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Newton's Method Calculator
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Newton's method is the method used to get the root of an equation of the form:
ax2 + bx + c = 0
It gives the root value to any desired level of accuracy.
Newtons Method Calculator calculates the root for any given function for the desired number of decimal places.
Default function with initial point is given in the calculator below. Roots are calculated by differentiating the function and substituting the initial value in the formula given in the steps. You will see the values of the function when you click on "Solve".
 

Steps for Newton's Method Calculator

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Step 1 :  

Read the given problem. Note down the given equation, initial point xo and no of decimal places n.



Step 2 :  

Check where exactly the root lies by find the values of f(x). Start from initial point, give different values of f(x), if the values of given function f(x) are such that one is positive and other is negative. It means that the root lies between those two numbers.



Step 3 :  

Differentiate f(x) to get f'(x).



Step 4 :  

Find the value of f(x) and f'(x) for x0.


Substitute x0 in the formula:
xn+1 = xn - $\frac{f(x)}{f'(x)}$


We get the value of x1.


Again check where the root lies by substituting the value of x1 in x2 and also substitute the value of x1 get the value of x2. Carry on the same process till the last decimal place. List out all the values for the given equation as x0, x1, x2......xn, which gives you the answer.



Problems on Newton's Method Calculator

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  1. Find the roots between 0 and 1 of the equation x3− 5x + 2 = 0 correct to three decimal places.


    Step 1 :  

    The given function is: x3 - 5x + 2 = 0,


    Initial point xo = 0,


    no of decimal places n = 3.



    Step 2 :  

    f(0) = 03 - 5(0) + 2 = 2


    f(1) = 13 - 5(1) + 2 = -2


    The Root lies between 0 and 1.


     



    Step 3 :  

    For xo = 1,  f(1) = 13 - 5 (1) + 2 = -2


                       f'(1) = 3(1)2 - 5 = -2


    x1 = xo - $\frac{f(x)}{f'(x)}$


         = 1 - $\frac{-2}{-2}$


         = 0.



    Step 4 :  

    for x1 = 0, f(0) = 03 - 5(0) + 2 = 2


    f'(0) = 0-5 = - 5


    x2 = x1 - $\frac{f(x)}{f'(x)}$


         = 0 - $\frac{2}{-5}$


         = 0 - $\frac{2}{-5}$


         = 0.4.


    f(0.4) = (0.4)3 - 5(0.4) + 2


              = 0.064


    f'(0.4) = 3(0.4)2 - 5


              = - 4.52


    x3 = x2 - $\frac{0.064}{- 4.52}$


         = 0.4 - 0.014159.


         = 0.414159.



    Answer  :  

    The root is 0.414159 correct to three decimal numbers.

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

    The root of x3−6x+4 = 0 are

     

     

     

    xo = 0

     

     

     

    x1 = 0.4

     

     

     

    x2 = 0.414159.



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