Table 1 Test equations and range of initial point
From: A new Newton-like method for solving nonlinear equations
Equation | Range of initial |
|---|---|
\(f_1(x) = \exp (x)\sin (x)+\ln (1+x^2)=0\) | \(x_0\in [-0.1,1]\) |
\(f_2(x) = \exp (x)\sin (x)+\cos (x)\ln (1+x)=0\) | \(x_0\in [-1,1]\) |
\(f_3(x) = \exp (\sin (x))-x/5 -1 =0\) | \(x_0\in [-0.5,1]\) |
\(f_4(x) = (x+1)\exp (\sin (x))-x^2\exp (\cos (x))=0\) | \(x_0\in [-1.5,1]\) |
\(f_5(x) = \sin (x)+\cos (x)+\tan (x)-1=0\) | \(x_0\in [-1,1]\) |
\(f_6(x) = \exp (-x)-\cos (x)=0\) | \(x_0\in [-1,0.5]\) |
\(f_7(x) = \ln (1+x^2)+\exp (x^2-3x)\sin (x)=0\) | \(x_0\in [-0.2,1]\) |
\(f_8(x) = x^3+\ln (1+x)=0\) | \(x_0\in [-0.5,1]\) |
\(f_9(x) = \sin (x)-x/3=0\) | \(x_0\in [-0.5,1]\) |
\(f_5(x) = (x-10)^6-10^6=0\) | \(x_0\in [-1,1]\) |