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I have a function $ f $ and it's derivative $f'$ that are both continuous on $ [0,2\pi] $ and for any $x$ in this range :

$$ f'(x) = a_0 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx))$$

I've been asked to:

  1. Determine a formula for $b_8$ in terms of $f$ and I'm completely stuck on what to do.

  2. (Assuming we now know the function is even) Justify whether $a_{12} = 0$ or not? My initial idea here is that since we know the function is even, all $b_n = 0$ and $a_n$ are not equal to $0$ so this statement must be false and $a_{12}$ cannot equal $0$. I'm not sure if this is correct or if there is more I need to show for this to be sufficient.

Any help would be much appreciated, especially for the first part of the question.

Bernard
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Charlie P
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    The whole point of Fourier series is that it is possible to find coefficients fairly straightforwardly; you multiply by the appropriate sine or cosine and integrate, to make every term in the series vanish except the one you want. – RobertTheTutor May 17 '21 at 15:25

1 Answers1

1

Assume f and f' are continuous on [0,2π].

Assume further $a_0 = 0$ and

$f(-π) = f(π)$ and $f ' (- π) = f ' (π)$.

If Fourier Series for f' is
$$ f'(x) = a_0 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx))$$ then its corresponding Fourier Series for f becomes $$ f (x) = b'_0 + \sum_{n=1}^\infty(b'_n\cos(nx) +a'_n\sin(nx))$$ Where ${a'_n} * n = a_n ~\text{and} -{b'_n} * n = b_n$

Define the inner product of $f$ and $g$ as $$\langle f,g\rangle = \int_{0}^{2π} {f(x) g(x)}\mathrm dx$$

$$ b'_n = -b_n/n = 1/π \cdot \langle f(x),\cos(nx)\rangle, a'_n = a_n/n = 1/π \cdot \langle f(x), \sin(nx)\rangle$$

or $$b'_0 = 1/π \cdot \langle f(x),1\rangle, b'_n = 1/π \cdot \langle f(x),\cos(nx)\rangle, a'_n = 1/π \cdot \langle f(x), \sin(nx)\rangle$$

Conclusion:

  1. $b_8 = -8/π \cdot \langle f(x), \cos(nx)\rangle $

  2. If $f$ is even, then $a_n = 0$ for all n, in particular $a_{12} = 0.$

Anyone interested to get a Fourier Series for $f$ or $f'$, let me know and please

indicate whether they are for $f(x)$ or $f'(x)$. We will try to get its Fourier

coefficients $a_n, b_n$ from the Internet!

Have fun and enjoy a nice day!

cdeamaze
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  • Ok, I get your working out but am I allowed to assume that $a_0 = 0$? Also, for part 1 of the question, could I leave the answer as you have or is there more I can do? – Charlie P May 17 '21 at 17:38
  • And just for clarification, it's $f$ that's even, so I would use 2a – Charlie P May 17 '21 at 17:39
  • For part 1 there is no ambiguity. If you start to get Fourier series for f then you a0 = 0 where a0 is the DC term of f" – cdeamaze May 17 '21 at 17:55
  • Oh ok, thanks a lot – Charlie P May 17 '21 at 18:21
  • One more question, sorry, how did you obtain the inner products? – Charlie P May 18 '21 at 12:01
  • we have the Fourier Coefficients , $bn=−n/π∗(f(x),cos(nx)),an=n/π∗(f(x),sin(nx))$ – cdeamaze May 18 '21 at 14:30
  • we have the Fourier Coefficients : a_n=n/π∗(f(x),sin(nx)), b_n=−n/π∗(f(x),cos(nx)), So $a_n = \frac{n \π} \int_0^2\π {f(x) sin(x)}, b_n = -\frac{n \π} \int_0^2\π {f(x) cos(x)}$ – cdeamaze May 18 '21 at 15:50
  • we have the Fourier Coefficients : a_n=n/π∗(f(x),sin(nx)), b_n=−n/π∗(f(x),cos(nx)), So $a_n = \frac{n π} \int_0^2π {f(x) sin(x)}, b_n = -\frac{n π} \int_0^2π {f(x) cos(x)}$ – cdeamaze May 18 '21 at 16:01
  • Any one interested to get a Fourier Series for f or f', Please give f(x) or f'(x) We will try to get Fourier coeff a_n, b_n from the Internet! Have fun!! – cdeamaze May 18 '21 at 17:49
  • I tidied up the formatting, as per your request. Check to make sure it is correct. – K.defaoite May 19 '21 at 03:13
  • It looks great! How do you make it so presentable? Can you show me the code?

    I'm glad that I am making progress toward using Mathjax as my essential tools.

    Thanks again!

     – cdeamaze May 19 '21 at 03:32