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So I am trying to use differentiation to find the power series representations for a series of variations of a function. They are f(x)=1/(8+x)^2 g(x)=1/(8+x)^3 h(x)=x^2/(8+x)^3. I was able to get the first representation to be Sum of x^n(-1)^n((8^-1-n)*(1+n)). I am struggling where to go from here, any help is appreciated.

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  • $\begingroup$ To make the question more readable please use mathjax $\endgroup$ – ℋolo Aug 8 '18 at 16:31
  • $\begingroup$ Can you give the numerical values of the first couple of terms of your expansion of $1/(8+x)^2$? I think you may be out by a factor of $8$ $\endgroup$ – Henry Aug 8 '18 at 16:53
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Hint:

When you have the first one $f(x)=\dfrac{1}{(8+x)^2}$ then for $g(x)$ and $h(x)$ you should compute $f'(x)$ and $x^2f'(x)$.

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As a pattern (when $|x| <1$) you have the expansions with binomial coefficients:

  • $\frac{1}{1+x} = 1 - x +x^2 -x^3 + x^4 - \ldots$
  • $\frac{1}{(1+x)^2} = 1 - 2x +3x^2 -4x^3 + 5x^4 - \ldots$
  • $\frac{1}{(1+x)^3} = 1 - 3x +6x^2 -10x^3 + 15x^4 - \ldots$
  • $\frac{1}{(1+x)^4} = 1 - 4x +10x^2 -20x^3 + 35x^4 - \ldots$

and the coefficient of $x^n$ in the expansion of $\frac{1}{(1+x)^k}$ is $(-1)^n {n+k-1 \choose n}$

But you are actually interested in the expansion of $\frac{1}{(8+x)^k} = \frac{1}{8^k}\times \frac{1}{\left(1+\frac{x}8\right)^k}$, so the coefficient of $x^n$ in its expansion is $$\frac{1}{(8+x)^k} = \sum_{n=0}^{\infty} x^n \frac{(-1)^n}{8^{n+k}} {n+k-1 \choose k-1}$$ for $|x| \lt 8$

Note when $k=2$ as in your example this would give $\frac{(-1)^n}{8^{n+2}} (n+1)$ which is almost what you have: you might want to check whether your $8^{-1-n}$ should really be $8^{-2-n}$

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