fourier coefficients

**hunkydory19** · February 23rd 2008, 05:26 AM

Find the Fourier cosine coefficients of $e^x$

$e^x = \frac{1}{2}a_0 + \displaystyle\sum_{n=1}^\infty a_n cos \frac{n \pi x}{L}$

Differentiating yields:

$e^x = - \displaystyle\sum_{n=1}^\infty \frac{n \pi}{L}a_n sin \frac{n \pi x}{L}$ ,

the Fourier sine series of e^x. Differentiating again yields

$e^x = - \displaystyle\sum_{n=1}^\infty (\frac{n \pi}{L})^2 a_n cos \frac{n \pi x}{L}$ ,

Since equations 1 and 3 both give Fourier cosine series of e^x, they must be identical. Thus,

$a_o = 0$ and $a_n = 0.$

Can anyone please explain step by step what is wrong with this? I'm supposed to correct the mistakes and then find $a_n$ without using the typical technique but I'm so confused!

Any mistakes at all you can see, please point them out!

Thanks in advance

**CaptainBlack** · February 23rd 2008, 07:47 AM

Originally Posted by hunkydory19

Find the Fourier cosine coefficients of $e^x$

$e^x = \frac{1}{2}a_0 + \displaystyle\sum_{n=1}^\infty a_n cos \frac{n \pi x}{L}$

Differentiating yields:

$e^x = - \displaystyle\sum_{n=1}^\infty \frac{n \pi}{L}a_n sin \frac{n \pi x}{L}$ ,

the Fourier sine series of e^x. Differentiating again yields

$e^x = - \displaystyle\sum_{n=1}^\infty (\frac{n \pi}{L})^2 a_n cos \frac{n \pi x}{L}$ ,

Since equations 1 and 3 both give Fourier cosine series of e^x, they must be identical. Thus,

$a_o = 0$ and $a_n = 0.$

Can anyone please explain step by step what is wrong with this? I'm supposed to correct the mistakes and then find $a_n$ without using the typical technique but I'm so confused!

Any mistakes at all you can see, please point them out!

Thanks in advance

What are the conditions under which you can differentiate a series term by term?

Have you considered integrating twice (and then expressing the integral as
a cosine series)?

RonL

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