Why in the world would one use a trig substitution to find a Taylor's series??
Of course, you could do this "the old-fashioned way": use the definition of Taylor's series. (By the way, you say "Taylor Series" but don't say about what point. I assume you mean the McLaurin series: about x_{0}= 0.)
f(x)= 1/(1-x^{2}) so f(0)= 1
f'(x)= ((1- x^{2})^{-1})'
= -(1-x^{2})^{-2}(-2x)
= (2x)(1-x^{2})^{-2} so f'(0)= 0
f"(x)= 2(1-x^{2})^{-2}-2(2x)(1-x^{2})^{-3} so f"(0)= 2, etc.
So far we would have f(0)+ f'(0)(x)+ f"(0)/2(x^{2}= 1+ x^{2} but the derivatives quickly become very difficult to find!
I would be inclined to remember the formula for sum of a geometric series: If |r|< 1, then Σr^{n}= 1/(1-r).
The right hand side looks a lot like 1(1- x^{2}) if we were to take r= x^{2}!
I believe if yo do it with the trig substitution, you wind up with a series like this:
1/(1-x^2)=SUM[ 1/(1-x^2) + other non-zero terms]
This indicates that the sum of the "other non-zero terms" converges to zero. It is more complicated, but I like the effect.
Also, whenever I'm working with 1/(1-x^2) I just feel like I'm using the wrong coordinate system. I realize that there is no "wrong" coordinate system, but it feels that way to me.
I think you missed out the spuare root in my equation when differentiating. Any chance you could show me the working again with the square root in there? ie the differentiation etc.
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