# Find the power series expansion of $f(z)=z^2$ around $z = 2$

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?

• 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.
– anon
Apr 26, 2012 at 4:12
• For polynomials, synthetic division (Horner) is an excellent technique for constructing Taylor expansions... Apr 26, 2012 at 4:13

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.

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:

$$z^n=\big(u+(z-u)\big)^n=\sum_{k=0}^n \left[\binom{n}{k}u^{n-k}\right](z-u)^k.$$

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, $$f(z)=4+4(z-2)+(z-2)^2.$$

• Oh, same as above!
– hkju
Apr 26, 2012 at 5:50

Yet anouther way is to actually work it through directly through the definition of a taylor series.

• Why don't you do it here? Maybe it will serve the OP.
– Pedro
May 1, 2012 at 15:51