# Why e?

1. Apr 21, 2004

### Zurtex

I'm revising over my maths for my exams and I just came across something I didn't understand. How do we know that:

$$\frac{d}{dx} \left( e^x \right) = e^x$$

I've seen the infinite series for $e^x$ but in our maths class we derived it by assuming the above statement . Preemptive thanks .

2. Apr 21, 2004

### matt grime

Assuming derivative commutes with sum you can prove it thusly, but seeing as e^x is *defined* to be the function that satisfies y'=y, i think you can rest easy.

3. Apr 21, 2004

### Zurtex

Sorry but I don't understand the first half of your sentence. How do you know that something to the power of x is going to satisfy y'=y?

4. Apr 21, 2004

### matt grime

differentiate k^x from first principles and see why, assuming something reasonable we can prove if we have to, that its derivative is proportional to it, k=e is just the number where the ratio is 1.

5. Apr 21, 2004

### Zurtex

How? We never have to differentiate from first principles in the course I take, it is only due to my own interest that I know how to differentiate polynomials from first principles.

6. Apr 21, 2004

### uart

What I think Matt means is that if you take the defn of e^x as the function that satisfies y'=y then it;s not hard to deduce that the power series for y must be 1 + x + x^2/2! + x^3/3! + ...

Last edited: Apr 21, 2004
7. Apr 21, 2004

### matt grime

Let's do it for exponentials then

let f(x) = k^x

f'(x) is defined to be the limit, when it exists, as d tends to zero, of {f(x+d)-f(x)}/d, which here is the same as f(x)*((k^d-1)/d

now it is reasonable, as the top tends to zero 'exponentially' and the bottom linearly, that this limit will exist, and it can only depend on k after all, so for all exponentials the derivative is a constant times the original function. Now we suppose that there might be such number such that f'=f, and we find we can actually construct this number by working out the talyor series for the function and letting x=1.

8. Apr 21, 2004

### Zurtex

Erm I've used Maclaurin Series to show the infinite series for e^x, but that assumes that e^x satisfies y'=y. I wanted to know how you show that e^x satisfies y'=y.

9. Apr 21, 2004

### matt grime

It would be clearer to say, ok we define e^x to be the unique solution to that diff eqn, but how do know that its actually the same as the function (2.whatever)^x

hopefully if you look back you'll see why.

10. Apr 21, 2004

### Zurtex

Right thanks . I don't feel 100% about that but certainly a lot better.

11. Apr 21, 2004

### matt grime

get a graphical calcluator, plot a numerical approx to the derivatives for 2 to the x and 3 to the x and see that they really do look like you hope. that;s how we were introduced to it in high school, it's quite illuminating.

12. Apr 22, 2004

### Zurtex

:tongue: Some are lucky, didn't know graphical calculators existed when I was in High School in England (only a couple of years ago).

I know what you'll get as I know:

$$\frac{d}{dx} \left( a^x \right ) = (\ln{a}) a^x$$

However is there anywhere with some direct mathematical proof that this works? I don't like just being given things in maths as facts without seeing full proof of it.

13. Apr 22, 2004

### matt grime

once you've accepted that e works as it does, then we can work out the constants because

a^x = exp^{log a^x} = exp{xloga} so differentiate this and we get

log(a) exp{xloga} = log(a) a^x

(and i was in high achool in england 10 years ago)

14. Apr 22, 2004

### Zurtex

I understand all the maths but one bit, you say that for $f(x) = k^x$ that:

$$f'(x) = \lim_{\substack{\delta x \rightarrow 0}} f(x) \frac{k^{\delta x} - 1}{\delta x}$$

I'm not sure how you can assume that there will be some limit .

P.S. When I was in high school most complex thing we did was quadratic formulae

15. Apr 22, 2004

### pig

Zurtex,

e is the limit of (1+1/n)^n as n tends to infinity, right?

= limit of (1+n)^(1/n) as n tends to zero. Remember this for now.

