MATH 447: Real Variables - Homework 10
Jerich Lee
December 22, 2024
Problem.
[26.6] Let $ s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots{} $ and $ c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots{} $ for $ x \in{} \mathbb{R} $.
[26.6] Let $ s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots{} $ and $ c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots{} $ for $ x \in{} \mathbb{R} $.
- Prove $ s' = c $ and $ c' = -s $.
- Prove $ (s^2 + c^2)' = 0 $.
- Prove $ s^2 + c^2 = 1 $.
Solution.
- Proof. For $x\in{}\mathbb{R}$: $$\begin{align} s(x) &= x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\ldots \\[10pt] c(x) &= 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\ldots \\[10pt] \lim_{t \to x} \phi(t)&=\lim_{t \to x} \frac{f(t)-f(x)}{t-x}\\[10pt] s(x)&=\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)}! \\[10pt] &=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{\left( 2n+1 \right)! } \\[10pt] c(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{(2x)}}{\left( 2n \right)! }\\[10pt] s^\prime (x)&=\sum_{n=0}^{\infty} \frac{(-1)^{n}(2n+1)x^{2n}}{\left( 2n+1 \right) !} \\[10pt] &= \frac{(-1)^{n}x^{2n}}{\left( 2n \right) !} \\[10pt] &= c(x) \end{align}$$∎
- Proof. $$\begin{align} c(x)&=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n}}{\left( 2n \right) !}\\[10pt] c^\prime (x)&=\sum_{n=0}^{\infty} \frac{(-1)^{n}(2n)x^{2n-1}}{(2n)!}\\[10pt] &= \frac{(-1)^{n}x^{2n-1}}{(2n-1)!}\\[10pt] &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!}\\[10pt] &= -s(x) \end{align}$$∎
- Proof. $$\begin{align} \left( s^{2}+c^{2} \right)^\prime &=0 \\[10pt] &= \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k+1}}{(2k+1)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}\\[10pt] &= \sum_{k=0}^{\infty} \left( \frac{(-1)^{k}x^{2k+1}}{(2k+1)!} + \frac{(-1)^{k}x^{2k}}{(2k)!} \right) \\[10pt] &= 0 \end{align}$$∎
- todo
Problem.
[33.3] A function $ f $ on $[a, b]$ is called a step function if there exists a partition $$\begin{align} P = \{a = u_0 < u_1 < \cdots < u_m = b\} \end{align}$$ of $[a, b]$ —not $ P = \{a = u_0 < u_1 < \cdots{} < c_m = b\} $, as stated in the textbook— such that $ f $ is constant on each interval $(u_{j-1}, u_j)$, say $ f(x) = c_j $ for $ x $ in $(u_{j-1}, u_j)$.
[33.3] A function $ f $ on $[a, b]$ is called a step function if there exists a partition $$\begin{align} P = \{a = u_0 < u_1 < \cdots < u_m = b\} \end{align}$$ of $[a, b]$ —not $ P = \{a = u_0 < u_1 < \cdots{} < c_m = b\} $, as stated in the textbook— such that $ f $ is constant on each interval $(u_{j-1}, u_j)$, say $ f(x) = c_j $ for $ x $ in $(u_{j-1}, u_j)$.
- Show that a step function $ f $ is integrable and evaluate $ \int_a^b f $.
Solution.
Proof.
If $f$ is constant on every interval, then $M_i = m_i$. Then, $f$ is monotone and bounded on $( u_{j-1}, u_{j} ) $, (in fact it is constant from $( u_{j-1},u_{j} ) $), so therefore it is uniformly continuous, which implies that
there exists a continuous extension to the closed set $[u_{j-1}, u_{j}] $. Invoking Theorem (3.38) from Ross (Piecewise Monotone), we show that $f\in{} \mathcal{R}$.
∎
Problem.
[33.7] Let $ f $ be a bounded function on $[a, b]$, so that there exists $ B > 0 $ such that $ |f(x)| \leq{} B $ for all $ x \in{} [a, b] $.
[33.7] Let $ f $ be a bounded function on $[a, b]$, so that there exists $ B > 0 $ such that $ |f(x)| \leq{} B $ for all $ x \in{} [a, b] $.
