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

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Solution of Term Exam 1

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Problem 1. All that is known about the angle $\alpha$ is that $\tan\frac{\alpha}{2}=\sqrt{2}$. Can you find $\sin\alpha$ and $\cos\alpha$? Explain your reasoning in full detail.

Solution. (Graded by C.-N. (J.) Hung) In class we wrote the formula $\sin2\beta=2\sin\beta\cos\beta$. Also using $\sin^2\beta+\cos^2\beta=1$ and taking $\beta=\frac{\alpha}{2}$ we get

$$\sin\alpha = \frac {2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} {\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}.$$

Dividing the numerator and denominator by $\cos^2\frac{\alpha}{2}$ this becomes

$$\frac{2\tan\frac{\alpha}{2}}{\tan^2\frac{\alpha}{2}+1} = \frac{2\sqrt{2}}{\sqrt{2}^2+1} = \frac{2\sqrt{2}}{3}.$$

Likewise using $\cos2\beta=\cos^2\beta-\sin^2\beta$ we get

$$\cos\alpha =\frac {\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}... ...}}{1+\tan^2\frac{\alpha}{2}} =\frac{1-\sqrt{2}^2}{1+\sqrt{2}^2} =-\frac{1}{3}.$$

Problem 2.

1. State the definition of the natural numbers.
2. Prove that every natural number $n$ has the property that whenever $m$ is natural, so is $m+n$.

Solution. (Graded by V. Tipu)

1. The set of natural numbers ${\mathbb{N}}$ is the smallest set of numbers for which
   • $1\in{\mathbb{N}}$,
   • if $n\in{\mathbb{N}}$ then $n+1\in{\mathbb{N}}$.
   Alternatively, the set of natural numbers ${\mathbb{N}}$ is the intersection of all sets $I$ of numbers satisfying
   • $1\in I$,
   • if $n\in I$ then $n+1\in I$.

2. Let $P(n)$ be the assertion ``whenever $m$ is natural, so is $m+n$''. We prove $P(n)$ by induction on $n$:
   1. $P(1)$ asserts that ``whenever $m$ is natural, so is $m+1$''. This is true by the second bullet in the definition of ${\mathbb{N}}$.
   2. Assume $P(n)$, that is, assume that whenever $m$ is natural, so is $m+n$. Let $m$ be a natural number. Then $m+(n+1)=(m+n)+1$ is a natural number because by $P(n)$ the number $m+n$ is natural and because adding one to a natural number gives a natural number by the second bullet in the definition of ${\mathbb{N}}$. So we have shown that whenever $m$ is natural so is $m+(n+1)$, and this is the assertion $P(n+1)$.

Problem 3. Recall that a function $g$ is called ``even'' if $g(x)=g(-x)$ for all $x$ and ``odd'' if $g(-x)=-g(x)$ for all $x$, and let $f$ be some arbitrary function.

1. Find an even function $E$ and an odd function $O$ so that $f=E+O$.
2. Show that if $f=E_1+O_1=E_2+O_2$ where $E_1$ and $E_2$ are even and $O_1$ and $O_2$ are odd, then $E_1=E_2$ and $O_1=O_2$.

Solution. (Graded by C. Ivanescu)

1. Set $E(x)=\frac12(f(x)+f(-x))$ and $O(x)=\frac12(f(x)-f(-x))$. Then $E(x)+O(x)=\frac12(f(x)+f(-x)+f(x)-f(-x))=\frac12(2f(x))=f(x)$ while $E(-x)=\frac12(f(-x)+f(-(-x)))=\frac12(f(x)+f(-x))=E(x)$ (so $E$ is even) and $O(-x)=\frac12(f(-x)-f(-(-x)))=-\frac12(f(x)-f(-x))=-O(x)$ (so $O$ is odd).

