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?

[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 ]
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

[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)

[ QUOTE ]
ln, is the natural log (base 2)

[/ QUOTE ]
I assume you mean base e (2.71828.....)

[Hiding]

yep, sorry about that.

[VivaLaViking]

Can't anyone offer any suggestions on how variable, a, may be isolated.

[evil_twin]

[ QUOTE ]
yep, sorry about that.

[/ QUOTE ]

Nothing to be sorry about, ln is the natural logarithm.

[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)

[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)).

Well using what I know of math (decent knowledge but rarely used) I get to.

ln(a) = c-ac

you exp each side you get.
e^(ln(a)) = e^(c-ac)

Simplify you get
a = (e^c)/(e^ac)

That is as far as I can get, it can't be simplified any further by me and I doubt it can be solved.

I'll join with the rest, where does that problem come from anyway?

[VivaLaViking]

See above Post:

Prolem Arises From (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=3271318&amp;page=0&amp;view=c ollapsed&amp;sb=5&amp;o=&amp;vc=1)

This is as far as I get, Note I'll call e ^ c, c', it still a constant.


a = c' * e^(1 - a)


a must be able to be isolated.

[Cooker]

[ QUOTE ]
a must be able to be isolated.

[/ QUOTE ]

This is not true. I haven't worked on this one bit, but I suspect this is a transcendental equation which means that a cannot be isolated. c=a e^a cannot be solved analytically for a and is an example of a transcendental equation.

Well I doubt you can isolate "a" as it's more complex than simple algebra. You start with the premise that (a) != 1 since you can't divide by 0. If you simplify you will see that (a) will tend toward 1 without reaching it...

BIG WARNING, I haven't touched math in over 3 years...I'm rusty and to avoid confusing things further I'll just read some post here to see what other post, lol. I'm pretty sure it doesn't tend toward 1, even though it's what I find...

/images/graemlins/confused.gif /images/graemlins/confused.gif /images/graemlins/confused.gif

Could those functions be what my teachers called non constant analytical functions? I'm sure a can't be isolated too, yet I figured I would give it a try...who knows had I found a way I could have been a celebrity /images/graemlins/cool.gif

[Cooker]

I believe the OPs original problem can be solved in terms of the Lambert W-function which is defined as the inverse of f(x) = x e^x. There are many more functions than most people bother to learn about, Bessel functions, Jacobi functions, and on and on. Just close your eyes and pretend it is sin or something you feel comfortable with. Read more about it here (http://mathworld.wolfram.com/LambertW-Function.html)

[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&lt;x&lt;=1

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

So for -1&lt;(a-1)&lt;=1, ie. for 0&lt;a&lt;=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

[Jcrew]

It looks like a transcendental equation and cannot be solved by algebraic means. The function looks like it can be solved using the Lambert W function.

http://mathworld.wolfram.com/LambertW-Function.html