# Week 8 hw

20.1 1: b) Discontinuities at $$0$$ and $$\pi$$. $$\lim_{n\to\infty} (\sin x)^{\frac{1}{n}} = f(x) = 1$$. $$\{f_n (x)\}$$ is uniformly convergent for $$a>0$$ and $$b<\pi$$.

c) No discontinuities. $$\lim_{n\to\infty}\frac{1}{n}e^{-n^2x^2} = f(x) = 0$$ $\forall \epsilon > 0, \exists N : \forall n,m : N < n<m$ $|f_n(x) - f_m(x)| = \frac{1}{n}e^{-n^2x^2} - \frac{1}{m}e^{-m^2x^2}$ $= (\frac{1}{ne^{n^2}} - \frac{1}{me^{m^2}})\frac{1}{e^{x^2}} < \frac{1}{ne^{n^2}} - \frac{1}{me^{m^2}} < \epsilon$ Therefore $$f_n(x)$$ is uniformly convergent for all x by Theorem I.

d) No discontinuities. Uniformly convergent for $$x \ge \delta$$ by a similar argument to part c.

e) Discontinuity at $$0$$. $$\lim_{n\to\infty} \arctan nx = f(x) = \frac{\pi}{2}$$. Uniformly convergent for $$a>0$$.

3) a) $$\lim_{n\to\infty}x^{a_n} = f(x) = x$$ b) $|f_n(x) - f(x)| = x-x^{a_n} =$

20.2 5: a) When $$x\ge0, \sum_{n=1}^{\infty}{\frac{x^n}{n^2(1+x^n)}}$$ is uniformly convergent because $$\frac{x^n}{n^2(1+x^n)} \le \frac{x^n}{n^2}$$which is a multiple of the convergent p-series $$\frac{1}{n^2}$$.

b) When $$|x| \le a$$ where $$0<a<1, \frac{x^n}{n^2(1+x^n)} \le \frac{1}{n^2}$$ because $$\frac{x^n}{n^2(1+x^n)} \le \frac{1}{n^2}$$ since $$\frac{x^n}{1+x^n} = 1 - \frac{1}{1+x^n} \le 1$$ for all x in the range.

c) Covered by part a and b.

d) When $$x \le -c$$ where $$c>1, \frac{x^n}{1+x^n}$$ is eventually bounded by 2. This is obvious for even $$n$$ and for odd n we have $$\frac{x^n}{1+x^n} =1 - \frac{1}{1+x^n} < 2 <=> \frac{1}{1+x^n} > -1 <=> -1 > 1+x^n <=> -2 > x^n$$ which is obviously true for sufficiently large $$n$$. So $$\sum_{n=1}^{\infty}{\frac{x^n}{n^2(1+x^n)}}$$ is uniformly convergent because, except for a finite number of terms, $$\frac{x^n}{n^2(1+x^n)} \le 2\frac{1}{n^2}$$, which is a multiple of a term of the convergent p-series $$\frac{1}{n^2}$$.

20.1 6:If $$\{f_n(x)\}$$ is uniformly convergent then for any $$\epsilon > 0$$, there exists an N such that $$|f_n(x) - f(x)| \le \frac{\epsilon}{2}$$ and $$|f_m(x) - f(x)| \le \frac{\epsilon}{2}$$ for $$n$$ and $$m$$ greater then $$N$$ so $$|f_n(x) - f_m(x)| \le \epsilon$$ by the triangle inequality.