Why e?

[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 :confused:. Preemptive thanks :smile: .

[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.

[Zurtex]

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.
:frown: 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?

[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.

[Zurtex]

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.
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.

[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! + ...

[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.

[Zurtex]

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[b] must be [b]1 + x + x^2/2! + x^3/3! + ...
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.

[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.

[Zurtex]

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

[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.

[Zurtex]

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.
: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.

[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)

[Zurtex]

:smile: 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 :frown:.

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

[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

[matt grime]

:smile: 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 :frown:.

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


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.

[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.

[matt grime]

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.


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)

[Zurtex]

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.
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?

[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.

[Zurtex]

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.
kk thanks I think I understand now :biggrin: 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.

[matt grime]

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

[uart]

:smile: 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 :frown:.

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

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.

[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.

[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.

[Zurtex]

Thanks :smile:, I feel much better seeing the maths of it. I would normally ask my teacher but he is looking rather stressed out from the amount of work we have left with only 3 weeks till study leave :wink: