Numerical Analysis Problem?

Question

Given the polynomial f(z)= z^3 - 7z^2 + 14z - 6

a. use Horner's algorithm to find p(4).

b. Use repeated Horner's algorithm to find the Taylor expansion of p(z) about the point z0 = 1. Compute the Taylor expansion without using Horner's method and compare the result.

c. Loate an approximate root z3 using Newton's and Horner's method when z0 = 2.

Answer 1

p(z) = ((z - 7)z + 14)z - 6

4 - 7 = 3
3*4 + 14 = 26
26*4 - 6 = 98

p(4) = 98

b.
f(z) = z^3 - 7z^2 + 14z - 6
f'(z) = 3z^2 - 14z + 14
f''(z) = 6z - 14
f'''(z) = 6

1 - 7 = - 6
- 6*1 + 14 = 8
8*1 - 6 = 2
f(1) = 2

3*1-14 = - 11
- 11*1 + 14 = 3

f'(1) = 3
f''(1) = 6*1-14 = - 8
f'''(1) = 6

f(z) = 2 + 3(z - 1) - 4(z - 1)^2 + (z - 1)^3

f(1) = 2 + 3*0 - 4*0 + 0 = 2

c.
2 - 7 = - 5
- 5*2 + 14 = 4
4*2 - 6 = 2
f(2) = 2

x = x0 + (z^3 - 7z^2 + 14z - 6 - 0)/(3z^2 - 14z + 14)

3*2 - 14 = - 8
-8*2 + 14 = 2

x = 2 + 2/2 = 3
3 - 7 = - 4
- 4*3 + 14 = 2
2*3 - 6 = 0

root = 3

Answer 2

The Rhind Mathematical Papyrus is a reproduction from 1650 BCE of a amazing until eventually eventually now artwork and shows us how the Egyptians extracted sq. roots. In historic India, the truth of theoretical and utilized aspects of sq. and sq. root replaced right into a minimum of as previous by using very reality the Sulba Sutras, dated round 800-500 B.C. (likely a lot until eventually eventually now). a way for gaining knowledge of very solid approximations to the sq. roots of two and three are given contained in the Baudhayana Sulba Sutra. Aryabhata contained in the Aryabhatiya (section 2.4), has given a way for gaining knowledge of the sq. root of numbers having many digits. D.E. Smith in heritage of mathematics, says, concerning the cutting-edge subject in Europe: "In Europe those procedures (for gaining knowledge of out the sq. and sq. root) did now not seem until eventually eventually now Cataneo (1546). He gave the approach of Aryabhata for determining the sq. root".