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Solution of the Final Exam

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Problem 1. Let $ f$ and $ g$ denote functions defined on some set $ A$.

  1. Prove that

    $\displaystyle \sup_{x\in A}(f(x)+g(x)) \leq \sup_{x\in A}f(x) + \sup_{x\in A}g(x). $

  2. Find an example for a pair $ f$, $ g$ for which

    $\displaystyle \sup_{x\in A}(f(x)+g(x)) = \sup_{x\in A}f(x) + \sup_{x\in A}g(x). $

  3. Find an example for a pair $ f$, $ g$ for which

    $\displaystyle \sup_{x\in A}(f(x)+g(x)) < \sup_{x\in A}f(x) + \sup_{x\in A}g(x). $

Solution.

  1. For any $ x\in A$, $ f(x)\leq\sup_{x\in A}f(x)$ and $ g(x)\leq\sup_{x\in A}g(x)$ and hence $ f(x)+g(x)\leq\sup_{x\in A}f(x) + \sup_{x\in A}g(x)$. Thus $ \sup_{x\in A}f(x) + \sup_{x\in A}g(x)$ is an upper bound for $ f(x)+g(x)$ on $ A$, and hence it is no smaller than the least upper bound for $ f(x)+g(x)$ on $ A$, which is $ \sup_{x\in A}(f(x)+g(x))$.
  2. Take say $ f$ and $ g$ to be the constant functions 0, and then $ \sup_{x\in A}(f(x)+g(x))$ and $ \sup_{x\in A}f(x) + \sup_{x\in A}g(x)$ are both 0.
  3. Take say $ f(x)=x$ and $ g(x)=-x$ on $ A=[0,1]$. Then $ f(x)+g(x)=0$ and hence $ \sup_{x\in A}(f(x)+g(x))=0$ while $ \sup_{x\in A}f(x)=1$ and $ \sup_{x\in A}g(x)=0$ and hence $ \sup_{x\in A}f(x) + \sup_{x\in A}g(x)=1$. Thus $ \sup_{x\in A}(f(x)+g(x))=0<1=\sup_{x\in A}f(x) + \sup_{x\in A}g(x)$ as required.

Problem 2. Sketch the graph of the function $ \displaystyle y=f(x)=\frac{x^2}{x^2-1}$. Make sure that your graph clearly indicates the following:

Solution. $ f(x)$ is defined for $ x\neq\pm 1$, and the following limits are easily computed: $ \lim_{x\to\pm\infty}f(x)=1$, $ \lim_{x\to-1^-}f(x)=\lim_{x\to 1^+}f(x)=\infty$ and $ \lim_{x\to-1^+}f(x)=\lim_{x\to 1^-}f(x)=-\infty$. The only solution for $ f(x)=0$ is $ x=0$, hence the only intersection of the graph of $ f(x)$ with the axes is at $ (0,0)$. Other than at $ x=0$, the numerator of $ f$ is always positive, hence the sign of the function is determined by the sign of the denominator $ x^2-1$. Thus $ f(x)\leq 0$ for $ \vert x\vert<1$ and $ f(x)>0$ for $ \vert x\vert>1$. Finally $ f'(x)=\frac{2x(x^2-1)-x^22x}{(x^2-1)^2}=-\frac{2x}{(x^2-1)^2}$ and thus $ f'$ is positive and $ f$ is increasing (locally) for $ x<0$ and $ f'$ is negative and $ f$ is decreasing (locally) for $ x>0$. Thus overall the graph is:

Plot of f(x)

Problem 3. Compute the following integrals:

  1. $ \displaystyle \int\frac{x^2+1}{x+1}dx$

    Solution. By long division of polynomials, $ x^2+1=(x+1)(x-1)+2$. Thus we can rewrite our integral as a sum of two terms as follows

    $\displaystyle \int\frac{(x+1)(x-1)}{x+1}dx + \int\frac{2}{x+1}dx
= \int (x-1)dx + 2\int\frac{1}{x+1}dx
$

    $\displaystyle = \frac{x^2}{2} - x + 2\log(x+1).
$

  2. $ \displaystyle \int\frac{x+1}{x^2+1}dx$

    Solution. Again we rewrite the integral as a sum of two terms. On the first we perform the substitution $ u=x^2$; the second is elementary:

    $\displaystyle \frac12\int\frac{2x dx}{x^2+1} + \int\frac{dx}{x^2+1}
= \frac12\int\frac{du}{u+1} + \arctan x
= \frac12\log(u+1) + \arctan x
$

    $\displaystyle = \frac12\log(x^2+1) + \arctan x.
$

  3. $ \displaystyle \int x^2\sin x dx$

    Solution. We integrate by parts twice, as follows:

    $\displaystyle \int x^2\sin x dx
= x^2(-\cos x) - \int 2x(-\cos x)dx
$

    $\displaystyle = -x^2\cos x -2x(-\sin x) - \int 2(-\sin x)dx
= 2x\sin x - x^2\cos x + 2\cos x.
$

  4. $ \displaystyle \int\frac{dx}{\sqrt{1+e^x}}$

    Solution. Set $ u=\sqrt{1+e^x}$ and then $ e^x=u^2-1$ and $ du=\frac{e^xdx}{2\sqrt{1+e^x}}=\frac{(u^2-1)dx}{2u}$ and so $ dx=\frac{2udu}{u^2-1}$ and

    $\displaystyle \int\frac{dx}{\sqrt{1+e^x}}
= \int\frac{2udu}{u(u^2-1)}
= \int\frac{du}{u-1}-\int\frac{du}{u+1}
$

    $\displaystyle = \log(u-1)-\log(u+1)
= \log\frac{u-1}{u+1}
= \log\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}.
$

  5. $ \displaystyle \int_0^\infty e^{-x}dx$

    Solution.

