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  1. 26 Aug '15 13:21 / 7 edits
    I want the function of x ( f(x) ) from just one mathematical clue in the form of just one rule about the definite integrals of the curve of the graph for f(x).

    All variables/constants here are positive real numbers only.

    On the graph for f(x), for some arbitrary constant K where K is some specific x value and where K>0, for any x value that is smaller than K i.e. x<K, then f(x) = 0. But I am not interested here in f(x) for x<K but rather f(x) for x≥K.

    The only clue we got what that function is this; if we have a=x1 and b=x2 where a≥K and b≥K and a<b, then the integral of the interval of x from x=a to x=b on that graph is given by:

    ʃ [a, b] = (K/a) – (K/b)

    The only things I have deduce from that clue so far is that the curve for f(x) for x≥K is wholly above the x-axis (thus making all integrals positive ) and that f(x) tends to zero as x tends to +infinity.
    But now I am stuck.

    So what is the formula for that f(x)?

    Any insight will be greatly appreciated.

    Not sure but I think I have also deduced, from non-mathematical deduction from its application I have in mind, that:

    ʃ [K, +∞] = 1

    Is that right?
  2. 26 Aug '15 13:39
    f(x) = K/x^2 if x>K
    f(x) = 0 otherwise

    Proof: the anti-derivative of 1/x^2 is -1/x + constant.
  3. 26 Aug '15 13:47 / 3 edits
    Originally posted by KazetNagorra
    f(x) = K/x^2 if x>K
    f(x) = 0 otherwise

    Proof: the anti-derivative of 1/x^2 is -1/x + constant.
    Thanks for that!
    and I now understand the proof.
    Greatly appreciated.
  4. Standard member DeepThought
    Losing the Thread
    26 Aug '15 14:05 / 1 edit
    Originally posted by humy
    Although I don't understand that proof ( "-1/x"? cannot see how that relates but I bet I am being completely stupid here ), I greatly appreciate your insight. "K/x^2" somehow intuitively looks about right to me even if I can't see why.
    Edit: posted in reply to a post that was later edited to say he understood it...this is a little redundant now.

    Identity 1:
    d/dx(x^p) = p*x^(p-1), where p is some parameter independent of x.

    Identity 2:
    ʃ [a, b] dx f(x) = F(a) – F(b) <=> dF(x)/dx = f(x)
    F(x) is the antiderivative that Kazet referred to.

    We are given:
    ʃ [a, b] dx f(x) = (K/a) – (K/b)

    so using F(x) = K/x in identity 2 we get d/dx(K/x) = f(x)

    The minus sign comes from putting p = -1 into the identity at the start:
    d/dx(1/x) = d/dx(x^-1) = -x^-2 = -1/x^2

    giving us:

    f(x) = -K/x^2
  5. 26 Aug '15 14:18 / 2 edits
    Originally posted by DeepThought
    Edit: posted in reply to a post that was later edited to say he understood it...
    Yes, I didn't get it at first but then suddenly got it!
  6. Standard member sonhouse
    Fast and Curious
    26 Aug '15 19:28
    Originally posted by humy
    Yes, I didn't get it at first but then suddenly got it!
    Is this part of your statistics study?
  7. 26 Aug '15 20:51 / 1 edit
    Originally posted by sonhouse
    Is this part of your statistics study?
    Yes. It is part of yet another "cav" probability distribution I am working on and only very recently and its a very important one for it indirectly helps to solve the problem of induction.
  8. Standard member sonhouse
    Fast and Curious
    27 Aug '15 12:22
    Originally posted by humy
    Yes. It is part of yet another "cav" probability distribution I am working on and only very recently and its a very important one for it indirectly helps to solve the problem of induction.
    When you get the Fields metal, remember your friends here
  9. 27 Aug '15 15:01
    Originally posted by sonhouse
    When you get the Fields metal, remember your friends here
    I will
    And I will make a thread all about it here as soon as I publish my work in about very roughly ~one years time.
  10. 02 Sep '15 18:50
    I normally am good at basic algebra. But, while working on something, I accidentally found that:

    ( x/(x + 1) ) – ( x/(x + 2) ) = x/( (x + 1)(x + 2) )

    and I cannot see why the above is true with the simple rules of algebra I know of.
    Can somebody break it down for me with several algebraic intermediate steps so I can see how you can simplify ( x/(x + 1) ) – ( x/(x + 2) ) to x/( (x + 1)(x + 2) ) ?
  11. 02 Sep '15 19:01 / 9 edits
    Originally posted by humy
    I normally am good at basic algebra. But, while working on something, I accidentally found that:

    ( x/(x + 1) ) – ( x/(x + 2) ) = x/( (x + 1)(x + 2) )

    and I cannot see why the above is true with the simple rules of algebra I know of.
    Can somebody break it down for me with several algebraic intermediate steps so I can see how you can simplify ( x/(x + 1) ) – ( x/(x + 2) ) to x/( (x + 1)(x + 2) ) ?
    Arr, after not getting it for ages, I suddenly got it just literally only one min after I posted this!

