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.

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