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Sagot :
The power series for given function [tex]g(x)=\frac{4x}{(x-1)(x+3)}[/tex] is [tex]g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)[/tex]
For given question,
We have been given a function g(x) = 4x / (x² + 2x - 3)
We need to find a power series for the function, centered at c, for c = 0.
First we factorize the denominator of function g(x), we have:
[tex]\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}[/tex]
We can write g(x) as,
[tex]\Rightarrow g(x)=\frac{1}{x-1}+\frac{3}{x+3}\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1+\frac{x}{3} }\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1-(-\frac{x}{3} )}\\[/tex]
We know that, [tex]\frac{1}{1-x}=\sum{_{n=0}^\infty}~{x^n}[/tex] if |x| < 1
and [tex]\frac{1}{1-(-\frac{x}{3} )}=\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n[/tex] if [tex]|\frac{x}{6}| < 1[/tex]
[tex]\Rightarrow g(x)=-\sum{_{n=0}^\infty}~x^n+\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n\\[/tex] if |x| < 1 and if [tex]|\frac{x}{6}| < 1[/tex]
[tex]\Rightarrow g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)[/tex] if |x| < 1
Therefore, the power series for given function [tex]g(x)=\frac{4x}{(x-1)(x+3)}[/tex] is [tex]g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)[/tex]
Learn more about the power series here:
https://brainly.com/question/11606956
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