Evaluate, thus Is 27- 5) Calculate the Taylor polynomials T2 (a) and T3 (x) centered at a = a of the function f(x) = sin a for the given %3D value of a. (a) a = 0 T (b) a = « Previous Next

"Let T_2(x) be the Taylor polynomial of f(x)=\sqrt(x) centered at a=4. Apply the Error Bound to find the maximum value of the error if we approximate \sqrt(x) with T_2(x) for x in [3.9,4]."

Find G′(w) if G(w)=3/8w^6+6^4*squareroot of w

Let 8 = {(1, 1, 0), (0, 1, 1), (1, 0, 1)} and 8' = {(1, 0, 0), (0, 1, 0), (0, 0, 1)) be bases for R², and let 3 -1 A = 2 be the matrix for 7: R³ R³ relative to 8. (a) Find the transition matrix P from 8' to 8. 1/2 1/2 -1/2 -1/2 1/2 1/2 P= 1/2 -1/2 1/2 11 (b) Use the matrices and A to find [v] and [7(v)], where [v]=[-1 10]. 0 1 [v] = -1 1/2 [7(v)]= -1/2 1/2 (c) Find and A' (the matrix for 7 relative to 8). 1 0 1 p=d= 1 0 0 1/2 -1/2 1/2 A'= N/~ N/a N/W 1 -1 2 1 ↓1 x (d) Find [7(v)] two ways. 0 [7(v)]g. =P~¹[7(v)]g -1 1 -3/2 5/2 1/2 ↓t 11 11

The square root is over the (x^2+1)

10. (a) Produce the linear Taylor polynomials to f(x) = in (x) on 1 < x < 2, expanding about x, = }. Graph the error. Produce the linear minimax approximation to f(x) = ln (x) on [1, 2]. (b) Graph the error, and compare it with the Taylor approximation.

please  very soon

Find the first four of the Taylor the function 1/x about the point a = -1 (Your answers should include the variable x when appropriate) 1/x = ?+?+?+? I got 1/x= -1/0!+ (x+1)/1!+-2(x+1)^2/2!+(6(x+1))/3! However, (x+1)/1! And (6(x+1))/3! Is incorrect. Please help me.

We are interested in the first few Taylor Polynomials for the function f(x) = 7e* + 9e¯* centered at a = 0. To assist in the calculation of the Taylor linear function, T1(x), and the Taylor quadratic function, T2(x), we need the following values: f(0) = f'(0) f''(0) = Using this information, and modeling after the example in the text, what is the Taylor polynomial of degree one: Ti (x) What is the Taylor polynomial of degree two: T3(x) :

Find the value of f)(3) if f(x) is approximated near x = 3 by the 10 Taylor polynomial p(x) = f(4) (3) n=0 (x-3)" n!