Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :


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

share|cite|improve this question
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 $f^{(k)}(a) = a_kk!$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.