Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I:

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Solution of the Final Exam

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Problem 1. We say that a set $A$ of real numbers is dense if for any open interval $I$, the intersection $A\cap I$ is non-empty.

1. Give an example of a dense set $A$ whose complement $A^c=\{x\in{\mathbb{R}}: x\not\in A\}$ is also dense.
2. Give an example of a non-dense set $B$ whose complement $B^c=\{x\in{\mathbb{R}}: x\not\in B\}$ is also not dense.
3. Prove that if $f\colon{\mathbb{R}}\to{\mathbb{R}}$ is an increasing function ($f(x)<f(y)$ for $x<y$) and if the range $\{f(x): x\in{\mathbb{R}}\}$ of $f$ is dense, then $f$ is continuous.

Solution.

1. Take for example $A={\mathbb{Q}}$, the set of rational numbers. Then $A^c$ is the set of irrational numbers. We've seen in class that between any two (different) numbers (i.e., within any open interval) there is a rational number and there is an irrational number. Hence both $A$ and $A^c$ are dense.
2. Take for example $B=[0,\infty]$, the set of non-negative numbers. Then $B^c=(-\infty,0)$ is the set of negative numbers. The set $B$ is not dense because, for example, it's intersection with the interval $(-2,-1)$ is empty. The set $B^c$ is not dense because, for example, it's intersection with the interval $(1,2)$ is empty.
3. We have to show that for every $a\in{\mathbb{R}}$ and for every $\epsilon>0$ there is a $\delta>0$ so that $\vert x-a\vert<\delta$ implies $\vert f(x)-f(a)\vert<\epsilon$. So let $\epsilon>0$ be given. By the density of $A:=\{f(x): x\in{\mathbb{R}}\}$ we know that we can find an element of $A$ in the interval $(f(a)-\epsilon,f(a))$ and another element of $A$ in the interval $(f(a),f(a)+\epsilon)$. That is, we can find $x_1$ and $x_2$ so that $f(a)-\epsilon<f(x_1)<f(a)$ and $f(a)<f(x_2)<f(a)+\epsilon$. It follows from the monotonicity of $f$ that $x_1<a$ and that $a<x_2$. Now set $\delta=\min(a-x_1,x_2-a)$ (this is a positive number because $x_1<a$ and $a<x_2$). Finally if $\vert x-a\vert<\delta$ then $x$ is in the interval $(a-\delta,a+\delta)\subset(a-(a-x_1),a+(x_2-a))=(x_1,x_2)$. By the monotonicity of $f$ it follows that $f(x)$ is in the interval $(f(x_1),f(x_2))\subset(f(a)-\epsilon,f(a)+\epsilon)$, and so $\vert f(x)-f(a)\vert<\epsilon$, as required.

Problem 2. Sketch the graph of the function $y=f(x)=xe^{-x^2/2}$. 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. Our function is defined for all $x$. As $x$ goes to $\pm\infty$ exponentials dominate polynomials, and so certainly $e^{x^2/2}$ gets much bigger than $x$. So $\lim_{x\to\pm\infty}f(x)=0$. Solving the equation $xe^{-x^2/2} = 0$ we see that the only intersection of the graph of $f$ with the axes is at $(0,0)$. We can compute $f'(x) = x'e^{-x^2/2}+x\left(e^{-x^2/2}\right)' = e^{-x^2/2}-x^2e^{-x^2/2} = (1-x^2)e^{-x^2/2}$ and $f''(x) = (1-x^2)'e^{-x^2/2}+(1-x^2)\left(e^{-x^2/2}\right)' = -2xe^{-x^2/2}-x(1-x^2)e^{-x^2/2}=x(x^2-3)e^{-x^2/2}$. Solving $f'(x)=0$ we see that the only critical points are when $1-x^2=0$. That is, at $x=\pm 1$. As $f''(1)=-2e^{-1/2}<0$, the point $(1,f(1))=(1,e^{-1/2})$ is a local max. As $f''(-1)=2e^{-1/2}>0$, the point $(-1,f(-1))=(-1,-e^{-1/2})$ is a local min. As there are no other critical points and the behaviour of $f$ near the ends of its domain of deifnition is mute (as determined before), $(1,e^{-1/2})$ is actually a global max and $(-1,-e^{-1/2})$ is actually a global min. Thus overall the graph is:

Problem and Solution 3. Compute the following derivative and the following integrals:

