1
$\begingroup$

Let function $f(t)$ is represented by Fourier series, $$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}}),$$ where $a$ and $b$ are lower and upper boundary, $$a_0=\frac{2}{b-a}\int_{a}^{b}f(t)dt,$$ $$a_n=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt,$$ $$b_n=\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt.$$

My question is, what conditions must be met so I can find derivative as (term by term) $$\frac{d}{dt}\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}a_n\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}b_n\sin{\frac{2n\pi t}{b-a}})=\sum_{n=1}^{\infty}\frac{d}{dt}(a_n\cos{\frac{2n\pi t}{b-a}})+\sum_{n=1}^{\infty}\frac{d}{dt}(b_n\sin{\frac{2n\pi t}{b-a}})?$$

$\endgroup$
2
$\begingroup$

A sufficient condition for differentiability of the series and commuting of sum and derivative is that $$ \sum_{n=1}^\infty n\big(\lvert a_n\rvert+\lvert b_n\rvert\big)<\infty. $$ See: Rudin, Principles of Mathematical Analysis, Theorem 7.17, p. 152.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

We need the Fourier Series to converge uniformly on the interval $ [A,B] $ (which may be open or closed). Then we can integrate and differentiate the Fourier Series term by term. To check for uniform convergence, we can use the Weierstrass M Test:

If $$ |x| \leq M_n \forall \space t \in I $$

Where $M_n$ is independent of t, and $$ \sum_{n=0}^\infty M_n$$ converges, then $$ \sum_{k=0}^\infty u_k(t)$$ converges uniformly on the interval $I$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot !! $\endgroup$ – etf Sep 28 '14 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.