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Proving MacLaurin's theorem
By Alexandra on March 16, 1998:

What is the proof, please, for MacLaurin's theorum

f(x) = f(0) + f'(0)x + (f''(0)x2)/2! + (f'''(0)x3)/3! etc

and how did MacLaurin come to discover it?

Thank you!

By David Sanders on March 22, 1998:

One way of proving MacLaurin's theorem is as follows. This assumes that the function you are considering actually can be expanded as a power series at all. This is a valid assumption at
A-level, since basically all the functions you will meet can be so expanded, but at degree level, you will meet some 'nastier' functions which cannot.

So let's assume that
f(x) = a0 + a1x + a2x2 + a3x3 + ... + anxn + ...
going on forever.

What is the justification for this? Well, consider the sum
g(x) = 1 + x + x2 + x3 + ...

This is just a G.P., and you know how to sum those to infinity - if you use the normal formula, you get g(x) = 1/(1-x). So the
function 1/(1-x) has a power series expansion. You can do similar things, it turns out, with lots of other series. MAYBE you can do it for a general function. This is a GUESS. You
don't KNOW that this is true, but you guess that it is, and then try and prove your guess. I'll come back to this after I've proved the result.

So we have ASSUMED that we can write, as before,
f(x) = a0 + a1x + ...

Then consider f(0). Since 01 = 02 = 03 = ... = 0, for all higher powers too, we have just f(0) = a0.

Now differentiate once:
f'(x) = a1 + 2a2x + 3a3x2 + ...

Then in the same way, f'(0) = a{1}.

Differentiate again:
f''(x) = 2a2 + 6a3x + ...

So f''(0) = 2a2, so a2 = f''(0) / 2.

Each time you differentiate you can see that another factor gets brought down.

Substituting these formulae for a0, a1 etc. back into the formula, gives:
f(x) = f(0) + f'(0)x + f''(0)x2/2 + ...
as we wanted.

But I left some questions unanswered. CAN I always do this? No, I can't. For example, if I can't differentiate the function f at 0, I certainly can't use this result. So you can't use this e.g. for the function f(x) = modulus(x), which means x, if x>0 and -x, if x<0, since this has a sharp point at x=0, where there is no well-defined gradient. Other things can also go wrong, for example 1/x is not defined at 0, so this also goes wrong. If you do a maths degree, some of the cryptic things I've said will become a bit clearer!

But basically, IF you assume that you can do all the things I've done, which you certainly can e.g. for sines, cosines and exponentials, then the result and the proof are fine!

I'm afraid I don't know the answer to how MacLaurin discovered his result, but power series were certainly being used a lot at the time, and he probably just happened upon it. In fact, it's quite possible that lots of people found this result, but didn't publish it - it just happens that for some reason MacLaurin got his name attached to it (I don't think he did much else of interest, although I may be wrong).

Hope this helps,

David.

By The Editor:

There is a biography of MacLaurin on the St Andrew's History of Maths website. He seems to have done quite a bit. Newton thought well of him, and he shared a prize with Bernoulli and Euler. MacLaurin's series, however, are just a special case of Taylor's series, as MacLaurin acknowledged himself.

By Alexandra on April 4, 1998:

Dear David,

Thanks. I follow the proof and it's really neat. But I'm going to ask you a question about it which you will probably think really strange. It's to do with the way my mind works trying to understand things.

Is there any way you can follow this proof conceptually? By that I mean on a graph or in any kind of 'picture language.' You see, while I believe the proof because there's no flaw in the logic I still can't understand the theorem in my head. Maybe it's the why behind the how that I'm missing.

I should warn you that when I first came across the proof that root two is irrational, I spent about 2 weeks trying to prove it wrong because I could not believe that such a simple set of steps
led to such a true thing.

As I said, it's an odd question. If you know any way of thinking about it that might help, please tell me even if it doesn't directly answer the query.

Thank you!

Alexandra