1. Using the fundamental theorem of calculus in the form $\frac{d}{du}\int_0^u f(t)dt=f(u)$ and the chain rule with $u=\sin x$ we get
   $$\frac{d}{dx}\left(\int_0^{\sin x}\sqrt{\arcsin t} dt\right) = \sqrt{\arcsin\sin x}\cdot(\sin x)' = \sqrt{x}\cos x.$$

2. We make the substitution $u=\sqrt{x}$ (and thus $x=u^2$ and $dx=2udu$) to compute
   $$\int\frac{e^{\sqrt{x}}}{\sqrt{x}}dx = \int\frac{e^u}{u}2udu = 2\int e^udu = 2e^u+C = 2e^{\sqrt{x}}+C.$$

3. Integrating by parts twice we get
   $$\int x^2e^xdx = x^2e^x-\int 2xe^x dx = x^2e^x-2xe^x+\int 2e^x dx = x^2e^x-2xe^x+2e^x+C.$$

4. We make the substitution $u=2^x$ (and thus $x=\log_2u$ and $dx=\frac{du}{u\log 2}$) to compute
   $$\int\frac{4^xdx}{2^x+1} = \int\frac{u^2\frac{du}{u\log 2}}{u+1} =... ...log 2}\int\frac{udu}{u+1} = \frac{1}{\log 2}\int\left(1-\frac{1}{u+1}\right)du$$
   $$= \frac{1}{\log 2}(u-\log\vert u+1\vert)+C = \frac{1}{\log 2}(2^x-\log\vert 2^x+1\vert)+C.$$

5. We use the factorization $x^2-3x+2=(x-1)(x-2)$ to get
   $$\int\frac{dx}{x^2-3x+2} = \int\frac{dx}{(x-1)(x-2)} = \int\left(\frac{dx}{x-2}-\frac{dx}{x-1}\right)$$
   $$= \log\vert x-2\vert-\log\vert x-1\vert+C = \log\left\vert\frac{x-2}{x-1}\right\vert+C.$$

Problem 4. In solving this problem you are not allowed to use any properties of the exponential function $e^x$.

1. Two differentiable functions, $e_1(x)$ and $e_2(x)$, defined over the entire real line ${\mathbb{R}}$, are known to satisfy $e_1'(x)=e_1(x)$, $e_2'(x)=e_2(x)$, $e_1(x)>0$ and $e_2(x)>0$ for all $x\in{\mathbb{R}}$ and also $e_1(0)=e_2(0)$. Prove that $e_1$ and $e_2$ are the same. That is, prove that $e_1(x)=e_2(x)$ for all $x\in{\mathbb{R}}$.
2. A differentiable function $e(x)$ defined over the entire real line ${\mathbb{R}}$ is known to satisfy $e'(x)=e(x)$ and $e(x)>0$ for all $x\in{\mathbb{R}}$ and also $e(0)=1$. Prove that $e(x+y)=e(x)e(y)$ for all $x,y\in{\mathbb{R}}$.

Solution.

1. Set $f(x):=e_1(x)/e_2(x)$ (this is well defined because $e_2(x)$ is never 0) and compute
   $$f' =\left(\frac{e_1}{e_2}\right)' =\frac{e_1'e_2-e_1e_2'}{e_2^2} =\frac{e_1e_2-e_1e_2}{e_2^2} =0.$$
   So $f$ is a constant. But $f(0)=e_1(0)/e_2(0)=1$, so that constant is 1 and $e_1(x)/e_2(x)=1$ for all $x$. This means that $e_1=e_2$.
2. Fix $y$ and set $e_1(x)=e(x+y)$ and $e_2(x)=e(x)e(y)$. Then $(e_1(x))'=(e(x+y))'=e(x+y)=e_1(x)$ and $(e_2(x))'=(e(x)e(y))'=(e(x))'e(y)=e(x)(e(y)=e_2(x)$ and $e_1(0)=e(0+y)=e(y)=1e(y)=e(0)e(y)=e_2(0)$. All the other conditions of the first part of this question are even easier to verify, and so the conclusion of that part holds. Namely, $e_1=e_2$, which means $e(x+y)=e(x)e(y)$.

Problem 5. In solving this problem you are not allowed to use any properties of the trigonometric functions.

1. A twice-differentiable function $c(x)$ defined over the entire real line ${\mathbb{R}}$ is known to satisfy $c''(x)=-c(x)$ for all $x\in{\mathbb{R}}$ and also $c(0)=c'(0)=0$. Write out the degree $n$ Taylor polynomial $P_{n,a,c}(x)$ of $c$ at $a=0$.
2. Write a formula for the remainder term $R_{n,0,c}(x):=c(x)-P_{n,0,c}(x)$. (To keep the notation simple, you are allowed to assume that $n$ is even or even that $n$ is divisible by 4).
3. Prove that $c$ is the zero function: $c(x)=0$ for all $x\in{\mathbb{R}}$.

