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- anonymous

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

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- anonymous

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- anonymous

hey u found me lol

- anonymous

Haha yeah.

- anonymous

Can you find the derivative of arcsinx for me?

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- anonymous

I'll help you with the idea of this question, and then you try and let me know if you face any problem.

- anonymous

\[{d \over dx}\sin^{-1}x={1 \over \sqrt{1-x^2}}=(1+(-x^2))^{1/2}\]
Which is in the given form, after substituting with p=1/2

- anonymous

Now 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

You following?!

- anonymous

Hello?! :)

- anonymous

so do i plug in (-x)^2 in for x in the formula

- anonymous

Exactly, and 1/2 for p.

- anonymous

simple. right?!

- anonymous

then i should find the antiderivative to get the orginal

- anonymous

Yep!

- anonymous

Does that make sense?

- anonymous

yah just a general question when do first use the derivative to find the taylor series

- anonymous

Well, 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

okay fantastic ty you for your help.

- anonymous

You're welcome!

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