Question

Anyone know if there's a function in Mathematica that will give numerical answers to the inequality:
\[2^n 14 years ago

Answer 1

Could you explain a bit more with one example ?

Answer 2

Actually, I only really need a function to solve:
\[2^n = n!\]
I'm looking for something similar to the reduce function. (Reduce doesn't seem to like factorials.)
For example I can use Reduce[(x + 1)/(x + 2) < (x + 3)/(x + 4)] and Mathematica gives the output x<-4||x>-2.
Does that make more sense?

Answer 3

Well, just try some values here

n = 1: 2^n = 2 > 1 = 1!
n = 2: 2^n = 4 > 2 = 2!
n = 3: 2^n = 8 > 6 = 3!
n = 4: 2^n = 16 < 24 = 4!
n = 5: 2^n = 32 < 120=5!

and it's not hard to convince yourself that for all n > 5, 2^n < n!

Answer 4

I understand that it is, and I know it's easy to prove by induction, but unfortunately I have to find a solution specifically through Mathematica and have been searching for quite a while.

Answer 5

But the equation \[ 2^n =n! \] is only solvable for n = 0,and \[ 2^n \gt n! \] for \[ n \in [1,3]\], and after that \[ 2^n \lt n! \] for any \[ n \ge 4 \]isn't ?

Answer 6

I see; I'm sorry, I don't know Mathematica commands.

Answer 7

Thanks for the help anyway.

Answer 8

Well then you have to write this specific function yourself,since I am not aware of any function that could do it by default.

Answer 9

Ok, I was hoping one existed. Thanks for the help guys.

Answer 10

are you supposed to use the gamma function? that is the only way to solve n!=2^n

Answer 11

Yes, was thinking that too, but I assumed that because he was using n the values are integral.

Answer 12

I doubt it. It's only undergraduate level and I think it would be overcomplicating it. And solutions are over natural n.

Answer 13

For the record, the Gamma function is very much an undergraduate concept, but if n is integral, it's integral.

Answer 14

Oh ok, I assumed it was a higher level probably being a tad naive.