compact, we must have $F_n=\emptyset$ for some $n$. Why? And which one of the bolded definitions it's using? I ask because in the beggining, it says that the chain of descending closed sets has the property of finite intersection, that is, any subfamily of them has a not empty intersection. How is it possible to conclude that the intersection IS empty? – Guerlando OCs Jun 24 '16 at 20:14


# verifying solution to first order differential equation

#### find_the_fun

##### Active member
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.

$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$

So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$

$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$

and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.

#### MarkFL

Staff member
I agree that if:

$$\displaystyle y=x+4\sqrt{x+2}$$

then:

$$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$

Now, let's look at the left side of the ODE:

$$\displaystyle (y-x)y'=\left(x+4\sqrt{x+2}-x \right)\left(1+\frac{2}{\sqrt{x+2}} \right)=4\sqrt{x+2}+8$$

And next, let's look at the right side:

$$\displaystyle y-x+8=x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$

This then shows that:

$$\displaystyle (y-x)y'=y-x+8$$

#### Prove It

##### Well-known member
MHB Math Helper
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.

$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$

So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$

$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$

and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.
\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \end{align*}

If \displaystyle \begin{align*} y = x + 4\sqrt{x+2} \end{align*} then \displaystyle \begin{align*} \frac{dy}{dx} = 1 + \frac{2}{\sqrt{x + 2}} \end{align*} and so substituting into the DE we have:

\displaystyle \begin{align*} LHS &= \left( y - x \right) \, \frac{dy}{dx} \\ &= \left( x + 4\sqrt{x + 2} - x \right) \left( 1 + \frac{2}{\sqrt{x + 2}} \right) \\ &= x + \frac{2x}{\sqrt{x + 2}} + 4\sqrt{x + 2} + 8 - x - \frac{2x}{\sqrt{x + 2}} \\ &= \left( x + 4\sqrt{x + 2} \right) - x + 8 \\ &= y - x + 8 \\ &= RHS \end{align*}

#### Prove It

##### Well-known member
MHB Math Helper
Another possibility is brute force, actually solving the DE:

\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \\ \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \end{align*}

Make the substitution \displaystyle \begin{align*} u = y - x \implies \frac{du}{dx} = \frac{dy}{dx} - 1 \implies \frac{dy}{dx} = \frac{du}{dx} + 1 \end{align*} and the DE becomes

\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \\ \frac{du}{dx} + 1 &= \frac{u + 8}{u} \\ \frac{du}{dx} &= \frac{u + 8}{u} - 1 \\ \frac{du}{dx} &= \frac{8}{u} \\ u\,\frac{du}{dx} &= 8 \\ \int{ u\,\frac{du}{dx}\,dx} &= \int{8\,dx} \\ \int{u\,du} &= 8x + C_1 \\ \frac{1}{2}u^2 + C_2 &= 8x + C_1 \\ \frac{1}{2}u^2 &= 8x + C_1 - C_2 \\ u^2 &= 16x + C \textrm{ where } C = 2C_1 - 2C_2 \\ \left( y - x \right) ^2 &= 16x + C \\ y - x &= \sqrt{16x + C} \\ y &= x + \sqrt{16x + C} \end{align*}

Of course, your given function \displaystyle \begin{align*} y = x + 4\sqrt{x + 2} \end{align*} is the case where \displaystyle \begin{align*} C = 32 \end{align*}, thus your given function is a solution to the DE.


# Evaluate the triple integral.

Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\:dy\:dz$.

I want to calculate the volume of this surface. I solved it for $dz\:dy\:dx$ and it was:

$$V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx$$

And for $dz\:dx\:dy$ would be this:

$$V=\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{x}dz\:dx\:dy$$

I tried to solve it and the result is this:

$$V=\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{z}^{1}dx\:dy\:dz + \int_{0}^{1}\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{y^2}^{1}dx\:dy\:dz$$

But i think its wrong please advice me the best solution .

I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.

-
What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$ –  Ross Millikan Aug 5 '12 at 21:18

## 2 Answers

Integrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}x\:dy\:dx=\int_0^12x\sqrt{x}\:dx=\frac 45 x^{\frac 52}|_0^1=\frac 45$. This is the triple integral of $1$ over that volume.

