The 2012 STEP Results Discussion Thread

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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  1. ben-smith's Avatar
    2241 22-04-2012  21:15
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    Re: The 2012 STEP Prep Thread!
    (Original post by hassi94)
    wcp don't look here
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    I believe farhan means doing \dfrac{a^{n+1}}{(n+1)(a-1)} = \dfrac{1 + a + a^2 + ... + a^n}{n+1} which is the same as what you were suggesting; except you're spotting that division by it being the formula for arithmetic progression. Correct me if I assumed what either of you were saying incorrectly.
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    I meant geometric
  2. Intriguing Alias's Avatar
    2242 22-04-2012  21:16
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    I was thinking of doing this.

    \displaystyle \frac{\alpha^{n-1}-1}{\alpha-1}=\alpha^n +...\alpha^k+....+1



    Geometric series....
    Yep that's right. Bit extra;
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    What can you say the coefficient of a^k is for the first integral then?

    and a tad more:
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    And how does this compare to the second?
    Last edited by Intriguing Alias; 22-04-2012 at 21:17.
  3. Farhan.Hanif93's Avatar
    2243 22-04-2012  21:16
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    I was thinking of doing this.

    \displaystyle \frac{\alpha^{n-1}-1}{\alpha-1}=\alpha^n +...\alpha^k+....+1



    Geometric series....
    Barring the small sign error (which I assume is a typo), that would be a good idea.
  4. Intriguing Alias's Avatar
    2244 22-04-2012  21:17
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    Re: The 2012 STEP Prep Thread!
    (Original post by ben-smith)
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    I meant geometric
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    sorry, so did I haha
  5. bananarama2's Avatar
    2245 22-04-2012  21:24
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    Re: The 2012 STEP Prep Thread!
    (Original post by Farhan.Hanif93)
    Barring the small sign error (which I assume is a typo), that would be a good idea.

    (Original post by hassi94)
    Yep that's right. Bit extra;
    Spoiler:
    Show
    What can you say the coefficient of a^k is for the first integral then?

    and a tad more:
    Spoiler:
    Show
    And how does this compare to the second?

    Ermmm.... Equating the coefficients of alpha to the k.

    \displaystyle \frac{1}{n+1}= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{1}{n+1}= \frac{n!}{k!(n-k)!} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{k!(n-k)!}{n!(n+1)}=\frac{k!(n-k)!}{(n+1)!} = \int_0^1 x^k(1-x)^{n-k} dx
  6. Intriguing Alias's Avatar
    2246 22-04-2012  21:25
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    Ermmm.... Equating the coefficients of alpha to the k.

    \displaystyle \frac{1}{n+1}= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{1}{n+1}= \frac{n!}{k!(n-k)!} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{k!(n-k)!}{n!(n+1)}=\frac{k!(n-k)!}{(n+1)!} = \int_0^1 x^k(1-x)^{n-k} dx
    Yep
  7. bananarama2's Avatar
    2247 22-04-2012  21:28
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    Re: The 2012 STEP Prep Thread!
    Wooo. I didn't even look at the spoilers
  8. Intriguing Alias's Avatar
    2248 22-04-2012  21:30
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    Wooo. I didn't even look at the spoilers
    Good stuff
  9. ben-smith's Avatar
    2249 22-04-2012  21:31
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    Ermmm.... Equating the coefficients of alpha to the k.

    \displaystyle \frac{1}{n+1}= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{1}{n+1}= \frac{n!}{k!(n-k)!} \int_0^1 x^k(1-x)^{n-k} dx

    \displaystyle \frac{k!(n-k)!}{n!(n+1)}=\frac{k!(n-k)!}{(n+1)!} = \int_0^1 x^k(1-x)^{n-k} dx
    as a marker, I would struggle to understand what is going on here.
    to be clear, where has the second equality come from in the last step?
  10. Intriguing Alias's Avatar
    2250 22-04-2012  21:35
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    Re: The 2012 STEP Prep Thread!
    (Original post by ben-smith)
    as a marker, I would struggle to understand what is going on here.
    to be clear, where has the second equality come from in the last step?
    I don't see the problem really. Okay some implication signs would be nice but it seems pretty clear to me.
  11. Farhan.Hanif93's Avatar
    2251 22-04-2012  21:36
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    Re: The 2012 STEP Prep Thread!
    (Original post by ben-smith)
    as a marker, I would struggle to understand what is going on here.
    to be clear, where has the second equality come from in the last step?
    There doesn't appear to be too many issues IMO, except the implication signs. You shouldn't need to explain why n!(n+1) = (n+1)!.
  12. ben-smith's Avatar
    2252 22-04-2012  21:38
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    Re: The 2012 STEP Prep Thread!
    (Original post by Farhan.Hanif93)
    There doesn't appear to be too many issues IMO, except the implication signs. You shouldn't need to explain why n!(n+1) = (n+1)!.
    I'm seriously not on it today! I thought he was making a stronger statement on the last line because, as you say, I didn't see why he had the n!(n+1) = (n+1)! bit at the end. My apologies.
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  13. bananarama2's Avatar
    2253 22-04-2012  21:42
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    Re: The 2012 STEP Prep Thread!
    (Original post by hassi94)
    I don't see the problem really. Okay some implication signs would be nice but it seems pretty clear to me.

    (Original post by Farhan.Hanif93)
    There doesn't appear to be too many issues IMO, except the implication signs. You shouldn't need to explain why n!(n+1) = (n+1)!.

