0
$\begingroup$

Finding $f^{(11)}(2)$ from taylor series $\sum (-\frac{2^n}{3^{n+1}})(x-2)^{2n+2}$

Given $g(x)=xf(x)$. But how does this hint help?

The answer looks like :

$g(x)=(x-2)f(x)+2f(x)$

but how does this translate to $xf(x)$?

The rest of the answer

enter image description here

UPDATE: full question, but I am on part ii

enter image description here

  • 1
    $(x-2)f(x)+2f(x)=xf(x)-2f(x)+2f(x)=xf(x)$.2012-05-01

1 Answers 1

1

If you have the Taylor series for $f$ around $a$, then any terms $a_n(x-a)^n$ with $n\lt k$ will vanish when you compute the $k$th derivative; any term $a_n(x-a)^n$ with $n\gt k$ will still have a factor of $(x-a)$, so when you evaluate the $k$th derivative at $a$ it will evaluate to $0$. So the only term that will matter when you compute $f^{(k)}(a)$ is $a_{k}(x-a)^{k}$; the $k$th derivative of this is $a_{k}(k!)$. So if $f(x) = \sum_{n=0}^{\infty}a_n(x-a)^n$ is the Taylor series expansion for $f(x)$ around $a$, and the radius of convergence is positive, then $f^{(k)}(a) = a_kk!$.