Deriving the taylor expansion from f(x)dx = f(x) - f(x+ dx)

I am fed up with this. Google, Yahoo AND Altavista all return a bunch of pages about the taylor expansion on how to do it with uggly mutliple integrals, or alternatively, by simply assuming the expansion exists. But no page whatsoever derives it in the nice simple way using the definition of the derivative.

Ok, so far I have got.

f '(x) = lim [ (f(x + dx) - f(x)) / dx ]

So as dx -> 0

f '(x) = ( f(x + dx) - f(x) )/ dx

f '(x)dx = f(x + dx) - f(x)

f(x + dx) = f(x) + f '(x)dx

Now, any tutorial on the internet that gets this far (including the worksheet we were given) just sais that from this it can be shown: and then immediately throws out the taylor expansion. I know I have seen how to apply the same argument again from here, I just can't remember how on earth it was done...

Edit: Many typoes in here, this is just what happens as you try to do maths 5:00AM in the morning....

Reply 1

Jonny W

8

Let x be fixed.

Let E(d) = f(x + d) - [f(x) + d*f'(x)] be the error at (x + d) of the Taylor expansion about x.

--

We want to show that, for small d, the error (E(d)) is much smaller than the final term in the Taylor expansion (d*f'(x)).

f'(x) doesn't change as we vary d. So our task is to show that, for small d, E(d) is much smaller than d.

For any nonzero d, E(d)/d = [f(x + d) - f(x)]/d - f'(x).

But [f(x + d) - f(x)]/d -> f'(x) as d -> 0.

So E(d)/d -> 0 as d -> 0. QED.

Reply 2

Jonatan

OP

13

Jonny W

Let x be fixed.

Let E(d) = f(x + d) - [f(x) + d*f'(x)] be the error at (x + d) of the Taylor expansion about x.

--

We want to show that, for small d, the error (E(d)) is much smaller than the final term in the Taylor expansion (d*f'(x)).

f'(x) doesn't change as we vary d. So our task is to show that, for small d, E(d) is much smaller than d.

For any nonzero d, E(d)/d = [f(x + d) - f(x)]/d - f'(x).

But [f(x + d) - f(x)]/d -> f'(x) as d -> 0.

So E(d)/d -> 0 as d -> 0. QED.

Thank you, though I was not looking for the error. I was looking for the rigorous derivation that actually shows that a polynomial aproximation exists. I know how to find teh coefficients once that is proved. You can do this by using integration by parts, but I was looking for the derivation using the derivative definition together with something that looks very similar to a proof by induction (I dont think you woudl write is as an induction proof, but essentially you just keep applying the definition of teh derivative over and over again together with an aproximation true for small dx.

Deriving the taylor expansion from f(x)dx = f(x) - f(x+ dx)

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