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    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.
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    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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    (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....
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    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.
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    (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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    (Original post by 16characterlimit)
    ..
    Sorry, a bit busy so can't give a proper explanation, but reading through this and this might help.
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    (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"
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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!
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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...
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    (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!
    Thank you! That's ii down.
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    (Original post by 16characterlimit)
    Thank you! That's ii down.
    Do you still need help with (i)?
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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 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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    (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.
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    (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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    (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!
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    Yeah I still am confused about i.
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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,

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

    You with me so far?

    But the gradient of the tangent at x_1 is the derivative of the curve evaluated at x_1, no? You with me?

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

    Now re-arrange.
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    (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,

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

    You with me so far?

    But the gradient of the tangent at x_1 is the derivative of the curve evaluated at x_1, no? You with me?

    So: \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.
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    (Original post by 16characterlimit)
    Once again thank you very much.
    No problemo!
 
 
 
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