Differentiation and integration of Taylor series

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cosmic

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Mar 3, 2014
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Hi guys, any help on this question would be hugely appreciated.

The Taylor series about 0 for the function f(x)=(1/4+x)-3/2 is

f(x)=8 - 48x + 240x^2 - 1120x^3 + ...

used differentiation to find the Taylor series about 0 for the function g(x)=(1/4+x)-5/2

I'm clueless here, don't know where to begin. I'd be greatful if someone could shed some light on how I go about approaching this.

Thanks in advance. :)
 
Well, the whole point of the problem is that g(x) is the derivative of f(x)! So find the Taylor's series for g(x) by differentiating the series for f(x) "term by term"- that is differentiate each of "8", "- 48x", "240x^2", "-1120x^3", etc.
 
HallsofIvy said:
Well, the whole point of the problem is that g(x) is the derivative of f(x)! So find the Taylor's series for g(x) by differentiating the series for f(x) "term by term"- that is differentiate each of "8", "- 48x", "240x^2", "-1120x^3", etc.
Click to expand...

Thanks for the reply. :)

I've tried that and I got 0-48+480x-3360x^2 but that can't be right. I used a software to calculate the Taylor series of g(x)=(1/4+x)^-5/2 to be 32-320x+2240x^2....

Any ideas what I'm doing wrong?
 
cosmic said:
Thanks for the reply. :)

I've tried that and I got 0-48+480x-3360x^2 but that can't be right. I used a software to calculate the Taylor series of g(x)=(1/4+x)^-5/2 to be 32-320x+2240x^2....

Any ideas what I'm doing wrong?
Click to expand...

Well g(x) isn't the exact differential of f(x). There's the matter of that (-3/2) hanging around.
 
Ishuda said:
Well (-3/2) times (-2/3) is equal to one.
Click to expand...

I managed to finally do it. The problem was like you said I hadn't differentiated the left-hand side properly and I was using the wrong set of derivatives I had managed to mix up with my notes. Fair to say that I feel like a total idiot having spent hours bashing my head against the wall with only myself to blame! :D

Anyway once again thanks for your help. :)
 
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