Series Thrm 1 .Claim For \(S_n(x)= \sum_{k=1}^n a_k(x)\) , \((S_n,\|\cdot\|_\infty)\) Cacuhy IFF \(S_n \rightarrow^{u.c.}S\)
proof . \(\Rightarrow\) \(S(x)\) s.t. \(S_k(x)\rightarrow^{p.w.}S(x)\) .
\[\begin{align*} |S_k(x)-S(x)| &\leq |S_k(x)-s_n(x)| + |s_n(x)-S(x)| \\ &\leq \epsilon + |S_n(x)-S(x)| &\text{by p.w. convergence} \\ &\leq \epsilon + \sup_{x\in\mathbb{R}} |S_n(x)-S(x)|\\ \end{align*}\]
Apply limit on \(n\rightarrow\infty\) , since \(n\) is independent on \(k\) , \(|S_k(x)-S(x)|\leq \epsilon + \lim_{n\rightarrow\infty}\sup_{x\in\mathbb{R}} |S_n(x)-S(x)| = \epsilon + 0\Rightarrow \|S_k-S\|_\infty \leq \epsilon\)
Example 1 \(S_n :=\sum_{k=1}^n \frac{x^k}{k!}\rightarrow^{u.c.}\sum^\infty \frac{x^k}{k!},x\in [-R,R]\)
\[\begin{align*} \|S_k - S_n\|_\infty &= \sup_{|x|\leq R} |\sum^{n+m}\frac{x^k}{k!} - \sum^n \frac{x^k}{k!}| \\ &= \sup|\sum_{n+1}^{n+m}\frac{x^k}{k!}| \\ &\leq \sum \frac{R^k}{k!} = e^R - \sum^n \frac{R^k}{k!}\rightarrow 0 \end{align*}\]
Example 2 \(S_n = \sum^n \frac{x^2}{k(k+x^2)}, x\in \mathbb R\)
proof . \(S_n\rightarrow^{p.w.}S\) but \(S_n \not\rightarrow^{u.c.}S\) \(x_0\) , \(S_n \leq x^2 \sum_{k=1}^n k^{-2} \leq x^2\pi^2/6 < \infty\)
\[\begin{align*} \|S_n-S\|_\infty &= \|\sum_{k\geq n}\frac{x^2}{k(k+x^2)}\|_\infty \\ &= \sup_{x\in\mathbb R}|\sum_{k\geq n}\frac{x^2}{k(k+x^2)}| \\ &\geq \sup_{|x|\leq N}|\sum_{k\geq n} \frac{1}{k}\frac{x^2}{k+x^2}|\\ \text{take N large so that }\frac{x^2}{k+x^2}&\geq 1-\epsilon \text{ and k is fixed} \\ &\geq \sum_{k\geq n}\frac{1-\epsilon}{k} = \infty \end{align*}\]
Thrm 2 . M-TestFor \(S_n(x) = \sum_{k=1}^n a_k(x), \|a_k\|_\infty \leq M_k \land \sum M_k < \infty \Rightarrow S_n \rightarrow^{u.c.}S < \infty\) .
proof . We'll show Cauchy so that uniform convergent.
\[\begin{align*} \|S_n - S_{n+m}\|_\infty &= \sup|\sum_{n+1}^{n+m}a_k(x)| \\ &\leq \sum_{n+1}^{n+m} \sup|a_k(x)| \\ &\leq \sum_{n+1}^{n+m} M_k < \epsilon \text{ for N large} \end{align*}\]
Example 1 \(S_n = \sum_{k=1}^n (-x^2)^k\rightarrow^{u.c.}S, x\in [-r,r]\subset (-1,1)\) but not for \(x\in(-1,1)\)
proof . For \(x\in [-r,r]\) .
