How do you find a power series converging to $f (x) = \frac{x}{{(1 + x)}^{4}}$ $f(x)=x/(1+x)^4$ and determine the radius of convergence?

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How do you find a power series converging to $f (x) = \frac{x}{{(1 + x)}^{4}}$ $f(x)=x/(1+x)^4$ and determine the radius of convergence?

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Answer 1 · 2017-05-20T16:53:39

$\frac{x}{{(1 + x)}^{4}} = \frac{1}{6} \infty \sum ν = 0 {(- 1)}^{ν + 1} ν (ν + 1) (ν + 2) x^{ν}$ $x/(1+x)^4=1/6 sum_(nu=0)^oo(-1)^(nu+1) nu(nu+1)(nu+2)x^nu$ for $| x | < 1$ $absx < 1$

Explanation:

There are infinite realizations, each depending to the point in which is done. We will develop a realization for the set $| x | < 1$ $abs x < 1$ centered at $x = 0$ $x=0$

We know that

$\frac{d^{3}}{{d x}^{3}} (\frac{1}{1 + x}) = - \frac{6}{{(1 + x)}^{4}}$ $d^3/(dx^3)(1/(1+x)) = -6/(1+x)^4$

and also that

$\frac{1}{1 + x} = \infty \sum k = 0 {(- 1)}^{k} x^{k}$ $1/(1+x) = sum_(k=0)^oo (-1)^k x^k$ for $| x | < 1$ $abs x < 1$ then

$\frac{x}{{(1 + x)}^{4}} = - \frac{x}{6} \frac{d^{3}}{{d x}^{3}} (\frac{1}{1 + x}) = - \frac{1}{6} \infty \sum k = 0 {(- 1)}^{k} k (k - 1) (k - 2) x^{k - 2}$ $x/(1+x)^4 = -x/6 d^3/(dx^3)(1/(1+x)) = -1/6sum_(k=0)^oo(-1)^k k(k-1)(k-2)x^(k-2)$

and making $ν = k - 3$ $nu=k-3$

$\frac{x}{{(1 + x)}^{4}} = - \frac{x}{6} \infty \sum ν = 0 {(- 1)}^{ν + 3} (ν + 1) (ν + 2) (ν + 3) x^{ν} =$ $x/(1+x)^4 =-x/6 sum_(nu=0)^oo (-1)^(nu+3)(nu+1)(nu+2)(nu+3)x^nu =$

$= \frac{1}{6} \infty \sum ν = 0 {(- 1)}^{ν + 1} ν (ν + 1) (ν + 2) x^{ν}$ $=1/6 sum_(nu=0)^oo(-1)^(nu+1) nu(nu+1)(nu+2)x^nu$

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How do you find a power series converging to $f (x) = \frac{x}{{(1 + x)}^{4}}$ $f(x)=x/(1+x)^4$ and determine the radius of convergence?

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How do you find a power series converging to f(x)=x(1+x)4f(x)=x/(1+x)^4 and determine the radius of convergence?

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How do you find a power series converging to $f (x) = \frac{x}{{(1 + x)}^{4}}$ $f(x)=x/(1+x)^4$ and determine the radius of convergence?