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    (Original post by atsruser)
    Where is this coming from? I don't see how you know this unless you know how to evaluate the integral with a=1 in the usual fashion.
    That's exactly how you do know it.
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    (Original post by james22)
    That's exactly how you do know it.
    In that case, the dimensional analysis approach seems to be less useful than the standard approach, doesn't it? I'm not sure I see the point, apart from the fact that it's a slightly unusual application of the technique.
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    (Original post by atsruser)
    In that case, the dimensional analysis approach seems to be less useful than the standard approach, doesn't it? I'm not sure I see the point, apart from the fact that it's a slightly unusual application of the technique.
    I'm not sure if the standard approach works in the more general case though.
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    (Original post by james22)
    I'm not sure if the standard approach works in the more general case though.
    You mean for I= \int_{-\infty}^\infty e^{-ax^2}dx? If so, yes. You can calculate I^2 as a double integral and convert to polars. It's only a change of variable from x \rightarrow \sqrt{a}x, after all.
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    Hey guys,
    new to this thread, but love maths. Hopefully I'll contribute a bit to discussion!
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    *!

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    Are you serious ?

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    Is that gcse?

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    Continuing on from Arithmeticae's 469, Tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

    Problem 473 **

    Show that:

    \displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

    Spoiler:
    Show
    The Beta function (and by extension, the Gamma function) might be of some use here.
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    (Original post by Khallil)
    Continuing on from Arithmeticae's 469, Tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

    Problem 473 **

    Show that:

    \displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

    Spoiler:
    Show
    The Beta function (and by extension, the Gamma function) might be of some use here.
    Solution 473**

    No need for any complicated functions. Note that, when integrating by parts, the extra terms are all 0 and teh 2 minus terms cancel so we get

    \displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x=\frac{4!}{21 \times 22 \times 23 \times 24}\displaystyle \int_{0}^{1}  \left( 1 - x \right)^{24} \text{ d}x=\frac{4!}{21 \times 22 \times 23 \times 24 \times 25}=\dfrac{1}{25 \binom{24}{4}}
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    (Original post by james22)
    ...
    I didn't think of integrating by parts but I think the solution to do with the Beta and Gamma functions is more elegant.

    Edit: Somebody, use the \Gamma hammer! :teeth:
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    (Original post by khallil)
    continuing on from arithmeticae's 469, tarquin's one is 470 and jjpneed1's problems are 471 and 472 respectively.

    problem 473 **

    show that:

    \displaystyle \int_{0}^{1} x^4 \left( 1 - x \right)^{20} \text{ d}x = \dfrac{1}{25 \binom{24}{4}}

    Spoiler:
    Show
    the beta function (and by extension, the gamma function) might be of some use here.
    \displaystyle \int_0^1 x^4 (1-x)^{20} \, \mathrm d x = \mathrm B (5,21) = \frac {\Gamma (5) \Gamma (21)} {\Gamma (26)} = \frac {4! \, 20!} {25!} = \frac 1 {25 \binom {24} 4}
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    (Original post by Khallil)
    I didn't think of integrating by parts but I think the solution to do with the Beta and Gamma functions is more elegant.

    Edit: Somebody, use the \Gamma hammer! :teeth:
    Why use complicated functions when there is a perfectly good (and not even messy or long) elementary solution.
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    (Original post by james22)
    Why use complicated functions when there is a perfectly good (and not even messy or long) elementary solution.
    Because I said so.
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    (Original post by Khallil)
    Why don't you add fractions, fam?
    I can't do maths when tired.
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    That's pretty harsh Khallil I'm sure you've made those kinds of mistakes at some point To be fair your problem isn't very interesting, in fact if the beta function was taught at A-level I'm sure I'd see it in exercise 1A

    Fun one but fairly simple:

    Problem 474 *

    I can write any (positive) number as a sum of arbitrarily many (positive) real numbers, e.g. 16 = 10 + 6 = 3.4 + 6.6 + 6, etc. When is the product of all elements in this sum maximised?
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    (Original post by Khallil)
    when ur anoos quivers
    I arrived at a similar answer. :—)

    (Mine also quivered. :—))
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    (Original post by jjpneed1)
    That's pretty harsh Khallil I'm sure you've made those kinds of mistakes at some point To be fair your problem isn't very interesting, in fact if the beta function was taught at A-level I'm sure I'd see it in exercise 1A

    Fun one but fairly simple:

    Problem 474 *

    I can write any (positive) number as a sum of arbitrarily many (positive) real numbers, e.g. 16 = 10 + 6 = 3.4 + 6.6 + 6, etc. When is the product of all elements in this sum maximised?
    I can do this by Lagrange multipliers (assuming a finite sum, of course). That wouldn't be a (*) kind of question, though, so presumably there is a nicer way?
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    (Original post by Smaug123)
    I can do this by Lagrange multipliers (assuming a finite sum, of course). That wouldn't be a (*) kind of question, though, so presumably there is a nicer way?
    A very nice way. Though I would be interested to see how you do it with Lagrange multipliers as I initially tried that??

    This was on a trinity admissions test by the way
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    (Original post by jjpneed1)
    A very nice way. Though I would be interested to see how you do it with Lagrange multipliers as I initially tried that??

    This was on a trinity admissions test by the way
    Hmm. I can Lagrange-multipliers it down to "Maximise a^i subject to a>0, i a natural number, a i = 16", anyway, because Lagrange multipliers on a b c subject to a+b+c=16 yields a=b=c. Then the only problem is to find how many terms are in the sum.

    That is, maximise (\frac{16}{i})^i for integer i. Let's move to the more general x as our sum (so x=16 above). Now, if i>x then we start raising something less than 1 to a high power, so that can't be a maximum. Our answer must therefore be less than x. Using i=1 gives us x straight off.

    Suppose we'd found i. What would we know? We'd have that \frac{x}{i}^i > \frac{x}{i+1}^{i+1}, or \frac{1}{i}^i > \frac{1}{i+1}^{i+1}x, or that x < \frac{i+1}{i}^i (i+1). That fraction term tends to e from below as i increases; in particular, we need i+1 > \frac{x}{e}, or i > \frac{x}{e}-1. That suggests using i = roundup(x/e - 1).

    How tight is this bound? Not sure. It is conceivable that the "fraction term tends to…" line makes the bound slacker than it needs to be.
 
 
 
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