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  1. 04 Mar '15 18:16 / 2 edits
    OK, I have already asked a similar question before but what I want is something a bit different here that I have been frustratingly been completely stuck on but hope someone here can show me the solution:

    It concerns the graph for:

    f(x) = (x^e)( (1 – x) ^ (c – e) )

    where 0 < x < 1 (x is not allowed to be less than 0 nor greater than 1 ) and c and e are both natural numbers (i.e. positive and whole numbers) and e cannot be larger than c (so the (c – e) exponent is never negative ).

    I know that the area under the curve (its integral ) of this graph from x=0 to x=1 is exactly:

    e!(c – e)! / (c + 1)!

    But what I want to know is, what is the real number M that is such that the area under the curve from x=0 to x=M is exactly HALF of that e!(c – e)! / (c + 1)! i.e. e!(c – e)! / 2(c + 1)!
    ?
    In other words, if you cut that area under the curve for f(x) = (x^e)( (1 – x) ^ (c – e) ) between x=0 and x=1 exactly in half with a perfectly vertical line so that there is exactly equal area under the curve either side of that vertical line, exactly where along the x-axis will that vertical line bisect? At x=...?
    ( Obviously, 0 < M < 1 )

    Anyone?
  2. 04 Mar '15 18:48
    What you need to do is find the solution for the definite integral from 0 to M and equate it to the desired value. I doubt there is an analytical expression for M, though. If you have a specific purpose in mind you can do it numerically quite easily for fixed choices of the parameters.
  3. Subscriber joe shmo
    Strange Egg
    04 Mar '15 19:01
    Originally posted by humy
    OK, I have already asked a similar question before but what I want is something a bit different here that I have been frustratingly been completely stuck on but hope someone here can show me the solution:

    It concerns the graph for:

    f(x) = (x^e)( (1 – x) ^ (c – e) )

    where 0 < x < 1 (x is not allowed to be less than 0 nor greater than 1 ) and c and e ...[text shortened]... along the x-axis will that vertical line bisect? At x=...?
    ( Obviously, 0 < M < 1 )

    Anyone?
    In theory you have to find the indefinite integral; call it F(x) of your function f(x)

    Then solve:

    F(x) = e!(c – e)! / 2(c + 1)!

    for "x"

    Simple...
  4. 04 Mar '15 19:09 / 1 edit
    If e is a true variable and not the number e, then I'd suggest using a different variable. Would you use pi? If not, then why use e?

    One other question, why bother explaining what the inequalities mean? A person who can answer your question will understand inequalities.
  5. 04 Mar '15 20:18 / 1 edit
    Originally posted by Eladar
    If e is a true variable and not the number e, then I'd suggest using a different variable. Would you use pi? If not, then why use e?
    e in the particular application I want it for represents the whole number of so-far observed cases c that have an observed event of type e happen to them (but some observed cases c may not have that event happen to them hence this is why said e can be less than c but cannot greater than c ) thus e must necessarily be a natural number.
  6. 04 Mar '15 20:19 / 4 edits
    Originally posted by joe shmo
    In theory you have to find the [b]indefinite integral; call it F(x) of your function f(x)

    Then solve:

    F(x) = e!(c – e)! / 2(c + 1)!

    for "x"

    Simple...[/b]
    but I don't know how to find the indefinite integral for that (although at least I know what an indefinite integral is) and that is the problem. The second part of then solving it for x by making x the subject of the equation I would imagine to be the easy part.
  7. Subscriber joe shmo
    Strange Egg
    04 Mar '15 20:51 / 2 edits
    Originally posted by humy
    but I don't know how to find the indefinite integral for that (although at least I know what an indefinite integral is) and that is the problem. The second part of then solving it for x by making x the subject of the equation I would imagine to be the easy part.
    Your question is interpreted as though you were looking for the method on how to find "M". (This should be apparent from the fact that Kazet and I both answered in the same manner)

    It is probably interpreted this way because the "reader" assumes that if you knew how to get to your result in theory, you would see that a generalized result for the integral is most likely out of the question. Also, if it could be integrated definitely from 0 to M, solving it for "M" would most likely be just as difficult...no part of this is the "easy part".
  8. Subscriber joe shmo
    Strange Egg
    04 Mar '15 20:57
    Originally posted by humy
    e in the particular application I want it for represents the whole number of so-far observed cases c that have an observed event of type e happen to them (but some observed cases c may not have that event happen to them hence this is why said e can be less than c but cannot greater than c ) thus e must necessarily be a natural number.
    He is saying to change your variable "e" to avoid confusing it with Euler's constant "2.71828..." to some other dummy variable that is like "a,b,c,d..." that doesn't have any mathematical significance.
  9. Standard member DeepThought
    Losing the Thread
    04 Mar '15 20:59 / 1 edit
    Originally posted by humy
    but I don't know how to find the indefinite integral for that (although at least I know what an indefinite integral is) and that is the problem. The second part of then solving it for x by making x the subject of the equation I would imagine to be the easy part.
    Here is a partial result, I haven't got the whole way. I don't like c and e because e is the symbol for the base of natural logs, and (c - e) as an exponent makes the integral look harder than it is. So I'll replace them with a and b

    let
    g(x) = x^a
    h(x) = (1 - x)^b

    f(x) = g(x)h(x)

    a, b ∈ ℕ

    a = e
    b = c - e

    The cases where a or b are zero are trivial (there is no reason to exclude the case a = 0, your formula still comes out right).