2. Assume $f=E+O$ where $E$ is even and $O$ is odd. Then
   $$f(x)+f(-x)=E(x)+O(x)+E(-x)+O(-x)=E(x)+O(x)+E(x)-O(x)=2E(x).$$
   So necessarily $E(x)=\frac12(f(x)+f(-x))$. Now if $f=E_1+O_1=E_2+O_2$ as above, then both $E_1$ and $E_2$ can play the role of $E$ in this argument, so they are both equal to $\frac12(f(x)+f(-x))$ and in particular they equal each other. Likewise,
   $$f(x)-f(-x)=E(x)+O(x)-E(-x)-O(-x)=E(x)+O(x)-E(x)+O(x)=2O(x)$$
   and arguing like before, $O_1(x)=\frac12(f(x)-f(-x))=O_2(x)$.

Problem 4. Sketch, to the best of your understanding, the graph of the function

$$f(x)=\frac{1}{x^2-1}.$$

(What happens for $x$ near 0? Near $\pm 1$? For large $x$? Is the graph symmetric? Does it appear to have a peak somewhere?)

Solution. (Graded by C. Ivanescu)

If $\vert x\vert>1$ then $x^2-1>0$ and so $\frac{1}{x^2-1}>0$; furthermore, the larger $\vert x\vert$ is (while $\vert x\vert>1$), the larger $x^2-1$ is and hence the smaller $\frac{1}{x^2-1}$ is. When $\vert x\vert$ approaches $1$ from above, $x^2-1$ approaches 0 from above and hence $\frac{1}{x^2-1}$ becomes larger and larger. If $\vert x\vert<1$ the $x^2-1<0$ and so $\frac{1}{x^2-1}<0$. When $x=0$, $f(x)=-1$ and when $\vert x\vert$ approaches $1$ from below, $x^2$ approaches $1$ from below and $x^2-1$ approaches 0 from below, and so $\frac{1}{x^2-1}$ becomes more and more negative. In summary, the graph looks something like:

Problem 5.

1. Suppose that $f(x)\leq g(x)$ for all $x$, and that the limits $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist. Prove that $\lim_{x\to a}f(x)\leq\lim_{x\to a}g(x)$.
2. Suppose that $f(x)<g(x)$ for all $x$, and that the limits $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist. Is it always true that $\lim_{x\to a}f(x)<\lim_{x\to a}g(x)$? (If you think it's always true, write a proof. If you think it isn't always true, provide a counterexample).

Solution. (Graded by C.-N. (J.) Hung)

1. Let $l=\lim_{x\to a}f(x)$ and $m=\lim_{x\to a}g(x)$ and assume by contradiction that $l>m$; that is, that $\epsilon:=\frac{l-m}{2}>0$. Use the existence of the two limits to find $\delta_1>0$ and $\delta_2>0$ so that
   $$0<\vert x-a\vert<\delta_1 \Longrightarrow \vert f(x)-l\vert<\epsilon$$
   and
   $$0<\vert x-a\vert<\delta_2 \Longrightarrow \vert g(x)-m\vert<\epsilon.$$
   Now choose some specific $x\neq a$ for which both $\vert x-a\vert<\delta_1$ and $\vert x-a\vert<\delta_2$. But then $\vert f(x)-l\vert<\epsilon$ and so $f(x)>l-\epsilon$ while $\vert g(x)-m\vert<\epsilon$ and so $g(x)<m+\epsilon$. Therefore remembering that $f(x)\leq g(x)$ for all $x$ we get
   $$l-\epsilon<f(x)\leq g(x)<m+\epsilon$$
   or
   $$l-\frac{l-m}{2} < m+\frac{l-m}{2}$$
   or
   $$\frac{m+l}{2} < \frac{m+l}{2}$$
   which is a contradiction. Thus the assumption that $l>m$ must be incorrect and thus $m\leq l$.

2. Take $f(x)=0$ for all $x$ and $g(x)=x^2$ for all $x\neq 0$ and $g(0)=157$. Then $f(x)<g(x)$ for all $x$ but $\lim_{x\to 0}f(x)=0=\lim_{x\to 0}g(x)$. So it isn't always true that if $f(x)<g(x)$ for all $x$ and the limits exist, then $\lim_{x\to a}f(x)<\lim_{x\to a}g(x)$.

The results. 105 students took the exam; the average grade was 67.19, the median was 70 and the standard deviation was 21.12.

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