Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: (263) Next: Homework Assignment 24
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Solution of Term Exam 4

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Problem 1. Is there a non-zero polynomial $ p(x)$ defined on the interval $ [0,\pi]$ and with values in the interval $ [0,\frac12)$ so that it and all of its derivatives are integers at both the point 0 and the point $ \pi$? In either case, prove your answer in detail. (Hint: How did we prove the irrationality of $ \pi$?)

Solution. There isn't. Had there been one, we could reach a contradiction as in the proof of the irrationality of $ \pi$. Indeed we would have that $ 0<\int_0^\pi p(x)\sin x dx<\frac12\int_0^\pi \sin x dx=1$, hence the integral $ I=\int_0^\pi p(x)\sin x dx$ is not an integer. But repeated integration by parts gives

$\displaystyle I
=\left(\parbox{18mm}{\centering boundary terms}\right)
...{18mm}{\centering boundary terms}\right)
\pm\int_0^\pi p''(x)\sin x dx

$\displaystyle =\left(\parbox{18mm}{\centering boundary terms}\right)
\pm\int_0^\pi p^{(2n)}(x)\sin x dx.

The assumptions on $ p^{(k)}(0)\in{\mathbb{Z}}$ and $ p^{(k)}(\pi)\in{\mathbb{Z}}$ along with the fact that $ \sin 0$, $ \sin\pi$, $ \cos 0$ and $ \cos\pi$ are all integers imply that the boundary terms are all integers. If $ n$ is large enough, $ p^{(2n)}=0$ and hence the remaining integral is 0. So $ I$ is an integer, and that's a contradiction.

Black Pawn Problem 2. Compute the volume $ V$ of the ``Black Pawn'' on the right -- the volume of the solid obtained by revolving the solutions of the inequalities $ 4x^2\leq y+3-(y-3)^3$ and $ y\geq 0$ about the $ y$ axis (its vertical axis of symmetry). (Check that $ 5+3-(5-3)^3=0$ and hence the height of the pawn is $ 5$).

Solution. This is the area of the rotation solid with radius $ r(y)=\frac12\sqrt{y+3-(y-3)^3}$ bounded by $ y=0$ and $ y=5$. Thus

$\displaystyle V=\pi\int_0^5 r(y)^2dy = \frac{\pi}{4}\int_0^5(y+3-(y-3)^3)dy $

$\displaystyle =\frac{\pi}{4}\left.\left(
\right)\right\vert _0^5 = \frac{175\pi}{16}.

Problem 3.

  1. Compute the degree $ n$ Taylor polynomial $ P_n$ of the function $ f(x)=\frac{1}{1-x}$ around the point 0.
  2. Write a formula for the remainder $ f-P_n$ in terms of the derivative $ f^{(n+1)}$ evaluated at some point $ t\in[0,x]$.
  3. Show that at least for very small values of $ x$, $ f(x)=\lim_{n\to\infty}P_n(x)$.


  1. $ f'=\frac{1}{(1-x)^2}$, $ f''=\frac{2}{(1-x)^3}$, $ f'''=\frac{2\cdot 3}{(1-x)^4}$ and so it can be shown by induction that $ f^{(k)}=\frac{k!}{(1-x)^{k+1}}$. Thus $ f^{(k)}(0)=k!$ and hence $ P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k=\sum_{k=0}^nx^k=1+x+x^2+\dots+x^n$.
  2. Cauchy's formula for the remainder is $ R_n(x)=f(x)-P_n(x)=\frac{f^{(n+1)}(t)}{(n+1)!}x^{n+1}=
\frac{x^{n+1}}{(1-t)^{n+2}}=\frac{1}{1-t}\left(\frac{x}{1-t}\right)^{n+1}$ for some $ t\in[0,x]$.
  3. If $ \vert x\vert<\frac12$ then $ \vert t\vert<\frac12$ and $ \vert 1-t\vert>\frac12$ and hence $ \left\vert\frac{x}{1-t}\right\vert<2\vert x\vert<1$ and thus $ \vert R_n(x)\vert<\frac{1}{1-t}(2\vert x\vert)^{n+1}\to 0$. Therefore $ f(x)-P_n(x)\to 0$, as required.

