Asked by Blair
Use Newton's method to approximate a root of the equation 5sin(x)=x as follows.
Let x1=2 be the initial approximation.
The second approximation x2 is:
and the third approximation x3 is:
Let x1=2 be the initial approximation.
The second approximation x2 is:
and the third approximation x3 is:
Answers
Answered by
Damon
5 sin x = x
let y = x - 5 sin x, search for y = 0
dy/dx = y' = 1 - 5 cos x
Xn+1 = Xn + y(Xn)/y'at Xn
X1 = 2
y = 2 - 5 sin 2 = 2 - 4.54 = -2.54
y'=1 - 5 cos 2 = 3.08
X2 = 2 -2.54/3.08 = 1.17
y = 1.17 - 5 sin 1.17 = -3.43
y' = 1 - 5 cos 1.17 = -.951
X3 = 1.17 -3.43/-.951 = 4.77
This is unlikely to work the way you want because you are jumping from cycle to cycle of the original sine wave
let y = x - 5 sin x, search for y = 0
dy/dx = y' = 1 - 5 cos x
Xn+1 = Xn + y(Xn)/y'at Xn
X1 = 2
y = 2 - 5 sin 2 = 2 - 4.54 = -2.54
y'=1 - 5 cos 2 = 3.08
X2 = 2 -2.54/3.08 = 1.17
y = 1.17 - 5 sin 1.17 = -3.43
y' = 1 - 5 cos 1.17 = -.951
X3 = 1.17 -3.43/-.951 = 4.77
This is unlikely to work the way you want because you are jumping from cycle to cycle of the original sine wave
Answered by
Blair
The answers are wrong for this one.
Answered by
Damon
sorry, sign wrong. I drew my picture wrong
5 sin x = x
let y = x - 5 sin x, search for y = 0
dy/dx = y' = 1 - 5 cos x
Xn+1 = Xn - y(Xn)/y'at Xn
X1 = 2
y = 2 - 5 sin 2 = 2 - 4.54 = -2.54
y'=1 - 5 cos 2 = 3.08
X2 = 2 + 2.54/3.08 = 2.82
y = 2.82 - 5 sin 2.82 = 1.24
y' = 1 - 5 cos 2.82 = 5.74
X3 = 2.82 -1.24/5.74 = 2.60
5 sin x = x
let y = x - 5 sin x, search for y = 0
dy/dx = y' = 1 - 5 cos x
Xn+1 = Xn - y(Xn)/y'at Xn
X1 = 2
y = 2 - 5 sin 2 = 2 - 4.54 = -2.54
y'=1 - 5 cos 2 = 3.08
X2 = 2 + 2.54/3.08 = 2.82
y = 2.82 - 5 sin 2.82 = 1.24
y' = 1 - 5 cos 2.82 = 5.74
X3 = 2.82 -1.24/5.74 = 2.60
Answered by
Blair
The answer is still wrong :(
Answered by
Damon
check my arithmetic carefully
Answered by
lauren
the correct answers are
x2=2.82658
x3=2.60457
x2=2.82658
x3=2.60457
Answered by
April
How did you solve this is this the same process?