# Generalised mean value theorem?

1. Aug 18, 2007

### pivoxa15

Does anyone know the mean value theorem associated with the Taylor series. Representing the Taylor series a finite sum and an end term? I don't get how they get it to look that way.

2. Aug 18, 2007

### quasar987

3. Aug 18, 2007

### quetzalcoatl9

i have seen the taylor series derived that way..you consider the n-th antiderivative of the n derivative. when you apply the fundamental thm of calculus you will get a series of terms that define the taylor series, with a remainder term. you use the mean value theorem to show that the remainder will go to 0 as long as your function grows no faster than n!

4. Aug 18, 2007

### pivoxa15

This could be it but I don't follow it very well.

5. Aug 18, 2007

### quetzalcoatl9

i know, but i dont feel like typing out all of the latex. just write out taking N definite integrals over some interval for a function that has been differentiated N times.

Arfken does exactly this derivation in his book (pg. 260 in the 2nd edition)

6. Aug 19, 2007

### jostpuur

http://en.wikipedia.org/wiki/Taylor's_theorem

Quite good explanations. This stuff was left unclear to me when I was sleeping in the calculus classes, but now I feel like understanding the integral trickery that is done in the Wikipedia's article.

7. Aug 19, 2007

### quetzalcoatl9

that is exactly the method that i was referring to. as long as the function doesn't grow faster than n! you will be fine and the remainder term will go to zero

<happy 600 posts to me>

8. Aug 20, 2007

### pivoxa15

THe problem is they still didn't mention how the mean value theorem works in the proof.

9. Aug 20, 2007

### quetzalcoatl9

you invoke the mean value thm in order to convert the integral into the remainder term. in other words,

$$\int_a^b f(x) dx = (b - a)f(y)$$

where y is an element of the interval (a,b)

10. Aug 21, 2007

### pivoxa15

I've seen a proof of the genearlised mean value theorem in books and its long but it finally explained it fully in terms of Taylor series.

11. Aug 21, 2007

### quetzalcoatl9

i seem to recall that the MVT can be proven all by itself, and that the fundamental theorem of calculus can be derived from it

12. Aug 22, 2007

### pivoxa15

But the generalised MVT is something else. It involves using induction.