How to find all the intersection points of the two functions log(x!) and x? I am trying to find where log(x!) and x intersect, and am unable to do so rigorously.I eventually have 2^x=x!, but I am unsure how to proceed from here. Any input as to how to go about solving this would be great.

Layton Park

Layton Park

Answered question

2022-11-05

How to find all the intersection points of the two functions log ( x ! ) and x?
I am trying to find where log ( x ! ) and x intersect, and am unable to do so rigorously.
I eventually have 2 x = x !, but I am unsure how to proceed from here.
Any input as to how to go about solving this would be great.

Answer & Explanation

jennasyliang4tr

jennasyliang4tr

Beginner2022-11-06Added 15 answers

If you look for a continuous function
f ( x ) = log ( x ! ) x
and search for the x such that f(x)=0, I do not think that there is any analytical solution (beside the trivial solution x=0).
If you plot the function, you should notice that there is a root close to x = 5 and numerical method should be used, such as Newton. Starting from a "reasonable" guess x 0 , it will be updated according to
x n + 1 = x n f ( x n ) f ( x n )
The problem is that the derivative is quite complex since
f ( x ) = ψ ( 0 ) ( x + 1 ) 1
where appears the digamma function.
Nevertheless, if you start at x 0 = 5, Newton iterates will be 5.30095, 5.29033, 5.29032 which is the solution for six significant figures.
If you look for a "good" approximate solution, you could replace x! by Stirling, Gosper or Burnisde approximation of the factorial. This means, that, using Gosper for example, you should solve
g ( x ) = log [ π ( 2 x + 1 3 ) ( x e ) x ] x = 0
that is to say
g ( x ) = log ( π ) + 1 2 log ( 2 x + 1 3 ) + x log ( x ) 2 x = 0
g ( x ) = 1 2 x + 1 3 + log ( x ) 1
This transform makes the problem much simpler. Starting at x 0 = 5, Newton iterates will then be 5.30127, 5.29063, 5.29061 which is not too bad.

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