f(x)=x$^3$ln(1+2x)
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.
Best Answer
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.