Which is bigger???

Which is larger? 2^(100!) or (2^100)! - a really fun problem that you don't require a calculator to solve!
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[@BunnyKhatri-pd8zm]

2^100! be like: Nah I'd win

[@GrifGrey]

which one 2^(100!) or (2^100)!

[@mt180extras]

@@GrifGrey that's probably the joke

[@JPiMaths]

😂😂😂

[@barbo866]

"Throughout exclamation marks and the power of 100, I alone am the bigger one" (sorry if it doesn't make sense, I have never studied math in english)

[@marcelino69420]

as (2)^100! started opening his domain J pi Maths asked "are you bigger than 2^100! because you are a factorial, or are you a factorial because you are bigger than 2^100!" Then (2)^100! calmly responded "Nah i'd grow"

[@shehannanayakkara4162]

If anyone is curious, (2^100)! has about 10^31 digits, whereas 2^(100!) has about 10^157 digits.

[@svxtn]

how did you get these numbers? they seem far too small.

[@shehannanayakkara4162]

@@svxtn 10^31 and 10^157 are not the actual numbers, but the number of digits that the actual numbers have. The larger number has 10^157 digits, there are less atoms in the known universe meaning that we could not write down this number even if we used atomic-sized font in a universe-sized book. And if we can't even write down the number, imagine how incomprehensibly huge the actual number is. In comparison, a billion is a huge number, but the number is very easy to write down (1,000,000,000).

[@alexanderdaum8053]

​@@svxtnYou can get the number of digits by computing the logarithm of the expressions. Using the logarithmic identities, it is possible to calculate the numbers on a normal computer. If you're interested, here are my solutions (using log base 10): log(2^(100!)) = 100! * log(2) ~ 2.8*10^157 For the left hand side, we also need an approximation for the factorial. n! ~ sqrt(2πn) * (n/e)^n. For this example, I'm just going to drop the sqrt() term, as it doesn't make a big difference with big numbers. log((2^100)!) ~ log((2^100/e)^(2^100)) = 2^100 * log(2^100/e) = 2^100 * (100 * log(2) - log(e)) ~ 3.8 * 10^31

[@adrianfireblaze9556]

Actually it is 10^1571 digits and 10^311 digits, so the numbers are even larger@@shehannanayakkara4162

[@blazoraptor3392]

​@shehannanayakkara4162 so in other words the real number is n×10^(10^157)?

[@justintroyka8855]

Ah, too bad, I got it wrong because I figured the one with the bigger exclamation point would be the bigger number.

[@JPiMaths]

I must admit it's not super clear which one should be bigger (or at least that was the case when I looked at it first). At one point I was convinced it was the other way around

[@ChepparaBabai]

​@@JPiMaths I just did the same with 3 instead of 100 and found the answer. Does that work?? 😅 (for context I'm a biology major)

[@sushimrexx]

​@@ChepparaBabai You dont get the same result tho. 2^(3!) is less than (2^3)!

[@ianweckhorst3200]

Big number calculator?

[@ianweckhorst3200]

I hereby retract that question

[@cbarnes2160]

I got faked out by the exact property you mentioned at the end, namely that factorials grow way faster than exponents. So I figured that the RHS would win. But the factorial in the exponent on the LHS essentially gets even more power because it's in the exponent. So it wins!

[@JPiMaths]

I thought the same too!!

[@CorrectHorseBatteryStaple472]

I thought the same as well. I sort of remember learning f(g(x)) > g(f(x)) for large x, if f grows faster than g, but maybe that's just not true...

[@rianantony]

I suppose its like having your slowest for loop inside of another for loop, In a sense

[@bestopinion9257]

Factorials have a rate of growth bigger than exponents. But these are not quite factorials and exponents, instead a combination of both.

[@qeuhvm2552]

⁠@@bestopinion9257Exponent has a rate of growth faster than Factorial. The LHS is doing factorial first and exponent later. That’s why it’s larger than RHS

[@ChrisMMaster0]

I really thought the right one was gonna be bigger since you are taking 2^100 (MASSIVE NUMBER) and applying the factorial compared to 2^(100!) (Not As Big Number) and factorial grows faster than exponentials.

[@Alex-mh4lt]

Exact same

[@JPiMaths]

It seems from the comments that most had the same idea!

[@offshack]

I started small and just compared (2^5)! vs 2^(5!) and got: 2^5 is 32, 32! is 2x10^35, meanwhile 2^(5!) is 2^120, that turns out to be 1.3x10^36 ----- so - the form of 2^(N!) was already beating (2^N)! for small N, just figured the pattern would hold through 100 if it held through 5. They were still surprisingly close there though (only one magnitude of order out!) -- BUT - By N=6 the 2^(N!) grows like crazy (2^720 is 5*10^216, meanwhile, (2^6)! is 64! which is only 1.2*10^89 -- my calculator can't even handle 2^(7!) = (2^5040) but 2^7 is only 128 - and 128! is STILL smaller than (2^720) from the last step!

