taylor's theorem: real analysis

Question

state and proof taylor's theorem?

 

Answer 1

Let f be a function, and c some value of x (“the center”). Denote, as usual, the degree n Taylor approximation of f with center x = c by Pn(x). As discussed before, this is the unique polynomial of degree n (or less) that matches f(x) and it first n derivatives at x = c. It is given by the expression below.

P_n (x)=f(c)+ f^' (c)(x-c)+ (f^'' (c))/2! *(x-c)^2+(f^''' (c))/3! *(x-c)^3+?+(f^((n) ) (c))/n! *(x-c)^n

Taylor’s theorem states that the difference between Pn (x) and f(x) at some point x (other than c) is governed by the distance from x to c and by the (n + 1)st derivative of f. More precisely, here is the statement.

Taylor’s Theorem. Assume that f is (n + 1)-times differentiable, and P_n is the degree n Taylor approximation of f with center c. Then if x is any other value, there exists some value b between c and x such that
f(x)=P_n (x)+ f^(n+1) (b)/(n+1)! *(x-c)^(n+1)

Proof. Since f(c) = 0 and f(x) = 0, the mean value theorem shows that there is some b1 between c and x such that f’(b1) = 0. Now, since this f’(c) = 0 and f’(b1) = 0, the mean value theorem shows that there is some b2 between c and b1 such that f’’(b2) = 0. Continue in this fashion, using the equations at c to construct a sequence of values b3, b4, … bn. Eventually, this shows that there is some bn between c and x such that f^(n+1) (bn )=0. Now b = bn is the desired value.
Although it may not appear that this lemma is related, we can use it to complete the proof of the mean value theorem, as follows. Let Pn(x) be the degree n Taylor approximations of f(x) with center x = c. Then the function g(x) = f(x) − Pn(x) has the property that g(c) = 0, and also g'(c) = g''(c) = ... = g(n)(c) = 0. . In effect, we have constructed Pn specifically to make these equations true.
Now let y be some value other than c. The error f(y) − Pn(y) of the Taylor approximation at y is simply g(y). Let C be the unique constant such that g(y) = −C*(y − c) n+1. In other words, let C = −g(y)/(y − c)^(n+1). Now define another function, h(x) = g(y) + C(x − c)^(n+1). The function h now has the properties described in the lemma.

h(c) = 0

h'(c) = 0
h''(c) = 0
…
h^(n)(c) = 0
h(y) = 0

It follows from this that for some value b between c and y, it is the case that h(n+1)(b) = 0. But now observe that h(n+1)(x) = f(n+1)(x) − (n + 1)!C, by an easy calculation from the definitions. Therefore, at x = b, we have h(n+1)(b) = f(n+1)(b) − (n + 1)!C = 0, so C = (f^((n+1)) (b))/((n+1)!) . But from the definition of C, it follows that
f(y)=Pn (y)+f^(n+1) (b) )/((n + 1)!)* (y - c)^(n+1). This is precisely Taylor’s theorem.

 

Contact me for a pdf of this which makes it more clear!

 

Jeff N.