- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
Which is larger, $2005!$ or $2^{18000}$?
Which is Larger: $2005!$ or $2^{18000}$?
- MHB
- Thread starter anemone
- Start date
-
- Tags
- Value
In summary, when comparing the values of $2005!$ and $2^{18000}$, we can use the Stirling approximations to find that $\ln 2005! \sim 13244.536$ and $\ln 2^{18000} = 12476.649$. Therefore, $2005!$ is larger than $2^{18000}$ since $\ln 2005! > \ln 2^{18000}$. Another approach using induction also confirms this result.
- #2
chisigma
Gold Member
MHB
- 1,628
- 0
anemone 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
$\chi$ $\sigma$
- #3
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
chisigma said:
Thank you chisigma for participating in this challenge problem! And thank you for your nice solution using the Stirling approximations route!
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.
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.
1. Is $2005!$ larger than $2^{18000}$?
Yes, $2005!$ is larger than $2^{18000}$. The factorial function grows much faster than the exponential function, so even though $2^{18000}$ is a very large number, it is still smaller than $2005!$.
2. How do you calculate $2005!$ and $2^{18000}$?
To calculate $2005!$, you would multiply all of the numbers from 1 to 2005 together. For $2^{18000}$, you would raise 2 to the power of 18000. This can be done using a calculator or by using mathematical formulas.
3. Can you give an example to illustrate the difference in size between $2005!$ and $2^{18000}$?
Sure, let's take a smaller number, like 5. $5!$ is equal to 120, which is much larger than $2^5$ (32). This shows that even with smaller numbers, the factorial function grows much faster than the exponential function.
4. Are there any real-world applications for such large numbers?
Yes, these types of large numbers are often used in cryptography, computer science, and other fields that deal with data and computations. For example, in cryptography, large numbers are used to create secure encryption keys.
5. How do mathematicians compare the sizes of extremely large numbers?
Mathematicians use a variety of methods to compare large numbers, such as logarithms, approximations, and mathematical proofs. They also use mathematical notation, such as the symbols $<$ (less than) and $>$ (greater than), to represent the relative sizes of numbers.
Similar threads
-
Introductory Physics Homework Help
-
Nuclear Engineering
-
Introductory Physics Homework Help
-
General Math
-
Computing and Technology
-
General Math
-
Nuclear Engineering
-
General Math
-
General Math