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Define $f_1$, $f_2$ $:$ $[0,1] \to R$ by $f_1(x)=\sum_{n=1}^{\infty}\frac{xsin(n^2x)}{n^2}$ and $f_2(x)=\sum_{n=1}^{\infty}x^2(1-x^2)^{n-1}$. Then which of the following is true?

a) $f_1$ is continuous but $f_2$ is NOT continuous.

b) $f_2$ is continuous but $f_1$ is NOT continuous.

c) both $f_1$ and $f_2$ are continuous.

d) neither $f_1$ nor $f_2$ is continuous.

My attempt:

At n tending to infinity, both $f_1$ and $f_2$ are zero for all x. But what does that tell us about continuity?

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aarbee
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1 Answers1

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For the first, you have

$\forall \,x\in[0,1] \;\;\forall n>0 $

$$|u_n(x)|=|\frac{x\sin(n^2x)}{n^2}|\leq \frac{1}{n^2}$$

thus the series of functions $\sum u_n(x)$ converges normally and uniformly at $[0,1]$. so $f_1$ is continuous at $[0,1]$ cause the functions $u_n$ are continuous.

for the second, it is a geometric sum.

if $x\neq 0$, $$f_2(x)=\frac{x^2}{1-(1-x^2)}=1$$

$f_2(0)=0 \implies f_2$ is not continuous at $[0,1]$.

The answer to your question is $a)$.

hamam_Abdallah
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  • Hi. I got ur point. Thanks. However, if we do it differently. Say, when x is lying b/w 0 and 1. $(1-x^2)$ would be lying b/w 0 and 1. So, its power infinity would be zero. But with ur method, we are getting 1. Can you explain please? – aarbee Dec 28 '16 at 17:14