Two functions $f_1$,$f_2$ from $[0,1]$ to $R$ are given by $$f_1(x)=\sum \ \frac{x\sin(n^2x)}{n^2}$$ and $$f_2(x)= \sum \frac{x^2(1-x^2)^n}{1-x^2}.$$ Then show that $f_1$ is continuous and $f_2$ is not continuous. (Here $\sum$ is from n=1 to infinity)
$f_1(x)=\sum \ \frac{x\sin(n^2x)}{n^2}$ is continuous and $f_2(x)= \sum \frac{x^2(1-x^2)^n}{1-x^2}$ is not continuous.
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real-analysis
sequences-and-series
1 Answers
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Since $|\frac{xsin(n^2x)}{n^2}| \le \frac{1}{n^2}$ for all $x \in [0,1]$ and all $n$, the Weierstrass M - test shows that $\sum \ \frac{xsin(n^2x)}{n^2}$ converges uniformly on $[0,1]$ . Since $\frac{xsin(n^2x)}{n^2}$ is continuous, $f_1$ is continuous.
In connection with $f_2$ something went wrong:
$f_2$ is not defined for $x=1$,
We have $\sum_{n=1}^{\infty} \frac{x^2(1-x^2)^n}{1-x^2}=1$ for $x \in (0,1)$. Since $f_2(0)=0$, $f_2$ is not continuous on $[0,1)$.
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0No, it is defined for x=1. first term of the series is $x^2$ , second is $x^2(1-x^2)$, third is $x^2(1-x^2)^2$, fourth is $x^2(1-x^2)^3$ and so on, we see each one is defined for x=1. – 2017-01-30
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0The expression $\frac{x}{x}$ is not defined for $x=0$. $\frac{x}{x}=1 $ for $x \ne 0$. – 2017-01-30
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0how to prove that the sum is 1 for x in (0,1) – 2017-01-30
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0$\sum_{n=1}^{\infty}x^2(1-x^2)^{n-1}=x^2 \frac{1}{1-(1-x^2)}=1$ – 2017-01-30