$x\in \mathbb{R}$ $$ f(x) = \sum_{n=1}^{\infty} {{\sin(nx^2)}\over 1+n^3} $$ will this be continuously differentiable?

I know that Uniform convergence does not confirm the fact. That's why though the aforesaid series is convergent we can not ensure the fact that $f(x)$ will be continuously differentiable.

Can anyone help me out by giving a rigorous proof?

  • $\begingroup$ I assume that you have already shown that the series converges uniformly? What would you need to be true in order for the derivative of $f$ to be continuous? Can you phrase that condition in terms of the series? or maybe the term-by-term derivative of the series? $\endgroup$ – Xander Henderson Aug 6 '17 at 14:43
  • $\begingroup$ I b/w can anyone tell me what exactly f(x) will be? $\endgroup$ – Pranita Gupta Aug 6 '17 at 15:48
  • $\begingroup$ I doubt that $f(x)$ has a closed form in terms of elementary functions. Wolfram Alpha cannot find any, and the partial sums formula is a somewhat ugly expression involving the Lerch transcendent. Source: wolframalpha.com/input/?i=sum+sin(nx%5E2)%2F(1%2Bn%5E3) $\endgroup$ – Boaz Moerman Aug 6 '17 at 16:24

If $f(x) = \sum_{n=1}^{\infty} {{\sin(nx^2)}\over 1+n^3}$, then $f(x)$ converges absolutely for all $x\in \mathbb{R}$ by the comparison test, as $|\frac{sin(nx^2)} {1+n^3}|\leq \frac{1}{1+n^3}$. And also $f'(x)=\sum_{n=1}^{\infty}{{\ 2nx \cos(nx^2)}\over 1+n^3}$. This is because for all $n \in \mathbb{N}$ the derivative of $f_m(x) = \sum_{n=1}^{m} {{\sin(nx^2)}\over 1+n^3}$ is $f_m'(x)=\sum_{n=1}^{m}{{\ 2nx \cos(nx^2)}\over 1+n^3}$ And since $|\frac{2nx \cos(nx^2)}{1+n^3}| \leq |\frac{2xn}{1+n^3}| \leq |\frac{2x}{n^2}$| and $\sum_{n=1}^{\infty}|{{2x}\over n^2}|$ converges, it follows from the Weierstrass M-test that the sequence $(f'_m)$ converges absolutely uniformly on any bounded interval and $\lim_{n\to\infty} f_n'(x)=f'(x)$. Thus, by the uniform limit theorem it follows that $f'(x)=\lim_{n\to\infty} f_n'(x)$ is continuous, and thus it follows that $f(x)$ is continuously differentiable.

  • $\begingroup$ Very nice.can u tell what is uniform limit thm u mentioned? You said on any bounded interval but here x belongs to R will this affect your ans? And how? $\endgroup$ – Pranita Gupta Aug 6 '17 at 15:17
  • $\begingroup$ The uniform limit theorem states that the uniform limit of a sequence of continuous functions is continuous. It follows that $f'(x)$ is continuous on any bounded interval, and since every real number has an non-degenerate bounded interval containing it, it follows that $f'$ is continuous in all $x\in \mathbb{R}$. Thus it follows that $f'$ is continuous. $\endgroup$ – Boaz Moerman Aug 6 '17 at 15:27
  • $\begingroup$ Oh okk.u mean uniform converges theorem.umm..one more thing "non degenerate bounded interval"? $\endgroup$ – Pranita Gupta Aug 6 '17 at 15:41
  • 2
    $\begingroup$ @Ikuyuki. A non-degenerate interval means an interval with more than one point in it, as opposed to the interval $(0,0)$, which is the empty set, or the interval $[1,1]=\{1\}$. $\endgroup$ – DanielWainfleet Aug 6 '17 at 16:00
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    $\begingroup$ You should explain why your formula for $f'(x)$ is valid. $\endgroup$ – zhw. Aug 6 '17 at 17:14

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