Question

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.

Answer 1

the function was d theta/dx and dx was taken to the other side after being implicitly differentiated.

Answer 2

the function after differentiated implicitly dtheta/dx=-1/sqrt 1-x^2

then dtheta=1/sqrt1-x^2

Answer 3

sorry dx should be at the end

Answer 4

@.Sam.

Answer 5

hi am only receiving this message @.Sam. and when i open the link its leading to another question

Answer 6

ok

Answer 7

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.

Answer 8

and the question has been answered

Answer 9

\[ \frac{1}{\sqrt{1 - 4x^2}} = \sum_{n=0}^\infty \binom{2n}{n} x^n\]

Answer 10

make appropriate substitution

Answer 11

ok

Answer 12

so how would we make the first substitution

Answer 13

x^2 = 4 (x/2)^2
1/sqrt(1-x^2) = summation of 2nCn (x/2)^n

Answer 14

ok am assuming the n is the first term

Answer 15

\[ \frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty \binom{2n}{n} (\frac{x}{2})^2\]
that 2nCn is called central binomail coefficient and the LHS is called generating function.
http://planetmath.org/centralbinomialcoefficient

Answer 16

woops!! sorry .. that's wrong.
you need to change x to x^2
\[ \frac{1}{\sqrt{1-x^2}} =\frac{1}{\sqrt{1-4(x^2/4)}} = \sum_{n=0}^\infty \binom{2n}{n} (\frac{x^2}{4})^n \]

Answer 17

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

Answer 18

so one final thing how would we obtain the first two terms -(1+x^2/2+3x^4/8

Answer 19

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.

Answer 20

using Binomail expansion you can evaluate it as
\[ (1 - x^2)^{-1/2} = 1 + (-1/2) (-x^2) + (-1/2)(-1/2-1)/2! (-x^2)^2 +\\ (-1/2)(-1/2-1)(-1/2-2)/3! (-x^2)^3 + ... \]