Year 13 Maths Help Thread

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    (Original post by Blake Jones)
    Does anyone have any links to any good A-level maths resources for the quotient rule etc?
    This website has some nice resources
    http://www.mathcentre.ac.uk/students/topics/
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    (Original post by Sal.Tek_ 〔サルテック〕)
    This website has some nice resources
    http://www.mathcentre.ac.uk/students/topics/
    Thanks!
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    Managed to finish STEP I 2007 Q5 !

    One area of maths which I'm particularly weak at is recurrence relations, particularly the types of questions which define u_{n+1} in terms of u_n and then ask you to define u_n in terms of n.

    What are some good questions that I can do to practice?
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    (Original post by Palette)
    Managed to finish STEP I 2007 Q5 !

    One area of maths which I'm particularly weak at is recurrence relations, particularly the types of questions which define u_{n+1} in terms of u_n and then ask you to define u_n in terms of n.

    What are some good questions that I can do to practice?
    What part of it are you weak at? Making sense of breaking down the recurrence system? If so 15, 17 and 35 of advanced problems in mathematics are good

    http://www.mathshelper.co.uk/110501_...athematics.pdf
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    (Original post by ValerieKR)
    What part of it are you weak at? Making sense of breaking down the recurrence system? If so 15, 17 and 35 of advanced problems in mathematics are good

    http://www.mathshelper.co.uk/110501_...athematics.pdf
    It's questions like 10b) here
    http://madasmaths.com/archive/iygb_p...apers/c1_t.pdf

    which I find difficult. Out of interest, do you prefer the old Siklos booklet or the new one?
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    (Original post by Palette)
    It's questions like 10b) here
    http://madasmaths.com/archive/iygb_p...apers/c1_t.pdf

    which I find difficult. Out of interest, do you prefer the old Siklos booklet or the new one?
    That question's similar to the loans and savings STEP ones (in the late 90s there's a few of those)

    I didn't use them too much when I was studying for STEP because I got tempted to scroll down - I'm working through them both now and haven't really gone through the other one properly yet
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    (Original post by Palette)
    It's questions like 10b) here
    http://madasmaths.com/archive/iygb_p...apers/c1_t.pdf

    which I find difficult. Out of interest, do you prefer the old Siklos booklet or the new one?
    If I give you the answer to that one the method becomes clear, so i've put it in the spoiler
    Spoiler:
    Show
    it's 2^(n-1) +5
    if you're stuck you can spot most common ones by taking the difference between terms (here the difference was the sequence 1,2,4,8 etc)
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    (Original post by Blake Jones)
    Does anyone have any links to any good A-level maths resources for the quotient rule etc?
    Examsolutions
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    Let L be the value of \sum^{\infty}_{r=1} \frac{1}{r^2}.

    Consider the series S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}-\frac{8}{9^2}+....

    I wrote that S is equal to \sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0; am I right?
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    (Original post by Palette)
    Let L be the value of \sum^{\infty}_{r=1} \frac{1}{r^2}.

    Consider the series S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}-\frac{8}{9^2}+....

    I wrote that S is equal to \sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0; am I right?
    Yeah, looks it
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    (Original post by Palette)
    Let L be the value of \sum^{\infty}_{r=1} \frac{1}{r^2}.

    Consider the series S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}-\frac{8}{9^2}+....

    I wrote that S is equal to \sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0; am I right?
    Your justification is very fishy, you'd need to prove that S converges absolutely.
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    (Original post by Zacken)
    Your justification is very fishy, you'd need to prove that S converges absolutely.
    http://madasmaths.com/archive/iygb_p...apers/sp_t.pdf

    I think the question (Q16) assumes convergence of S] otherwise the 'limit of the following infinite series' would not be asked. If I am wrong on this, I may try and exploit the direct comparison test when I'm not in a half-asleep state.
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    (Original post by Palette)
    http://madasmaths.com/archive/iygb_p...apers/sp_t.pdf

    I think the question (Q16) assumes convergence of S] otherwise the 'limit of the following infinite series' would not be asked. If I am wrong on this, I may try and exploit the direct comparison test when I'm not in a half-asleep state.
    It's not too hard to prove it anyway, just note that \displaystyle S < \sum_{\geq 1} \frac{1}{k^2} and \displaystyle s_m = \sum_{1 \leq r \leq m} \frac{1}{r^2} is a bounded monotone sequence or apply Cauchy's Condensation Test to get convergence of \displaystyle \sum_{\geq 1}\frac{1}{k^2}.
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    (Original post by Zacken)
    It's not too hard to prove it anyway, just note that \displaystyle S < \sum_{\geq 1} \frac{1}{k^2} and \displaystyle s_m = \sum_{1 \leq r \leq m} \frac{1}{r^2} is a bounded monotone sequence or apply Cauchy's Condensation Test to get convergence of \displaystyle \sum_{\geq 1}\frac{1}{k^2}.
    Would the direct comparison test work too?
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    (Original post by Palette)
    Would the direct comparison test work too?
    That is the comparison test?
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    (Original post by Zacken)
    That is the comparison test?
    Oops, I failed to spot that you just wrote it out in a different form to what I learnt it by; my mistake!
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    (Original post by metrize)
    Examsolutions
    Ooh thanks!
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    Can you be deducted a mark by writing \int x^{-1} dx as \ln x +C instead of \ln |x|+ C?
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    (Original post by Palette)
    Can you be deducted a mark by writing \int x^{-1} dx as \ln x +C instead of \ln |x|+ C?
    You could but I don't think it matters too much. It could be important for a definite integral though where the argument inside the log is negative because of one of the limits. It's best to put it when it's needed.
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    Consider I=\int^b_a f(x) dx, b>a. How can one intuitively explain why \int^a_b f(x) dx=-I?
 
 
 
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