    (Original post by ben-smith)
    I'm seriously not on it today! I thought he was making a stronger statement on the last line because, as you say, I didn't see why he had the n!(n+1) = (n+1)! bit at the end. My apologies.
    I do put the implications in the written working but Latex becomes cumbersome.
  14. Farhan.Hanif93's Avatar
    2254 22-04-2012  21:43
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    Re: The 2012 STEP Prep Thread!
    (Original post by ben-smith)
    I'm seriously not on it today! I thought he was making a stronger statement on the last line because, as you say, I didn't see why he had the n!(n+1) = (n+1)! bit at the end. My apologies.
    Happens. No need to apologise. :p:
  15. Intriguing Alias's Avatar
    2255 22-04-2012  21:44
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    Re: The 2012 STEP Prep Thread!
    (Original post by wcp100)
    I do put the implications in the written working but Latex becomes cumbersome.
    That's what I assumed to be honest; I do the same
  16. Xero Xenith's Avatar
    2256 23-04-2012  14:33
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    Re: The 2012 STEP Prep Thread!
    (Original post by gff)
    I may claim it works for n = -1.
    (Assume we'll add the proper constant of integration this time)
    Love what you did here! Seems like you were ignored, but it does indeed work for n = -1. Very nicely done

    Spoiler:
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    with a spoonful of L'Hopital's Rule

    EDIT: Though I can see why you were uneasy about the \frac{-1}{0}... now I'm starting to think it requires a better argument than "the constant of integration scoops it up".
    Last edited by Xero Xenith; 23-04-2012 at 14:37.
  17. Intriguing Alias's Avatar
    2257 23-04-2012  14:43
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    Re: The 2012 STEP Prep Thread!
    (Original post by Xero Xenith)
    Love what you did here! Seems like you were ignored, but it does indeed work for n = -1. Very nicely done

    Spoiler:
    Show
    with a spoonful of L'Hopital's Rule

    EDIT: Though I can see why you were uneasy about the \frac{-1}{0}... now I'm starting to think it requires a better argument than "the constant of integration scoops it up".
    Excuse my ignorance but I'm pretty confused here

    Why does it work for n = -1? That would say that the integral of 1/x = 0/0 surely?

    I don't know much about L'hopital's Rule but I don't see the relevance?
  18. Xero Xenith's Avatar
    2258 23-04-2012  14:49
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    Re: The 2012 STEP Prep Thread!
    (Original post by hassi94)
    Excuse my ignorance but I'm pretty confused here

    Why does it work for n = -1? That would say that the integral of 1/x = 0/0 surely?

    I don't know much about L'hopital's Rule but I don't see the relevance?
    Here's the argument, as best I understand it. Please forgive the terrible TeX...

    Spoiler:
    Show

    \int x^{-1} dx = \lim_{a \to 0} (\dfrac{x^a-1}{a}) + c

    = \lim_{a \to 0} ( \frac{d}{da} x^a) + c

    = lnx \lim_{a \to 0} (e^{alnx}) + c

    = lnx + c


    L'Hopital's Rule is beyond both A-level and STEP. It says that if the top and the bottom of a limit come out the same - infinity or zero - then you can differentiate the top and the bottom, with respect to the 'limit variable' (in this case a), and the limit stays the same.

    The thing that seems bad to me is that you've essentially added a constant into the limit which equals negative infinity (\frac{-1}{a}).
    Last edited by Xero Xenith; 23-04-2012 at 14:52.
  19. Intriguing Alias's Avatar
    2259 23-04-2012  14:55
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    Re: The 2012 STEP Prep Thread!
    (Original post by Xero Xenith)
    Here's the argument, as best I understand it. Please forgive the terrible TeX...

    Spoiler:
    Show

    \int x^{-1} dx = \lim_{a \to 0} (\dfrac{x^a-1}{a}) + c

    = \lim_{a \to 0} ( \frac{d}{da} x^a) + c

    = lnx \lim_{a \to 0} (e^{alnx}) + c

    = lnx + c


    L'Hopital's Rule is beyond both A-level and STEP. It says that if the top and the bottom of a limit come out the same - infinity or zero - then you can differentiate the top and the bottom, with respect to the 'limit variable' (in this case a), and the limit stays the same.

    The thing that seems bad to me is that you've essentially added a constant into the limit which equals negative infinity (\frac{-1}{a}).
    Ah okay I see thanks. Is that sort of logic allowed where instead of taking a=0 (which it is) we take a tending to zero (which technically it is not)?
    Last edited by Intriguing Alias; 23-04-2012 at 14:57.
  20. ben-smith's Avatar
    2260 23-04-2012  15:04
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    Re: The 2012 STEP Prep Thread!
    (Original post by Xero Xenith)
    Here's the argument, as best I understand it. Please forgive the terrible TeX...

    Spoiler:
    Show

    \int x^{-1} dx = \lim_{a \to 0} (\dfrac{x^a-1}{a}) + c

    = \lim_{a \to 0} ( \frac{d}{da} x^a) + c

    = lnx \lim_{a \to 0} (e^{alnx}) + c

    = lnx + c


    L'Hopital's Rule is beyond both A-level and STEP. It says that if the top and the bottom of a limit come out the same - infinity or zero - then you can differentiate the top and the bottom, with respect to the 'limit variable' (in this case a), and the limit stays the same.

    The thing that seems bad to me is that you've essentially added a constant into the limit which equals negative infinity (\frac{-1}{a}).
    There are some caveats to this though e.g. lim[g'(x)/f'(x)] must exist for the rule to apply.
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