# Approximating $e^x$ with a polynomial of degree $n$

Would the cubic equation that is the best approximation for $e^x$ just be the first 4 terms of its Taylor Series expansion? $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$$ I guess it would depend on the bounds you are inspecting though? I'm not well-versed in dealing with infinities. Is there a cubic equation that is the best approximation on $[-\infty,\infty]?$

• This is indeed the best approximation you can do near $x = 0$. If you wanted an approximation near $x = 2$ for example, you would have a different cubic. Commented Sep 28, 2017 at 0:33
• The answer is usually "yes". Unless specified otherwise, I assume they meant "Maclaurin expansion" i.e. expansion near $x=0$, not near $\pm\infty$. Commented Sep 28, 2017 at 0:37

How close a function is to another definitely depends on how you define that notion. The $n$-th Taylor polynomial $T_n$ of $f$ at $x_0$ is the best approximation of $f$ among all degree $n$ polynomials, precisely in the sense that this is a unique polynomial satisfying

$$f(x) = T_n(x) + \mathcal{O}((x-x_0)^{n+1}) \qquad \text{as} \quad x \to x_0.$$

So this a priori concerns only what happens near $x_0$, and there is no reason that it should serve good approximation outside a neighborhood of $x_0$.

For instance, the following is a comparison of the degree 8 Taylor polynomial of $f(x) = \sin(2x)$ at $x = 0$ and the degree 8 polynomial interpolating 9 points $(k, f(k))$ for $k = -4, -3,\cdots, 4$.

$\hspace{6em}$

It depends what is it you try to minimize ?

Taylor expansion is the one that optimize the fitness of the polynomial for $||\cdot||_{\infty}$ to the exponential in a neighbourhood of $0$.

But you can try to minimize other things, by having other norms and/or trying to minimize locally or globally.

Here is an example of a linear approximation minimizing $||\cdot||_1$ on the interval $[-1,1]$ :

Linear approximation of exponential for the $L^1$ norm. Find $a,b$ that minimize $||ax+b-e^x||_{1}$

This is $l(x)=2x\sinh(\frac 12)+\cosh(\frac 12)$

It is different from Taylor expansion because we tried to minimize a global property (i.e. the fitness of the integral).