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 Posted: Wed, 16 May 2012 11:31:10 UTC
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Joined: Thu, 15 Dec 2011 16:54:36 UTC
Posts: 16
Is the below equation solvable analytically? -or- Must it be solved using iterative numerical methods?

exp(W*(1/(x*(x+1))^.5))=(2*J+1)*(2*x+1)

or

this may also written as W/(x^2+x)^.5=ln((2*J+1)*(2*x+1))

In this equation, "W" and "J" are constants, "ln" is the natural logarathm, "exp" is the exponential function - more commonly known as "e", * is the multiplication sign, "^" denotes raised to a power - so "^.5" denotes the square root of the expression x*(x+1) or x^2+x as I have written. How do we solve for x?

This equation attempts to model an entropy relationship of the nucleus of an atom, where X is attempting to model a quantity called nuclear spin.

Any algebraic insites would be most appreicated,

Jeff

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 Posted: Wed, 16 May 2012 13:44:38 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 4188
Location: Ottawa Ontario
jjbyram wrote:
exp(W*(1/(x*(x+1))^.5))=(2*J+1)*(2*x+1)

So above same as:
eW / SQRT(x^2 + x) = (2J + 1)(2x + 1)

Let A = eW and B = 2J +1; then:
A / SQRT(x^2 + x) = B(2x + 1)
Rewrite:
A / [B(2x + 1)] = SQRT(x^2 + x)
Square both sides:
A^2 / [B^2 (4x^2 + 4x + 1)] = x^2 + x
Rewrite:
A^2 / B^2 = (4x^2 + 4x + 1)(x^2 + x)
Do right side multiplication; simplify:
A^2 / B^2 = 4x^4 + 8x^3 + 5x^2 + x

I guess you can "nicely" wrap up this way:
Let K = A^2 / B^2
4x^4 + 8x^3 + 5x^2 + x - K = 0

Well, what can I say?! No cute quadratic for you today

Anyway, seems to work: tested it "guessed at" data;
why don't you try it out...

http://www.numberempire.com/equationsolver.php

_________________
I'm just an imagination of your figment...

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 Posted: Wed, 16 May 2012 19:21:55 UTC
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Joined: Thu, 15 Dec 2011 16:54:36 UTC
Posts: 16
Perhaps I wrote the wrong question. Shouldn't

exp(W*(1/(x*(x+1))^.5))=(2*J+1)*(2*x+1)

be equivalent to:

e^(W/(x^2+x)^.5)=(2J+1)*(2x+1) where "e" is euhlers constant?

How is the first term equivalent to: eW SQRT(x^2+x)?

Respectfully,

Jeff

P.S. You have helped me with quartic equations in the past and I am most grateful for your help.

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 Posted: Wed, 16 May 2012 20:29:31 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 4188
Location: Ottawa Ontario
jjbyram wrote:
e^(W/(x^2+x)^.5)=(2J+1)*(2x+1) where "e" is euhlers constant?

Of course....was on my 1st coffee....apologies...

Then:
e = [(2J+1)(2x+1)] ^ [SQRT(x^2+x) / W]
or
W / SQRT(x^2+x) = LOG[(2J+1)(2x+1)]

I believe that requires numeric solving.
Hope one of the experts here confirms that.

_________________
I'm just an imagination of your figment...

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