Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (75) Next: Homework Assignment 19
Previous: Term Exam 3

Solution of Term Exam 3

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Problem 1. In a very condensed form, the definition of integration is as follows: For $ f$ bounded on $ [a,b]$ and $ P:
a=t_0<t_1<\dots<t_n=b$ a partition of $ [a,b]$ set $ m_i=\inf_{[t_{i-1},t_i]}f(x)$, $ M_i=\sup_{[t_{i-1},t_i]}f(x)$, $ L(f,P)=\sum_{i=1}^n m_i(t_i-t_{i-1})$ and $ U(f,P)=\sum_{i=1}^n
M_i(t_i-t_{i-1})$. Then set $ L(f)=\sup_P L(f,P)$ and $ U(f)=\inf_P
U(f,P)$. Finally, if $ U(f)=L(f)$ we say that ``$ f$ is integrable on $ [a,b]$'' and set $ \int_a^b f=\int_a^b f(x)dx=U(f)=L(f)$.

From this definition alone, without using anything proven in class about integration, prove that the function $ f$ given below is integrable on $ [-1,1]$ and compute its integral $ \int_{-1}^1f$:

$\displaystyle f(x)=\begin{cases}0 & x\neq 0  1 & x=0. \end{cases} $

Solution. (Graded by Cristian Ivanescu) Let $ P:-1=t_0<t_1<\dots<t_n=1$ be an arbitrary partition of $ [-1,1]$. Then for any $ i$ the infimum $ m_i=\inf_{[t_{i-1},t_i]}f(x)$ is 0 and so $ L(f,P)=\sum_{i=1}^n
m_i(t_i-t_{i-1})=0$. Thus $ L(f)=\sup_P L(f,P)=0$. At the same time, for any $ i$ the supremum $ M_i=\sup_{[t_{i-1},t_i]}f(x)$ is $ \geq 0$, and hence $ U(f,P)=\sum_{i=1}^n M_i(t_i-t_{i-1})\geq 0$ and so $ U(f)=\inf_P
U(f,P)\geq 0$. Now let $ 0<\epsilon<1$ be given and let $ P_\epsilon$ be the partition $ -1=t_0<t_1=-\frac{\epsilon}{2}<\frac{\epsilon}{2}=t_2<1=t_3$. Then $ M_1=M_3=0$ while $ M_2=1$ and so $ U(f,P_\epsilon) =
0(1-\frac{\epsilon}{2}) + 1(\frac{\epsilon}{2} + \frac{\epsilon}{2}) +
0(1-\frac{\epsilon}{2})=\epsilon$. Thus $ U(f)=\inf_P
U(f,P)\leq\epsilon$. But this is true for any $ 0<\epsilon<1$ and we already know that $ U(f)\geq 0$. So it must be that $ U(f)=0$. Thus $ U(f)=L(f)=0$ and hence $ f$ is integrable on $ [-1,1]$ and its integral is $ \int_{-1}^1 f=U(f)=L(f)=0$.

Problem 2. Prove that the function

$\displaystyle g(x):=\int_0^x\frac{dt}{1+t^2} + \int_0^{1/x}\frac{dt}{1+t^2} $

is a constant function.

Solution. (Graded by Julian C.-N. Hung) Differentiate $ g$ using the first fundamental theorem of calculus. The first summand yields $ \frac{1}{1+x^2}$. The second summand is the first summand pre-composed with the function $ x\mapsto
1/x$. So by the chain rule, the derivative of the second summand is $ \frac{1}{1+(1/x)^2}(1/x)'=-\frac{1}{1+(1/x)^2}(1/x^2)=-\frac{1}{1+x^2}$. $ g'$ is the sum of these two terms, $ g'=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$. Hence $ g$ is a constant.

Problem 3. In class we have proven that a twice-differentiable function $ f$ satisfying the equation $ f''=-f$ is determined by $ f(0)$ and $ f'(0)$. Use this fact and the known formulas for the derivatives of $ \cos x$ and $ \sin x$ to derive a formula for $ \cos(\alpha+\beta)$ in terms of $ \cos\alpha$, $ \cos\beta$, $ \sin\alpha$ and $ \sin\beta$.

Solution. (Graded by Julian C.-N. Hung) Let $ \beta$ be a constant and consider the functions $ f_1(\alpha)=\cos(\alpha+\beta)$ and $ f_2(\alpha)=\cos\alpha\cos\beta - \sin\alpha\sin\beta$. Then $ f_1'' =
(\cos(\alpha+\beta))'' = (-\sin(\alpha+\beta))' =
-\cos(\alpha+\beta)=-f_1$ and $ f_2''=(\cos\alpha\cos\beta -
\sin\alpha\sin\beta)'' = (-\sin\alpha\cos\beta-\cos\alpha\sin\beta)' =
-\cos\alpha\cos\beta + \sin\alpha\sin\beta=-f_2$ so both $ f_1$ and $ f_2$ satisfy $ f''=-f$. We also have $ f_1(0)=\cos\beta=\cos
0\cos\beta-\sin 0\sin\beta=f_2(0)$ and $ f'_1(0)=-\sin\beta=(-\sin
0)\cos\beta-\cos 0\sin\beta=f_2'(0)$. So by what we have proven in class $ f_1=f_2$ or $ \cos(\alpha+\beta)=\cos\alpha\cos\beta -
\sin\alpha\sin\beta$.

Problem 4. The function $ F$ is defined by $ F(x):=x^x$.

  1. Compute $ F'(x)$ for all $ x>0$.
  2. Explain why $ F(x)$ has a differentiable inverse for $ x>\frac{1}{e}$.
  3. Let $ S$ be the inverse function of $ F$ (with the domain of $ F$ considered to be $ (\frac{1}{e},\infty)$). Compute $ S'(x)$ and simplify your result as much as you can. Your end result may still contain $ S(x)$ in it, but not $ S'$, $ F$ or $ F'$.

Solution. (Graded by Vicentiu Tipu)

  1. $ \displaystyle
F'(x) = \left(e^{x\log x}\right)'
= e^{x\log x}\left(\log x+x\frac{1}{x}\right)=x^x(1+\log x)
= F(x)(1+\log x)
$.
  2. For $ x>\frac{1}{e}$ we have that $ \log x>\log\frac{1}{e}=-1$ and hence $ 1+\log x>0$ and $ F'(x)>0$. So $ F$ is increasing on $ (\frac{1}{e},\infty)$. It is also differentiable on that interval, so by a theorem proven in class, it has a differentiable inverse.
  3. $ \displaystyle
S'(x)=\frac{1}{F'(S(x))}
= \frac{1}{F(S(x))(1+\log S(x))}
= \frac{1}{x(1+\log S(x))}
$.

The results. 82 students took the exam; the average grade was 69.3, the median was 78 and the standard deviation was 26.76.

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Dror Bar-Natan 2004-10-18