Find the power series expansion of $f(z)=z^2$ around $z = 2$.
My result was $z^2 = 4 + 4(z − 2) + (z − 2)^2$ But I need different ways to solve Could someone help me through this problem?
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3$\begingroup$ Actually, since you never said how you got your result, how can anyone know if their way is different from your way? I assumed you used derivatives. $\endgroup$– anonApr 26, 2012 at 4:12
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1$\begingroup$ For polynomials, synthetic division (Horner) is an excellent technique for constructing Taylor expansions... $\endgroup$– J. M. ain't a mathematicianApr 26, 2012 at 4:13
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4 Answers
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Division with an appropriate linear divisor works nicely here. We have, as a start, the identity
$$\frac{z^2}{z-2}=z+2+\frac4{z-2}$$
A second division yields
$$\frac{z^2}{(z-2)^2}=1+\frac4{z-2}+\frac4{(z-2)^2}$$
Rearrange and you obtain
$$z^2=(z-2)^2+4(z-2)+4$$
As I mentioned in the comments, Horner's method, a.k.a synthetic division, is a good way to perform the divisions needed to extract your Taylor coefficients.
answered Apr 26, 2012 at 10:14
J. M. ain't a mathematicianJ. M. ain't a mathematician
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Here's one way; just use $(a+b)^2=a^2+2ab+b^2$ after rewriting
$$z^2=\big(2+(z-2)\big)^2=4+4(z-2)+(z-2)^2.$$
More generally, the power series of $z^n$ around $z=u$ can be derived via the Binomial Theorem:
$$z^n=\big(u+(z-u)\big)^n=\sum_{k=0}^n \left[\binom{n}{k}u^{n-k}\right](z-u)^k. $$
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Taylor series of $f(z)$ at $z=2$ says: $$f(z)= \sum_{n \ge 0} \frac{f^{(n)}(2)}{n!} (z-2)^n$$ $f(2)=4, f'(2)=4, f''(2)=2, f^{(n)} (2)=0 $ for $n \ge 3.$ Hence, $$f(z)=4+4(z-2)+(z-2)^2.$$
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Yet anouther way is to actually work it through directly through the definition of a taylor series.
answered Apr 26, 2012 at 5:07
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$\begingroup$ Why don't you do it here? Maybe it will serve the OP. $\endgroup$– Pedro ♦May 1, 2012 at 15:51