Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-l] Can Anyone Find a Solution For This Equation?



In a message dated 5/13/2010 3:44:16 PM Eastern Daylight Time,
David_Bowman@georgetowncollege.edu writes:

Bob, what kinds of quantities are k_1 & k_2 supposed to be? Since the
equation is trancendental we don't expect to find a general solution formula.
But we can make some general observations about the solutions when k_1 &
k_2 are real-valued.
)))))))))))))))))))))))))))))))))

k_1 and k_2 are real vales. The actual equation is


t= { 8*piG*R(0)^5*exp[5*t/t_d]*rho/(3*c^4)}^(1/3)


Trying to find a solution for t.


Actually I can get rid of one of the constants


Define a new variable y=k_2*x then solve

1/(k_1*k_2) = exp(y)/y

This can be solved numerically


)))))))))))))))))))))))))






If k_1 & k_2 are supposed to be real numbers then the equation has no real
solution for x when k_1*k_2 > 1/e.
If k_1*k_2 = 1/e then x = 1/k_2.
If 0 < k_1*k_2 < 1/e then there are *two* real solutions for x of the same
algebraic sign as k_2 (one larger than 1/k_2 and the other one between 0 &
1/k_2).
If k_1 = 0 then x = 0.
If k_1*k_2 < 0 then there is one real solution for x whose algebraic sign
is opposite of k_2.

Suppose we define the real-valued function f(u) of the real argument u
according to:

f(u) = = u*exp(-u). Then in general for real k_1 & k_2 the value(s) of
the solution x can be formaly written as:

x = (1/k_2)*f^-1(k_1*k_2) where f^-1(y) means the inverse function of f,
i.e. y = f( f^-1(y)).

The function f^-1(y) is single-valued and negative for a negative
argument. It is zero when its argument u is zero. It is double-valued with both
values being positive for positive arguments that are also less than 1/e (in
which case one value is between zero & one and the other value is greater
than 1). It has the value 1 when its argument is exactly 1/e. And it has
no real values at all for arguments larger than 1/e.


Thanks very clever and helpful.

Bob Zannelli



David Bowman