Approximation of $26!$

Question

Peltzl's Cryptology states on page 8 that $26!$ is approximately $2^{88}$. I have tried different variations of Stirling's formula to confirm this but no luck. I know the argument is hiding in there somewhere and a countable infinity of heads are better than one. Any help out there?

Answer 1

Write it as

$$26!=2^{23}\cdot3^{10}\cdot5^6\cdot7^3\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\;,$$

and take the log base $2$:

$$23+10\lg3+6\lg5+3\lg7+2\lg11+2\lg13+\lg17+\lg19+\lg23$$

is approximately $88.382$, so $88$ is the nearest integer exponent.

Answer 2

As suggested by vadim123: $$\log_2(26!) = \sum_{n=2}^{26}\log_2n$$ Approximating brutally with $\log_2n \approx \lfloor\log_2n\rfloor$ we get: $$\log_2(26!)\approx 2\cdot1+4\cdot2+8\cdot3+11\cdot4=78$$ Thus with this (extremely inefficient) approximation: $$26!\approx2^{78}$$ Summing the "exact" logarithms with Wolfram Alpha, we obtain $$\log_2(26!)\approx88.4$$ $$\Rightarrow 26!\approx2^{88}$$

Answer 3

$ \ln\left(26!\right) \sim \left(26 + 1/2\right)\ln\left(26\right) - 26 + \ln\left(2\pi\right)/2 \sim n\ln\left(2\right)$ $$ n \sim {26.5\ln\left(26\right) -26 + \ln\left(2\pi\right)/2\over \ln\left(2\right)} = 88.3773\ldots $$

$$ \color{#ff0000}{\large\ln\left(26!\right)} \sim 2^{0.3373} \times 2^{88} \sim \color{#ff0000}{\large 1.2633 \times 2^{\color{#0000ff}{88}}} $$

Answer 4

Stirling's Formula works: $$ n! \sim (\frac{n}{e})^n\sqrt{2\pi n}$$ and so $$\log_2{n!}=\frac{\ln{n!}}{\ln2} \sim \frac{n(\ln n-1)}{\ln 2}+\frac{\ln \pi +\ln n}{2 \ln 2}+\frac{1}{2}$$

This gives the approximation $$\log_2 26! \sim 88.377$$

for $$\log_2 26! = 88.381$$