Mary Boas: Mathematical Methods Problem 1.13.8

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SUMMARY

The discussion focuses on finding the Maclaurin series for the function \( \frac{1}{\sqrt{1-x^2}} \). The user, Chris Maness, initially derived the series incorrectly by substituting \( 1+x^2=t+1 \) without solving for \( t \) first. The correct series representation is \( \sum_{n=0}^{\infty} \binom{-1/2}{n} (-x^2)^{n} \), as provided in the solution manual. The discussion emphasizes the importance of proper substitution in series expansion.

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Homework Statement



Using the methods of this section:
(a) Find the first few terms of the Maclaurin series for each of the following functions.
(b) Find the general term and write the series in summation form.
(c) Check your results in (a) by computer.
(d) Use a computer to plot the function and several approximating partial sums of the
series.

#8 [tex]\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }[/tex]

Homework Equations



[tex](1+x)^{P}=\sum _{ n=0 }^{ \infty }{ \binom {P} {n}} x^n[/tex]

The Attempt at a Solution



I got: [tex]\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }=\sum _{ n=0 }^{ \infty }{ \binom {-1/2} {n}} (1-x^{2})^{n}[/tex]

Solution manual's solution is: [tex]\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }=\sum _{ n=0 }^{ \infty }{ \binom {-1/2} {n}} (-x^{2})^{n}[/tex]

What went wrong?

Thanks,
Chris Maness
 
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I see my mistake. I made a bad assumption when I subbed in 1+x^2=t+1. I needed to solve for t before I subbed it back in.

Thanks,
Chris Maness
 

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