Find the Taylor series for $f(x) = \ln (1+x)$ centered at $x = 0$ using the formula for Taylor Series.
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Disregard the answer below if it is still visible. – John W. Smith Feb 20 '17 at 05:06
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http://math.stackexchange.com/questions/878374/taylor-series-of-ln1x and – John Wayland Bales Feb 20 '17 at 05:17
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$$\ln(1+x)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac {x^4}4+\&\text{c}=\sum\limits_{r=0}^{\infty}\dfrac {(-1)^{r}x^{r+1}}{r+1}$$
The Taylor Series for $f(x)$ at the point $x=a$ is
$$f(x)=f(a)+f^1(a)(x-a)+\dfrac {f^2(a)}{2!}(x-a)^2+\dfrac {f^3(a)}{3!}(x-a)^3+\cdots+\dfrac {f^r(a)}{r!}(x-a)^r$$
Since we want it at $x=0$, $a=0$ and simply plug them into the formula to get the expansion!
Note that another way is to start with the exponential theorem$$a^x=1+cx+\dfrac {c^2x^2}{2!}+\dfrac {c^3x^3}{3!}+\&\text{c}$$ Where $$c=(a-1)-\dfrac 12(a-1)^2+\dfrac 13(a-1)^3-\&\text c$$ And by setting $x=1,\ x=c$, we obtain$$a=e^c\implies c=\ln a$$ And therefore,$$\begin{align*} & \ln a=(a-1)-\dfrac 12(a-1)^2+\dfrac 13(a-1)^3-\&\text{c}\\ & \implies \ln(x+1)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac {x^4}4+\&\text c\end{align*}$$
Frank
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