Many students have asked me why it is important to for $x$ to be close to zero for the Maclaurin’s expansion to be a good approximation. So here, I plot it in the 4 curves:
$y =e^x$ which is our actual curve.
$y = 1+x$ which is the estimation of $e^x$, up to and including $x$.
$y = 1+x+\frac{x^2}{w!}$ which is the estimation of $e^x$, up to and including $x^2$.
$y = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$ which is the estimation of $e^x$, up to and including $x^3$.

We can observe from the graphs, that as we increase the degree of order, the estimated curves become more like that of $e^x$, although it still tend to deviate a lot. The idea of maclaurin’s is that it provides us a way to interpolate and write the humble $e^x$ equation out in a polynomial that actually never ends. So as we continue to consider more terms, our estimation will get close and closer to the actual curve.

And clearly, we see that when $x=0$, we found the actual value. This is simply because the maclaurin’s expansion is centred about zero. 🙂

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