anonymous
  • 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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
hey u found me lol
anonymous
  • anonymous
Haha yeah.
anonymous
  • anonymous
Can you find the derivative of arcsinx for me?

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anonymous
  • 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
  • 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
  • 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
  • anonymous
You following?!
anonymous
  • anonymous
Hello?! :)
anonymous
  • anonymous
so do i plug in (-x)^2 in for x in the formula
anonymous
  • anonymous
Exactly, and 1/2 for p.
anonymous
  • anonymous
simple. right?!
anonymous
  • anonymous
then i should find the antiderivative to get the orginal
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
yah just a general question when do first use the derivative to find the taylor series
anonymous
  • 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
  • anonymous
okay fantastic ty you for your help.
anonymous
  • anonymous
You're welcome!

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