anonymous
  • anonymous
how do i expand the equation into an infinite series: -(1/√(1-x^2)) dx the answer should be -(1-x^2/2+3x^4/8+5x^6/16+...) if you can show how we get the first three terms i would really appreciate thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the function was d theta/dx and dx was taken to the other side after being implicitly differentiated.
anonymous
  • anonymous
the function after differentiated implicitly dtheta/dx=-1/sqrt 1-x^2 then dtheta=1/sqrt1-x^2
anonymous
  • anonymous
sorry dx should be at the end

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Loser66
  • Loser66
@.Sam.
anonymous
  • anonymous
hi am only receiving this message @.Sam. and when i open the link its leading to another question
anonymous
  • anonymous
ok
anonymous
  • anonymous
this is the question which has been answered: hey I need some chain rule help *advanced chain rule* f(x)=x^5sec(1/x) answer is: -x^3sec(1/x)tan(1/x)+5x^4sec(1/x) I need to knw how to get the answer. This is so I can study for my Ap Calc AB final on monday. PLEASE HELP!! Thanks.
anonymous
  • anonymous
and the question has been answered
experimentX
  • experimentX
\[ \frac{1}{\sqrt{1 - 4x^2}} = \sum_{n=0}^\infty \binom{2n}{n} x^n\]
experimentX
  • experimentX
make appropriate substitution
anonymous
  • anonymous
ok
anonymous
  • anonymous
so how would we make the first substitution
experimentX
  • experimentX
x^2 = 4 (x/2)^2 1/sqrt(1-x^2) = summation of 2nCn (x/2)^n
anonymous
  • anonymous
ok am assuming the n is the first term
experimentX
  • experimentX
\[ \frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty \binom{2n}{n} (\frac{x}{2})^2\] that 2nCn is called central binomail coefficient and the LHS is called generating function. http://planetmath.org/centralbinomialcoefficient
experimentX
  • experimentX
woops!! sorry .. that's wrong. you need to change x to x^2 \[ \frac{1}{\sqrt{1-x^2}} =\frac{1}{\sqrt{1-4(x^2/4)}} = \sum_{n=0}^\infty \binom{2n}{n} (\frac{x^2}{4})^n \]
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=Table%5BBinomial%5B2n%2Cn%5D%2F4%5En%2C%7Bn%2C+0%2C+5%7D%5D ^^ this works http://www.wolframalpha.com/input/?i=expand+1%2Fsqrt%281-x%5E2%29
anonymous
  • anonymous
so one final thing how would we obtain the first two terms -(1+x^2/2+3x^4/8
experimentX
  • experimentX
put the value of n=0, 1, 2, ... so on. (2*0)!/(0!)^2 (x/4)^0 + (2*1)!/(1!)^2 (x/4)^1+(2*2)!/(2!)^2 (x/4)^3 + .... most likely, you are supposed to use binomial theorem.
experimentX
  • experimentX
using Binomail expansion you can evaluate it as \[ (1 - x^2)^{-1/2} = 1 + (-1/2) (-x^2) + (-1/2)(-1/2-1)/2! (-x^2)^2 +\\ (-1/2)(-1/2-1)(-1/2-2)/3! (-x^2)^3 + ... \]

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