This is the only bit I'd question: the Taylor series for e^x is quite a big "extra thing" to your original "main thing".

(But you certainly need to assume *something* more, because you can find (*) functions E with E(a+b) = E(a)E(b) for all a, b in R, but E is not continuous, let alone differentiable).

(*) You need to use the Axiom of Choice, so "find" here is certainly not constructive...

EDIT: arkanm, what if

\log a

is not an integer?

Yes, this is a fairly fundamental problem with the argument IMHO...

Reply 22

DFranklin

Original post by arkanm

See above. It reduces to proving it for |a|<1 EDIT: which is ̶e̶a̶s̶y̶ left as an exercise to the reader. :wink:

So, as I said, there is a fundamental problem with your argument.

Reply 23

atsruser

Original post by DFranklin

This is the only bit I'd question: the Taylor series for e^x is quite a big "extra thing" to your original "main thing"

Would I be alone in thinking that any argument from "first principles" that appeals to use of the Taylor series for another function is a *heck* of a long way from the spirit of a first principles argument?

Someone unpleasant could start asking questions about convergence, derivation of the properties of e^x from the Taylor series defn, etc.

Reply 24

DFranklin

Original post by atsruser

Well yes, but from pre-uni people I'm happy enough with an argument that actually recognizes what results it's assuming, even if they are not "first principles".

Someone unpleasant could start asking questions about convergence, derivation of the properties of e^x from the Taylor series defn, etc.

Nah, I'm still being unpleasant about arguments like "1 + 1 + 1 + ... + 1 (log a terms)" which apparently still work when log a isn't an integer :rolleyes:

Reply 25

atsruser

Original post by DFranklin

Well yes, but from pre-uni people I'm happy enough with an argument that actually recognizes what results it's assuming, even if they are not "first principles"

Nah, I'm still being unpleasant about arguments like "1 + 1 + 1 + ... + 1 (log a terms)" which apparently still work when log a isn't an integer :rolleyes:

Well, I think Euler made a few arguments of that quality, and it didn't do him any harm :-)

Reply 26

Desk-Lamp

Original post by atsruser

That's a fair point. The trouble with arguments about derivatives of log and exponential functions that I have seen is that they can easily end up being circular...

I think we've already demonstrated that at least twice in the space of this thread :smile:

Reply 27

DFranklin

Original post by atsruser

That's a fair point. The trouble with arguments about derivatives of log and exponential functions that I have seen is that they can easily end up being circular, unless you carefully demonstrate the results that you're assuming.The trouble with "the trouble with arguments like this is the danger of being circular" is that there are a lot of possible ways to build the foundations, so for any particular question, you can usually found a way of building the foundations that largely trivialises the problem.

(For example, in this case I'd probably start by defining exp(x) to be the unique function with d/dx (exp(x)) = exp(x) and exp(0) = 1).

Reply 28

Desk-Lamp

Original post by DFranklin

(For example, in this case I'd probably start by defining exp(x) to be the unique function with d/dx (exp(x)) = exp(x) and exp(0) = 1).

So the next step here is to find a way to show that exp(x) is equivalent to a function

c^x

where c is a constant, I suppose? once you have done that you're entitled to use log laws with exp(x) and you can solve the problem fairly trivially.

Reply 29

DFranklin

One you've got exp'(x) = exp(x), fix y, and show exp(x)exp(y)/exp(x+y) is constant and then exp(x)exp(y) = exp(x+y). The exp(x) = exp(1)^x is fairly quick to show from there (at least for rational x, and you might as well then define exp(1)^x as exp(x) for x irrational).

Reply 30

davros

Original post by arkanm

Yes, unfortunately I had to use the expansion of ln(x), which makes the problem trivial (just use it at the beginning). So there doesn't appear to be an "elementary" approach, i.e. from first principles.

The problem with trying to produce any sort of "elementary" approach to differentiating something like

a^x

is that you first have to understand what the function means, given that x can take real values. I mean, we can all agree what

2^3

means or

2^7

, and we can easily extend our understanding to

2^{-2}

2^{3/7}

, but how do you make sense of something like

2^{\sqrt{2}}

2^{\pi}

? It's a bit ambitious to talk about the derivative of something if you don't really know how the "something" works :smile:

My personal starting point would be to define exp(x) by its power series - this gives you continuity and differentiability, and the ability to define an inverse function ln x, THEN you can define

a^x = exp(alnx)

and continue from there, but as DFranklin says, you can build up the foundations in different ways if you prefer.

Reply 31

davros

Original post by arkanm

That's what I thought of at first but it seemed to "high-powered" by introducting e^x and stuff, but then I ended up using the power series for ln (1+x) anyway. :frown:

Can we perhaps use like a make-shift power series for a^x? (analogously with the e^x approach)

for example, since

1+a+a^2+\cdots +a^{x-1}=\frac{a^x-1}{a-1}

we get

a^x=1+(a-1)(1+a+a^2+\cdots +a^{x-1})

EDIT: Thanks for the explanation by the way! I tried +1 but it didn't work, maybe tomorrow

I don't quite follow what you're trying to do there, but my first comment is that the power series for ln(1+x) has a very small interval of convergence (basically for |x| < 1) so any argument you base on that series will only be valid in that interval!

The other problem I think you're going to have is that once you start introducing infinite series you have to be very careful about assuming what happens when you try to apply limits to those series - limits do not always work intuitively as you have already seen!

