Question

So I have encountered during a problem-solving excercise a function with the form:
\[h(x)= 3e ^{3x}-8x\]
And I am interested to see if this is solvable algebraically, which I tried and got the solution \(x=\frac{ W(-\frac{ 9 }{ 8 }) }{ -3 }\) wanted to check if I got it right and would gladly appreciate if I got some feedback.

Answer 1

That is correct, and you can read more about the function here: http://mathworld.wolfram.com/LambertW-Function.html

Answer 2

One way would be to plug it back in:

\[3e^{-W(-9/8)} + \frac{8}{3} W(-9/8)\]

Then simplify that left term with this rearrangement:

\[W(x)e^{W(x)}=x\] into \[\frac{W(x)}{x}=e^{-W(x)}\]

that'll give you

\[3\frac{8}{9}W(-9/8) + \frac{8}{3} W(-9/8) \]

Which clearly equals 0 (I assume you were looking for the roots of h(x) I guess?)

Answer 3

Yes, So I got the correct answer.
How would you go about approximating the value of the solution using Newtons Method?

Answer 4

Is that what your assignment is to do, or is this what you believe is the best way to evaluate the product log?

Answer 5

Oh, I thought it could've been the best way to approximate it, but is there one which is not possibly restricted (Newtons method sometimes does fail).

Answer 6

Depends, like I'm asking, what's the purpose here. If it was up to me, I would just throw it in maple or mathematica and be done with it at this point if I needed some actual physical approximation to something. Otherwise leaving it like this is just as well as \(\sqrt{ \ln x } \) to me.

Answer 7

One cheap trick to evaluating it could be \[W(x)e^{W(x)} = x\] take the logarithm:
\[\ln W(x) + W(x) = \ln x \]
Now you have a recursive way to calculate it simply that's dead easy to implement with no trouble:

\[W(x) = \ln x - \ln W(x)\]

At least that's what I've done in the past, I think just using Newton's method just like you would normally should work too, I think I have some papers on evaluating it actually with other numerical methods if I look.

Here's another fun problem to try to evaluate exactly while we're on the subject:

\[W(-\tfrac{\pi}{2})\]

Answer 8

I don't quite get the whole
\[W(x)=lnx-lnW(x)\]
How does that simplify the work?

Answer 9

https://repl.it/B3bI

I'm busy helping out a friend right now, but that's what I'm talking about.

Answer 10

?? closed? :(

Answer 11

https://en.wikipedia.org/wiki/Lambert_W_function#Numerical_evaluation

Answer 12

Your program won't work for a value that gives a complex answer, though.

Didn't know about repl.it, thanks!

Answer 13

It wasn't really supposed to work for complex answers, Java doesn't support complex values as a primitive type

Answer 14

Yeah, was just pointing it out in case the OP decided to try using it to evaluate either the original problem or the one you suggested...