Solving equation using Newton Raphson's method

Solving f(x) equation using Newton Raphson's method First you shall assume initial x₁ and solve f(x₁). Then check f(x₁) whether is 0(zero) or not. If not 0(zero), you assume x₂. x₂ is solved as following equation \[x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]

 

 

\[f'(x)=\frac{f(x+dx)-f(x)}{dx}\]

 

\[f'(x_1)=\frac{f(x_1+dx)-f(x_1)}{dx}\]

 

Check f(x₂) wheter is 0 or not. If not 0(zero), you assume x₃ using above method If you continue to solve it this way, the answer will be calculated.

 

For example,

\[f(x)=2x^3+3x^2+5x+850\]

 

x₁ = 30 (assume)

\[f(x_1)=2(30)^3+3(30)^2+5(30)+850 = 57700.00000\]

not 0(zero) In order to obtain x₂, you solve f'(x₁) , f(x₁+dx) dx=0.00001

\[f(x_1+dx)=f(30+0.00001)\]

\[=2(30.00001)^3+3(30.00001)^2+5(30.00001)+850 = 57700.055850\]

\[f'(x_1)=\frac{f(x_1+dx) - f(x_1)}{dx}=\frac{57700.055850 - 57700.00000}{0.00001}=5585.001830\]

\[x_{2}=x_1 - \frac{f(x_1)}{f'(x_1)}=30 - \frac{57700.055850}{5585.001830}=19.6687590\]

Then f(x₂) is

\[f(x_2)=f(19.6687590)=17327.03937\]

not zero.  For x₃ ...

\[x_3=12.5796411\]

\[f(x_3)=f(12.5796411)=5369.03054\]

... ... ...

\[x_{10}=-7.9350620\]

\[f(x_{10})=f(-7.9350620)=-0.04534\]

\[x_{11}=-7.9349267\]

\[f(x_{11})=f(-7.9349267)=0.00000\]

Finding 0,  x=-7.9349267 You can check the answer following excel. Excel link