-
yes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it –  shahin Aug 6 '12 at 0:37
for $dx \:dy \:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\frac 12$'s around. –  Ross Millikan Aug 6 '12 at 0:50
yes i think it would be $$\int_0^1\int_{-1}^{1 }\int_1^{y^2}dxdydz$$ –  shahin Aug 6 '12 at 1:02
but its still wrong –  shahin Aug 6 '12 at 1:28

(to follow up on where this was left in August 2012):

Here is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $\ x = y^2 \$ is a parabolic cylinder extending indefinitely in the $\ z-$ directions; the surface $\ z = x \$ is an oblique plane; and $\ z = 0 \$ is, of course, the $\ xy-$ plane. The volume is then a sort of wedge with a tilted "roof" and a parabolic "wall".

The integrals for the integration orders $\ dz \ dy \ dx \$ , which is sort of the "natural" order many people would use, and $\ dz \ dx \ dy \$ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with.

In this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any "crisis". For the order $\ dx \ dy \ dz \$ , we are able to make use of the fact that one of the boundary surfaces is $\ z \ = \ x \$ . Since we want to "herd" the integration toward working in the variable $\ z \$ , we can carry out the integration in $\ x \$ from $\ 0 \$ to $\ z \$ . For the integration in $\ y \$ , we can therefore also "replace" $\ x \$ by $\ z \$ to express the relevant portion of the parabolic surface as $\ z \ = \ y^2 \$ . The integration limits on $\ z \$ become $\ 0 \$ to $\ 1 \$ , as they were for $\ x \$ .

We can now write the "re-ordered" integration as

$$\int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ \ \text{or} \ \ \ 2 \ \int_0^1 \int_0^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ ,$$

by exploiting the symmetry of the region about the $\ xz-$ plane. Since this new integral simply looks like a "relabeled" version of the first integral in the original post, using the order $\ dz \ dy \ dx \$ , this will plainly give the same result for the volume of $\ \frac{4}{5} \$ .

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# distance formula: simplifying sqrt(148)

#### marshall1432

##### Banned
i have a small question...regarding the distance formula...

(-7, 1) (5,3)

using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)

How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...

I don't need it in a decimal

#### Mrspi

##### Senior Member
Re: distance formula

marshall1432 said:
i have a small question...regarding the distance formula...

(-7, 1) (5,3)

using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)

How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...

I don't need it in a decimal
You need to see if you can simplify sqrt(148)

Does 148 have any perfect square factors? If so, you can remove the square root of any perfect square factor from under the radical sign.

Here's an example:

Simplify sqrt(60)

Note that 60 = 4*15, and that 4 is a perfect square. So,

sqrt(60) = sqrt(4*15), or sqrt(4)*sqrt(15)

Now, you know that sqrt(4) is 2, so sqrt(4)*sqrt(15) can be written as 2 sqrt(15).

And sqrt(60 = 2 sqrt(15).

Since 15 does not have any perfect square factors, this is the simplest form for sqrt(60).

Try the same process on sqrt(148).

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: distance formula

marshall1432 said:
i have a small question...regarding the distance formula...

(-7, 1) (5,3)

using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)

How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...

I don't need it in a decimal
This is good enough in my view.

However, you can reduce it one step further by factorizing148 and taking some factor/s out-of-the radical sign. That is

$$\displaystyle \sqrt{50} \,=\, \sqrt{5^2\cdot2} \, = \,5\sqrt{2}$$

#### marshall1432

##### Banned
Re: distance formula

well there is a perfect square being sqrt 4 * sqrt 37

so would you consider the answer to be 2 sqrt 37?

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: distance formula

marshall1432 said:
well there is a perfect square being sqrt 4 * sqrt 37

so would you consider the answer to be 2 sqrt 37<<<<< Correct
?


# Probability of dying from smallpox?

A family of four is infected with Variola major. There is a fatality rate of 30%. Calculate the probability that...

Here are my attempts,

The probability that nobody dies, $$0.7^4\cdot100\%=24.01\%$$

The probability that everybody dies, $$0.3^4\cdot100\%=0.81\%$$

The probability that at least one person dies is equal to $$(1-3^4)\cdot100\%=99.19\%$$ or the probability that 1 person dies + 2 people + 3 people + 4 people

The probability that one person dies, two people die, three people die are what I'm having trouble with. For instance, for the probability 1 person dies: $0.3\cdot\binom{4}{1}\cdot n$

I know I have to multiple it by some value n, but I just can't figure out the logic behind what I need to do. I figure I have to find the probability that 1 person dies and multiply that by the probability that the other 3 people live.

