Does anyone have the math skills to isolate the variable, a.

[VivaLaViking]

c is an arbitrary constant, a is a real number.

ln(a)
------- = c
(1 - a)

Isolate a.

[fishsauce]

[ QUOTE ]
c is an arbitrary constant, a is a real number.

ln(a)
------- = c
(1 - a)

Isolate a.

[/ QUOTE ]

The only equivalence I can find with only one occurence of a is

c = \int_1^a (x+1)/x dx

where \int_1^a is the definite integral from 1 to a. This is derived from the original equation by first multplying bth sides by (1-a), then exponentiating both sides to get
a e^a = e^(c+1).
Then taking the natural log of both sides you get
c+1 = ln(a e^a) = a + ln(a).
Since ln(a) = \int_1^a 1/x dx and a = \int_1^a 1 dx + 1,
we have
c+1 = \int_1^a (1 + 1/x) dx + 1.
But I think this is all you can do.

Also, from the second FTC, since c is constant,
0 = (a+1)/a
for which the only solution is a = -1, for which the original problem statement is undefined.

Anyhow, where does this problem come from?

[PairTheBoard]

[ QUOTE ]
fishsauce --

multplying bth sides by (1-a), then exponentiating both sides to get
a e^a = e^(c+1).


[/ QUOTE ]

You have an error here. You turned c(1-a) into c+1-a.

PairTheBoard

[fishsauce]

[ QUOTE ]
[ QUOTE ]
fishsauce --

multplying bth sides by (1-a), then exponentiating both sides to get
a e^a = e^(c+1).


[/ QUOTE ]

You have an error here. You turned c(1-a) into c+1-a.

PairTheBoard

[/ QUOTE ]

My bad. I was working it out and made the mistake of
e^(c(1-a)) = (e^c)(e^(1-a)).

[VivaLaViking]

A power series. If you entered a number of poker tournaments with a variable number of entrants you may ask yourself, "How often do I finish in the upper quarter". You would probably normaize your finish by simply taking the percent of the place finished over the number of entrants.

Probability, p, of finishing in the top x percent, x, is p = x ^ a (eg Prob finishing in the top ½ is p = ½ ^ a).



For N trials, the data points of the percent finishes are used to find c with the data points d1 to dN.


ln(d1 * d2 * … dN)
------------------ = c
. . . . N

Then we can find a.


ln(a)
------ = c
(1 - a)

[PairTheBoard]

I don't see how how that probabilty situation relates. Although it would be cool if there was a solution that came from looking at a probabilty problem.

The power series looks like an idea but here's where I go with it.

Using the expansion for ln(1+x), For -1<x<=1

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...

So for -1<(a-1)<=1, ie. for 0<a<=2

ln(a) = ln(1+(a-1) = (a-1) - (a-1)^2/2 + (a-1)^3/3 - ...

and dividing by (1-a) we get,

ln(a)/(1-a) = -1 + (a-1)/2 - (a-1)^2/3 + (a-1)^3/4 - ...

But I don't see how to come back out of the series.


PairTheBoard

[gumpzilla]

Answer: <font color="white"> Differentiate both sides, leading to 1 / (a * (1-a)) + ln a / (1 - a)^2 = 0. Now substitute in our relationship from before and manipulate to get a = - 1 / c. </font>

EDIT: I don't think this makes a lick of sense. At the very least the result I got doesn't make sense. It's time to go to bed.

[PairTheBoard]

[ QUOTE ]
Answer: Differentiate both sides, leading to 1 / (a * (1-a)) + ln a / (1 - a)^2 = 0. Now substitute in our relationship from before and manipulate to get a = - 1 / c.
<font color="white">.
</font>
EDIT: I don't think this makes a lick of sense. At the very least the result I got doesn't make sense. It's time to go to bed.

[/ QUOTE ]

Taking the derivative makes no sense. On the left hand side you have a function in the variable a which is Not constant. Just because you're trying to solve for when the function equals c does not mean the derivative of the function is zero.

It would be like trying to solve the equation x^2=3 by saying the derivative of x^2 must be zero. ie. 2x=0 so x=0. It makes no sense.

PairTheBoard

[BluffTHIS!]

Doesn't the logarithmic base here need to be specified?

[Hiding]

ln, is the natural log (base 2)