    The integral is

    M
    ∫dx f(x)
    0

    f(x) = (x^a)((1 - x)^b) = Σ(-1^r) * C(b,r) * x^(a + r)

    the sum is from 0 to b and C(b, r) is the binomial coefficient. The anti-derivative (ignoring constants of integration) is:

    S = Σ(-1^r) * C(b,r) * x^(a + r + 1) / (a + r + 1)

    If x = 1, we get:

    S = Σ(-1^r) * C(b,r) / (a + r + 1)

    which you have in closed form. I don't instantly see how to sum the polynomial, but you might find it in a table of standard polynomials.
  10. 04 Mar '15 20:59 / 1 edit
    Originally posted by KazetNagorra
    What you need to do is find the solution for the definite integral from 0 to M and equate it to the desired value. I doubt there is an analytical expression for M, though. If you have a specific purpose in mind you can do it numerically quite easily for fixed choices of the parameters.
    you just made me think of a new approach; first give c and e specific examples of values, simplify the equation for those specific values (so get rid of those horrible messy factorials ) and only then try and do integration by parts. I will give that a go in due course and see if I can go anywhere with that.
    If that works, I would give my result here and that result would be good enough for my purposes.
  11. 04 Mar '15 21:07 / 2 edits
    Originally posted by joe shmo
    He is saying to change your variable "e" to avoid confusing it with Euler's constant "2.71828..." to some other dummy variable that is like "a,b,c,d..." that doesn't have any mathematical significance.
    oh I see. In that case, I would change e to t ( t for trait -sort of close enough to the right meaning )

    If c is a problem (sometimes c is used to mean "a constant" ) then could always replace that with i for "instance" -close enough to what I want it to mean.
  12. 04 Mar '15 21:14 / 3 edits
    Originally posted by DeepThought
    Here is a partial result, I haven't got the whole way. I don't like c and e because e is the symbol for the base of natural logs, and (c - e) as an exponent makes the integral look harder than it is. So I'll replace them with a and b

    let
    g(x) = x^a
    h(x) = (1 - x)^b

    f(x) = g(x)h(x)

    a, b ∈ ℕ

    a = e
    b = c - e

    The cases where a or b are ze ...[text shortened]... stantly see how to sum the polynomial, but you might find it in a table of standard polynomials.
    That looks good to me so far! I will study this and then get back to this thread. Thanks.
  13. Subscriber joe shmo
    Strange Egg
    04 Mar '15 22:47
    Originally posted by humy
    oh I see. In that case, I would change e to t ( t for trait -sort of close enough to the right meaning )

    If c is a problem (sometimes c is used to mean "a constant" ) then could always replace that with i for "instance" -close enough to what I want it to mean.
    Well if we are going to nit pick, "t" is usually reserved for time in physical equations, and i could be confused with " imaginary"...I think he was just trying to point out that letters like "e" are specific numbers in mathematics, and the misinterpretation of that "e" in your equation being Euler's number( or "i" being imaginary √(-1)) has specific consequences regarding the mechanics of your equation. The misinterpretation of "c" as "constant" or "t" as time is less of an offense, as it does not change the functions form. Call the variables whatever you want, just try to avoid the symbolic representation of numbers, and maybe variable that are used regularly to denote physical quantities, its no big deal, just some thoughts on the matter.
  14. Standard member DeepThought
    Losing the Thread
    05 Mar '15 21:18 / 5 edits
    There's a big difficulty with the last step, this doesn't solve the problem. But I don't think integration by parts really helps (at first I thought I'd got it and then spotted the problem):

    Using my notation from earlier (a, b) rather than (c, e) we have:

    F(X; a, b) = ∫x^a (1 - x)^b dx

    With the limits of the integral being [0, X]. We can integrate by parts in two ways to produce two recurrence relations:

    (1 + b)F(X; a, b) = a F(X; a - 1, b + 1) - X^a(1 - X)^(b + 1)
    (1 + a)F(X; a, b) = b F(X; a + 1, b - 1) - X^(a + 1) (1 - X)^b

    Now for the trick, rearrange each equation so that we have X^a(1 - X)^b on the left:

    X^a(1 - X)^b = [a F(X; a - 1, b + 1) - (1 + b) F(X;a, b)]/(1 - X)
    X^a(1 - X)^b = [b F(X; a + 1, b - 1) - (1 + a) F(X;a, b)]/X

    the left hand sides are equal, so we can equate them and rearrange to get:

    X/(1 - X) = [b F(X; a + 1, b - 1) - (1 + a)F(X;a, b)]/[(a F(X; a - 1, b + 1) - (1 + b) F(X; a, b)]

    Getting this in terms of X is straightforward, but tedious I'll leave it to you.

    You also have:

    F(X; a, b) = F(1; a, b)/2

    Now we get to the catch. That formula is only right for X, a and b. For F(X, a+1, b - 1) = 1/2F(1; a + 1, b - 1) for a different value of X than F(X, a, b). If we try using the recurrence relations again to get the right value of F(X, a + 1, b - 1) and F(X; a - 1, b + 1):

    (1 + b)F(1; a, b)/2 = a F(X; a - 1, b + 1) - X^a(1 - X)^(b + 1)
    (1 + a)F(1; a, b)/2 = b F(X; a + 1, b - 1) - X^(a + 1) (1 - X)^b

    you get X/(1 - X) = X/(1 - X). So not much help. Basically you can only solve specific cases. You have to do the iterations by hand, or find a way of summing the polynomial in my previous post.
  15. 06 Mar '15 00:14
    Originally posted by humy
    e in the particular application I want it for represents the whole number of so-far observed cases c that have an observed event of type e happen to them (but some observed cases c may not have that event happen to them hence this is why said e can be less than c but cannot greater than c ) thus e must necessarily be a natural number.
    I don't think you understood what I was saying.