3. Example `f(x) = 3x-cos(yz)-3/2=0,4x^2-625y^2+2z-1=0,20z+e^(-xy)+9=0` and Initial guesses `= 1,1,1`
Find root using Modified Newton Raphson method (Multivariate Newton Raphson method)
`f(x) = 3x-cos(yz)-3/2=0,4x^2-625y^2+2z-1=0,20z+e^(-xy)+9=0` and Initial guesses `= 1,1,1`
Solution:
`f_1=3x-cos(yz)-(3/2)=0 ->(1)`
`f_2=4x^2-625y^2+2z-1=0 ->(2)`
`f_3=20z+exp(-xy)+9=0 ->(3)`
In order to use Modified Newton Raphson method (Multivariate Newton Raphson method), we must first determine the functional form of the partial derivatives
`J_(1,1) = (del f_1)/(del x)=3`
`J_(1,2) = (del f_1)/(del y)=sin(yz)z`
`J_(1,3) = (del f_1)/(del z)=sin(yz)y`
`J_(2,1) = (del f_2)/(del x)=8x`
`J_(2,2) = (del f_2)/(del y)=-1250y`
`J_(2,3) = (del f_2)/(del z)=2`
`J_(3,1) = (del f_3)/(del x)=-exp(-xy)y`
`J_(3,2) = (del f_3)/(del y)=-exp(-xy)x`
`J_(3,3) = (del f_3)/(del z)=20`
`J = [[3,sin(yz)z,sin(yz)y],[8x,-1250y,2],[-exp(-xy)y,-exp(-xy)x,20]]`
with `x_0=[1,1,1]^T`, we get
Formula
`[[x_(n+1)],[y_(n+1)],[z_(n+1)]] = [[x_(n)],[y_(n)],[z_(n)]] - [[(del f_1)/(del x),(del f_1)/(del y),(del f_1)/(del z)],[(del f_2)/(del x),(del f_2)/(del y),(del f_2)/(del z)],[(del f_3)/(del x),(del f_3)/(del y),(del f_3)/(del z)]]^(-1) xx f([[x_(n)],[y_(n)],[z_(n)]])`
`n=0,x=1,y=1,z=1`
`[[1],[1],[1]] - [[3,0.8415,0.8415],[8,-1250,2],[-0.3679,-0.3679,20]]^(-1) xx f([[1],[1],[1]])`
`=[[1],[1],[1]] - [[3,0.8415,0.8415],[8,-1250,2],[-0.3679,-0.3679,20]]^(-1) xx [[0.9597],[-620],[29.3679]]`
`=[[1.2327],[0.5031],[-0.4733]]`
`n=1,x=1.2327,y=0.5031,z=-0.4733`
`[[1.2327],[0.5031],[-0.4733]] - [[3,0.1116,-0.1187],[9.8616,-628.9151,2],[-0.2706,-0.663,20]]^(-1) xx f([[1.2327],[0.5031],[-0.4733]])`
`=[[1.2327],[0.5031],[-0.4733]] - [[3,0.1116,-0.1187],[9.8616,-628.9151,2],[-0.2706,-0.663,20]]^(-1) xx [[1.2263],[-154.082],[0.0728]]`
`=[[0.8326],[0.2518],[-0.4906]]`
`n=2,x=0.8326,y=0.2518,z=-0.4906`
`[[0.8326],[0.2518],[-0.4906]] - [[3,0.0605,-0.031],[6.6607,-314.7581,2],[-0.2042,-0.6751,20]]^(-1) xx f([[0.8326],[0.2518],[-0.4906]])`
`=[[0.8326],[0.2518],[-0.4906]] - [[3,0.0605,-0.031],[6.6607,-314.7581,2],[-0.2042,-0.6751,20]]^(-1) xx [[0.0054],[-38.8375],[-0.0019]]`
`=[[0.8332],[0.1284],[-0.4947]]`
`n=3,x=0.8332,y=0.1284,z=-0.4947`
`[[0.8332],[0.1284],[-0.4947]] - [[3,0.0314,-0.0082],[6.6659,-160.5074,2],[-0.1154,-0.7487,20]]^(-1) xx f([[0.8332],[0.1284],[-0.4947]])`
`=[[0.8332],[0.1284],[-0.4947]] - [[3,0.0314,-0.0082],[6.6659,-160.5074,2],[-0.1154,-0.7487,20]]^(-1) xx [[0.0017],[-9.5173],[0.0045]]`
`=[[0.8333],[0.0691],[-0.4971]]`
`n=4,x=0.8333,y=0.0691,z=-0.4971`
`[[0.8333],[0.0691],[-0.4971]] - [[3,0.0171,-0.0024],[6.6662,-86.3523,2],[-0.0652,-0.7867,20]]^(-1) xx f([[0.8333],[0.0691],[-0.4971]])`
`=[[0.8333],[0.0691],[-0.4971]] - [[3,0.0171,-0.0024],[6.6662,-86.3523,2],[-0.0652,-0.7867,20]]^(-1) xx [[0.0004],[-2.1996],[0.0011]]`
`=[[0.8333],[0.0436],[-0.4982]]`
`n=5,x=0.8333,y=0.0436,z=-0.4982`
`[[0.8333],[0.0436],[-0.4982]] - [[3,0.0108,-0.0009],[6.6662,-54.4818,2],[-0.042,-0.8036,20]]^(-1) xx f([[0.8333],[0.0436],[-0.4982]])`
`=[[0.8333],[0.0436],[-0.4982]] - [[3,0.0108,-0.0009],[6.6662,-54.4818,2],[-0.042,-0.8036,20]]^(-1) xx [[0.0001],[-0.4063],[0.0002]]`
`=[[0.8333],[0.0361],[-0.4985]]`
`n=6,x=0.8333,y=0.0361,z=-0.4985`
`[[0.8333],[0.0361],[-0.4985]] - [[3,0.009,-0.0007],[6.6663,-45.1459,2],[-0.035,-0.8086,20]]^(-1) xx f([[0.8333],[0.0361],[-0.4985]])`
`=[[0.8333],[0.0361],[-0.4985]] - [[3,0.009,-0.0007],[6.6663,-45.1459,2],[-0.035,-0.8086,20]]^(-1) xx [[0],[-0.0349],[0]]`
`=[[0.8333],[0.0353],[-0.4985]]`
`n=7,x=0.8333,y=0.0353,z=-0.4985`
`[[0.8333],[0.0353],[-0.4985]] - [[3,0.0088,-0.0006],[6.6663,-44.1789,2],[-0.0343,-0.8091,20]]^(-1) xx f([[0.8333],[0.0353],[-0.4985]])`
`=[[0.8333],[0.0353],[-0.4985]] - [[3,0.0088,-0.0006],[6.6663,-44.1789,2],[-0.0343,-0.8091,20]]^(-1) xx [[0],[-0.0004],[0]]`
`=[[0.8333],[0.0353],[-0.4985]]`
Approximate root using Newton Raphson method is
`x=0.8333,y=0.0353,z=-0.4985`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
