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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

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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 $g^{(11)}$....2012-05-01
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    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.2012-05-01
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    @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 $$g(x) = x \sum (-\frac{2^n}{3^{n+1}})(x-2)^{2n+2}$$2012-05-01
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    $(x-2)f(x)+2f(x)=xf(x)-2f(x)+2f(x)=xf(x)$.2012-05-01

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