The Unapologetic Mathematician

Mathematics for the interested outsider

The Exponential Series

What is it that makes the exponential what it is? We defined it as the inverse of the logarithm, and this is defined by integrating \frac{1}{x}. But the important thing we immediately showed is that it satisfies the exponential property.

But now we know the Taylor series of the exponential function at {0}:

\displaystyle\exp(x)=\sum\limits_{k=0}^\infty\frac{x^k}{k!}

In fact, we can work out the series around any other point the same way. Since all the derivatives are the exponential function back again, we find

\displaystyle\exp(x)=\sum\limits_{k=0}^\infty\frac{\exp(x_0)}{k!}(x-x_0)^k

Or we could also write this by expanding around a and writing the relation as a series in the displacement b=x-a:

\displaystyle\exp(a+b)=\sum\limits_{l=0}^\infty\frac{\exp(a)}{l!}b^l

Then we can expand out the \exp(a) part as a series itself:

\displaystyle\exp(a+b)=\sum\limits_{l=0}^\infty\left(\sum\limits_{k=0}^\infty\frac{a^k}{k!}\right)\frac{b^l}{l!}

But then (with our usual handwaving about rearranging series) we can pull out the inner series since it doesn’t depend on the outer summation variable at all:

\displaystyle\exp(a+b)=\left(\sum\limits_{k=0}^\infty\frac{a^k}{k!}\right)\left(\sum\limits_{l=0}^\infty\frac{b^l}{l!}\right)

And these series are just the series defining \exp(a) and \exp(b), respectively. That is, we have shown the exponential property \exp(a+b)=\exp(a)\exp(b) directly from the series expansion.

That is, whatever function the power series \sum\limits_{k=0}^\infty\frac{x^k}{k!} defines, it satisfies the exponential property. In a sense, the fact that the inverse of this function turns out to be the logarithm is a big coincidence. But it’s a coincidence we’ll tease out tomorrow.

For now I’ll note that this important exponential property follows directly from the series. And we can write down the series anywhere we can add, subtract, multiply, divide (at least by integers), and talk about convergence. That is, the exponential series makes sense in any topological ring of characteristic zero. For example, we can define the exponential of complex numbers by the series

\displaystyle\exp(z)=\sum\limits_{k=0}^\infty\frac{z^k}{k!}

Finally, this series will have the exponential property as above, so long as the ring is commutative (like it is for the complex numbers). In more general rings there’s a generalized version of the exponential property, but I’ll leave that until we eventually need to use it.

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October 8, 2008 - Posted by | Analysis, Calculus, Power Series

3 Comments »

  1. [...] Now we’ve got another definition, using a power series, which is its Taylor series at . And we’ve shown that this definition also satisfies the exponential [...]

    Pingback by The Exponential Differential Equation « The Unapologetic Mathematician | October 10, 2008 | Reply

  2. need example

    Comment by soemoe | November 10, 2008 | Reply

  3. An example of what, exactly?

    Comment by John Armstrong | November 10, 2008 | Reply


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