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math104-f21:hw15 [2021/12/03 12:54] pzhou |
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This is for your practice only, not going to be graded. The solution will be release next Wednesday. | This is for your practice only, not going to be graded. The solution will be release next Wednesday. | ||
- | 1. Ross Ex 33.9 (uniform convergence and integral) | + | === 1. Ross Ex 33.9 === |
+ | (uniform convergence and integral). | ||
- | 2. Ross Ex 33.5 (bound an integral by replacing an part of the integrand by something nice) | + | See Rudin Thm 7.16 (page 151) |
- | 3. Ross Ex 33.14 (a). | + | === 2. Ross Ex 33.5 === |
+ | (bound an integral by replacing an part of the integrand by something nice) | ||
- | 4,5. Ross Ex 34.5, 34.6. Fundamental theorem of calculus, and chain rule | + | We want to show |
+ | $$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq 16 \pi^3 /3 $$ | ||
+ | since $|\sin^8(e^x)| \leq 1$, we have | ||
+ | $$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq | ||
- | 6, 35.3 | ||
- | 7, 35.4 | + | === 3. Ross Ex 33.14 (a). === |
+ | For any continuous $g(x) \geq 0$ on $[a,b]$ and continuous $f(x)$, we want to show | ||
+ | $$ \int_a^b f(t) g(t) dt = f(x) \int_a^b | ||
+ | for some $x \in [a,b]$. | ||
+ | |||
+ | Proof: If $g(t) =0 $ for all $t \in [a,b]$ then both sides are zero, and there is nothing to prove. Otherwise, assume $g(t) \neq 0$ for some $t \in [a,b]$. By continuity, $g(t)$ is non-zero hence positive on an open subset, hence $\int_a^b g(t) dt > 0$. | ||
+ | |||
+ | let $M = \sup \{f(t): t \in [a,b]\}$ and $ m= \inf \{f(t): t \in [a,b]\}$, then since $g(t) \geq 0$, we have $m g(t) \leq f(t) g(t) \leq M g(t)$, thus | ||
+ | $$ m \int_a^b | ||
+ | Thus, $$ A = \frac{\int_a^b f(t) g(t) dt}{ \int_a^b | ||
+ | by intermediate value theorem, there is $x \in [a,b]$, such that $f(x) = A$. This finishes the proof. | ||
+ | |||
+ | |||
+ | === note=== | ||
+ | For the following two problems, if $a > b$, then $\int_a^b f(t) dt = - \int_b^a f(t) dt $. | ||
+ | |||
+ | === 4. Ross Ex 34.5,=== | ||
+ | |||
+ | |||
+ | Let $f$ be continuous on $\R$, and let $F(x)$ be defined by | ||
+ | $$ F(x) = \int_{x-1}^{x+1} f(t) dt $$ | ||
+ | show that $F$ is differentiable on $\R$ and compute $F' | ||
+ | |||
+ | Solution: we prove by definition. For any $\epsilon \neq 0$, we have | ||
+ | $$ F(x+\epsilon) - F(x) = \int_{x+\epsilon-1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+1} f(t) dt | ||
+ | = \int_{x+1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+\epsilon-1} f(t) dt $$ | ||
+ | Divide by $\epsilon$ and taking limit, we have | ||
+ | $$ \lim_{\epsilon \to 0} \frac{F(x+\epsilon) - F(x)}{\epsilon} = \lim_{\epsilon \to 0^+} \epsilon^{-1} \int_{x+1}^{x+\epsilon+1} f(t) dt - \epsilon^{-1} \int_{x-1}^{x+\epsilon-1} f(t) dt = f(x+1) - f(x-1) $$ | ||
+ | where we used fundamental theorem of calculus. | ||
+ | |||
+ | Hence $$F' | ||
+ | |||
+ | === 34.6. === | ||
+ | Let $f$ be continuous function on $\R$, let | ||
+ | $$ G(x) = \int^{sin(x)}_0 f(t) dt $$ | ||
+ | prove that $G(x)$ is differentiable, | ||
+ | |||
+ | Define $H(u) = \int^{u}_0 f(t) dt $, for any $u \in \R$, then $G(x) = H(\sin(x))$. Since $H(u)$ is differentiable, | ||
+ | $$ G'(x) = H' | ||
+ | |||
+ | === 6, 35.3 === | ||
+ | See definition 35.2 for Stieljes integral. In particular, if $F$ has a jump on the integration domain' | ||
+ | $$ \int_a^b f(t) d F(t) = \sum_{n \in [a,b] \cap \Z} f(n) $$ | ||
+ | For example, | ||
+ | $$ \int_0^6 x dF(x) = 0 + 1 + \cdots + 6 = .. $$ | ||
+ | |||
+ | === 7, 35.4 === | ||
+ | Since $F(t)$ is differentiable and monotone over that range, we have $dF(x) = F'(x) dx$, with $F'(x) = \cos(x)$ for $t \in [-\pi/2, \pi/2]$. | ||
+ | $$ \int_0^{\pi/ | ||
+ | |||
+ | Alternatively, | ||
+ | $$ \int_a^b f dF = \int_a^b d(f F) - F df = f F|^b_a - \int_a^b F df(x) $$ | ||
+ | Here, in this problem, we have $f(x) = x$, thus | ||
+ | $$ \int_0^{\pi/ | ||