MATH SOLVE

3 months ago

Q:
# Find a power series representation for the function. Determine the interval of convergence. (Give your power series representation centered at x = 0.) f(x) = 5 2 − x

Accepted Solution

A:

I'm guessing the function is[tex]f(x)=\dfrac5{2-x}=5(2-x)^{-1}[/tex]whose first few derivatives are[tex]\dfrac{\mathrm df}{\mathrm dx}=-5(2-x)^{-2}(-1)=5(2-x)^{-2}[/tex][tex]\dfrac{\mathrm d^2f}{\mathrm dx^2}=-2\cdot5(2-x)^{-3}(-1)=2!\cdot5(2-x)^{-3}[/tex]with a general pattern of[tex]\dfrac{\mathrm d^nf}{\mathrm dx^n}=n!\cdot5(2-x)^{-(n+1)}=\dfrac{5n!}{(2-x)^{n+1}}[/tex]Then [tex]f[/tex] has power series (centered at 0)[tex]\displaystyle\sum_{n=0}^\infty\frac{\frac{5n!}{2^{n+1}}}{n!}x^n=\frac52\sum_{n=0}^\infty\left(\frac x2\right)^n[/tex]which is a geometric series, converging only for[tex]\left|\dfrac x2\right|<1\implies|x|<2\implies-2<x<2[/tex](this is the interval of convergence)###If you know that for [tex]|x|<1[/tex][tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]then we can derive the same series by manipulating this result:[tex]\displaystyle\frac5{2-x}=\frac52\frac1{1-\frac x2}=\frac52\sum_{n=0}^\infty\left(\frac x2\right)^n[/tex]