Question

Suppose
\[
f(x)=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots+a_n(x-c)^n .
\]
Find \(a_k\).

Answer 1

\[
a_k=\frac{f^{(k)}(c)}{k !}
\]
The steps are given as below.
\[
\begin{gathered}
f(c)=a_0 . \\
f^{\prime}(x)=a_1+2 a_2(x-c)+3 a_3(x-c)^2+\cdots+n a_n(x-c)^{n-1}
\end{gathered}
\]
So
So
\[
\begin{gathered}
f^{\prime}(c)=a_1 . \\
f^{\prime \prime}(x)=2 a_2+6 a_3(x-c)+\cdots+n(n-1) a_n(x-c)^{n-2} .
\end{gathered}
\]
\[
f^{\prime \prime}(c)=2 a_2 .
\]
Similarly
\[
f^{(k)}(x)=k ! a_k+(2 \cdots(k+1))(x-c)+(3 \cdots(k+2))(x-c)^2+\cdots+
\]
So
\[
f^{(k)}(c)=k ! a_k .
\]
Or
\[
a_k=\frac{f^{(k)}(c)}{k !}
\]