Example solution in a way Newoton-Raphson Example 1: Find the root(s) of the function f (x) =x² – 2.5x +1.5625 to ɛ=1×10-°.Solution:Check the sign of f(x) at different values of x:- 11234f(x) 5.06..1.56..0.06..0.56.. 3.06.. 7.56..

Given: f(x)=x2-2.5x+1.5625 To find: Root of the given function using Newton-Raphson method.

Example 1: Find the root(s) of the function f(x)=x²-2.5x+1.5625 to s=1x10-6. Solution: Check the sign of f(x) at different values of .x: - 1 0 1 f(x) 5.06.. 1.56.. 0.06.. X 2 0.56.. 3 4 3.06.. 7.56..

Example 1: Find the root(s) of the function f(x)=x²-2.5x+1.5625 to s=1x10-6. Solution: Check the sign of f(x) at different values of .x: - 1 0 1 f(x) 5.06.. 1.56.. 0.06.. X 2 0.56.. 3 4 3.06.. 7.56..

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Example solution in a way Newoton-Raphson

Example 1: Find the root(s) of the function f (x) =x² – 2.5x +1.5625 to ɛ=1×10-°.
Solution:
Check the sign of f(x) at different values of x:
- 1
1
2
3
4
f(x) 5.06..
1.56..
0.06..
0.56.. 3.06.. 7.56..

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