The Unapologetic Mathematician

Mathematics for the interested outsider

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

5 Comments »

  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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