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Calculus Level 1

Which is greater?

$100! \text{ or } 10 ^{162} ?$

You may use the fact that $e^ 5 > 148$ .

100!

10^{162}

1 solution

Calvin Lin Staff
Aug 14, 2015

There is no need to resort to wolfram alpha to prove that the $10 ^ {162}$ is greater.

Let's factor out 100 from both sides, and then take logarithms. We thus want to compare
$\sum_{i=1} ^ {99 } \ln x \text{ VS } 160 \ln 10$

Observe that the sum can be over-approximated by the Riemann sum of $\int_1 ^ {100} \, \ln x \, dx$ , which is equal to $[ x ( \ln x - 1) ]_{1}^{100} = 100 ( \ln 100 - 1)$ .

We thus want to compare

$\begin{array} { l l l } 200 \ln 10 - 100 & ?? & 160 \ln 10 \\ 40 \ln 10 & ?? & 100 \\ \ln 10 & ?? & \frac{5}{2} \\ 100 & ?? & e^ 5 \\ \end{array}$

And since $e ^ 5 = 148$ , hence we see that the RHS is larger.

Note: According to Wolfram Alpha, $100 ! < 10 ^ {158}$ . How can we improve to this bound?

To improve on the bound, we need to know the values of $\log_{10}(2), \log_{10} (e), \log_{10} (\pi)$ beforehand up to certain desired accuracy.

Apply the asymptotic formula for Stirling series :

$n! = \sqrt{2\pi n} \left(\frac ne\right)^2 \left(1 + \frac1{12n} + \frac{1}{288n^2} - \frac{139}{51840n^3} - \frac{571}{2488320n^4} + \ldots \right)$

Computing with $\ln$ instead of $\log_{10}$ then just divide the result by $\ln(10)$ :

$\ln(n!) = (\ln(n) - 1) n + \ln(\sqrt{2\pi}) + \frac12 \ln(n) + \frac1{12n} - \frac1{360n^3} + O(n^{-5})$ .

For $n = 100$ , we just need to show that the following inequality is true.

$201 - \frac{100}{\ln(10)} + \frac{\ln(2\pi)}{2 \ln(10)} = 201 - 100 \log_{10} e + \frac12 \log_{10} (2\pi) < 158$

And we're done.

Pi Han Goh - 5 years, 10 months ago

Good. I was also going to say this.

Satyajit Mohanty - 5 years, 10 months ago

Yes, unfortunately in order to get a tighter bound, we have to start obtaining the value of more constants.

Another approach that I was thinking of, is to bound the first few terms tighter, since they contribute the most error. For example, if we can bound $\sum_{i=1}^9 \ln i$ 13 because, $9! < 2.7^{13} < e^{13}$ , then the LHS would be bounded by

$100 ( \ln 100 - 1 ) - 10 ( \ln 10 - 1 ) + 13 < 190 \ln 10 -77$

And since $190 \ln 10 - 77 < 157 \ln 10$ , we can conclude that $100! < 10^{159 }$ (remember to add back the 100 that we factored out initially).

To tighten this further using this method, we have to hunt down much more terms.

Calvin Lin Staff - 5 years, 10 months ago

Yes, we need more information for tighter bounds. It's best left for a computer to solve this. Even knowing the inequality $9! < 2.7^{13} < e^{13}$ looks very out of place.

A better follow up question would probably be:

i) What is the largest integer $n$ such that $n! < 10^{162}$ ?

ii) What is the smallest integer $n$ such that $n! > 10^{162}$ ?

iii) What is the first 6 (leftmost) digits of $(100!)$ ?

iii) After removing the trailing zeroes, what is the last 6 digits of $(100!)$ ?

Pi Han Goh - 5 years, 10 months ago

My way js - sknce 100!= 100x99x98x97x96x95x94x93...3x2x1x0! & 10^162=100^81 which means 100x100x100x100x100x100x100.... Clearly, 100>99 or 98 or97 , 100^162> 100! .

Yajur Phullera - 5 years, 10 months ago

Nope. It doesn't work. You missed out on a lot of crucial steps. What you have left is:

$100^{80} \qquad ? ? \qquad 99!$

That's all.

Pi Han Goh - 5 years, 10 months ago

Yeah thanks i realised my mistake there. 😊😊

Yajur Phullera - 5 years, 10 months ago

But what about the other 19 numbers that you didn't get to compare against?

I agree that $100 ! < 100 ^ {100} = 10^{200}$ . But it's not immediately clear what to do with the rest.

Calvin Lin Staff - 5 years, 10 months ago

Got my mistake thanks .

Yajur Phullera - 5 years, 10 months ago

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