Sorry: Browser does not support Graphics Canvas Sorry: Browser does not support Graphics Canvas

$\displaystyle{p_0(x) = 1}$

$\displaystyle{p_1(x) = 1 + x}$

$\displaystyle{p_2(x) = 1 + x + \frac{1}{2}x^2}$

$\displaystyle{p_3(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3}$

$\displaystyle{p_4(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4}$

$\displaystyle{p_5(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5}$

$\displaystyle{p_6(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + \frac{1}{720}x^6}$

$\displaystyle{p_7(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + \frac{1}{720}x^6 + \frac{1}{7!}x^7}$

$\displaystyle{p_8(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{7!}x^7 + \frac{1}{8!}x^8}$

$\displaystyle{p_9(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{8!}x^8 + \frac{1}{9!}x^9}$

$\displaystyle{p_{10}(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{9!}x^9 + \frac{1}{10!}x^{10}}$

$\displaystyle{p_{11}(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{10!}x^{10} + \frac{1}{11!}x^{11}}$

$\displaystyle{p_{12}(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{1!}x^{11} + \frac{1}{12!}x^{12}}$

$\displaystyle{p_n(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{(n-1)!}x^{n-1} + \frac{1}{n!}x^n}$

The black line shows the function $f(x) = e^x$. The red line shows the polynomial

(We've used the factorial notation, $n! = 1\cdot2\cdot3\cdot\dots\cdot n$)

Notice how the $p_n(x)$ approaches $e^x$ as the degree increases. The low degree polynomials do a fairly good job of approximating $e^x$ if $x$ is close to 0, whereas higher degree is needed to approximate $e^x$ if $x$ is not so close to 0. Beyond degree $n \gtrapprox 12$, the polynomial looks like it is almost overlayed on $e^x$. But beware: it is only on this fixed domain, $[-4,2]$, that $p_n(x)$ looks very close to $e^x$. If you enlarge the domain, you will see that the polynomial isn't close to $e^x$ if $x$ is far away from 0, and you will need to increase the degree of the polynomial further to make it close to $e^x$. In fact, no matter how high the polynomial degree is, it won't do a good job of approximating $e^x$ if $x$ is really far from 0, and you will need to increase the degree further still.

What if you increase the degree of the polynomial to infinity? It's not hard to believe that, for each $x$, $$ \lim_{n \rightarrow \infty} p_n(x) = e^x . $$