The coefficient of the Taylor series is given by?
By John Campbell •
what is considered the coefficient in a taylor series? and how would i solve this
The coefficient of $(x-1)^3$ in the Taylor series of $ f(x) = ln x$ about $a=1$ is given by ?
$\endgroup$2 Answers
$\begingroup$If $\displaystyle \sum_{n=1}^\infty c_n(x-\alpha)^n$ is the Taylor series of $f(x)$ about $\alpha$ then $c_n=\dfrac{f^{(n)}(\alpha)}{n!}$.
$\endgroup$ $\begingroup$The coefficient of order $n$ is given by $\frac{f^{(n)}(x_0)}{n!}$, where $f$ is the function you are going to express as a power series, $f^{(n)}$ denotes the $n$-th derivative of $f$ and $x_0$ is the center point of expansion. In your case $$\log(x)=\log(1+(x-1))$$ hence $x_0=1$ and the coefficient of $(x-1)^3$ is the third derivative of $\log x$, divided by $6=3!$, evaluated in $x_0=1$
$\endgroup$