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The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. It is about $0.76$, or if you prefer percent (I don't) it is about $76\%$. – André Nicolas Aug 26 '14 at 4:57
What about exactly one person dying, exactly two people dying, exactly three people dying? Sorry, I must have worded my question poorly, but that's what I meant. I thought about this after watching an episode of House, M.D., and my curiosity sparked. – Jason Aug 26 '14 at 5:00
Andre is pointing out that your third answer is incorrect. You found the complement probability to everyone dying, but you need the complement probability to no one dying. – alex.jordan Aug 26 '14 at 5:01
Ah yes, I see how I was wrong. Thank you for pointing that out. – Jason Aug 26 '14 at 5:04
But hasnt smallpox been eradicated ? – Rene Schipperus Aug 26 '14 at 5:05

The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier.

Now let us compute the probability exactly one person dies. Call the people A, B, C, D. The event exactly one person dies can happen in $4$ ways: (i) A dies and the rest survive; (ii) A survives, B dies, C and D survive; (iii); (iv).

If we assume independence, the probability of (i) is $(0.3)(0.7)(0.7)(0.7)$. The probability of (ii) is $(0.7)(0.3)(0.7)(0.7)$, the same. We also get the same answer for (iii) and (iv). So the probability exactly $1$ dies is $(4)(0.3)(0.7)^3$.

Now let us find the probability exactly $2$ die. Again, there are several ways this can happen. How many ways? As many as there are ways to choose the $2$ unlucky people. There are $6$ ways to do that. Each of them has probability $(0.3)^2(0.7)^2$. So the probability exactly $2$ die is $(6)(0.3)^2(0.7)^2$.

The probability exactly $3$ die is calculated in a similar way. There are $4$ "patterns."

In general, if the probability of "success" (in this case death) is $p$, and the experiment is repeated independently $k$ times, then the probability of exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, please see Wikipedia, Binomial distribution. There are many other web sources. The Khan Academy stuff is good.

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For choosing 2 people to die, why do we count each person 3 times? i.e., if people are named ABCD, we can choose AB to die, AC to die, AD to die; then we also add BC, BD and CD. Sure, this is six. But we're counting each person three times. Doesn't this cause problems? – Jason Aug 26 '14 at 6:25
The two people to die could be A and B, A and C, A and D, B and C, B and D, or C and D. Then for example the probability A and C die and the other two survive is $(0.3)(0.7)(0,3)(0.7)$, that is, $(0.3)^2(0.7)^2$. Same for the other five. No problems, there is no double-counting involved. – André Nicolas Aug 26 '14 at 6:31

The probability that say, exactly one person dies is $(0.3)(0.7)^3\binom{4}{1}$. The $0.3$ is the chance that the first person will die, and then the $(0.7)^3$ is the chance that the second, third, and fourth survive. The $\binom{4}{1}$ counts for the other "combinations" where exactly one person dies: maybe it was the first person, maybe it was the second, maybe it was the third, and maybe it was the fourth.

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Let $N$ be the number of people who die. Then we have a binomial distributution. $$N\sim\mathcal{Bin}(n, p), n=4, p=0.3$$

Hence the probability that exactly $k$ people die is: $$\mathsf P(N=k)= {n\choose k}\cdot p^k \cdot (1-p)^{n-k} = \frac{4! \cdot 0.3^k \cdot 0.7^{4-k}}{k! \cdot (4-k)!}$$

The binomial coefficient counts the ways to select the required number of people. The exponentials on the probability and its complement measure the probability of a each of those people either dying or not, as selected.

\begin{align} \mathsf P(N=0) & = 1 \cdot (0.7)^4 & = 24.01\% \\ \mathsf P(N=1) & = 4 \cdot (0.3) \cdot (0.7)^3 & = 41.16\% \\ \mathsf P(N=2) & = 6 \cdot (0.3)^2 \cdot (0.7)^2 & = 26.46\% \\ \mathsf P(N=3) & = 4 \cdot (0.3)^3 \cdot (0.7) & = 7.56 \% \\ \mathsf P(N=4) & = 1 \cdot (0.3)^4 & = 0.81 \% \\ \end{align}

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