how do you solve e^x = x
by taking ln for both sides
Use Newton-raphson method
what is the newton-raphson method?
taking the ln does not anything
please help nowhereman
there is no analytic solution, so you should use an approximation method like Newton's tangent method
how do you know there is no analytic solution?
you're right. ln does not solve the problem
oh wait, there is no solution at all, because \[ e^x >= 1 + x \]
ya, e^x always larger than x, so e^x = x does not have any solution
you are wrong. The answer is nonreal, which i am well aware of. I know what the number is to many decimal places. (Just take the ln of any number over and over again), but i have no idea how to find the number.
you have to iterate using computer methods to find this value
can you prove that?
If you use the moivre's theorem you know that e^(ix) = isinx + cosx or something like that
I used software program and found that x=-W(-1), where W(z) is the product log function
what software did you use? did u use wolfram alpha? what is the product log function?
yes.. here is the definition of the function http://www.wolframalpha.com/input/?i=product+log+function
ok
Join our real-time social learning platform and learn together with your friends!