Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I:

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

2. Find an example for a pair $f$, $g$ for which
   $$\sup_{x\in A}(f(x)+g(x)) = \sup_{x\in A}f(x) + \sup_{x\in A}g(x).$$

3. Find an example for a pair $f$, $g$ for which
   $$\sup_{x\in A}(f(x)+g(x)) < \sup_{x\in A}f(x) + \sup_{x\in A}g(x).$$

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

• The domain of definition of $f(x)$.
• The behaviour of $f(x)$ near the points where it is not defined (if any) and as $x\to\pm\infty$.
• The exact coordinates of the $x$- and $y$-intercepts and all minimas and maximas of $f(x)$.

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:

Problem 3. Compute the following integrals:

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

   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

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

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

   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:

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

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

   Solution. We integrate by parts twice, as follows:

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

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

   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

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

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

   Solution.

   $$\int_0^\infty e^{-x}dx = \lim_{X\to\infty}-\left.e^{-x}\right\vert _0^X = \lim_{X\to\infty} e^{-0}-e^{-X} = 1.$$
   Or using a shorter and less precise notation, but good enough --
   $$\int_0^\infty e^{-x}dx = -\left.e^{-x}\right\vert _0^\infty = e^{-0}-e^{-\infty} = 1.$$

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

• $e(x+y)=e(x)e(y)$ for all $x,y\in{\mathbb{R}}$.
• $e(0)=1$
• $e$ is differentiable at 0 and $e'(0)=1$.

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
   $$\lim_{h\to 0} \frac{e(x+h)-e(x)}{h} = \lim_{h\to 0} \frac{e(x)e(h)-e(x)}{h} = e(x) \lim_{h\to 0} \frac{e(h)-1}{h} = e(x).$$
   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
   $$q'(x) = e'(x)e^{-x}+e(x)(e^{-x})' = e(x)e^{-x}-e(x)e^{-x} = 0.$$
   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:

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

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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