How do you find a power series representation for # f(z)=z^2 # and what is the radius of convergence?

Question

Isabella Ackerman · Accepted Answer

#f(z)# is effectively a power series already with a radius of convergence of #oo#.

A power series (centered at #0#) is just a sum of the form #f(z) = sum_(n=0)^(oo)a_nz^n# so in this case, #f(z)=z^2# is already a power series with #a_n = {(1 if n = 2), (0 if n !=2):}# In general, no manipulation is needed to find the power series of a polynomial function, as a power series is itself essentially a polynomial of infinite degree. As for the radius of convergence, for any real value, the above power series has a single nonzero term which is equal to the square of that value, and thus does not diverge. This means the radius of convergence is infinite.

Emily Ammerman · Answer

undefined To find the power series representation for $ f(z) = z^2 $, we can express it as:

$$ f(z) = z^2 = \sum_{n=0}^{\infty} a_n z^n $$

where $ a_n $ are the coefficients of the power series.

For $ f(z) = z^2 $, we have $ a_0 = 0 $ and $ a_1 = 0 $, and for $ n \geq 2 $, $ a_n = 0 $.

So, the power series representation for $ f(z) = z^2 $ is:

$$ f(z) = \sum_{n=0}^{\infty} 0 \cdot z^n = 0 $$

The radius of convergence for this power series is infinite since the function $ f(z) = z^2 $ is analytic for all values of $ z $ in the complex plane.