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Use the power series for f(x) = 1/(1-x) to write the function g(x) = x^2/(2 - x^2) as a power series. Write your answer as a summation with a lower index of n = 0, and simplify the summation to a single variable.

Would really appreciate the help

Respuesta :

LammettHash
[tex]\dfrac{x^2}{2-x^2}=\dfrac{x^2}2\cdot\dfrac1{1-\frac{x^2}2}[/tex]

Since

[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]

for [tex]|x|<1|[/tex], the power series for the rightmost rational expression is

[tex]\dfrac1{1-\frac{x^2}2}=\displaystyle\sum_{n=0}^\infty\left(\frac{x^2}2\right)^n=\sum_{n=0}^\infty \frac{x^{2n}}{2^n}[/tex]

which is applicable for [tex]\left|\dfrac{x^2}2\right|<1[/tex], or [tex]|x|<\sqrt2[/tex].

Distributing the other rational expression gives

[tex]\dfrac{x^2}{2-x^2}=\displaystyle\frac{x^2}2\sum_{n=0}^\infty\frac{x^{2n}}{2^n}=\sum_{n=0}^\infty\frac{x^{2n+2}}{2^{n+1}}[/tex]

I'm not sure what is meant by "simplify the summation to a single variable", but you can rewrite this as

[tex]\displaystyle\sum_{n=0}^\infty\frac{x^{2(n+1)}}{2^{n+1}}=\sum_{n=0}^\infty\left(\frac{x^2}2\right)^{n+1}[/tex]

which could be interpreted as a simpler form, but that's really in the eye of the beholder...

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