Iteration Formula (Can someone please help me with part ii? I dont know the initial value)

[cired2002]

 

[topsquark]

Please show us whatever work you've been able to do with this. That will help us identify where you need the help.

And posting a clearer picture would help too.

-Dan

 

[cired2002]

Ac

Please show us whatever work you've been able to do with this. That will help us identify where you need the help.

And posting a clearer picture would help too.

-Dan

Actually, theres no working. Ive only been able to prove part (i) and the answer is there already. Part (ii) is about finding p but i just dont have the initial value to subs it in

 

[lev888]

Ac

Actually, theres no working. Ive only been able to prove part (i) and the answer is there already. Part (ii) is about finding p but i just dont have the initial value to subs it inView attachment 18860

It's been a while since I took numerical methods, but I think any value close enough to the root should work if the process converges. Try it.

 

[JeffM]

I think the whole point is that you are not given an initial value.

You do not say what topic that this problem relates to. My guess is that the topic is iterative methods of approximation. It has literally been many decades since I studied numerical methods, but my recollection is that some iterative methods do not always work unless you start the process with a "good enough" initial guess and that almost all methods converge more rapidly if the process is started with a reasonable initial guess.

You are, I suspect, being asked to provide your own "good enough" initial guess

 

[HallsofIvy]

I don't see that this problem has anything to do with "iteration" or requires an initial value. The problem does not ask you to find p.
Clearly at that value of p, since it is a maximum, y'= 0. Differentiate \(\displaystyle y(x)= x^2cos(2x\), set the derivative equal to 0 and show that x must satisfy that equation.

 

[lev888]

I don't see that this problem has anything to do with "iteration" or requires an initial value. The problem does not ask you to find p.
Clearly at that value of p, since it is a maximum, y'= 0. Differentiate \(\displaystyle y(x)= x^2cos(2x\), set the derivative equal to 0 and show that x must satisfy that equation.

(ii) is quite explicit about iteration.

 

[JeffM]

I don't see that this problem has anything to do with "iteration" or requires an initial value. The problem does not ask you to find p.
Clearly at that value of p, since it is a maximum, y'= 0. Differentiate \(\displaystyle y(x)= x^2cos(2x\), set the derivative equal to 0 and show that x must satisfy that equation.

Part ii of the problem says quite explicitly "Use the iteration formula ... to determine the value of p correct to 2 decimal places."

That seems sufficient to me to suggest that the problem has something to do with iteration. Of course, it would greatly help if posters gave us a clue what topic they are currently studying, but we do not live in an ideal world.

 

[Cubist]

I just tried twice, once with the starting value 0.1 and then with pi/4. Both converged within 8 iterations (to 2dp). A better starting guess, somewhere in the middle, would obviously be a bit quicker.

Just take a rough guess and start! You'll see a pattern develop.