Problem 4.

  1. Prove that if $ \lim_{n\to\infty} a_n=l$ and the function $ f$ is continuous at $ l$, then $ \lim_{n\to\infty}f(a_n)=f(l)$
  2. Let $ b>1$ be a number, and define a sequence $ a_n$ via the relations $ a_1=1$ and $ a_{n+1}=\frac12(a_n+b/a_n)$ for $ n\geq 1$. Assuming that this sequence is convergent to a positive limit, determine what this limit is.


  1. See the ``easy'' part of Theorem 1 of Spivak's Chapter 22.
  2. Assume $ \lim a_n=l>0$. Then $ l=\lim a_{n+1}=\lim\frac12(a_n+b/a_n)$. Using the first part of this question on the function $ x\mapsto\frac12(x+b/x)$, which is continuous at $ l$, we find that $ \lim\frac12(a_n+b/a_n)=\frac12(l+b/l)$. Hence $ l$ satisfies $ l=\frac12(l+b/l)$. Dividing by $ l$ we get $ 1=\frac12+\frac{b}{2l^2}$ which is $ 1=\frac{b}{l^2}$ which along with $ l>0$ implies that $ l=\sqrt{b}$.

Problem 5. Do the following series converge? Explain briefly why or why not:

  1. $ \displaystyle \sum_{n=1}^\infty \frac{n}{2n+1}$.

    Solution. $ \lim_{n\to\infty}\frac{n}{2n+1}=\frac12$ hence by the vanishing test the series cannot converge.

  2. $ \displaystyle \sum_{n=1}^\infty \frac{\sqrt{n}}{n\sqrt{n}+1}$.

    Solution. $ \frac{\sqrt{n}}{n\sqrt{n}+1}>\frac{\sqrt{n}}{2n\sqrt{n}}=\frac{1}{2n}$. The latter is a multiple of the harmonic series which doesn't converge, hence the original series doesn't converge either.

  3. $ \displaystyle \sum_{n=1}^\infty \frac{n^2}{n!}$.

    Solution. Ignoring the first two terms of the series, which don't change convergence anyway,

    $\displaystyle \frac{n^2}{n!}=\frac{n^2}{n(n-1)(n-2)!}<\frac{n^2}{2n^2(n-2)!}

    The latter sequence is summable as we have shown in class, hence the original series is convergent.

  4. $ \displaystyle \sum_{n=1}^\infty \frac{\log n}{n^2}$.

    Solution. The function $ f(x)=\sqrt{x}-\log x$ is positive at $ x=1$ and simple differentiation shows that $ f'(x)>0$ for $ x\geq 1$, hence it is increasing, and hence it is positive for all $ x\geq 1$. Thus $ \frac{\log n}{n^2}<\frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}$ which is summable as was shown in class.

  5. $ \displaystyle \sum_{n=2}^\infty \frac{1}{n\log n}$.

    Solution. That's a tough one. Here's a solution inspired by the solution to Problem 20 of Spivak's Chapter 23, which by itself is inspired by the proof of the divergence of the harmonic series:

    $\displaystyle \sum_{n=2}^{2^K}\frac{1}{n\log n}
\sum_{n\colon 2^{k-1}<n\leq 2^k}\frac{1}{n\log n}
\right) = \char93 .

    If we replace each of the inner sums here by the number of terms in it times the smallest of those, which is the last of those, it only becomes smaller. Hence

    $\displaystyle \char93 >\sum_{k=1}^K 2^{k-1}\frac{1}{2^k\log 2^k}
=\sum_{k=1}^K \frac{2^{k-1}}{2^kk\log2}

    The latter are partial sums of a divergent positive series, hence they approach infinity. Therefore the partial sums $ \sum_{n=2}^{2^K}\frac{1}{n\log n}$ approach infinity and our series is divergent.

The results. 75 students took the exam; the average grade is 47.4, the median is 46 and the standard deviation is 23.55.

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Dror Bar-Natan 2004-03-19