Re: i need help, im stuck

"pfecik@xxxxxxxxx" <pfecik@xxxxxxxxx> writes:

hey hey hey,
i have tried and tried and i am very curious how to solve this. can
anyone help me with this one. thanx (ans. = .394)
1,065*r = 525/[2.73+1.5915*ln(r)]
how do you isolate "r"?

You can't do it with elementary functions, but you can do it with
Lambert's W function. Lambert's W(z) is the solution for W in the
equation

z = W*exp(W)

If you have the equation

r*ln(r) + a*r = b

you can rewrite this as

ln(r) + a = b/r

a = b/r - ln(r)

a + ln(b) = b/r + ln(b) - ln(r)

a + ln(b) = b/r + ln(b/r)

exp(a+ln(b) = exp(b/r + ln(b/r))

b*exp(a) = (b/r)*exp(b/r)

This is now in the form z=W*exp(W) where we can invert it using Lambert's
W function.

b/r = W(b*exp(a))

Therefore

r = b/W(b*exp(a))

This is as close to closed form as you're going to get. Now, if what you
really want to do is to calculate r (you actually asked about isolating
r) then one way would be to calculate successive approximations that get
closer and closer to the answer you seek. Here are web descriptions of
Newton's method:

en.wikipedia.org/wiki/Newton's_method
mathworld.wolfram.com/NewtonsMethod.html

If you are seeking the root of f(r) = r*ln(r) + a*r - b = 0, then start
with an initial guess a_0. Then given an approximation a_k, you can
calculate the next approximation:

f(a_k) a_k*ln(a_k) + a*a_k - b
a_{k+1} = a_k - ------- = a_k - -----------------------
f'(a_k) ln(a_k) + a + 1

If your initial guess isn't close enough, Newton's method may not converge
to the desired root. Once it is close enough, the number of correct
decimal places doubles with each new approximation.

Scott
--
Scott Hemphill hemphill@xxxxxxxxxxxxxxxxxx
"This isn't flying. This is falling, with style." -- Buzz Lightyear
.

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