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?
Originally posted by humyIn theory you have to find the indefinite integral; call it F(x) of your function f(x)
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?
Then solve:
F(x) = e!(c – e)! / 2(c + 1)!
for "x"
Simple...
Originally posted by Eladare 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.
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?
Originally posted by joe shmobut 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.
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]
Originally posted by humyYour 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)
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.
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".
Originally posted by humyHe 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.
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.
Originally posted by humyHere 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
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.
Originally posted by KazetNagorrayou 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.
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.
If that works, I would give my result here and that result would be good enough for my purposes.
Originally posted by joe shmooh I see. In that case, I would change e to t ( t for trait -sort of close enough to the right meaning )
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.
Originally posted by DeepThoughtThat looks good to me so far! I will study this and then get back to this thread. Thanks.
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.
Originally posted by humyWell 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.
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.
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.
Originally posted by humyI don't think you understood what I was saying.
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.