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FP2 Newton Raphson

https://goo.gl/S3bldn

3 i and ii

I have a vague idea of the first part, probably constructing some equation of a line with the info I'm given, I couldn't get it to work, perhaps I'm rusty with coordinate geometry.

ii is really weird for me though. I thought I could draw y=x and f(x) and show it staircase or cobweb to a, but f(x) is not in x=F(x) form so I can't do that. I've seen several of these types of questions where its something to do with tangents and convergence but I don't get them, I would like to know. Thanks in advance.
If the initial approximation and the second, with their corresponding y value where to be sketched on a graph, the line connecting these two point is very close to the tangent of the curve.
EDIT: whoops wrong chapter :/

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(edited 8 years ago)
Reply 2
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TeeEm
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Original post by 16characterlimit
https://goo.gl/S3bldn

3 i and ii

I have a vague idea of the first part, probably constructing some equation of a line with the info I'm given, I couldn't get it to work, perhaps I'm rusty with coordinate geometry.

ii is really weird for me though. I thought I could draw y=x and f(x) and show it staircase or cobweb to a, but f(x) is not in x=F(x) form so I can't do that. I've seen several of these types of questions where its something to do with tangents and convergence but I don't get them, I would like to know. Thanks in advance.


@16Characters....
Part one involves replacing f(x+e) with its first degree Taylor expansion, which gives f(X) +f,(X)e is approximately 0


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Original post by drandy76
If the initial approximation and the second, with their corresponding y value where to be sketched on a graph, the line connecting these two point is very close to the tangent of the curve.


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I'm not sure I get what you mean.

Draw a tangent at (x2, f(x2) ) and it gets very close to a ?

That seems to make sense, but I don't know if that's what you meant.
Original post by 16characterlimit
I'm not sure I get what you mean.

Draw a tangent at (x2, f(x2) ) and it gets very close to a ?

That seems to make sense, but I don't know if that's what you meant.


Sorry I was thinking of the approximations and errors chapter, I'm not sure if this still applies to Newton-raphson, but yeah that's what I meant


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Zacken
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Original post by 16characterlimit
..


Sorry, a bit busy so can't give a proper explanation, but reading through this and this might help.
Original post by Zacken
Sorry, a bit busy so can't give a proper explanation, but reading through this and this might help.


Thank you, was useful for my understanding however Taylor series aren't in the spec so I don't know how OCR would want me to answer.

https://goo.gl/YseIKK

Mark scheme just mumbles something about "attempt gradient"
Reply 8
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Zacken
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Original post by 16characterlimit
Thank you, was useful for my understanding however Taylor series aren't in the spec so I don't know how OCR would want me to answer.

https://goo.gl/YseIKK

Mark scheme just mumbles something about "attempt gradient"


I didn't want you to look at the taylor series part, scroll down a few pages and look at the "geometric derivation part" - that applies to both PDF's, there's a useful picture in the second link! :smile:
Reply 9
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Zacken
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Original post by 16characterlimit


Mark scheme just mumbles something about "attempt gradient"


That would be for part (i), which is basically just the difference in y-values f(x_1) - 0 over the change in the x value x_2 - x_1, etc...
Original post by Zacken
I didn't want you to look at the taylor series part, scroll down a few pages and look at the "geometric derivation part" - that applies to both PDF's, there's a useful picture in the second link! :smile:


Thank you! That's ii down.
Reply 11
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Zacken
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Original post by 16characterlimit
Thank you! That's ii down.


Do you still need help with (i)?
Original post by 16characterlimit
Thank you, was useful for my understanding however Taylor series aren't in the spec so I don't know how OCR would want me to answer.

https://goo.gl/YseIKK

Mark scheme just mumbles something about "attempt gradient"


I think I misread that actually, taylor series is how you establish the general formula, not how you use it as an iterative one


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Reply 13
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Zacken
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Original post by drandy76
I think I misread that actually, taylor series is how you establish the general formula, not how you use it as an iterative one


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The general formula is the iterative one, you derive them using the Taylor series.
Original post by Zacken
The general formula is the iterative one, you derive them using the Taylor series.


Ugh, doesn't appear to be my day today, at least I got to nab some extra notes for fp2 so it's not all bad


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Zacken
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Original post by drandy76
Ugh, doesn't appear to be my day today, at least I got to nab some extra notes for fp2 so it's not all bad


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I know the feeling. Bright side looks good, Yay! :biggrin:
Yeah I still am confused about i.
Reply 17
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Zacken
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Original post by 16characterlimit
Yeah I still am confused about i.


Okay, so you can see that the tangent on the diagram. Find it's gradient. You know two of the points on the tangent,

(x1,f(x1))(x_1, f(x_1)) and (x2,0)(x_2, 0). This means your gradient is: 0−f(x1)x2−x1\displaystyle \frac{0 - f(x_1)}{x_2 - x_1}.

You with me so far?

But the gradient of the tangent at x1x_1 is the derivative of the curve evaluated at x1x_1, no? You with me?

So: −f(x1)x2−x1=f′(x1)\frac{-f(x_1)}{x_2 - x_1} = f'(x_1) You cool with me here?

Now re-arrange.
Original post by Zacken
Okay, so you can see that the tangent on the diagram. Find it's gradient. You know two of the points on the tangent,

(x1,f(x1))(x_1, f(x_1)) and (x2,0)(x_2, 0). This means your gradient is: 0−f(x1)x2−x1\displaystyle \frac{0 - f(x_1)}{x_2 - x_1}.

You with me so far?

But the gradient of the tangent at x1x_1 is the derivative of the curve evaluated at x1x_1, no? You with me?

So: −f(x1)x2−x1=f′(x1)\frac{-f(x_1)}{x_2 - x_1} = f'(x_1) You cool with me here?

Now re-arrange.


Once again thank you very much.
Reply 19
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Zacken
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Original post by 16characterlimit
Once again thank you very much.


No problemo!

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