MHB Which is Larger: $2005!$ or $2^{18000}$?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Value
AI Thread Summary
The discussion centers on comparing the sizes of $2005!$ and $2^{18000}$. Using Stirling's approximation, it is calculated that $\ln 2005! \sim 13244.536$, while $\ln 2^{18000} \approx 12476.649$. This indicates that $2005!$ is significantly larger than $2^{18000}$. Additionally, an alternative solution method involving induction was mentioned, but the primary focus remains on the Stirling approximation results. Overall, the conclusion is that $2005!$ is the larger value.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Which is larger, $2005!$ or $2^{18000}$?
 
Mathematics news on Phys.org
anemone said:
Which is larger, $2005!$ or $2^{18000}$?

[sp]Using the Stirling approximations You have...

$\displaystyle \ln n! \sim (n + \frac{1}{2})\ \ln n - n + \frac{1}{2}\ \ln (2\ \pi)\ (1)$

... so that is $\displaystyle \ln 2005! \sim 13244,536...$

On the other side is...

$\displaystyle \ln 2^{n} = n\ \ln 2\ (2)$

... so that is $\displaystyle \ln 2^{18000} = 12476.649...$ [/sp]

Kind regards

$\chi$ $\sigma$
 
chisigma said:
[sp]Using the Stirling approximations You have...

$\displaystyle \ln n! \sim (n + \frac{1}{2})\ \ln n - n + \frac{1}{2}\ \ln (2\ \pi)\ (1)$

... so that is $\displaystyle \ln 2005! \sim 13244,536...$

On the other side is...

$\displaystyle \ln 2^{n} = n\ \ln 2\ (2)$

... so that is $\displaystyle \ln 2^{18000} = 12476.649...$ [/sp]

Kind regards

$\chi$ $\sigma$

Thank you chisigma for participating in this challenge problem! And thank you for your nice solution using the Stirling approximations route!:cool:

Solution proposed by other through induction method:
We show that for all $n$, $n!>\left(\dfrac{n}{3} \right)^n$. This is true for $n=1$.

Suppose it is true for $n$. Then $(n+1)!=(n+1)n!>(n+1)\left(\dfrac{n}{3} \right)$.

Now, $\left(\dfrac{n+1}{3} \right)^{n+1}\left(\dfrac{3}{n} \right)^n=\dfrac{n+1}{3}\left(1+\dfrac{1}{n} \right)^n<\dfrac{(n+1)e}{3}<(n+1)$, this proves that $(n+1)!>\left(\dfrac{n+1}{3} \right)^{n+1}$ which then justifies the claim by induction.

Therefore, $2005!>2004!>668^{2004}>512^{2004}=2^{9\cdot2004}=2^{18036}>2^{18000}$ and we can conclude by now that $2005!$ is the greater of the two.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top