How would one expand the following functions as power series around the given points:
- $f(x) = \frac{1}{x}$ around $2$
- $g(x) = \frac{x}{x-3}$ around $5$
I know how to do that using Taylor series. Is there another method?
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1If a function can be written as a power series $\sum_{n\geqslant 0} a_n x^n$, then the coefficients $a_n$ are uniquely determined... so they are necessarily equal to $f^{(n)}(0)/n!$. – Luke Collins Jul 07 '22 at 11:29
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Related https://math.stackexchange.com/questions/83951/uniqueness-of-power-series – DanielV Jul 07 '22 at 11:33
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Yes, @LukeCollins, but one still has to show that the Taylor series converges to the function – Seven9 Jul 07 '22 at 11:40
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The hint is: geometric series. – GEdgar Jul 07 '22 at 11:52
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I see! Would estimating the error of the Taylor polynomials using the Lagrange remainder work? – Seven9 Jul 07 '22 at 11:54
1 Answers
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For the first one, write $1/x=1/(2+x-2)$ and divide by $2$ getting:
$(1/2)/1-((2-x)/2)$ . Clearly $(2-x)/2$ < $1$ in a neighborhood of $2$
hence $1/(1-((2-x)/2)=1+(2-x)/2+((2-x)/2)^{2}+((2-x)/2)^{3}+.....$ and
multiplying by $1/2$ we get the expansion.
In the second case we write $1/((x-5)+2)$ and dividing by $2$ we
obtain $(1/2)(1/((x-5)/2)+1$ which is
$(1/2)(1-((x-5)/2)+((x-5)/2)^{2}-((x-5)/2)^{3}+......)$.
Multiplying this expansion by $x$ which can be written as $(x-5)+5$ we get the required expansion!