Can you solve this FACTORIAL problem?

[@mikemossey]

I did competition math in high school 40 years ago. Never got to Olympiad level, though. I can follow these proofs but I have no idea where a mathematician comes up with them. It would be nice if you motivated it showed some of your thought process to come up with it, so we can learn something about mathematical intuition. I had a teacher one time (who had a PhD in applied mathematics) say that math proofs are neat and orderly and depend on exact logic, but the real creativity is in coming up with the proof, which can depend on all sorts of "irrational" insights - hunches, guesses, intuition you can't put into words, and so on. He said that his PhD got back on track when he had some kind of vision of light rays that represented metaphorically the logic he needed for a final proof.

[@Hex2a]

You can also compare the natural log of each side, using the fact that log(n!) = n * log(n) - n + very-small-term Then using just the first term for log(n!): n * log(n), and disregarding constant multiplier log(2): log(LHS) = 100! log(RHS) = 100 * 2^100 Comparing between terms in each product, similarly to what you did, you get the result: log(LHS) > log(RHS)

[@lfoevuf340]

6:00 Obviously, we can get away with the 1 at the end. Now, we can write 8 as 2x2x2, which gives us the same number of terms in both expressions (one hundred). The three 2s are common to both of them and all of the remaining terms of 99! are bigger then 2 => 99! > 2^100.

[@italixgaming915]

My solution: ln[2^(100!)]=100!.ln(2). ln[(2^100)!]=ln(1)+ln(2)+...ln(2^100). But since the function ln is strictly rising then ln(x)

[@jadbridge]

Your proof is incorrect at one point because you do not account for the fact that 99! has only 99 terms but 2^100 has 100 terms. This flaw is easily fixed by combining three 2s into an eight in the comparison. This also eliminates the last term which is the other way around, avoiding any hand waving. Of course the result is correct.

[@balkeebalakrishnan493]

Brilliant catch! I didn’t get that but once you said it, I was thinking write the last 7 terms 2.2.2.2.2.2.2 at the bottom as 4.4.2.2.2.1 to compare with 6.5.4.3.2.1. Now both rows have 99 terms and EACH term at the top is >= corresponding term at the bottom.

[@ReasonableForseeability]

@@balkeebalakrishnan493 Yes! 2^7 < 6! since 128 < 720. We're left 93 *factors* each: 2x2x...x2 < 99x98x...x7

[@ReasonableForseeability]

2x2x2 vs 2x1 is the wrong way round. Try 7 factors: 2x2x2x2x2x2x2 vs 6x5x4x3x2x1 i.e. 128 vs 720 Also it's *factors* and not *terms*

[@PhilippeVerdy]

Never compare a series of multiplications or additions without looking at the number of terms, otherwise comparing term by term by just looking at the first few and the last few is ALWAYS flawed. So the video shows an incorrect proof. And the conclusion is flawed too. It demonstrates nothing. Instead of comparing the two numbers, I wuold have just compared their log2() here, because all what is in these huge integers numbers is a power of 2. So you had just to compare the base-2 exponents. Which is the same as seeing if the ratio of the two (strictly positive) numbers is larger or lower than 1, or seeing if the difference of their log2() is positive or negative. Then I would have used a recursive series definition of the remmining terms composing the exponents, to find that only the first few terms in each series had the result sign inverted, then computed them to set a lower bound, then how may terms just after were positive ot counter balance the negative terms, and demonstrating that all the remainging terms are positive. We would have then made a much more general solution about (2^n)! compared to 2^(n!) for any n above the demonstrated minimum threshold. (This is the general way to be used as well when comparing huge powers or factorials that may "look" similar on first thought, for example 50^50 and 49^51: explicit the series) And this is a result that should be know to all theoricians of computing complexity: O((b^n)!) > O(b^(n!)) for *any* constant base "b" when "n" is large enough (i.e. when "n" is higher than a finite constant "N(b)" that depends only on "b".

[@BalaSribhaskarImmidi]

Here we have 100 2"s and 99! , see 2*2*2*2*2*2*2* so on 100 times can be written as 4 *2*2*2*2*2*299 times which contains a four and 98 no of 2"s now you can compare easily Infact I too got the same doubt but I tried to explain by myself Hope you understood

[@knighttakes-817]

( 2^{100!} ):This represents the number 2 raised to the power of 100 factorial ((100!)).( (2^{100})! ):This represents the factorial of the number (2^{100}).Comparison:( 100! ) (100 factorial) is an extremely large number.( 2^{100} ) is also large, but much smaller than (100!). The factorial of (2^{100}) is then the product of all integers from 1 to (2^{100}).Given that factorial growth is much faster than exponential growth, ( (2^{100})! ) will be vastly larger than ( 2^{100!} ).Conclusion:( (2^{100})! ) is much bigger than ( 2^{100!} ).

[@huskypup3489]

For small values, like 3 and 4, instead of 100, the RHS is bigger. For numbers larger than 5, the LHS is bigger.

[@berateyn]

What is rhs and lhs?

[@huskypup3489]

@@berateyn right hand side and left hand side

[@portakal6929]

The videos just making me better and have fun of it.Keep it up i like ur videos and this type of question came when i was at school tomorrow im gonna give answer with explanation ty one of the best math channel subbed 😊

[@gnanadesikansenthilnathan6750]

Simple but needs little thinking n understanding

[@RealQinnMalloryu4]

2^10^10 2^10^10 2^2^5^2^5 2^2^52^5 1^1^12^1 2^1 (x ➖ 2x+1). 1^1^12^1 2^1 (x ➖ 2x+1) .

[@kosterix123]

I’d guess that RHS would be bigger but seems I was wrong.

[@GODCHESS.]

Uhhhh 😮 guys im in sixth and my brain just reprogrammed after watching this 🎥 😅😅😅😅 This is gonna eat my mind in high 🏫. Wow tbh it was so complicated that it was going a bouncer over my head😅

[@rabotaakk-nw9nm]

2^(100!)=2^(1•2•3•…•99•100)= =2^(100•2•3•…•98•99)= =(2^100)^(2^(3•4•5•…•98•99)> >(2^100)^(2^100)>(1•2•3•…•(2^100)= =(2^100)! => 2^(100!)>(2^100)! 😁

[@EUGEN093]

Is it possible to draw a graph and check what grows faster?

[@jamesbond_007]

Yeah, go see wolframalpha and ask it to plot the two expressions. I don't know the exact syntax, but it often accepts natural language formulations of expressions. Try this to see how it's done "graph (2^n)! vs 2^(n!) for n in 1 to 5"

[@BN-hy1nd]

Brill as usual

[@joso5554]

This proof is elegant but it looks not works because you already know the answer, or have a hint of it. If the answer was the opposite, the 1st part would have been totally useless !!! So it’s a dishonest trick to not first explain how you hint at the result in the first place.

[@ACE-wg8hq]

true

[@ReasonableForseeability]

Not dishonest, just badly phrased, maybe. "Hi guys, I claim that RHS

[@du6ttfe]

Так это недавно на канале "Эльмир math" было