Now, let L mean "limit as h tends to zero", and h mean "delta x":

ln(x)' =

= L of (ln(x+h) - ln(x))/h
= L of (ln((x+h)/x)/h)
= L of (ln(1+h/x)/h)
= L of x*(1/x) * ln(1+h/x)/h
= L of (1/x) * (x/h) * ln(1+h/x)
= L of (1/x) * ln((1+h/x)^(x/h))

As h tends to zero, h/x tends to zero, so (1+h/x)^(x/h) tends to e.

= L of (1/x)lne
= 1/x

So, ln(x)' = 1/x.

Now:

ln(a^x)=ln(a^x)
ln(a^x)=xlna
(ln(a^x))'=(xlna)'
(a^x)' * 1/(a^x) = lna
(a^x)' = a^x * lna

(e^x)' = e^x * lne
(e^x)' = e^x

16. Apr 22, 2004

### matt grime

the fact that the limit exists was the thing i said was reasonable to assume. if you want to see this proved rigorously then you're going to have to get used to lots of epsilon and delta arguments and it is messy and completely unhelpful unless you're about to teach yourself rigorous analysis. in which case no simple post in a forum like this will suffice. you'll need to get a text book and learn a lot of material.

17. Apr 22, 2004

### Zurtex

Thank you very much, still one problem though how do you know that "e is the limit of (1+1/n)^n as n tends to infinity"? Although I have seen that in my own study of maths I have never seen a proof for it or come across it in class.

18. Apr 22, 2004

### matt grime

the rigorous proof of that is very messy indeed (takes about 3 pages of work as far as i can remember). you will learn it when it is necessary, or get yourself some decent text (Tom Korner may have the exercise sheets on his web page at dpmms.cam.ac.uk where the proof is an exercise)

19. Apr 22, 2004

### Zurtex

What maths would I already need to know then to teach myself rigorous analysis and what books would you recommend for me to look at buying to teach myself?

20. Apr 22, 2004

### pig

Zurtex,

I don't know how to prove that the limit I used is equal to e. I thought that e was actually defined that way. The problem is - what is the actual definition of e? I don't really know.. For example, the sum from n=0 to infinity of 1/n! is also e, but to take that as a definition doesn't really make any more sense than what I used.

All those definitions can surely be proven equal, but not that they equal "e" - because "e" is simply what people decided to call that particular number, not the other way around :) Of course, I might be wrong.

21. Apr 22, 2004

### Zurtex

kk thanks I think I understand now had to write it out a bit to make sense of it but I get it.

But my question to matt still applies, I love learning maths particularly pure maths.

22. Apr 22, 2004

### matt grime

to be honest i've no idea what textbook you should buy, i never bought one as an undergraduate.

23. Apr 23, 2004

### uart

To prove that there is a limit consider the following very simple arguement. For convenience I'll just denote your above limit as L and also exclude the trivial case of k=1.

L = lim h->0 of (k^h - 1) / h

So L is not a function of x, it is therefore either a finite constant or is unbounded.

Now lets consider all possibilities.

1. L is finite and non zero

2. L is zero.

3. L is unbounded.

Proof by contradiction that L is a finite non zero constant :

Assume that L is either zero or unbounded. Note that we have previously shown that the derivative (d/dx) of y=k^x is equal to L * k^x.

Therefore if L=0 then the slope of y=k^x is everywhere zero, contradicted by inspection.

Similarly if L is unbounded then the slope of y=k^x is everywhere unbounded, again easily contradicted by inspection.

So that's it, y' = L y for a finite constant L.

Last edited: Apr 23, 2004
24. Apr 23, 2004

### matt grime

that doesn't hold uart. you are assuming it is differentiable to prove it's differentiable. inspection doesn't count i'm afraid.

25. Apr 23, 2004

### HallsofIvy

Let y(x)= ax where a is a positive number.

y(x+h)= ax+h = axah so

y(x+h)- y(x)= axah- ax= (aa-1)ax.

(y(x+h)-y(x)/h= ax (ah-1)/h)

Notice that we have one factor that depends upon x but not h, another that depends upon h but not x. Assuming that lim(h->0) (ah-1)/h exists (showing that it exists is the hard part) and Calling it Ca, we have
y'= Caax: a constant times the original function itself.

e is defined as the number such that Ce= 1.