Find n-th Taylor Polynomial of $f(x)$ around $0$.

Question

Let $f(x)=\ln(1+x)$. Find $n$-th Taylor Polynomial of $f(x)$ around $0$.

Recall $P_n (x)=\sum _{i=0}^n \dfrac {f^i (c)} {i!} (x-c)^i$.

My question is: in the question, ...''around $0$'' means that $c=0$, right?

Answer 1

"My question is: ...''around $0$'' means that $c=0$, right?"

Yes, you're absolutely right! In your case the $n$th term of the Taylor polynomial of $f(x)$ expanded around $0$ is:

$$\dfrac {f^n (0)} {n!} (x)^n \;=\; \pm \frac {x^n}n$$

If $n=2k+1$ (is odd) then, the $n$th term is positive and conversely, if $n=2k$ (is even) then, the $n$th term carries a negative sign.

The series expansion of $\ln(x+1)$ looks like this for the first $10$ terms:

$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^ 7}{7}-\frac{x^8}{8}+\frac{x^9}{9}-\frac{x^{10}}{10}$$

notice that the terms have alternating signs and the denominator of the $n$th term is always $n$ and there is no factorial left.

The formula for the series of $\ln(x+1)$ for $|x|\lt1$ is $$\ln(x+1)\;=\;-\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}$$ and for $|x|\gt1$ it is $$\ln(x+1)\;=\;\ln(x)-\sum_{k=1}^\infty\frac{(-1)^k}{kx^k}$$

Also, there is a very useful Wolfram Alpha widget called Taylor series calculator, very handy if you ever need to double-check your results quick when you are working with Taylor series.