# The Unapologetic Mathematician

## The Taylor Series of the Exponential Function

Sorry for the lack of a post yesterday, but I was really tired after this weekend.

So what functions might we try finding a power series expansion for? Polynomials would be boring, because they already are power series that cut off after a finite number of terms. What other interesting functions do we have?

Well, one that’s particularly nice is the exponential function $\exp$. We know that this function is its own derivative, and so it has infinitely many derivatives. In particular, $\exp(0)=1$, $\exp'(0)=1$, $\exp''(0)=1$, …, $\exp^{(n)}(0)=1$, and so on.

So we can construct the Taylor series at ${0}$. The coefficient formula tells us $\displaystyle a_k=\frac{\exp^{(k)}(0)}{k!}=\frac{1}{k!}$

which gives us the series $\displaystyle\sum\limits_{k=0}^\infty\frac{x^k}{k!}$

We use the ratio test to calculate the radius of convergence. We calculate $\displaystyle\limsup\limits_{k\rightarrow\infty}\left|\frac{\frac{x^{k+1}}{(k+1)!}}{\frac{x^k}{k!}}\right|=\limsup\limits_{k\rightarrow\infty}\left|\frac{x^{k+1}k!}{x^k(k+1)!}\right|=\limsup\limits_{k\rightarrow\infty}\left|\frac{x}{(k+1)}\right|=0$

Thus the series converges absolutely no matter what value we pick for $x$. The radius of convergence is thus infinite, and the series converges everywhere.

But does this series converge back to the exponential function? Taylor’s Theorem tells us that $\displaystyle\exp(x)=\left(\sum\limits_{k=0}^n\frac{x^k}{k!}\right)+R_n(x)$

where there is some $\xi_n$ between ${0}$ and $x$ so that $R_n(x)=\frac{\exp(\xi_n)x^n}{(n+1)!}$.

Now the derivative of $\exp$ is $\exp$ again, and $\exp$ takes only positive values. And so we know that $\exp$ is everywhere increasing. What does this mean? Well, if $x\leq0$ then $\xi_n\leq0$, and so $\exp(\xi_n)\leq\exp(0)=1$. On the other hand if $x\geq0$ then $\xi_n\leq nx$, and so $\exp(\xi_n)\leq\exp(x)$. Either way, we have some uniform bound $M$ on $\exp(\xi_n)$ no matter what the $\xi_n$ are.

So now we know $R_n(x)\leq\frac{Mx^n}{(n+1)!}$. And it’s not too hard to see (though I can’t seem to find it in my archives) that $n!$ grows much faster than $x^n$ for any fixed $x$. Basically, the idea is that each time you’re multiplying by $\frac{x}{n+1}$, which eventually gets less than and stays less than one. The upshot is that the remainder term $R_n(x)$ must converge to ${0}$ for any fixed $x$, and so the series indeed converges to the function $\exp(x)$.

October 7, 2008 - Posted by | Analysis, Calculus, Power Series

1. Is there a typo in the penultimate line? (gets less than and stays less than 1) Comment by Richard Roe | October 8, 2008 | Reply

2. You’re right, sorry. Comment by John Armstrong | October 8, 2008 | Reply

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