Finding the Lagrange Remainder Theorem from the Integral Form of a Taylor Series

[StellaLuna]

Homework Statement

I am looking for some help in finding the Lagrange Remainder Theorem from the integral form of the remainder of a Taylor series

Homework Equations

Integral form of Taylor Series:
R_n,a(x) = ^x∫_a [f(n+1)(t)]/n! *(x-t)dt

The Attempt at a Solution

We are given the hint to use the mean value theorem.

Thanks for any help!

 

[mighty2000]

Thank you for your post. The Lagrange Remainder Theorem is a very useful tool in analyzing the accuracy of Taylor series approximations. It states that the remainder term of a Taylor series can be expressed as the value of the next term in the series evaluated at some point c between the center of the series (a) and the point at which we are approximating (x). In other words, Rn,a(x) = f(n+1)(c)/n! *(x-a)^(n+1).

To derive this result from the integral form of the remainder, we can use the mean value theorem. This theorem states that for a continuous function f on a closed interval [a,b], there exists a point c in the interval such that f'(c) = (f(b)-f(a))/(b-a). In other words, the slope of the tangent line at c is equal to the slope of the secant line connecting (a,f(a)) and (b,f(b)).

Now, let's consider the integral form of the remainder: Rn,a(x) = x∫a [f(n+1)(t)]/n! *(x-t)dt. By applying the mean value theorem to the function f(n+1)(t), we can rewrite this integral as Rn,a(x) = f(n+1)(c)/n! * x∫a (x-t)dt. Simplifying this, we get Rn,a(x) = f(n+1)(c)/n! * [x^2/2 - x*a + a^2/2].

Now, we can use the fact that f(n+1)(c) is the value of the (n+1)th derivative of f at some point c between a and x. We can also note that the expression in the brackets is simply the (n+1)th term in the Taylor series for f centered at a. Therefore, we have Rn,a(x) = f(n+1)(c)/n! * (x-a)^(n+1), which is the Lagrange Remainder Theorem.

I hope this helps! Let me know if you have any further questions. Good luck with your studies!