Proof of Maclaurin series

Just doing Maclaurin series in FP2 and was wondering if there was a proof anwhere of the formula? It just seems sort of random... I can see it works when you try it on various functions, but I would like to know the reasoning behind it. Thanks :biggrin:

Edit: Also, is there a proof that functions can be written in the form a0 + a1x + a2x^2 .... etc

(edited 11 years ago)

Reply 1

dave_chapman

The proof you're looking for is a proof of Taylor's Theorem, which basically states that Taylor (and therefore Maclaurin) Series exist.

http://en.wikipedia.org/wiki/Taylor%27s_theorem#Proof_for_Taylor.27s_theorem_in_one_real_variable

Reply 2

jack.hadamard

4

I don't know how much of the proof you can actually take. However, this video offers the intuition behind it (not so random).

Original post by Conjecture

Edit: Also, is there a proof that functions can be written in the form a0 + a1x + a2x^2 .... etc

This is a very tough question and there is a proof. The result is called the Stone-Weierstrass theorem (link to simpler version).

Reply 3

esbo

11

Original post by jack.hadamard

I don't know how much of the proof you can actually take. However, this video offers the intuition behind it (not so random).

This is a very tough question and there is a proof. The result is called the Stone-Weierstrass theorem (link to simpler version).

Yes I have struggled to find a proof for this and have not found a satisfactory one.

The proof I (and no doubt you have seen) can follow starts with the answer with some unknown constants and shows you how to get the constants.

A bit like y=4+c

And show you how to get the constant c ie c=y-4.

So it kind of shows you the answer works but does not explain how it was dreamed up in the first place.

Seems to be like the fourier series in that respect.

I have seem some attempts as a proof with out starting with the answer
but none that I would follow.

I guess you could say if you know all the derivatives at a point match for two expressions then the expressions must be identical at all points
because they must all change (in the limit) in the same way to the next point and thus must get to the same next point in the limit.

I guess I will have to settled for that.

Reply 4

DFranklin

18

Original post by jack.hadamard

I don't know how much of the proof you can actually take. However, this video offers the intuition behind it (not so random).

This is a very tough question and there is a proof. The result is called the Stone-Weierstrass theorem (link to simpler version).

To my mind the Weierstrass theorem is a very different result. (Approximation with finite series v.s. equality with infinite series).

Reply 5

esbo

11

I think I might have proof now, I am still kind of working on it but it looks pretty good :smile:

Reply 6

esbo

11

Yep, my theory looks pretty good but I have never seen it anywhere else.

I was kind of surprise when I looked for a proof but could not find a satisfactory one, I think a few have felt the same!!

Reply 7

FrostShot

2

With A levels I think you're just meant to have an intuitive feel graphically of the way these things work. Discussing when they actually work and proving that an infinite number of approximations does equal perfection (finding the values of x for which the taylor approximation is valid) is probably part of Analysis.

Note that the approximations aren't always valid for all domains of all functions, but it's easy to see that it's always correct for a finite polynomial.

Reply 8

esbo

11

Original post by FrostShot

With A levels I think you're just meant to have an intuitive feel graphically of the way these things work. Discussing when they actually work and proving that an infinite number of approximations does equal perfection (finding the values of x for which the taylor approximation is valid) is probably part of Analysis.

Note that the approximations aren't always valid for all domains of all functions, but it's easy to see that it's always correct for a finite polynomial.

Yes I think that not every function has a Taylor series in which case there
would not be general proof, which is maybe why I can't find one, indeed
although I though I had a proof it would only be for certain types of functions
ie of polynomials.

I am not sure I ever got a proof at A level just some vague waffling, so I guess I find the Taylor series for sin x cos x etc unsatisfactory.

There is something here but it is a bit dense, I will take look when I have more time and energy!!!

http://math.feld.cvut.cz/mt/txte/3/txe3ea3e.htm

Reply 9

TheMagicMan

15

Original post by Conjecture

Edit: Also, is there a proof that functions can be written in the form a0 + a1x + a2x^2 .... etc

Well it is a bit old but yes there is. It follows from Zorn's Lemma, which is equivalent to the Axiom of Choice which is a basic axiom that we accept or reject.

Reply 10

DFranklin

18

Original post by TheMagicMan

Well it is a bit old but yes there is. It follows from Zorn's Lemma, which is equivalent to the Axiom of Choice which is a basic axiom that we accept or reject.

Possibly I shouldn't be resurrecting this (I found the thread when searching for a response to something more recent), but I'm rather confused by what you mean here.

For sure some functions can be written in the form a_0 + a_1 x + ... ; polynomials being the obvious example.

At the same time, almost all functions (in a sense that can be made precise) cannot be written as a power series.

Even when the Maclaurin series exists, it need not converge to the given function anywhere (other than the origin, where it converges by definition).

The canonical counterexample is f(x) = exp(-1/x^2), with f(0) defiinded = 0. f is infinitely differentiable and all deriviatives =0 at the origin. So Maclaurin series is 0, which obviously doesn't equal f(x) when x isn't 0.

And I'm not sure at which point the Axiom of Choice gets a look in.

Proof of Maclaurin series

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