Talk:Taylor's theorem
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Three expressions for R are available. Two are shown below
- It is a bit odd to say nothing about the 3rd. - Patrick 00:37 Mar 24, 2003 (UTC)
There ought to be a proof of this theorem.
I removed the following text from the article, which was added on Feb 5, 2003 by an anonymous contributor.
- Proof:
- Assume that :<math>f(x)<math> is a function that can be expressed in terms of a polynomial (it does not have to appear to be one). The n-th derivative of that function will have a constant term as well as other terms. The "zeroth derivative" of the function (plain old :<math>f(x)<math>) has what we will call the "zeroth term" (term with the zeroth power of x) as its constant term. The first derivative will have as a constant the coefficient of the first term times the power of the first term, namely, 1. The second derivative will have as a constant the coefficient of the second term times the power of the second term times the power of the first term: coefficient * 2 * 1. The next will be: coefficient * 3 * 2 * 1. The general pattern is that the n-th derivative's constant term is equal to the n-th term's coefficient times n factorial. Since a polynomial, and by extension, one of its derivatives, equals its constant term at x=0, we can say:
- <math>
f^{(n)}(0) = a_x{n!} <math>
- <math>
\frac{f^{(n)}(0)}{n!} = a_x <math>
- So we now have a formula for determining the coefficient of any term for the polynomial version of :<math>f(x)<math>. If you put these together, you get a polynomial approximation for the function.
I am not sure what this is supposed to prove, but it appears to be meant as a proof of Taylor's theorem. In that case, it does not seem quite right to me; in particular, the assumption in the first sentence ("f is a function that can be expressed in terms of a polynomial") is rather vague and appears to be just what needs to be proven. Hence, I took the liberty of replacing the above text with a new proof. -- Jitse Niesen 12:50, 20 Feb 2004 (UTC)
What is ξ?
In the Lagrange form of the remainder term, is ξ meant to be any number between a and x or is the theorem supposed to state that there exists such a ξ? I would guess the latter (because the proof uses the Mean Value Theorem), but the article doesn't make it totally clear. Eric119 15:51, 23 Sep 2004 (UTC)
- Thanks for picking this up. I rewrote that part to clarify (hopefully). -- Jitse Niesen 16:00, 24 Sep 2004 (UTC)