SHW 10, Calculus Exploration 4B:  Newton-Raphson Method

Use the Newton-Raphson Method to estimate all the zeroes of the following function f(t) and also to estimate the zeroes of the derivative of f(t). For f(t) and its derivative take the process out four steps.

f(t) = 10*t*exp(-2*t)-1

Warm Up on your calculator (TI-83, 84 Plus):

Step 1.

Do [Y = ] and [CLEAR] out all fuctions.

Then set [Y1 = 10Xe^(-2X) - 1] and set [Y2 = 10e^(-2X) - 20Xe^(-2X)]

Step 2.

Next [QUIT] and then [CLEAR] the calculating window.

Step 3.

Now test to make sure the functions are entered correctly by first setting X to 0 as follows:

do 0 [STO] X and 0 -> X will appear in the window.

Press [ENTER] and you should get the value of X, namely 0.

Step 4.

Next evaluate the functions Y1 and Y2 at X = 0 as follows:

press [VARS], select Y-VARS, select Function, press [ENTER], and then select Y1

and press [ENTER]. Y1 should now appear in the calculating window, press [ENTER]

and you should get the value of the function Y1 for X = 0, i.e. Y1 = -1.

To test Y2 repeat theabove for Y2 and you should get the value of the function Y2 for X = 0,

i.e. Y2 = 10.

f(t) = 10*t*exp(-2*t)-1, `f '`(t) = 10*exp(-2*t)-20*t*exp(-2*t)

`Estimate zeroes of `*f(t)*`using Newton-Raphson Method:`

``

L[i](t) = `f '`(t[i])*t+b[i], `  b`[i] = f(t[i])-`f '`(t[i])*t[i], i = 0 .. n

`Solve L`[i](t) = 0, ` or equivalently`, ` t`[i+1] = ` t`[i]-f(` t`[i])/`f '`(` t`[i])

`___________________ Iteration 0 ____________________`

` t`[0] = 2, ` L`[0](t) = -.5494691667*t+.465251111

`Solve L`[0](t) = 0, ` or equivalently`, ` t`[1] = ` t`[0]-f(` t`[0])/`f '`(` t`[0])

` t`[1] = .8467283320, ` f`(` t`[1]) = .556987792

Iteration 0 on your calculator (TI-83, 84 Plus):

Step 1. (You already did this in the warm up above. So, just do [Y = ])

Do [Y = ] and [CLEAR] out all fuctions.

Then set [Y1 = 10Xe^(-2X) - 1] and set [Y2 = 10e^(-2X) - 20Xe^(-2X)]

Step 2.

Set up Newton-Raphson formula as function Y3 as follows:

[Y3 = X - (Y1/Y2)]  

Note that in order to enter the Y1 function press [VARS],

select Y-VARS, select Function, press [ENTER], and then select Y1

and press [ENTER].

Step 3.

Next [QUIT] and then [CLEAR] the calculating window.

Step 4.

Begin calculations for Newton-Raphson Iteration 0 :

do 2 [STO] X and 2 -> X will appear in the window.

Press [ENTER] and you should get the value of X, namely 2.

Step 5.

Next evaluate the Newton-Raphson function Y3 at X = 2 as follows:

press [VARS], select Y-VARS, select Function, press [ENTER], and then select Y3

and press [ENTER]. Y3 should now appear in the calculating window, press [ENTER]

and you should get the value of the function Y3 for X = 2, i.e. Y3 = .8467283322.

Step 6.

Evaluate your original funtion at the estimate you just obtained

by setting X = .8467283322  as follows:

do Y3 [STO] X and Y3 -> X will appear in the window. (Note to get Y3 do [VARS] etc.)

Press [ENTER] and you should get the new value of X, namely X = .8467283322.

Now evaluate your original function Y1 (Note to get Y1 do [VARS] etc.)

and you should get the value of Y1 at X = .8467283322, namely Y1 = .5569877912.

That completes Iteration 0 !!

`___________________ Iteration 1 ____________________`

` t`[1] = .8467283320, ` L`[1](t) = -1.275147553*t+1.636691353

`Solve L`[1](t) = 0, ` or equivalently`, ` t`[2] = ` t`[1]-f(` t`[1])/`f '`(` t`[1])

` t`[2] = 1.283530952, ` f`(` t`[2]) = -.147520806e-1

Now Iteration 1 is dead easy!

Since the new estimate is already set up, X = .8467283322 (to double check type X and

press [ENTER]), just repeat Step 5 above by evaluating the Newton-Raphson function

Y3 at X = .8467283322 and you should get Y3 = 1.283503952.

Now do Y3 [STO] X and Y3 -> X will appear in the window.

Press [ENTER] and you should get the new value of X, namely X = 1.283503952.

Now evaluate your original function Y1 and you should get the value of Y1 at

X = 1.283503952, namely Y1 = -.0147520804.

That completes Iteration 1 !

`___________________ Iteration 2 ____________________`

` t`[2] = 1.283530952, ` L`[2](t) = -1.202888390*t+1.529192399

`Solve L`[2](t) = 0, ` or equivalently`, ` t`[3] = ` t`[2]-f(` t`[2])/`f '`(` t`[2])

` t`[3] = 1.271267070, ` f`(` t`[3]) = .65051e-4

Now Iteration 2 is dead easy!

Since the new estimate is already set up, X = 1.283503952 (to double check type X and

press [ENTER]), just repeat Step 5 above by evaluating the Newton-Raphson function

Y3 at X = X = 1.283503952 and you should get Y3 = 1.271267071.

Now do Y3 [STO] X and Y3 -> X will appear in the window.

Press [ENTER] and you should get the new value of X, namely X = 1.271267071.

Now evaluate your original function Y1 and you should get the value of Y1 at

X = 1.271267071 , namely Y1 = .0000650504462.

That completes Iteration 2 !