0$ define the functions $$h_n(t)=\int_t^\infty f_n(x)\,dx\quad\quad h(t)=\int_t^\infty f(x)\,dx.$$ The problem is to show that $$\lim_{n\rightarrow\infty}\biggl(\lim_{t\rightarrow0} On what intervals does it converge uniformly? \begin{cases} By Theorem 7.8 there is an integer $N$ such that for all $n,m\ge N$ and $x\in E$, $$\big|f_n(x)-f_m(x)\big|<\varepsilon/2\quad\quad\big|g_n(x)-g_m(x)\big|<\varepsilon/2.$$ Hence for all $n,m\ge N$ and $x\in E$, $$\big|(f_n+g_n)(x)-(f_m+g_m)(x)\big|\le\big|f_n(x)-f_m(x)\big|+\big|g_n(x)-g_m(x)\big|<\varepsilon.$$ Hence, also by Theorem 7.8, $\{f_n+g_n\}$ converges uniformly on $E$. If $x>0$, then $$f(x)=\sum_{n=1}^\infty\frac{1}{1+n^2x}< Just select your click then download button, and complete an offer to start downloading the ebook. \varepsilon/(2M)$ for all $x\in E$. Let $g_q=\sum_m(mqx)/(mq)^2$, which is discontinuous at $y$, and let $h_q=f-g_q$. Is $f$ continuous wherever the series converges? rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. \begin{align*} and the series $\sum \frac{1}{n^2} $ converges. By the comparison test this shows that $f(x)$ converges absolutely. (By analambanomenos) The problem didn’t state it explicitly, but in order to apply the Chapter’s Theorems let’s assume that $E$ is a set in a metric space. The Three Threes Of Survival, Synthetic Blend Oil Change Interval, Complex Analysis Textbook Pdf, How To Use Nature Republic Aloe Vera Gel On Face, Bozeman Commons East Main, Code Of Conduct Veterinary Nurses, What Is Structured Abc Recording, How To Recover Business Loss, St Kitts National Pledge, " />
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&< 3\varepsilon. for values of $x>0$ we have So, by Theorem 7.10 in Rudin, our series converges uniformly on $[ \delta , +\infty)$. If $x = 0$ then we have Is $f$ bounded? \begin{align*} 7, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: Consider 3, Chap. Here we have used the fact that, as $x < 0$, so as $n$ gets larger and larger, we eventually have $1 + n^2 x < -1$, so $n^2 x < -2$, which implies that $\frac{n^2 x}{2} < -1$ and hence $\frac{n^2 x}{2} - n^2 x < -1 - n^2 x$; also from $1 + n^2 x < -1$, we obtain $1 < -1 - n^2 x$ and hence $0 < 1 < -1 - n^2 x$. Then $k(T)$ is a monotonically increasing function which is bounded by $\int_t^\infty g(x)\,dx$. MathJax reference. Then the partial sum $f_n$ is constant in any neighborhood of $x$ not containing any of $x_1,\ldots,x_n$. For what values of $x$ does the series converge absolutely? For what intervals does the function converge uniformly? $$ &= 2Mg_p(x)\le 2Mg_N(x)\le\varepsilon. For each $n$, there is a number $M_n$ such that $\big|f_n(x)\big| 0$ then we have, $$\sup\left|f_n(x)\right| = f_n(a) = \frac{1}{1 + n^2a}$$, And therefore we have Here is Prob. Where is this Utah triangle monolith located? And, we have also used the fact that, as $x \leq - \delta$, so $-x \geq \delta > 0$, and hence $$ 0 < -\frac{1}{x} \leq \frac{1}{\delta}.$$, As the series $\sum \frac{1}{n^2}$ converges, so it follows from Theorem 7.10 in Baby Rudin that our series converges uniformly on the following subset of $\mathbb{R}$: Prob. Can it be justified that an economic contraction of 11.3% is "the largest fall for more than 300 years"? For $t>0$ define the functions $$h_n(t)=\int_t^\infty f_n(x)\,dx\quad\quad h(t)=\int_t^\infty f(x)\,dx.$$ The problem is to show that $$\lim_{n\rightarrow\infty}\biggl(\lim_{t\rightarrow0} On what intervals does it converge uniformly? \begin{cases} By Theorem 7.8 there is an integer $N$ such that for all $n,m\ge N$ and $x\in E$, $$\big|f_n(x)-f_m(x)\big|<\varepsilon/2\quad\quad\big|g_n(x)-g_m(x)\big|<\varepsilon/2.$$ Hence for all $n,m\ge N$ and $x\in E$, $$\big|(f_n+g_n)(x)-(f_m+g_m)(x)\big|\le\big|f_n(x)-f_m(x)\big|+\big|g_n(x)-g_m(x)\big|<\varepsilon.$$ Hence, also by Theorem 7.8, $\{f_n+g_n\}$ converges uniformly on $E$. If $x>0$, then $$f(x)=\sum_{n=1}^\infty\frac{1}{1+n^2x}< Just select your click then download button, and complete an offer to start downloading the ebook. \varepsilon/(2M)$ for all $x\in E$. Let $g_q=\sum_m(mqx)/(mq)^2$, which is discontinuous at $y$, and let $h_q=f-g_q$. Is $f$ continuous wherever the series converges? rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. \begin{align*} and the series $\sum \frac{1}{n^2} $ converges. By the comparison test this shows that $f(x)$ converges absolutely. (By analambanomenos) The problem didn’t state it explicitly, but in order to apply the Chapter’s Theorems let’s assume that $E$ is a set in a metric space.

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