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The equation x³ = cos x has one root. Choose ALL possible iteration formulae that can be used to find this root out of the following: Xn+1 = √√Cos Xn cos³ Xn xn Xn+1 cos² Xn Xn² Xn+1 = (cos-1 xn)3 Ăn+1 = cos-1 Xn+1 = Xn Xn+1 = Xn Xn+1 = Xn+1 = Xn+1 = - Xn+1 = X³+cOS X 3X²+COS X Xn³+cos Xn 3X²-cos Xn Cos³ Xn cos Xn+Xn²+Xn Xn²+Xn+1 cos Xn+Xn²-Xn Xn²+Xn-1 3 ·¹x³ Xn+1 = Xn+1 Xn+1 = (cos x)² Xn+1 = cos xn Xn+1 = Xn Xn+1 = (cos¯¹xñµ)² Xn+1=Xn — Xn+1 = -√√/cos³ Xn Xn+1 = cos Xn¬Xn²-Xn Xn²-Xn-1 cOS Xñ−Xñ ²+Xn In Xn+1 n³+cos Xn 3x²-sin xn Xn+1 = COS¯¹ Xn³-cos Xn 3Xn +sin Xn cos Xn 2 Xn

The equation x³ = cos x has one root. Choose ALL possible iteration formulae that can be used to find this root out of the following: Xn+1 = √√Cos Xn cos³ Xn xn Xn+1 cos² Xn Xn² Xn+1 = (cos-1 xn)3 Ăn+1 = cos-1 Xn+1 = Xn Xn+1 = Xn Xn+1 = Xn+1 = Xn+1 = - Xn+1 = X³+cOS X 3X²+COS X Xn³+cos Xn 3X²-cos Xn Cos³ Xn cos Xn+Xn²+Xn Xn²+Xn+1 cos Xn+Xn²-Xn Xn²+Xn-1 3 ·¹x³ Xn+1 = Xn+1 Xn+1 = (cos x)² Xn+1 = cos xn Xn+1 = Xn Xn+1 = (cos¯¹xñµ)² Xn+1=Xn — Xn+1 = -√√/cos³ Xn Xn+1 = cos Xn¬Xn²-Xn Xn²-Xn-1 cOS Xñ−Xñ ²+Xn In Xn+1 n³+cos Xn 3x²-sin xn Xn+1 = COS¯¹ Xn³-cos Xn 3Xn +sin Xn cos Xn 2 Xn

Advanced Engineering Mathematics
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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The equation x³ = cos x has one root. Choose ALL possible iteration formulae that can be used to find this root out of the following: Xn+1 = ³√√cos Xn cos³ Xn Xn+1 = cos² Xn Xn² Xn+1 = (cos-1Xn)3 Xn+1 = C0S-1 Xn+1 = Xn Xn+1=Xn Xn+1 = Xn+1 = Xn+1 = Xn+1 = Xn³+cos Xn 3X²+COS X Xn³+cos Xn 3Xn²-cos Xn cos³ Xn cos Xn+Xn²+Xn Xn²+Xn+1 cos Xn+Xn²-Xn Xn²+Xn-1 ¹Xn³ Xn+1 = Xn+1 = Xn+1 = (cos x) Xn+1 = cos Xn Xn² cos xn Xn³ Xn+1 = (cos-1xn) Xn+1 = cos-1 Xn+1 = Xn Xn+1 = Xn – Xn+1==√ Xn+1 = Xn²+cos Xn 3x²-sin Xn cos Xn-Xn²-Xn Xn²-Xn-1 COS Xn-Xn²+Xn Xn²-Xn+1 Xn-cos Xn 3x²+sin Xn cos3 Xn 2
Transcribed Image Text:The equation x³ = cos x has one root. Choose ALL possible iteration formulae that can be used to find this root out of the following: Xn+1 = ³√√cos Xn cos³ Xn Xn+1 = cos² Xn Xn² Xn+1 = (cos-1Xn)3 Xn+1 = C0S-1 Xn+1 = Xn Xn+1=Xn Xn+1 = Xn+1 = Xn+1 = Xn+1 = Xn³+cos Xn 3X²+COS X Xn³+cos Xn 3Xn²-cos Xn cos³ Xn cos Xn+Xn²+Xn Xn²+Xn+1 cos Xn+Xn²-Xn Xn²+Xn-1 ¹Xn³ Xn+1 = Xn+1 = Xn+1 = (cos x) Xn+1 = cos Xn Xn² cos xn Xn³ Xn+1 = (cos-1xn) Xn+1 = cos-1 Xn+1 = Xn Xn+1 = Xn – Xn+1==√ Xn+1 = Xn²+cos Xn 3x²-sin Xn cos Xn-Xn²-Xn Xn²-Xn-1 COS Xn-Xn²+Xn Xn²-Xn+1 Xn-cos Xn 3x²+sin Xn cos3 Xn 2
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