How can I write the taylor series expansion for the function (1+x^2)^(-1/2) in code

Hello. I am trying to get this taylor series expansion in code but I am having a hard time. I attached a picture of the function and its taylor series expansion. I thought of using the factorial function but I don't think that will work here. Thank you for the help.

2 Comments

Joshua - no, you probably don't need to use the factorial function here. But from what you have shown, what is the pattern? Once you figure that out then the expansion should be easy.

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 Accepted Answer

a0 = 1
an = -a(n-1) * (2*n-1)/(2*n)

4 Comments

Thanks. What is a? And what does n go from?
Just insert n = 1,2,3,...
a0 = 1
a1 = -1*1/2=-1/2
a2 = 1/2*3/4
a3 = -1/2*3/4*5/6
...
Now you see that the an's are the coefficients of the taylor expansion in front of x^(2*n).
Best wishes
Torsten.
Thank you Torsten, now to get 50 terms of it, I just need to put it in a for loop?
x=0.5;
f=1.0;
a=1.0;
xn=1.0;
n=25;
for i=1:n
a=-a*(2*i-1)/(2*i);
xn=xn*x^2;
f=f+xn*a;
end

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Asked:

on 5 Jul 2018

Commented:

on 6 Jul 2018

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