    Sorry about that.
    Perhaps this would be a nice bit of maths exercise for one of you if you don't look at my answer below first.

    The answer is:





    ( x/(x + 1) ) – ( x/(x + 2) )
    = ( x(x + 2) /((x + 1) (x + 2) ) ) – ( x(x + 1)/((x + 1) (x + 2) ) )
    = (x(x + 1) – x(x + 2)) / ((x + 1)(x + 2))
    = x( (x + 1) – (x + 2) ) / ((x + 1)(x + 2) ) (factorize the (x(x + 1) – x(x + 2)) part)
    = x( x + 1 – x + 2 ) / ((x + 1)(x + 2))
    = x( 1 ) / ((x + 1)(x + 2))
    = x / ((x + 1)(x + 2))
  12. Standard member DeepThought
    Losing the Thread
    02 Sep '15 19:28
    Originally posted by humy
    Arr, after not getting it for ages, I suddenly got it just literally only one min after I posted this!

    Sorry about that.
    Perhaps this would be a nice bit of maths exercise for one of you if you don't look at my answer below first.

    The answer is:





    ( x/(x + 1) ) – ( x/(x + 2) )
    = ( x(x + 2) /((x + 1) (x + 2) ) ) – ( x(x + 1)/((x + 1) (x + 2 ...[text shortened]...
    = x( x + 1 – x + 2 ) / ((x + 1)(x + 2))
    = x( 1 ) / ((x + 1)(x + 2))
    = x / ((x + 1)(x + 2))
    Or more simply just multiply each side of the equation by (x+1)(x+2):

    LHS * (x+1)(x+2) = x(x+2) - x(x + 1) = x
    RHS * (x+1)(x+2) = x

    When you have a problem like that where you are trying to prove a known result you don't need to derive it, showing it's true is enough. Incidentally this is a partial fractions problem and you should give the Wikipedia page on partial fractions a look.

    https://en.wikipedia.org/wiki/Partial_fraction_decomposition
  13. 04 Sep '15 21:05 / 1 edit
    I am stuck on another problem.
    I have produced a very complex mathematical algorithm that seems to output decimals that seem to be one natural number divided by another natural number in a form of a top-heavy fraction but I fail to see the pattern between input x (x>0, x ∈ ℕ) and output y (y ≥ 1).
    here are the first 10 inputs and outputs:

    input x | output y
    1 → 1
    2 → 3 / 2
    3 → 19 / 12
    4 → 29 / 18
    5 → 1169 / 720
    6 → 5869 / 3600
    7 → 41183 / 25200
    8 → 288731 / 176400
    9 → 128461 / 78400
    10 → 10413181 / 6350400
    ...

    Note I don't know how much each output fraction has been inadvertently cancelled down from the unknown general formula, whatever that unknown general formula is, so you must take that into account.

    Any ideas?
  14. 05 Sep '15 19:12 / 3 edits
    What I really want to know, far more than the equation for the above pattern (although that would be far better than nothing! ), is the general algebraic formula for:

    ∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) where (x>0, x ∈ ℕ)

    This is so, to work out that sum (which is for a very important probability model that indirectly helps to solve the problem of induction ), we don't have to use an inefficient iteration to get a good approximation of it but rather use vastly less computation time by deriving the sum directly from a general equation.

    So far, I tried in vain to obtain one. But I have worked out that, for arbitrary 'high' x input values, we can derive a reasonable approximation of it with:

    ∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) ≈ 1/( 2(x^2) )
    where (x>0, x ∈ ℕ)
    and the ~1/( 2(x^2) ) approximation is always at least just slightly too low and never too high. And the approximation gets better and better with every increase in x.

    Any ideas?
  15. 05 Sep '15 19:21
    Originally posted by humy
    What I really want to know, far more than the equation for the above pattern (although that would be far better than nothing! ), is the general algebraic formula for:

    ∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) where (x>0, x ∈ ℕ)

    This is so, to work out that sum (which is for a very important probability model that indirectly helps to solve the problem of in ...[text shortened]... r too high. And the approximation gets better and better with every increase in x.

    Any ideas?
    There doesn't seem to be a simple answer to the sum. Try the following Wolfram Alpha query:

    Sum[1/( (X^2)(X + 1) ),{X,x,\infty}]