Why, intuitively, does the Maclaurin series for $e^x$ but not $\ln(1+x)$ converge globally? So we all know that, $\forall x\in\mathbb{R}$,
$$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$$
And that
$$\ln (1+x) = \sum_{k=0}^{\infty} \frac{(-1)^{k-1}}{k}x^k$$
But that this only holds for $x\in(-1,1]$.
My question is, given that one is a simple, shape-preserving transformation of the other, why should we expect the former to be readily approximated globally by polynomials while this is only possible locally for the former?
I know that mathematically this comes basically from the discontinuity in the derivative of $\ln (1+x)$; I'm trying to understand how we get "global information" at $x=0$ from $e^x$, but somehow this global information is destroyed when we flip the curve about $y=x$.
What is it that allows polynomials to fit this curve "vertically" but not "horizontally"?
 A: let $L:x \mapsto \log(1+x)$ and $E:x \mapsto e^x$. $E$ we know to be entire.  suppose $L$ were entire. then the composition: $G = E \circ (-L)$ must also be entire. but $G: x \mapsto \frac1{1+x}$ which has a simple pole at $x=-1$
as Dan Shved noted in his comment on your question there is no actual loss of information because $\log(1+x)$ is a translate of $\log x$ and you will see that the graphs of $e^x$ and $\log x$ in $\mathbb{R}^2$ are mirror images in the line $y=x$
A: How can $\log(1+x)$ have a convergent Maclaurin series for $x\le-1$?
Your question actually hints at something deeper. Consider instead the function $f(x)=\frac1{1+x^2}$. This is defined on all of $\mathbb R$, and at every point it has a Maclaurin series with nonzero radius of convergence. However,
$$
f(x)=\sum_{n=0}^\infty(-1)^nx^{2n}$$
does not converge for $|x|>1$. Asking why this function does not have a globally convergent Maclaurin series is a more interesting question, and the answer lies in complex analysis. I'll leave you to think about it.
EDIT: It seems like you are really asking why intervals of convergence must be symmetric. Let's look at a simple example: $\sum_n x^n$. In this case, you can explicitly work out the partial sums and note that the series converges if $|x|<1$. However, if $|x|>1$ then we are summing larger and larger terms, so of course there will be no convergence.
Now for a general power series $\sum_na_nx^n$, the radius of convergence is, very roughly speaking, the number $\rho$ such that: if $|x|<\rho$, $|a_nx^n|\le y^n$ for all large enough $n$, where $0\le y<1$, and and if $|x|>\rho$, $|a_nx^n|\ge z^n$ for infinitely many $n$, where $z>1$. Comparing with our previous series gives appropriate convergence/divergence conclusions.
The point is, we are only ever interested in $|x|$. If the expansion is about $x_0$, i.e. $\sum_na_n(x-x_0)^n$, everything still holds only now we are interested in $|x-x_0|^n$, so the interval of convergence will be $(x_0-\rho,x_0+\rho)$ rather than $(-\rho,\rho)$. This (hopefully) explains why intervals of convergence must always be symmetric.
This extends to complex analysis as I alluded to earlier in the following way. Suppose now we are interested in the power series $\sum_na_nz^n$, where $z$ is now allowed to be a complex number (and if you like the sequence $(a_n)$ may also be complex). The exact same arguments above still work; convergence will be determined by $|z|$ (except on boundary cases, which we shall ignore). Why then should my function $f$ defined above have radius of convergence $1$ about the origin, even though it is analytic everywhere? Well, once we consider the complex function $f(z)$, it becomes rather clear; this function has a singularity at $z=\pm i$, so the radius of convergence can certainly be no more than $1$.
A: My best answer is that if we consider the derivative, we get $\frac{1}{1+x}$, which means that, for a series centered at $0$, the radius is at most $1$.
This question might help you see why the radius has to be symmetric on both sides.