    $\displaystyle \int_0^\infty e^{-x}dx
= \lim_{X\to\infty}-\left.e^{-x}\right\vert _0^X
= \lim_{X\to\infty} e^{-0}-e^{-X}
= 1.
$

    Or using a shorter and less precise notation, but good enough --

    $\displaystyle \int_0^\infty e^{-x}dx
= -\left.e^{-x}\right\vert _0^\infty
= e^{-0}-e^{-\infty}
= 1.
$

Problem 4. Agents of the CSIS have secretly developed a function $ e(x)$ that has the following properties:

Prove the following:
  1. $ e$ is everywhere differentiable and $ e'=e$.
  2. $ e(x)=e^x$ for all $ x\in{\mathbb{R}}$. The only lemma you may assume is that if a function $ f$ satisfies $ f'(x)=0$ for all $ x$ then $ f$ is a constant function.

Solution.

  1. The given fact that $ 1=e'(0)$ means that $ 1=\lim_{h\to 0}\frac{e(h)-e(0)}{h}=\lim_{h\to 0}\frac{e(h)-1}{h}$. Hence, using $ e(x+h)=e(x)e(h)$ we get

    $\displaystyle \lim_{h\to 0} \frac{e(x+h)-e(x)}{h}
= \lim_{h\to 0} \frac{e(x)e(h)-e(x)}{h}
= e(x) \lim_{h\to 0} \frac{e(h)-1}{h} = e(x).
$

    This proves both that $ e$ is differentiable at $ x$ and that $ e'(x)=e(x)$.
  2. Consider $ q(x)=e(x)e^{-x}$. Differentiating we get

    $\displaystyle q'(x) = e'(x)e^{-x}+e(x)(e^{-x})' = e(x)e^{-x}-e(x)e^{-x} = 0. $

    Hence $ q(x)$ is a constant function. But $ q(0)=e(0)e^0=1\cdot 1=1$, hence this constant must be $ 1$. So $ e(x)e^{-x}=1$ and thus $ e(x)=e^x$.

Problem 5.

  1. Prove that if a sequence of continuous functions $ f_n$ converges uniformly to a function $ f$ on some interval $ [a,b]$, then $ f$ is continuous on $ [a,b]$.
  2. Prove that the series $ \sum_{n=1}^\infty\frac{1}{2^n}\sin(3^nx)$ converges on $ (-\infty,\infty)$ and that its sum is a continuous function of $ x$.

Solution.

  1. See Spivak's Theorem 2 of Chapter 24.
  2. $ \vert\frac{1}{2^n}\sin(3^nx)\vert\leq\frac{1}{2^n}$ and $ \sum_{n=1}^\infty\frac{1}{2^n}$ converges. Hence by the Weierstrass M-Test the series $ \sum_{n=1}^\infty\frac{1}{2^n}\sin(3^nx)$ converges uniformly. As each of the terms $ \frac{1}{2^n}\sin(3^nx)$ is continuous, the first part of this question implies that so is the sum.

Problem 6. Prove that the complex function $ z\mapsto\bar{z}$ is everywhere continuous but nowhere differentiable.

Solution. The key point is that $ \vert w\vert=\vert\bar{w}\vert$ for every complex number $ w$. Let $ \epsilon>0$ and set $ \delta=\epsilon$. Now if $ \vert z-z_0\vert<\delta$ then $ \vert\bar{z}-\bar{z}_0\vert=\vert\overline{z-z_0}\vert=\vert z-z_0\vert<\delta=\epsilon$. This proves the continuity of $ z\mapsto\bar{z}$. Let us check if this function is differentiable:

$\displaystyle \lim_{h\to 0}\frac{\overline{z+h}-\bar{z}}{h}
= \lim_{h\to 0}\frac{\bar{z}+\bar{h}-\bar{z}}{h}
= \lim_{h\to 0}\frac{\bar{h}}{h}.
$

If we restrict our attention to real $ h$ then the latter quotient is always $ 1$, so the limit would be $ 1$. If we restrict our attention to imaginary $ h$, $ h=iy$ with real $ y$, then that quotient is $ \frac{\bar{h}}{h}=\frac{-iy}{iy}=-1$ so the limit would be $ -1$. Hence the limit cannot exist and $ z\mapsto\bar{z}$ is not differentiable at (an arbitrary) $ z$.

The results. 76 students took the exam; the average grade was 72.66/120, the median was 71.5/120 and the standard deviation was 25.5. The overall grade average for the course (of $ X=0.05T_1+0.15T_2+0.1T_3+0.1T_4+0.2HW+0.4\cdot 100(F/120)$) was 66.92, the median was 64.9 and the standard deviation was 17.16. Finally, the transformation $ X\mapsto 100(X/100)^\gamma$ was applied to the grades, with $ \gamma=0.82$. This made the average grade 71.55, the median 70 and the standard deviation 15.31. There were 25 A's (grades above 80) and 5 failures (grades below 50).

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Dror Bar-Natan 2003-05-01