Solution.

1. From $c''(x)=-c(x)$ it is clear that $c^{(2k)}=(-1)^kc$ and that $c^{(2k+1)}=(-1)^kc'$. So $c^{(2k)}(0)=(-1)^kc(0)=0$ and $c^{(2k+1)}(0)=(-1)^kc'(0)=0$ and hence all the coefficients of $P_{n,a,c}(x)$ are 0. In other words, $P_{n,a,c}=0$.
2. If $n$ is divisible by $4$ then $c^{(n+1)}=c'$ and so the remainder formula says that for any $x\neq 0$ there is a $t$ between 0 and $x$ for which
   $$R_{n,0,c}(x) = \frac{c^{(n+1)}(t)}{(n+1)!}x^{n+1} = \frac{c'(t)}{(n+1)!}x^{n+1}.$$

3. Factorials grow faster then exponentials, so in the remainder formula the denominator $(n+1)!$ grows faster then the term $x^{n+1}$, while the numerator $c'(t)$ is bounded (by the theorem that a continuous function on a closed interval is bounded). So the remainder goes to 0 when $n$ goes to $\infty$, and hence $\lim_{n\to\infty}P_{n,a,c}(x)=c(x)$. But $P_{n,a,c}(x)=0$ for all $n$, so necessarily $c(x)=0$.

Remark 1. Two alternative forms of the remainder formula are

$$\frac{c^{(n+1)}(t)}{n!}x(x-t)^n = \frac{c'(t)}{n!}x(x-t)^n$$   and$$\quad \int_0^x\frac{c^{(n+1)}(t)}{n!}(x-t)^ndt = \int_0^x\frac{c'(t)}{n!}(x-t)^ndt.$$

Either one of those could equaly well be used to solve part 3 of the problem.

Remark 2. There is an alternative approach to the whole problem; start with part 3 and go backwards. To do part 3, consider the function $f:=c^2+(c')^2$. We have $f'=2cc'+2c'c''=2cc'-2c'c=0$, so $f$ is a constant function. But $f(0)=c(0)^2+c'(0)^2=0^2+0^2=0$, so $f$ must be the 0 function. But $f$ is a sum of squares, and the only way a sum of squares can be 0 is if each summand is 0. So $c^2=0$ and hence $c=0$ as required in part 3. But if $c$ is the 0 function then its Taylor polynomials are all 0 and the remainder terms are also all 0, solving parts 1 and 2 as well. This is not the solution I had in mind when I wrote the problem, but people who solved the problem this way got full credit.

Problem 6. In solving this problem you are not allowed to use the irrationality of $\pi$, but you are allowed, indeed advised, to borrow a few lines from the proof of the irrationality of $\pi$.

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.

Solution. There is no such polynomial. Had there been one, we would have

$$0<\int_0^\pi p(x)\sin x dx<\int_0^\pi\frac12\sin x dx=1,$$

but also, by repeated integration by parts (an even number of times, for simplicity),

$$\int_0^\pi p(x)\sin x dx =\left.-p(x)\cos x\right\vert _0^\pi + \int_0^\pi p'(x)\cos x dx$$

$$=\left.-p(x)\cos x + p'(x)\sin x\right\vert _0^\pi - \int_0^\pi p''(x)\sin x dx =\ldots$$

$$= \left.\text{ (terms involving $\pm 1$, $p^{(k)}(x)$, $\sin x$ and $\cos x$) }\right\vert _0^\pi \pm p^{(2n)}(x)\sin x dx.$$

For any $n$ the first term in this formula involves only integers (as $p^{(k)}(0)$, $p^{(k)}(\pi)$, $\sin 0$, $\sin\pi$, $\cos 0$ and $\cos\pi$ are all integers), and if $2n$ is larger than the degree of $p$, the second term is 0. So $\int_0^\pi p(x)\sin x dx$ is an integer. But by the first formula it is in $(0,1)$. That can't be.

The results. 80 students took the exam; the average grade was 69.33/120, the median was 71.5/120 and the standard deviation was 26.51. 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 68.5, the median was 71.57 and the standard deviation was 18.64. Finally, the transformation $X\mapsto 100(X/100)^\gamma$ was applied to the grades, with $\gamma=0.92$. This made the average grade 70.41, the median 73.5 and the standard deviation 17.77. There were 30 A's (grades higher or equal to 80) and 12 failures (grades below 50).

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