Q. 73 - Calculus

Question

Prove that if the series $\sum _{\mathrm{k}=0}^{\infty }{\mathrm{a}}_{\mathrm{k}} {\mathrm{x}}^{\mathrm{k}}$ and $\sum _{\mathrm{k}=0}^{\infty }{\mathrm{b}}_{\mathrm{k}} {\mathrm{x}}^{\mathrm{k}}$ both converge to the same sum for every value of x in some nontrivial interval, then a_k = b_k for every nonnegative integer k.

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{series} \sum _{\mathrm{k}=0}^{\infty }\left({\mathrm{a}}_{\mathrm{k}}-{\mathrm{b}}_{\mathrm{k}}\right){\mathrm{x}}^{\mathrm{k}} \mathrm{is} \mathrm{a} \mathrm{maclaurin} \mathrm{series} \mathrm{for} \mathrm{the} \mathrm{zero} \mathrm{function}. \\ \mathrm{Thus}, \mathrm{each} \mathrm{coefficient} {\mathrm{a}}_{\mathrm{k}}-{\mathrm{b}}_{\mathrm{k}}=0 \mathrm{and} \mathrm{hence}, {\mathrm{a}}_{\mathrm{k}}={\mathrm{b}}_{\mathrm{k}}. \end{gathered}$$

Step 1. Given information is:

$$\begin{gathered} \sum _{\mathrm{k}=0}^{\infty }{\mathrm{a}}_{\mathrm{k}} {\mathrm{x}}^{\mathrm{k}} \mathrm{and} \sum _{\mathrm{k}=0}^{\infty }{\mathrm{b}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}} \mathrm{both} \mathrm{converge} \mathrm{to} \mathrm{the} \mathrm{same} \mathrm{sum} \mathrm{for} \mathrm{every} \mathrm{value} \mathrm{of} \mathrm{x} \\ \mathrm{in} \mathrm{some} \mathrm{nontrivial} \mathrm{interval}. \end{gathered}$$

Step 2. Proving a k = b k

$$\begin{gathered} \mathrm{The} \mathrm{two} \mathrm{series} \mathrm{are} \mathrm{the} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{some} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right). \\ \mathrm{Thus}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=0}^{\infty }\left({\mathrm{a}}_{\mathrm{k}}-{\mathrm{b}}_{\mathrm{k}}\right){\mathrm{x}}^{\mathrm{k}} \mathrm{is} \mathrm{a} \mathrm{maclaurin} \mathrm{series} \mathrm{for} \mathrm{the} \mathrm{zero} \mathrm{function}. \\ \mathrm{Thus}, \mathrm{each} \mathrm{coefficient} {\mathrm{a}}_{\mathrm{k}}-{\mathrm{b}}_{\mathrm{k}}=0. \\ \mathrm{Therefore}, \mathrm{we} \mathrm{can} \mathrm{say} \mathrm{that} {\mathrm{a}}_{\mathrm{k}}={\mathrm{b}}_{\mathrm{k}} \mathrm{for} \mathrm{every} \mathrm{value} \mathrm{of} \mathrm{k}. \end{gathered}$$