finding the series representation

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May 3rd 2010, 02:04 AM
slapmaxwell1

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finding the series representation

ok the first problem i had to construct a power series and i feel pretty good about that one, but please check my work anyways. the second problem is the one really giving me fits or maybe im jus tired. basically i was going to jus take the e to the x function and subtract 1 from it and then divide by x? well i was trying to simply it so i could take the integral of each term, can you tell me where i went wrong. thanks in advance.Attachment 16671
May 3rd 2010, 03:02 AM
simplependulum

Quote:

Originally Posted by slapmaxwell1

ok the first problem i had to construct a power series and i feel pretty good about that one, but please check my work anyways. the second problem is the one really giving me fits or maybe im jus tired. basically i was going to jus take the e to the x function and subtract 1 from it and then divide by x? well i was trying to simply it so i could take the integral of each term, can you tell me where i went wrong. thanks in advance.Attachment 16671

The second one you don't need to expand $\frac{1}{x}$ , just expand $e^x - 1 = ( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .... ) - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + ....$

Then the integral $\int_0^x \frac{e^x-1}{x}~dx =$

$\int_0^x 1 + \frac{x}{2!} + \frac{x^2}{3!} + .... ~dx$

$= x + \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} + ....$
May 3rd 2010, 03:55 AM
slapmaxwell1

why dont i need to expand 1/x? thanks again for your response...
May 3rd 2010, 04:14 AM
HallsofIvy

Because $\frac{1}{x}(x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \cdot\cdot\cdot)= 1+ \frac{x}{2!}+ \frac{x^2}{3!}+ \cdot\cdot\cdot$ . Leaving $\frac{1}{x}$ in that form makes the rest of the calculation easier.
May 3rd 2010, 04:48 AM
slapmaxwell1

ok i was thinking i had to integrate before i could write it in terms of x. ok i think im going to need to practice that more. now the problem never said when to stop, so how do i know how many terms to write out? and thank you for your response...