en+1=Xn+1-r=Xn- = en By Taylor's Theorem, we have f(xn) f'(xn) Putting this in Equation (2) leads to en+1 = - 0= f(r) = f(xn-en) = f(xn) - en f'(xn) + ½e²f" (En) where is a number between x, and r. A rearrangement of this equation yields en f'(xn) - f(x) = {/f" (§ne ²/2 1 f" (§n) 2 f'(xn) f(xn) f'(xn) en f'(xn)-f(xn) f'(xn) 1 f" (r)² = Cen Ce² 2 f'(r) (2) (3)
en+1=Xn+1-r=Xn- = en By Taylor's Theorem, we have f(xn) f'(xn) Putting this in Equation (2) leads to en+1 = - 0= f(r) = f(xn-en) = f(xn) - en f'(xn) + ½e²f" (En) where is a number between x, and r. A rearrangement of this equation yields en f'(xn) - f(x) = {/f" (§ne ²/2 1 f" (§n) 2 f'(xn) f(xn) f'(xn) en f'(xn)-f(xn) f'(xn) 1 f" (r)² = Cen Ce² 2 f'(r) (2) (3)
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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Question
I was wondering what would happen to equation 3 if the newton's formula was changed to Xn+1 = Xn - f(Xn)/f'(X0)? So the derivative is only taken of the starting value funciton.
Transcribed Image Text:en+1 = Xn+1 =r=Xn-
= en
By Taylor's Theorem, we have
-
f(xn)
f'(xn)
en+1 =
Putting this in Equation (2) leads to
f(xn)
f'(xn)
0= f(r) = f(xn-en) = f(xn) - en f'(xn) + 1/2e²f" (5₁)
where is a number between x, and r. A rearrangement of this equation yields
en f'(xn) — ƒ (xn) = 1½ ƒ" (§n)e²
1 f" (n)²
2 f'(xn)
r
en f'(xn)-f(xn)
f'(xn)
1 f"(r)
e² = Cen
2 f'(r)
(2)
(3)
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