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El_Arrow
 one year ago
Help! find the taylor series for f(x) center at the given value
El_Arrow
 one year ago
Help! find the taylor series for f(x) center at the given value

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El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0i already got the series i just need the sum dw:1437185617267:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal @Michele_Laino

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0i got 9  18/1! (x1) + 52/2! (x1)^2  216/3! (x1)^3 + 1080/4! (x1)^4  6480/5! (x1)^5

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing help with one more problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean? You have the series but need the sum? Saying you need a sum implies that there's a numerical answer when there isnt.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean you need the nth term?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0yeah thats what i meant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would suggest doing what we did before, actually. Consider integrating the function and then applying the new center. \[\int\limits_{}^{}\frac{ 9 }{ x^{2} }dx = \frac{ 9 }{ x }+C\] Now I want to shift this antiderivative. It would have been too difficult to do so while it was squared, though. So what we can do is write this in the form of a geoseries by doing this: \[\frac{ 9 }{ x } + C = \frac{ 9 }{ 1x1 } = \frac{ 9 }{ 1(x+1) } = 9\sum_{n=0}^{\infty} (x+1)^{n} + C\] Now proceed exactly like we did in the last problem. Since we'll be differentiating, we don't need to worry about the +C.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0okay so that would be (9x+1)^n right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, the 9 is just stuck outside. Not much you can do with it. Gotta head out, though. Good luck :)

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0@iambatman could you please help?
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