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    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    q3d

    can i just keep on going and do x4 x5 x6 and so on until i get a consistent answer instead of choosing the numbers 1.2905 and 1.2915?
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    (Original post by will'o'wisp)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    q3d

    can i just keep on going and do x4 x5 x6 and so on until i get a consistent answer instead of choosing the numbers 1.2905 and 1.2915?
    Yes, that's what I generally do.
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    (Original post by will'o'wisp)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    q3d

    can i just keep on going and do x4 x5 x6 and so on until i get a consistent answer instead of choosing the numbers 1.2905 and 1.2915?
    You could but I suspect it may take a while.

    The quickest way is substituting in the upper and lower bounds of alpha into the original equation, this should give you one positive and one negative answer. The change of sign indicates that the root is in between them, meaning that it is 1.291 to 3dp
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    (Original post by HapaxOromenon3)
    Yes, that's what I generally do.
    Good stuff. Can you also look at q4 and tell me if the range of g(x) is 1 \leq g(x) \leq 6??
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    (Original post by KaylaB)
    You could but I suspect it may take a while.

    The quickest way is substituting in the upper and lower bounds of alpha into the original equation, this should give you one positive and one negative answer. The change of sign indicates that the root is in between them, meaning that it is 1.291 to 3dp
    B-b-but all i have to do is press the equals button a few more times....
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    (Original post by will'o'wisp)
    B-b-but all i have to do is press the equals button a few more times....
    This way, your answer isn't dependant on whether your iteration values are correct. If for some reason you accidently messed up part c, using the upper and lower bound method you'd still be able to get the correct answer.

    M1: Choose suitable interval for x, e.g. [1.2905, 1.2915] or tighter and at least one
    attempt to evaluate f(x). A1: both values correct to awrt (or truncated) 1 sf, sign change and conclusion.
    The mark scheme shows that you need to prove that the sign changes
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    (Original post by will'o'wisp)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    q3d

    can i just keep on going and do x4 x5 x6 and so on until i get a consistent answer instead of choosing the numbers 1.2905 and 1.2915?
    No you would have got no marks for doing this.

    Quote from the examiners report:

    "A minority of candidates who attempted part (d) by using a repeated iteration technique received no credit because the question required the candidate to consider a change of sign.
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    (Original post by KaylaB)
    This way, your answer isn't dependant on whether your iteration values are correct. If for some reason you accidently messed up part c, using the upper and lower bound method you'd still be able to get the correct answer.



    The mark scheme shows that you need to prove that the sign changes
    (Original post by notnek)
    No you would have got no marks for doing this.

    Quote from the examiners report:

    "A minority of candidates who attempted part (d) by using a repeated iteration technique received no credit because the question required the candidate to consider a change of sign.
    :_____________________________:

    examiners pls...

    Ok then i'll do it that way.
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    (Original post by will'o'wisp)
    :_____________________________:

    examiners pls...

    Ok then i'll do it that way.
    I had the same issue when I did C3 so don't worry :lol::lol: I always kept wondering why I was being given 0 marks for those questions, so I learned to do them properly
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    (Original post by will'o'wisp)
    :_____________________________:

    examiners pls...

    Ok then i'll do it that way.
    There are certain questions where you can use continued iteration but in your question you are specifically told to use a change of sign.

    E.g. June 2011 Q2c can be done by continued iteration but you have to iterate far enough and have to give a clear explanation at the end referring to how this particular iterative sequence oscillates. It's easier to just get used to the change of sign method.
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    (Original post by HapaxOromenon3)
    Yes, that's what I generally do.
    (Original post by KaylaB)
    You could but I suspect it may take a while.

    The quickest way is substituting in the upper and lower bounds of alpha into the original equation, this should give you one positive and one negative answer. The change of sign indicates that the root is in between them, meaning that it is 1.291 to 3dp
    You cannot do it like this, because it doesn't preclude that your iteration is converging to what you think. To prove that there's a root, you need to exhibit a change of sign.

    Given that you need to show a change of sign (about a given point), the thing to do is simply plug in 1.2905 and 1.2915 and show f changes sign in between. The earlier parts of the question are completely irrelevant.

    Edit: thinking about this, I'm not sure whether it's too strict to say you can't do this at A-level (if the question doesn't ask you about a change of sign). Showing the change is definitely the most failsafe method, however.
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    (Original post by DFranklin)
    You cannot do it like this, because it doesn't preclude that your iteration is converging to what you think. To prove that there's a root, you need to exhibit a change of sign.

    Given that you need to show a change of sign (about a given point), the thing to do is simply plug in 1.2905 and 1.2915 and show f changes sign in between. The earlier parts of the question are completely irrelevant.

    Edit: thinking about this, I'm not sure whether it's too strict to say you can't do this at A-level (if the question doesn't ask you about a change of sign). Showing the change is definitely the most failsafe method, however.
    You're right; I didn't actually read the question that the OP linked to, and hence assumed it was the type that says "Find, using this iterative formula, the root to some given degree of accuracy", and there you can either do a large enough number of iterations to justify the given number of decimal places, or just do enough iterations to get an idea of what the root is and then do the justification using a change of sign over an appropriate interval. Apologies for any confusion.
 
 
 
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