- Show $$ U(f^2, P) - L(f^2, P) \leq 2B[U(f, P) - L(f, P)] $$ for all partitions $ P $ of $[a, b]$. \emph{hint:} $ f(x)^2 - f(y)^2 = [f(x) + f(y)] \cdot [f(x) - f(y)] $.
- Show that if $ f $ is integrable on $[a, b]$, then $ f^2 $ also is integrable on $[a, b]$.
Solution.
Proof.
$f$ is a bounded function on $[a,b]$, so there exists $B>0 \ s.t. \ \vert f(x) \vert \leq{} B$ for all $x\in{}[a,b]$.
$$\begin{align}
u(f^{2},p) &= \sum_{i=1}^{\infty} M_i^{2}\Delta x_{i} \\[10pt]
&= M_i^{2} = \sup f(x)^{2} \\[10pt]
&= \sum_{i=1}^{n} \left( M_i^{2}-m_i^{2} \right)\Delta x_i \\[10pt]
&= \sum_{i=1}^{n} \left( M_{i+m_i} \right)\left( M_{i} - m_i \right)\Delta x_{i} \\[10pt]
&\leq \sum_{i=1}^{n} 2B\left( M_{i}-m_{i} \right)\Delta x_{i} \\[10pt]
&= 2B \sum_{i=1}^{\infty} \left( M_i - m_i \right)\Delta x_{i} \\[10pt]
&= 2B \left( U(f,p)-L(f,p) \right)
\end{align}$$
∎
Problem.
[34.2] Calculate
[34.2] Calculate
- $ \lim_{h \to 0} \frac{1}{h} \int_3^{3+h} e^{t^2} \, dt $.
Solution.
Proof.
$$\begin{align}
\int_{3}^{3+h}e^{t^{2}} \,\mathrm{d}t \\[10pt]
&= \lim_{h \to 0} \frac{F(3+h)-F(3)}{h}\\[10pt]
\end{align}$$
By FTC I,
$$\begin{align}
F^\prime (3) = e^{9}
\end{align}$$
∎
Problem.
[34.5] Let $ f $ be a continuous function on $ \mathbb{R} $ and define $$\begin{align} F(x) = \int_{x-1}^{x+1} f(t) \, dt \quad \text{for } x \in \mathbb{R}. \end{align}$$ Show $ F $ is differentiable on $ \mathbb{R} $ and compute $ F' $.
[34.5] Let $ f $ be a continuous function on $ \mathbb{R} $ and define $$\begin{align} F(x) = \int_{x-1}^{x+1} f(t) \, dt \quad \text{for } x \in \mathbb{R}. \end{align}$$ Show $ F $ is differentiable on $ \mathbb{R} $ and compute $ F' $.
Solution.
Proof.
$$\begin{align}
F(x) &= \int_{x}^{0} f(t) \,\mathrm{d}t + \int_{0}^{x+1} f(t) \,\mathrm{d}t \\[10pt]
&= -\int_{0}^{x-1} f(t) \,\mathrm{d}t + \int_{0}^{x+1} f(t) \,\mathrm{d}t \\[10pt]
\end{align}$$
By FTC II, we get:
$$\begin{align}
F^\prime (x_0) &= f(x_{0}+1) - f(x_{0}-1)
\end{align}$$
∎
Problem.
A. [Bonus problem] Suppose $ f $ is a continuous non-negative function on $[a, b]$, with $$\begin{align} M = \max_{x \in [a, b]} f(x). \end{align}$$ For $ n \in{} \mathbb{N} $, let $$\begin{align} M_n = \left( \int_a^b f^n \, dt \right)^{1/n}. \end{align}$$ Prove that $ \lim{} M_n = M $.
A. [Bonus problem] Suppose $ f $ is a continuous non-negative function on $[a, b]$, with $$\begin{align} M = \max_{x \in [a, b]} f(x). \end{align}$$ For $ n \in{} \mathbb{N} $, let $$\begin{align} M_n = \left( \int_a^b f^n \, dt \right)^{1/n}. \end{align}$$ Prove that $ \lim{} M_n = M $.
Solution.
Proof.
$$\begin{align}
\left( \int_{a}^{b} f^{n} \right)^{\frac{1}{n}} &\leq \left( (b-a)(M^{n}) \right)^{\frac{1}{n}} \\[10pt]
&= \lim_{n \to \infty} \underbrace{(b-a)^{\frac{1}{n}}}_{1} M = M
\end{align}$$
∎