- 04 Mar '15 18:16 / 2 editsOK, 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? - 04 Mar '15 18:48What 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.
- 04 Mar '15 19:01

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

Then solve:

**F**(x) = e!(c – e)! /**2**(c + 1)!

for "x"

Simple... - 04 Mar '15 20:18 / 1 edit

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*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?***necessarily*be a natural number. - 04 Mar '15 20:19 / 4 edits

but I don't know how to find the indefinite integral for that (although at least I know what an indefinite integral is) and*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]*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. - 04 Mar '15 20:51 / 2 edits

Your question is interpreted as though you were looking for the method on*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.*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". - 04 Mar '15 20:57

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.*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. - 04 Mar '15 20:59 / 1 edit

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*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.

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. - 04 Mar '15 20:59 / 1 edit

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*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.***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. - 04 Mar '15 21:07 / 2 edits

oh I see. In that case, I would change e to t ( t for trait -sort of close enough to the right meaning )*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.**

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. - 04 Mar '15 21:14 / 3 edits

That looks good to me so far! I will study this and then get back to this thread. Thanks.*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. - 04 Mar '15 22:47

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.*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. - 05 Mar '15 21:18 / 5 editsThere'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. - 06 Mar '15 00:14

I don't think you understood what I was saying.*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.