\[\|(-x^2)^k\| \leq r^{2k} =:M_k\]
Since \(\sum^\infty r^{2k} = (1-r^2)^{-1}-1 < \infty\) , apply M-test, \(S_n\rightarrow^{u.c.}S\)
However, take \(x\in(-1,1)\)
\[\begin{align*} \|S_n - S_{m+n}\|_\infty &= \sup_{x\in(-1,1)}|\sum_n^{n+m}(-x^2)^k| \\ \text{take }x_0 = 0.5^{1/2N} &\in (-1,1) \text{ and }n + m\leq N \\ &\geq |\sum_n^{n+m} (-1)^k 0.5^{k/N}| \\ &= |0.5^{n/N} + (-1)^{n+1}0.5^{\frac{n+1}{N}}| \\ &\sim 0.5^{n/N} &\text{ for N large} \\ &> \epsilon &\text{since }N\rightarrow\infty, 0.5^{n/N}\rightarrow 1 \end{align*}\]
Alternatively, take \(\lim_{N\rightarrow\infty} = |-1+1-1+1...| = 1\) when \(n+m\) is odd
Example 2 For the series for \(\arctan(x)\)
\[\begin{align*} \arctan(x) &= \int_0^x \frac{1}{1+t^2}dt \\ &= \int_0^x \sum (-t^2)^k dt &|x|<1 \\ &= \sum \int_0^x (-t^2)^k dt &\text{ICT} \\ &= \sum (-1)^k (\frac{t^{2k+1}}{2k+1}\Big|_0^x) \\ &= \sum (-1)^k (\frac{x^{2k+1}}{2k+1}) \end{align*}\]
\(\arctan(1)\) is defined and \(\arctan(1) = \pi/4\)
if \(a_k \geq a_{k+1} > 0, |\sum^\infty a_k(-1)^k| \leq a_n\)
proof .
\[\begin{align*} \sup_{x\in[-1,1]}|S_n(x)-S(x)| &= \sup_{|x|\leq 1}|\sum^\infty_n (-1)^k \frac{x^{2k+1}}{2k+1}| \\ &= \sup_{|x|\leq 1} |\frac{x^{2n+1}}{2n+1}(-1)^n)|&\text{since }\frac{x^{2k+1}}{2k+1} \text{ decreasing} \\ &= \frac{1}{2n+1} \leq \epsilon \end{align*}\]
Claim For \(f_n \in C([a,b])\) If \(\exists c \in [a,b], f_n(c)\rightarrow \gamma, f'_n \rightarrow^{u.c.}g\) , then \(f_n \rightarrow f\) where \(f'=g\)
proof . By FTC,
\[\begin{align*} f_n(x)&=f_n(c) + \int_c^x f'_n(t)dt \\ &\rightarrow\gamma + \int_c^x g(t)dt =:f(x) \\ |f_n(x)-f(x)| &= |f_n(x)-\gamma -\int_c^x g(t)dt| \\ &\leq |f_n(x)-\gamma| + |\int_c^x f'_n(t)-g(t)dt| \\ &\leq \epsilon/2 + \|f'_n - g\|_\infty \int_c^x dt \\ &= \epsilon/2 + \|f'_n-g\|(x-c)\\ &\leq \epsilon/2 + \frac{\epsilon}{b-a} (x-c)\\ &leq\epsilon \end{align*}\]
Corollary Swap derivative and sum
\[S'_n \rightarrow^{u.c.}S\land\exists c. S_n(c)\rightarrow S(c) \Rightarrow d_x \sum a_k(x) = \sum d_x a_k(x)\]
Def'n. Power Series \(\sum a_k x^k\)
Thrm 3. Ratio test \(\lim_\infty \frac{a_k}{a_{k+1}}< 1\Rightarrow \sum a_k < \infty\)
Thrm 4. Root test \(\lim\sup(|b_k|)^{1/k} < 1\Rightarrow \sum b_k < \infty\)
Example \(\sum \frac{x^n}{n}\) is u.c. for \(x\in[-1,r], r< 1\)
\[\begin{align*} \sup_{x\in[-1,1)}|S_N - S| &= \sup_{x\in[-1,1)}|\sum_{n\geq N} x^n/n| \\ (i) \sup_{x\in[-1,0]}|\sum_{n\geq N}x_n /n | = \sup_{y\in[0,1]}&|\sum (-1)^n y^n/n| \leq \sup|y^N/N| \leq 1/N \quad\text{for large }N \\ (ii) \sup_{x\in[0,r]}|\sum_{n\geq N}x_n /n | = \sup_{x\in[0,r]}&\sum |x^n/n| \leq \sum r^n/n \leq \epsilon \quad\text{for large }N \end{align*}\]
Thrm 5. Hadamard's Theorem Let \(S(x)=\sum a_k x^k\) and \(R^{-1} = \lim\sup(|a_n|)^{1/n}\) , then \(S_k \rightarrow^{u.c.}S, x\in (-R,R)\) .
