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Iteration A level maths question help!!!

I’ve been struggling with this question..
Ive attached it below

I know i need to do 3=3 in my calculator and then put Ans whenever there is an x.. but i have no idea if im putting the Ans in the wrong places because im not getting close to the answer?!

This is what i put in my cal (where Ans=3)

-sin(4Ans)+e^(-Ans)+0.75/4cos(4Ans)-e^(-Ans)

And what is the Xn before it? Is that why im getting it wrong? But idk what Xn actually is so not sure.
Reply 1
Original post by dust bunny
I’ve been struggling with this question..
Ive attached it below
I know i need to do 3=3 in my calculator and then put Ans whenever there is an x.. but i have no idea if im putting the Ans in the wrong places because im not getting close to the answer?!
This is what i put in my cal (where Ans=3)
-sin(4Ans)+e^(-Ans)+0.75/4cos(4Ans)-e^(-Ans)
And what is the Xn before it? Is that why im getting it wrong? But idk what Xn actually is so not sure.



Kareem wants to solve the equation sin 4x+e^-x+ 0.75 = 0. He uses his calculator to create the
following table of values for f(x) = sin 4x+ e^-x + 0.75.

Use the Newton-Raphson method with
Xo= 3 to find a root of the equation f(x) = 0.

Show three iterations and give your answer to a suitable degree of accuracy.

The answers are
X1=2.920853
X2=2.908274
X3=2.907846

The root is 2.908 to 4sf
Reply 2
Original post by dust bunny
I’ve been struggling with this question..
Ive attached it below
I know i need to do 3=3 in my calculator and then put Ans whenever there is an x.. but i have no idea if im putting the Ans in the wrong places because im not getting close to the answer?!
This is what i put in my cal (where Ans=3)
-sin(4Ans)+e^(-Ans)+0.75/4cos(4Ans)-e^(-Ans)
And what is the Xn before it? Is that why im getting it wrong? But idk what Xn actually is so not sure.

Im presuming youve got a description of newton-raphson so something like page 75 in
https://www.lboro.ac.uk/media/media/schoolanddepartments/mlsc/downloads/HELM%20Workbook%2031%20Numerical%20Methods%20of%20Approximation.pdf
The key idea of iteration is that you use an old value to calculate a new value so something line
new value = old value - f(old value)/f'(old value)
then set
old value = new value
and repeat.

Here the initial old value = 3 so you calculate
new value = 3 - f(3)/f'(3)
so the new value is now 2.92 and then set
old value = 2.92
and keep repeating.

The key idea is that you guess value of the root. Then evaluate both the functions and the derivative value at that guess so you form the tangent line at your guess and if that is a good approximation of the function close to the root, then you can use where the tangent line crosses the x-axis (new root estimate) to improve it. The iterative scheme will converge when the estimate does not change.
(edited 1 month ago)
Reply 3
Original post by mqb2766
Im presuming youve got a description of newton-raphson so something like page 75 in
https://www.lboro.ac.uk/media/media/schoolanddepartments/mlsc/downloads/HELM%20Workbook%2031%20Numerical%20Methods%20of%20Approximation.pdf
The key idea of iteration is that you use an old value to calculate a new value so something line
new value = old value - f(old value)/f'(old value)
then set
old value = new value
and repeat.
Here the initial old value = 3 so you calculate
new value = 3 - f(3)/f'(3)
so the new value is now 2.92 and then set
old value = 2.92
and keep repeating.
The key idea is that you guess value of the root. Then evaluate both the functions and the derivative value at that guess so you form the tangent line at your guess and if that is a good approximation of the function close to the root, then you can use where the tangent line crosses the x-axis (new root estimate) to improve it. The iterative scheme will converge when the estimate does not change.


I think i understand this but what do i put in my calculator? As i think i am missing something when i put it in as i said above??
Reply 4
Original post by dust bunny
I think i understand this but what do i put in my calculator? As i think i am missing something when i put it in as i said above??

to get the first estimate of 2.92... you calcualte
3 - f(3)/f'(3)
where f(3) is the given function evaluated when x= 3 and f'(3) is the derivative of the function evaluated at x= 3.

Then the next estimate is given by
2.92 - f(2.92)/f'(2.92)
then ....

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