Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (27) Next: Homework Assignment 7
Previous: Term Exam 1

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

$\displaystyle \sin\alpha
= \frac

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

$\displaystyle \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

$\displaystyle \cos\alpha

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 Alternatively, the set of natural numbers $ {\mathbb{N}}$ is the intersection of all sets $ I$ of numbers satisfying

  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

    $\displaystyle 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,

    $\displaystyle 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

$\displaystyle 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:

Plot of f(x)

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

    $\displaystyle 0<\vert x-a\vert<\delta_1 \Longrightarrow \vert f(x)-l\vert<\epsilon $


    $\displaystyle 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

    $\displaystyle l-\epsilon<f(x)\leq g(x)<m+\epsilon $


    $\displaystyle l-\frac{l-m}{2} < m+\frac{l-m}{2} $


    $\displaystyle \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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Dror Bar-Natan 2004-10-18