in taylor theorem if h=x and a=0, then taylor theorem reduces in maclaurin's theoram
Answers
Answer:
One of the most important uses of infinite series is the potential for using an initial portion of the series for f to approximate f. We have seen, for example, that when we add up the first n terms of an alternating series with decreasing terms that the difference between this and the true value is at most the size of the next term. A similar result is true of many Taylor series.
Theorem 11.11.1 Suppose that f is defined on some open interval I around a and suppose f(N+1)(x) exists on this interval. Then for each x≠a in I there is a value z between x and a so that
f(x)=
N
∑
n=0
f(n)(a)
n!
(x−a)n+
f(N+1)(z)
(N+1)!
(x−a)N+1.
Proof.
The proof requires some cleverness to set up, but then the details are quite elementary. We want to define a function F(t). Start with the equation
F(t)=
N
∑
n=0
f(n)(t)
n!
(x−t)n+B(x−t)N+1.
Here we have replaced a by t in the first N+1 terms of the Taylor series, and added a carefully chosen term on the end, with B to be determined. Note that we are temporarily keeping x fixed, so the only variable in this equation is t, and we will be interested only in t between a and x. Now substitute t=a:
F(a)=
N
∑
n=0
f(n)(a)
n!
(x−a)n+B(x−a)N+1.
Step-by-step explanation: