# Use bisection method to find the root for: y = 1.5x/(1+x2)2 - 0.65*atan(1/x) + 0.65*x/(1+x2) Use xL = 0.1, xU = 2. Do five iterations.

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To find the root of y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) using xL =0.1 and xu = 2 by iteration method .

We substitute xL = 0.1 in y = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

y(XL) = y(0.1) = -0.7448 < 0.

Similarly we put x= 2 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

y(xu) = y(2) = 0.0786 > 0

So y(0.1) and y(2) are opposite sign. Therefore we take x = (0.1+2)/2 = 1.05 and find y(1.05) by putting x = 1.05 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2):

y(1.05) = 0.1862 > 0.

So y(1.05) > 0 and y(0.1) < 0. So we take x = (0.1+1.05)/2 = 0.0575.

Put x= 0.575 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

y(0.575) = 0.0862 which is > 0. y(0.1) = -0.7448 < 0. So we take x = (0.1+0.575)/2 = 0.3375.

We substitute x = 0.3375 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

y(0.3375) = -0.0285 which is < 0. y(0.575) = 0.0862> 0. So we now take x= ( 0.575+0.3375)/2 = 0.45625.

So we put x = 0.45625 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

y(0.45625) = -0.0285 < 0, y(0.575) = 0.0862 is the nearest positive value. So we take x = (0.45625+0.575)/2 = 0.515625.

We put x= 0.515625 in y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) and get:

We get y(0.515625) = 0.0358.

Therefore the the 5h estimate for x is 0.515625 . So x= 0.515625 is an estimate of the root of y(x) = 1.5x/(1+x^2)^2 - 0.65*atan(1/x) + 0.65*x/(1+x^2) .