proof . By root test, \(\lim\sup (a_kx^k)^{1/k} < 1\) \(|a_k x^k|^{1/k} = |x|\lim\sup |a_k|^{1/k} = |x|R^{-1} < 1\Rightarrow |x|< R\) \(M_k := |a_k|r^k\) by root test, \(\sum M_k < \infty\) , then by M-test, \(S_k \rightarrow^{u.c.}S, \forall x\in[-r,r]\)
Ratio test can do the same work
Example 1 \(\sum \frac{2^{2k}}{k^2} x^{2k}\) \(y := x^2\) ,
\[\begin{align*} \frac{1}{R} &= \lim\sup (|\frac{2^{2k}}{k^2}|^{1/k}) \\ &= \lim\sup 2^2 k^{-2/k} \\ &= 4 \lim\sup k^{-2/k} &(i)\\ &= 4 \\ \Rightarrow R &= 1/4 \\ \\ (i) \log L &= \lim\sup\log(k^{-2/k}) \\ \frac{2}{k}\log(\frac{1}{k}) \rightarrow 0 &\Rightarrow k^{-2/k}\rightarrow 1 \\ |y| \leq R = 1/4 &\Rightarrow |x|\leq 1/2 \end{align*}\]
So the interval of convergence \((-1/2, 1/2)\)
Example 2 Claim \(\sum a_k x^k\) with radius \(R_1\) , \(\sum b_k x^k\) with radius \(R_2\) . Then \(\sum (a_k + b_k)x^k\) has radius \((R_1^{-1}+R_2^{-1})^{-1}\) \(\sum_{n=1}^\infty (\sum_{k=1}^n a_k b_{n-k})x^n\)
Fix \(\epsilon > 0\) , take \(N = \max(N_1, N_2)\) s.t. \(\forall n \geq N_1. |a_n|^{1/n}\leq R_1^{-1} + \epsilon, \forall n \geq N_2. |b_n|^{1/n} \leq R_2^{-1}\) .
(i) proof .
\[\begin{align*} R^{-1} &= \lim\sup (|a_n| + |b_n|)^{1/n} \\ &\leq \lim\sup ((R_1^{-1}+\epsilon)^n + (R_2^{-1}+\epsilon)^n)^{1/n} \\ &\leq \lim\sup ((R_1^{-1}+\epsilon) + (R_2^{-1}+\epsilon))^{n/n} &\text{by }|a|^q + |b|^q \leq (|a|+|b|)^q, \forall q > 1 \\ & = R_1^{-1} + R_2^{-1} + 2\epsilon \\ &\rightarrow^{\epsilon\rightarrow 0} R_1^{-1} + R_2^{-1} \\ \Rightarrow R &= (R_1^{-1} + R_2^{-1})^{-1} \end{align*}\]
(ii) proof . Let \(M_1 = \max(\{1\}\cup \{\frac{|a_k|}{(R_1^{-1} + \epsilon)^k}\}), M_2 = \max(\{1\}\cup \{\frac{|b_k|}{(R_2^{-1} + \epsilon)^k}\})\) .\(|a_n|\leq M_1(R_1^{-1}+\epsilon)^n\land|b_n|\leq M_2(R_2^{-1}+\epsilon)^n. \forall n\geq 1\)
\[\begin{align*} R^{-1}&\leq \lim\sup (\sum^n M_1(R_1^{-1}+\epsilon)^kM_2(R_2^{-1} +\epsilon)^{n-k})^{1/n} \\ &= \lim\sup (M_1M_2)^{1/n}\lim\sup ((R_2^{-1}+\epsilon)^n \sum^n r^k)^{1/n} &r := \frac{R_1^{-1}+\epsilon}{R_2^{-1}+\epsilon} \\ \end{align*}\]
Take log limt for each lim sup, \(\lim\sup(M_1M_2)^{1/n} = 1\) .\(\lim\sup (R_2^{-1}+\epsilon)^{n/n}\lim\sup (\frac{1-r^{n+1}}{1-r})^{1/n} = R_2^{-1}+\epsilon\) \(R_1,R_2\) , similar conclusion can be made \(=R_1^{-1}+\epsilon\) , therefore \(R^{-1} \leq \max(R_1^{-1}+\epsilon, R_2^{-1}+\epsilon)\)
January 9, 2023 January 9, 2023