maclaurin series for this function

iDoof

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Oct 17, 2005
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i'm supposed to get the infinite series representation for the INTEGRAL of (sin(x))/x dx.

so i start by finding the maclaurin series for sin(x)/x right? then i need to integrate term by term, but when i try to get the first couple terms, i get undefined values (divide by 0)...does that make sense?

am i doing something wrong? if my specific issue doesn't make sense, could someone just walk through how to get the maclaurin series?

thanks so much
 
You could start with the MacLaurin series for sin(x).

Let's derive it:

\(\displaystyle f(x)=sin(x), f(0)=0\)

\(\displaystyle f'(x)=cos(x), f'(0)=1\)

\(\displaystyle f''(x)=-sin(x), f''(0)=0\)

\(\displaystyle f'''(x)=-cos(x), f'''(0)=-1\)

The pattern 0,1,0,-1 will repeat.

So we have, using the MacLaurin series definition:

\(\displaystyle x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+.....+(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}+.....\)

Divide by x to get:

\(\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)}=

1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+....+(-1)^{k}\frac{x^

{2k}}{(2k+1)!}+....\)

Now integrate the series. Is this what you were looking for?.
 
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