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

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?

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

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

UPDATE: full question, but I am on part ii

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What is $g(x)$? What is $f(x)$? Your question asks about $f^{(11)}$, the answer you quote gives $g^{(11)}$.... – Arturo Magidin May 1 '12 at 3:43
To see that $g(x)=xf(x)=(x-2)f(x)+2f(x)$, just simplify the right-hand side. You never did tell us what $g(x)$ was supposed to be. Incomplete problems $\,=\,$ no answers. – André Nicolas May 1 '12 at 3:47
@AndréNicolas, updated question with full problem... hmm but I still dont get $g(x)=xf(x)=(x-2)f(x)+2f(x)$, in a real exam, I cant simplify the RHS, I am only given $g(x)=xf(x)$. What I probably can think of is $$g(x) = x \sum (-\frac{2^n}{3^{n+1}})(x-2)^{2n+2}$$ – Jiew Meng May 1 '12 at 3:52
$(x-2)f(x)+2f(x)=xf(x)-2f(x)+2f(x)=xf(x)$. – André Nicolas May 1 '12 at 4:09

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 $f^{(k)}(a) = a_kk!$.