Bear in mind that mathematicians very rarely differentiate "from first principles" if they can help it - they're naturally very lazy :smile:

Most first year uni analysis is about setting up machinery that lets you derive general rules that you can then apply to numerous specific cases e.g.

this is how differentiation is defined - let's apply that definition to some "basic" functions
this is how the product / chain rules are proved - now we can differentiate a wider class of functions
this is how a power series is defined. Power series have something called a radius of convergence. Inside the radius of convergence a power series is not only continuous, but differentiable too. Now we can define an even more interesting class of functions purely from power series and get continuity and differentiability for free.

etc etc,

It's good that A level students want to explore the meaning and rationale behind topics like differentiation, but it shouldn't be taken to extremes :smile:

Reply 32

atsruser

Original post by DFranklin

(For example, in this case I'd probably start by defining exp(x) to be the unique function with d/dx (exp(x)) = exp(x) and exp(0) = 1).

I guess I'd start by seeing how

\displaystyle \lim_{h \rightarrow 0} \frac{a^h-1}{h}

is evaluated.

For

f(x) = \frac{a^x-1}{x}

we want to show that:

1. It's bounded below by 0
2. It's increasing
3.

\displaystyle \lim_{h \rightarrow 0} a^h = 1

so that

\displaystyle \lim_{h \rightarrow 0} \frac{a^h-1}{h}

exists.

Then we note that:

\displaystyle \frac{(ab)^x-1}{x} = \frac{a^xb^x -a^x + a^x-1}{x} = \frac{a^x(b^x-1)}{x} + \frac{a^x-1}{x}

So that:

\displaystyle \lim_{x \rightarrow 0} \frac{(ab)^x-1}{x} = \lim_{x \rightarrow 0}( \frac{a^x(b^x-1)}{x} + \frac{a^x-1}{x}) = \lim_{x \rightarrow 0} \frac{a^x-1}{x} + \lim_{x \rightarrow 0} \frac{b^x-1}{x}

so if we write

\displaystyle \ln(a) = \lim_{x \rightarrow 0} \frac{a^x-1}{x}

then we have

\ln(ab) = \ln(a)+\ln(b)

and this limit behaves like a log function (with

\ln(1) = 0

, too, obviously, from this definition).

Then define e such that

\ln(e)=1

(exists since this log function is increasing, continuous, and the IVT, I guess).

We'd still need to show that this function is bijective, and that it has the correct derivative. Not sure how we do that though.

After all that, we'd have

\displaystyle \frac{d(a^x)}{dx} = a^x \lim_{h \rightarrow 0} \frac{a^h-1}{h} = a^x\ln(a)

as required.

(And in the great tradition of this thread, missing details are left as an exercise for the reader :-)

Reply 33

XxKingSniprxX

CAN SOMEONE EXPLAIN TO ME -> IS THIS CORE 3/CORE 4 syllabus or FP modules?

Reply 34

Desk-Lamp

Alright folks, here's my full solution based on suggestions from DFranklin above ^.
It seems pretty concrete, the only thing that might cause difficulty is the definition of e^x for irrational numbers, but i'm not going to worry about that. :wink:

Spoiler

(edited 9 years ago)

Reply 35

davros

Original post by arkanm

Final flourish:

As before (my first approach) we find that it's equivalent to

a^x\lim_{h\rightarrow 0} \frac{(1+h)^{\ln a}-1}{h}

. But by l'Hopital's rule, this is just

a^x\lim_{h\rightarrow 0} d((1+h)^{\ln a})/dh = a^x\lim_{h\rightarrow 0} \ln a(1+h)^{\ln a-1} = a^x\ln a

, and we're done.

Don't let DFranklin catch you using l'Hopital's rule in a question about differentiation from first principles :smile:

Reply 36

DFranklin

Original post by davros

My personal starting point would be to define exp(x) by its power series - this gives you continuity and differentiability, and the ability to define an inverse function ln x, THEN you can define

a^x = exp(alnx)

and continue from thereThis was certainly the "standard approach" when I did IA (and seems common elsewhere as well).

In my experience, there's usually a bit of "hypocrisy" in this treatment though (which is also true for the "foundation" I suggested, to be clear): we're all very careful and rigorous about exp(x), differentiability, showing exp(a+b) = exp(a)exp(b), and then we just say "define

a^x = exp(alnx)

" without any kind of rigorous demonstration that this definition actually makes sense. [Not that it's particularly hard but it really ought to be done, and it seems very rare that it actually is done].

Reply 37

DFranklin

Original post by arkanm

Taking the limit as x goes to infinity we have

\lim_{x\rightarrow \infty} \left(1+\frac{1}{x}\right)^x =\lim_{x\rightarrow \infty} \sum_{r=0}^{\infty} \frac{1(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{r-1}{x})}{r!} = \sum_{r=0}^{\infty} \frac{1}{r!}=e

. Thus

\lim_{x\rightarrow 0} (1+x)^{1/x} = e

, and continue as before.

Note that

\sum_0^\infty a_r = \lim_{N \to \infty} \sum_0^N a_r

. So you've interchanged the order of two limits in going from the 2nd equation to the 3rd. This isn't valid without justification.

Reply 38

Zenarthra

Original post by atsruser

I have the impression that you are not yet totally familiar with differentiation from first principles. It may be useful to read up on that before trying it with such a tricky example.

How would you differentiate

y=x^2

from first principles? (Hint: you need to mimic algebraically the little right triangle that you would draw when working graphically, with a couple of points on a curve).

Sorry i was a little drunk when posting this thread haha.
I do understand differentiation from first principles.

So back to equation y=a^x

Let any point be (x,a^x) and another point be (x+h,a^(x+h))
Differentiating from first principles we have lim h>0 [a^(x+h) - a^x / x+h-x]

simplifies to lim h>0 [a^(x+h) - a^x / h]

And for the example y=x^2
General point of the curve (x,x^2) another point deltah along x axis (x+h, (x+h)^2)

as h>0 dy/dx > 2x