- #1
GreenPrint
- 1,196
- 0
New Question (Changed Old one) - Taylor Polynomial - Upper Bound for Absolute Error
(a) Find the 3-rd degree Taylor polynomial of sin(pix) centered at x=1.
(b) Use (a) to approximate sin(1.1*pi)
(c) Use the remainder term to find an upper bound for the absolute error in this approximation
Estimate of the Remainder
Let n be a fixed positive integer. Suppose there exists a number M such that |f(x)^(n+1)| <= M for all c between a and x inclusive. the remainder in the nth-order Taylor polynomial for f centered at a satisfies
|Rn(x)| = |f(x) - Pn(x)| <= (M|x-a|^(n+1))/(n+1)!
Remainder in a Taylor Polynomial
Let Pn be the Taylor polynomial of order n for f. The remainder in using Pn to approximate f at the point x is
Rn(x) = f(x) - Pn(x)
Taylor Polynomials
Let f be a function with f', f'', ..., f^(n) defined at a. The nth-order Taylor polynomial for f with its center a, denoted Pn, has the property that it matches f in value, slope and all derivatives up to the nth derivative at a; that is,
Pn(x) = sigma[k=0,n] ck (x-a)^k, where the coefficients are
ck = f(a)^k/k!, for k = 0, 1 2, ..., n
Taylor's Theorem
Let f have continuous derivatives up to f^(n+1) on an open interval I containing a. For all x in I,
f(x) = Pn(x) + Rn(x),
where Pn is the nth-order Taylor polynomial for f centered at a, and the remainder is
Rn(x) = ( f(x)^(n+1)*(x-a)^(n+1) )/(n+1)!,
for some point c between x and a
(a) Find the 3-rd degree Taylor polynomial of sin(pix) centered at x=1.
f(x) = sin(pix), df(x)/dx = pi cos(pix), d^2f(x)/dx^2 = -pi^2*sin(pix), d^3f(x)/dx^3 = -pi^3*cos(pix), f(1) = 0, df(1)/dx = -pi, d^2f(1)/dx^2 = 0, d^3f(1)/dx^3 = pi^3
P3(x) = -pi(x-1) + (pi^3*(x-1)^3)/6
(b) Use (a) to approximate sin(1.1*pi)
P3(1.1) is about -.309
(c) Use the remainder term to find an upper bound for the absolute error in this approximation
d^4f(x)/dx^4 = pi^4*sin(pix), |d^4f(x)/dx^4| <= pi^4 = M
R3(x) <= ( pi^4 |x-1|^4 )/4!
So I take it that sense the maximum value of R3(x) occurs at R3(inf) which has a value of infinity I say that the upper bound is infinity? I think I may be doing something wrong
I apologize for posting a post #2 I had to change some things around because I original had asked a different question but found the answer myself, so instead of creating another topic I just edited this one post for my new question that I seem to be struggling with.
Homework Statement
(a) Find the 3-rd degree Taylor polynomial of sin(pix) centered at x=1.
(b) Use (a) to approximate sin(1.1*pi)
(c) Use the remainder term to find an upper bound for the absolute error in this approximation
Homework Equations
Estimate of the Remainder
Let n be a fixed positive integer. Suppose there exists a number M such that |f(x)^(n+1)| <= M for all c between a and x inclusive. the remainder in the nth-order Taylor polynomial for f centered at a satisfies
|Rn(x)| = |f(x) - Pn(x)| <= (M|x-a|^(n+1))/(n+1)!
Remainder in a Taylor Polynomial
Let Pn be the Taylor polynomial of order n for f. The remainder in using Pn to approximate f at the point x is
Rn(x) = f(x) - Pn(x)
Taylor Polynomials
Let f be a function with f', f'', ..., f^(n) defined at a. The nth-order Taylor polynomial for f with its center a, denoted Pn, has the property that it matches f in value, slope and all derivatives up to the nth derivative at a; that is,
Pn(x) = sigma[k=0,n] ck (x-a)^k, where the coefficients are
ck = f(a)^k/k!, for k = 0, 1 2, ..., n
Taylor's Theorem
Let f have continuous derivatives up to f^(n+1) on an open interval I containing a. For all x in I,
f(x) = Pn(x) + Rn(x),
where Pn is the nth-order Taylor polynomial for f centered at a, and the remainder is
Rn(x) = ( f(x)^(n+1)*(x-a)^(n+1) )/(n+1)!,
for some point c between x and a
The Attempt at a Solution
(a) Find the 3-rd degree Taylor polynomial of sin(pix) centered at x=1.
f(x) = sin(pix), df(x)/dx = pi cos(pix), d^2f(x)/dx^2 = -pi^2*sin(pix), d^3f(x)/dx^3 = -pi^3*cos(pix), f(1) = 0, df(1)/dx = -pi, d^2f(1)/dx^2 = 0, d^3f(1)/dx^3 = pi^3
P3(x) = -pi(x-1) + (pi^3*(x-1)^3)/6
(b) Use (a) to approximate sin(1.1*pi)
P3(1.1) is about -.309
(c) Use the remainder term to find an upper bound for the absolute error in this approximation
d^4f(x)/dx^4 = pi^4*sin(pix), |d^4f(x)/dx^4| <= pi^4 = M
R3(x) <= ( pi^4 |x-1|^4 )/4!
So I take it that sense the maximum value of R3(x) occurs at R3(inf) which has a value of infinity I say that the upper bound is infinity? I think I may be doing something wrong
I apologize for posting a post #2 I had to change some things around because I original had asked a different question but found the answer myself, so instead of creating another topic I just edited this one post for my new question that I seem to be struggling with.
Last edited: