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Phoenix [80]
2 years ago
5

Find a power series for the function, centered at c. g(x) = 4x x2 2x − 3 , c = 0

Mathematics
1 answer:
BartSMP [9]2 years ago
5 0

The power series for given function g(x)=\frac{4x}{(x-1)(x+3)} is g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)

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:

\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}

We can write g(x) as,

\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} )}\\

We know that, \frac{1}{1-x}=\sum{_{n=0}^\infty}~{x^n} if |x| < 1

and \frac{1}{1-(-\frac{x}{3} )}=\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n  if |\frac{x}{6}| < 1

\Rightarrow g(x)=-\sum{_{n=0}^\infty}~x^n+\sum{_{n=0}^\infty}~x^n(-\frac{x}{3} )^n\\     if |x| < 1 and  if |\frac{x}{6}| < 1

\Rightarrow g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n) if |x| < 1

Therefore, the power series for given function g(x)=\frac{4x}{(x-1)(x+3)} is g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)

Learn more about the power series here:

brainly.com/question/11606956

#SPJ4

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