Newton Raphson Method

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Here is my written report for the Newton-Raphson Method for ChE Mathematics. It tackles what the method is all about, as well as the applications and the validity of the method.
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NEWTON’S METHOD   Also known as the Newton- Raphson Method, Newton’s Method is used to approximate real zeros of a function. It is done using the tangent lines to approximate the graph of a function near its x-intercept. Supposing that we have a continuous and differentiable function  f(x) , we find first the tangent line to f(x) to a certain point x 0 . Using the point-slope form, we solve for the equation of the tangent line to f(x) at that point.       (1) Knowing that the slope m  of the tangent line is the first derivative of the function f(x) at  point x 0 , we can replace it in the first equation:           (2) Since y o  can be solved when x o  is substituted in the function f(x), we can represent y 0  as f(x 0 ). Placing it in the equation, we can get:           (3) As seen in the figure, the tangent line passes through a point in the x-axis, labeled as x 1 . We now use this point in order to solve for the tangent line to f(x) at x 1 . Knowing the coordinates of the x-intercept of the first tangent line (depicted in green) to be (x 1 , 0), we solve for x 1 . Substituting the given coordinates in (3), we get:              (4) Transforming the equation in terms of x 1 , we now get:               (5) This process, called iteration , is done until the value obtained is approximately equal to that of the srcinal value. Therefore, the general equation for Newton’s Method is:              n = 1, 2, 3…  (6) In using Newton’s Method, however, one must take note that the function f(x) must be equal to 0, since Newton’s Method aims to approximate the real zeroes of a function. In contrast, f’(x n ) must not be equal to zero, since when plugged in to (6), an undefined answer will be obtained. Lastly, the zero of a function f on an interval must be able to satisfy the condition of convergence defined by the equation (7) Problem Solving: Example 1 : Calculate to four iterations of Newton ’s Method the root of x 2 =3, using x 1 =1. Step 1 : Transform x 2 =3 such that f(x)=0. x 2 =3 x 2 -3=0 f(x)=x 2 -3 Step 2: Get f’(x). f(x)=x 2 -3 f’(x)=2x  Step 3: Set- up a table. In using Newton’s Method, it is helpful to make a table that would summarize all the steps done.  First iteration: f(1)= (1) 2 -3 =-2.000000 f’(1 )=(2)(1)=2.000000    =1.000000     = 1-(1.000000)=2.000000 Second iteration: f(2)=(2) 2 -3=1.000000 f  ’ (2)=(2)(2)=4.000000    =  =0.250000     =1.750000 Third iteration: f(1.75)=(1.75) 2 -3=0.062500 f’(1.7 5)=(2)(1.75)=3.500000    =  =0.017857     =1.732143 Fourth iteration: f(1.732143)=(1.732143) 2 -3=0.000319 f’(1.73 2143)=2(1.732143)=3.464286    =  =9.21899x10 -5     =1.73205081After four iterations, the value obtained is 1.73205081, which is very close to the srcinal value of x= √   (1.732050808). The values are accurate until the 7 th  decimal place, therefore, the iterations made are correct. Example 2: Approximate the positive root of f(x)=    . Continue the iterations until two successive approximations differ by less than 0.001. Step 1:   Get f’(x). f(x)=       f’(x)= 2x-1 Step 2 : Plot the function approximately to determine an appropriate x 1 . N x n  f(x n ) f’x n )                        1.000000 -2.000000 2.000000 1.000000 2.000000 2 2.000000 1.000000 4.000000 0.250000 1.750000 3 1.750000 0.062500 3.500000 0.017857 1.732143 4 1.732143 0.000319 3.464286 9.21899x10 -5 1.73205081 5 1.73205081    Let x 1 =2.500000 First iteration: f(2.500000)=(2.500000) 2 -2.500000-3=0.750000 f’(2.500000)=2(2.500000) -1= 4.000000    =  =0.187500     =2.312500 Second iteration: f(2.312500)=(2.312500) 2 -2.312500-3= 0.03515625 f’(2.312500)=2(2.312500) -1=3.625    =  =0.009698276     =2.3028017 Third iteration: f(2.3028017)=(2.3028017) 2 -2.3028017-3= 0.00009405655 f’(2.3028017)=2(2.3028017) -1=3.6056034    =  =0.0000260822      =2.3027756 The obtained value is 2.3027756. The correct value of the root is 2.302775638, meaning that the third iteration is very close to the correct answer. Also, the iterations were also able to satisfy the condition that the two successive approximations, since it differed less than 0.001, making the procedure correct. N x n  f(x n ) f’x n )                        2.500000 0.750000 4.000000 0.187500 2.312500 2 2.312500 0.03515625 3.625000 0.009698276 2.3028017 3 2.3028017 0.000094057 3.6056034 0.0000260822 2.3027756 4 2.3027756

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