What is the significance of the number e and Euler's formula?

[Ben Niehoff]

Right. So, after some investigation, I have found that there are two looser definitions one might use:

Definition 1:
\exp(z) is the unique function over \mathbb{C} such that:

1. \frac{d}{dz}(\exp(z)) exists, and

2. \exp(x+i0) = e^x for all real numbers x.

Definition 2
\exp(z) is the unique function over \mathbb{C} such that:

1. \frac{d}{dz}(\exp(z)) = \exp(z), and

2. \exp(0) = 1.

Either definition is sufficient, and they are equivalent.

 

[jacobrhcp]

I was bored in class so I did this, don't really know if it's any use here, but I think so and found it very interesting:

d/dx[exp(x)]=lim h->0 \frac{exp(x+h)-exp(x)}{h} = exp(x) lim h->0 \frac{exp(h)-1}{h}=exp(x) lim h->0 \frac{\sum\frac{h^k}{k!}-1}{h}, with the sum from 0 to infinity,

= exp(x) lim h->0 \frac{\sum\frac{h^k}{k!}}{h}, with the sum from 1 to infinity

= exp(x) lim h->0 \sum\frac{h^{(k-1)}}{k!}

and all terms in the sum go to zero, except the first, which goes to 1... so:

= exp(x)

so, when you define exp(x) by \sum\frac{x^k}{k!}, the deravative of exp(x) is exp(x)

 

[Gib Z]

Yes, and that result would have been much easier if you differentiated the series directly, term by term. Or even if you used the limit definition, at the point \lim_{h\to 0} \frac{e^h -1}{h} you could have replaced with the series definition there, subtract one from it and then divide by h, you have the same series again.

 

[jacobrhcp]

well that's what I did >_>, at least the second thing.

 

[Gib Z]

My bad, I didn't pay attention to the intervals of summation.

 

[mathwonk]

have you studied linear algebra? an "eigenvector" for a linear operator T is a vector v such that Tv is a scalar multiple of v. These vectors provide the most natural coordinate system appropriate to the operator T. If one wants to solve an equation like TX = Y, for X, it is easy to do if Y is expanded in terms of eigenvectors of T.

The functions e^ax provide the eigenvectors for the linear operator D (differentiation). Using them, one gets the most natural expansion of a smooth function, its Fourier series. this makes it easy to solve differential equations like Df = g, if one can expand g in a Fourier series.

 

[HallsofIvy]

I will point out, again, that there is nothing all that magical about e. Any exponential function a^kx is an eigenfunction of the derivative operator.

 

[mathwonk]

well there is something special about the eigenvalue 1. or is your point that we should say "fixed points" of the operator D, to characterize ce^x?

i.e. e^x is the unique solution of the primordial ode: Dy = y, y(0) = 1.