hi Alex23,my duodecimal-system-loving friend

the sum of inverse factorials is done by getting the taylor series expansion around 0 (or in other words-the mclaurin's series) of e^x.so technically that can't exactly be used as a definition of e,but rather a way of computing it.

Sylvia104 wrote:

Definitions are not supposed be proved. It makes no sense to "prove" a definition. I think what you're looking for is a proof of the equivalence of two definitions of e.

Exactly! Also how does an exponent x of e transfer to the specific position in the definitions. How these connections where discovered?

hi Alex23 and everyone,

I agree with Sylvia104 that you don't have to prove definitions.  Any mathematical theory may be developed by starting with some definitions and deriving the rest.  My brother once started with the compound angle formulas as axioms and derived all the usual trig stuff from them.

For 'e' the way I like to start is this:

Is there a function of x that, when differentiated, returns the same function?

If there is then it is relatively easy to prove that any other function having the same property must be a numeric multiple of that function.

So, after my students have tried a bit to find such a function, I tell them that this function does exist and it has an 'x' as one of its terms.

As it must differentiate to give itself that implies

but that third term implies that the fourth term is

and that implies the next term is

........ and so on.

They quickly grasp that this series continues for ever.

Then I switch to considering functions of the form

We look at cases like a = 2, a= 3, a = 4, a = ½, a = -2 etc

The graphs all have properties in common (such as the shape and they go through x=0, y=1) so we look at those properties and especially consider the gradient at any point and the gradient at x = zero.

as this last limit is just the gradient at x = 0 and inspection of the family of graphs shows each clearly has some fixed gradient value at (0,1) for each 'a'.

Now this family of curves can clearly have every possible gradient at x = 0 by choosing a suitable 'a'.

So define e to be the value of a such that the gradient at zero is 1.

then

So this function is the one that has the property that it differentiates to give itself.

It is then just a matter of setting x equal to 1 and you have the series expansion for e.




Bob

Last edited by bob bundy (2012-03-02 08:37:39)

Have a look at

http://www.milefoot.com/math/calculus/l … nOfE10.htm

Bob

hi Stefy,

Thanks for the tip.  I've done that now.

What did you think of the method?

Bob

You've told me about \cdot before. 
I'm just too lazy to do all that extra typing, so I just use a dot.  ( Or \times if I'm trying to make it very clear.)
And I'm bound to make a mistake with it; I always do with Latex.



Bob

Yes, you have and I do use it a bit ... depends what I'm doing.

Sometimes I just type the code from my head and then use copy and paste quite a bit.

Bob