Write the first four non-zero terms of the Taylor Series for the above function with x centered at a=0.

Using this model:

ln(1+x) = Σ$\frac{(-1)^{k}(x)^{k}}{k}$

I get the following...

x$^{3}$ln(1+2x) => x$^3$Σ$\frac{(-1)^{k}(2x)^{k}}{k}$

k=0 -----> Cannot divide by zero.

k=1 -----> x$^3$$\frac{(-1)^{1}(2x)^{1}}{1}$$\frac{(x-a)}{1!}$ -----> (0)$^3$$\frac{(-1)^{1}(2(0))^{1}}{1}$$\frac{(x-0)}{1!}$ = 0

k=2 -----> x$^3$$\frac{(-1)^{2}(2x)^{2}}{2}$$\frac{(x-a)}{1!}$ -----> (0)$^3$$\frac{(-1)^{2}(2(0))^{2}}{1}$$\frac{(x-0)}{2!}$ = 0

and so on...

I keep getting zero for every value of k in the series so I never reach a non-zero term.

I don't even know if I am doing this wrong but I am assuming that I am supposed to get an answer that isn't, "there are no non-zero terms" for this expansion.

  • 1
    $\begingroup$ "with $x$ centered at $a=0$" means that your series will contain powers of $x$, not that you should put $x=0$ in the expression. $\endgroup$ – njguliyev Oct 10 '13 at 22:08

As njguliyev said in the comments, you should be substituting $a=0$ rather than $x=0$. Your use of the power series for $\ln$ is correct, however, so the first term is $-2x^4$, as your work suggests.


Here how you advance

$$ \ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^k x^k}{k}\implies \ln(1+2x)=\sum_{k=1}^{\infty}\frac{(-1)^k (2x)^k}{k}$$

$$ \implies x^3\ln(1+2x)=\sum_{k=1}^{\infty}\frac{(-1)^k 2^kx^{k+3}}{k}=\dots.. $$

Now, you can find the desired number of terms.

  • $\begingroup$ One little fix: the $2$ should not get an additional three powers when the $x$ moves into the sum. $\endgroup$ – Eric Stucky Oct 10 '13 at 22:14
  • $\begingroup$ @EricStucky: You are right. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Oct 10 '13 at 22:16

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