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anonymous
 5 years ago
use (1+x)^p= 1+px+p(p1)/2!x^2+ p(p1)(p2)/3!x^3+... to find the first four nonzero terms of the taylor series about 0 for arcsinx
anonymous
 5 years ago
use (1+x)^p= 1+px+p(p1)/2!x^2+ p(p1)(p2)/3!x^3+... to find the first four nonzero terms of the taylor series about 0 for arcsinx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you find the derivative of arcsinx for me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll help you with the idea of this question, and then you try and let me know if you face any problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{d \over dx}\sin^{1}x={1 \over \sqrt{1x^2}}=(1+(x^2))^{1/2}\] Which is in the given form, after substituting with p=1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now find the first 4 terms of the expansion of the derivative of arcsinx using the given formula you wrote above. After that, just integrate these four terms, and that would be your first five term of the taylor series of arcsinx.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so do i plug in (x)^2 in for x in the formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly, and 1/2 for p.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then i should find the antiderivative to get the orginal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah just a general question when do first use the derivative to find the taylor series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, that depends on the question. In this case, the question tells us what method to use to find the Taylor expansion of arcsinx. Here we should look for anything related to arcsin, that's similar to the given formula. Its derivative is the best choice, I believe.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay fantastic ty you for your help.
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