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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?
    A partial solution, if any 2 of the numbers are different, call them y-e and y+e, then (y-e)(y+e)=y^2-e^2 so making them equal will only increase the product, keeping the sum the same. Repeating this we see that we can continue to increase the product indefinately, and the terms tend to the same number. This isn't rigorous but it seems obvious (from the smoothness of the process) that having them all the same is the way to maximise the product. So if x is our number, we need to maximise (x/N)^N where N is an integer.
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    (Original post by Smaug123)
    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.
    A quick check with a computer suggests that your value is correct. For x=1000 I get the max occurs at 368 terms. I think this can be made rigorous by differentiating and noting that the function is nice enough that no other itnegers will be better.

    EDIT: Differentiation does work, and it is indeed what you said.
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    James and Smaug, you have the main parts even if it's a bit hand-wavey.
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    The fact that all the terms are the same follows from AM-GM, then you're left to maximise (k/x)^x which you can just do using calculus, giving the number of terms = k/e (where k is the sum of the parts). Though it appears my solution also isn't complete as I don't know which integer value for the number of terms (floor(k/e) or ceiling(k/e)) will be most favourable, any ideas?
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    (Original post by jjpneed1)
    James and Smaug, you have the main parts even if it's a bit hand-wavey.
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    The fact that all the terms are the same follows from AM-GM, then you're left to maximise (k/x)^x which you can just do using calculus, giving the number of terms = k/e (where k is the sum of the parts). Though it appears my solution also isn't complete as I don't know which integer value for the number of terms (floor(k/e) or ceiling(k/e)) will be most favourable, any ideas?
    AM-GM does work better.

    Calculus only gives the max for continuous n. To actually prove that the max is when n is one of floor(x/e) or cel(x/e) is a bit harder. You need to show that the function is decreasing either side of the maximum (not difficult, but not trivial either). I see no reason why the answer must be the same choice of floor or cel, I think it will depend of x as the maximum could be closer to one than the other for same values, but the other way round for others.
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    (Original post by james22)
    AM-GM does work better.

    Calculus only gives the max for continuous n. To actually prove that the max is when n is one of floor(x/e) or cel(x/e) is a bit harder. You need to show that the function is decreasing either side of the maximum (not difficult, but not trivial either). I see no reason why the answer must be the same choice of floor or cel, I think it will depend of x as the maximum could be closer to one than the other for same values, but the other way round for others.
    Well intuitively the function is strictly increasing initially and reaches a maximum, and clearly it will strictly decrease afterwards and go to zero pretty soon after x>k. I don't imagine it would be difficult to show with calculus, and as ceiling/floor are closest integers to the maximum one of them gives the maximum in the integers, right? And yeah I think you're right, I think a calculator will do the trick if needed
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    (Original post by jjpneed1)
    Well intuitively the function is strictly increasing initially and reaches a maximum, and clearly it will strictly decrease afterwards and go to zero pretty soon after x>k. I don't imagine it would be difficult to show with calculus, and as ceiling/floor are closest integers to the maximum one of them gives the maximum in the integers, right?
    Yes, it is actually much easier than I thought to show that it is decreasing (the sign of the derivative is trivial to calculate). This proves in a competely rigorous way that the answer is either floor or ceiling of x/e. Still need to work out which one though.
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    (Original post by jjpneed1)
    Well intuitively the function is strictly increasing initially and reaches a maximum, and clearly it will strictly decrease afterwards and go to zero pretty soon after x>k. I don't imagine it would be difficult to show with calculus, and as ceiling/floor are closest integers to the maximum one of them gives the maximum in the integers, right? And yeah I think you're right, I think a calculator will do the trick if needed
    Using matlab, it is floor for x=250 and ceiling for x=1000 (as examples). There may be a pattern to which one it is but I doubt that was part of the question.
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    (Original post by james22)
    Using matlab, it is floor for x=250 and ceiling for x=1000 (as examples). There may be a pattern to which one it is but I doubt that was part of the question.
    I doubt it as well. The question is as follows from the '99 Trinity admissions quiz:

    "I can divide the number 11 up in a number of ways, eg
    11 = 9 + 2
    11 = 3.1 + 4.5 + 3.2 + 0.2
    but what I want is the way of splitting it up so that the product of all the numbers I split it into is as large as possible. Eg for 11 = 9 + 2 the product concerned is 9 × 2 = 18. Can you find a general way of solving this puzzle so that I could have started with numbers other than 11?"

    Seeing as they'd go over the questions in the interview I suspect they'd want to see as much as we've done and perhaps they'd go over that subtlety. I'd be surprised if there was a pattern to be honest
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    (Original post by james22)
    ...
    (Original post by jjpneed1)
    ...
    Just to clarify, somebody I know IRL seized the opportunity to post the last few messages on this thread whilst my window was left open. Sorry about that!

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    Also, come on guys, when was the last time I said 'fam' and didn't use the correct grammar? 'Qtsie tootsie' should've been a dead giveaway that it wasn't me! :laugh:
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    (Original post by Khallil)
    Just to clarify, somebody I know IRL seized the opportunity to post the last few messages on this thread whilst my window was left open. Sorry about that!

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    Also, come on guys, when was the last time I said 'fam' and didn't use the correct grammar? 'Qtsie tootsie' should've been a dead giveaway that it wasn't me! :laugh:
    Seems they were excited about it :laugh:

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    (Original post by Khallil)
    Just to clarify, somebody I know IRL seized the opportunity to post the last few messages on this thread whilst my window was left open. Sorry about that!

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    Also, come on guys, when was the last time I said 'fam' and didn't use the correct grammar? 'Qtsie tootsie' should've been a dead giveaway that it wasn't me! :laugh:
    Oh, I assumed that you were drunk. That makes sense though.
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    Problem 475*

    f:\;\mathbb{R}\to\mathbb{R}. Must there exist x,y\in\mathbb{R} s.t. f\big( x-f(y)\big)>x+yf(x)?
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    (Original post by Lord of the Flies)
    ...
    Are you asking for us to give a counterexample for which there are no pairs x,y that satisfy the inequality?

    If x \mapsto 0, then that would imply that 0>0. Do you mean for the inequality to be strict?
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    (Original post by Khallil)
    Are you asking for us to give a counterexample for which there are no pairs x,y that satisfy the inequality?

    If x \mapsto 0, then that would imply that 0>0. Do you mean for the inequality to be strict?
    In the case f(x)=0, just consider x=-1, y=0.
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    (Original post by Khallil)
    ...
    The question is asking whether, given any real function f, there is a pair of real numbers that satisfies the inequality. If not we need to construct a clever counterexample, otherwise we need to prove the assertion.
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    Problem 476***

    Prove that, for integers n \geq 2

    \displaystyle \dfrac{e^2}{2} \left(\dfrac{n}{e} \right)^n \leq n! \leq \dfrac{ne^2}{4}\left(\dfrac{n}{e  } \right)^n
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    Problem 477***

    Evaluate the following product (note that p represents a prime number).

    \displaystyle \prod_p \left(1+\dfrac{1}{p(p+1)}\right)
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    (Original post by Indeterminate)
    Problem 477***

    Evaluate the following product (note that p represents a prime number).

    \displaystyle \prod_p \left(1+\dfrac{1}{p(p+1)}\right)
    Can you offer a hint?
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    (Original post by jjpneed1)
    Can you offer a hint?
    \displaystyle  \prod_p \left(1-\frac{1}{p^2}\right) \left(1+\frac{1}{p(p+1)}\right)=  \prod_p \left(1-\frac{1}{p^3}\right)
 
 
 
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