[@Xnoob545]

Factorial grows slower than exponentials though?

[@condorianonegdiffsgoku]

Except 100! is far greater than 2^100. Factorial grows faster doesn't apply.

[@red1bk190]

I love the way of explaing, you didn't leave out any details at all and guided the viewer through your thought process very well

[@toricon8070]

My first guess was that the bigger factorial was bigger, but I couldn't figure out a justification quickly enough before I got bored and tried a different tactic. For a second guess, I approximated n! as n^n, which is easier to work with, and reduced the problem to 100^100 vs 100*(2^100), which was easy enough. The lesson I'm taking from this is that even rough approximations are better than gut feelings.

[@JPiMaths]

Nice!!

[@asandax6]

My gut feeling told me the one with the bigger exponent is bigger and I was right.

[@MrLogistician]

My gut said the RHS was bigger, then i replaced 100 with 4 and LHS was bigger. I stuck with my choice like a shitty mathematician because i figured 4 was too small for the fast growth of factorial to catch up. I was too lazy to try to work it out on paper and i was wrong lol.

[@stickman_lore_official6928]

2^((100^100)!) Is over 2^^5 digits

[@kNightAkuma]

Two to the one hundred factorial.

[@JPiMaths]

lol nice 😂

[@robertveith6383]

What you wrote is ambiguous. You need to write it differently.

[@yeager593]

​@@robertveith6383 that's the joke

[@akshatpratapsingh5476]

the intuitive explanation was really beautiful

[@eniferxamm]

Very fun video, setting the bases equal and then comparing the exponents was a way to make this comparison very accessible to confused precalc students like myself!

[@JPiMaths]

Thank you - I think it's also the nicest way of comparing them!

[@ericsiegel1087]

interesting video. idk if this is possible but it may help if you use a more visible cursor, at times it can get a bit confusing what youre referring to when you say "this term" and "that term". hopefully thats helpful feedback to hear

[@JPiMaths]

Thanks for the feedback! This is a good shout - will see if I can make my cursor more visible!

[@aaasm4460]

Perhaps just mark one "a" and other "b" and don't use "this" at all..

[@Nartymer]

Maybe you could've used arrows rather than underlining, and then just delete them with CTRL+Z. At least that's what I usually do when explaining mathematical proofs in paint haha.@@JPiMaths

[@mapi5032]

Lol yeah, start the video at 2:08... This is bigger than this, thus this is bigger than that thus this is bigger than this showing that this is bigger than that. done. :)

[@tcoren1]

Cool! I just eyeballed it with the sterling approximation but getting a proper proof with that method is probably harder than what you did. Kudos!

[@frenchfries8164]

nice video, you really made it easy for me to understand by breaking down everything. Keep up the work, I wish you reach 10K subscribers soon!👍

[@usyan]

I had seen this in reddit and tried to solve but cannot. This is a brilliant solution. Thank you.

[@Renville80]

I had to think it through in general terms (particularly the part that factorial numbers grow more quickly than exponents) and was pleased to see that I chose correctly.

[@phoenix-cw8vf]

i think a more direct, although equivalent, approach to your last step would be to use the fact that the ratio between the two terms must be greater than one. then you simply divide each factor of each term respectively, notice that the only quotient which is less than 1 is 1/4 (the rightmost factor in your work), then conclude that the product of the other quotients are clearly greater than 4 thus the product is greater than 1, therefore 99! > 2^100.

[@JPiMaths]

Nice, although I don't see how this is different to what I did?

[@alien3200]

Thanks. This will improve my maths skills 😃❤

[@MeixingYip]

Great explanation! Liked!

[@kennethgee2004]

so i thought that this question has already been answered. if the numbers in question are > e than the one with the larger exponent is the winner. if both are e and the exponents are greater than 1. Maybe I misunderstood other videos that cover this type of evaluation.

[@kellymoses8566]

I guessed the left one because of my intuition on how fast exponentiation grows.

[@kujmous]

Well done. Please put more contrast on your on-screen pointer.

[@paolo_mrtt]

Awesome video thanks!!

[@blessedfragger5988]

Way better explanation of induction than I received through a whole CS education

[@minidreschi2]

it is true if the power is bigger than 4.974, but if lower (so 2^4.973), than the right value/formula is the bigger

[@Demonz2000]

"the numbers are far too big for calculators" Wolfram alpha: nah, I'd compute

[@Murk_Matter]

I remember using that for discord

[@Macieks300]

It's pretty easy to do this using Stirling's approximation.

[@xelspeth]

Woo, thought about this before actually clicking on the video and that was my exact thought process 😊

[@kepler_22b83]

Had you asked me to solve for this without external help, I wouldn't know where to start... But this is a simple and fast way to get the answer that may be extrapolated to many other dumb questions and other not so dumb ones... So thank you!

[@sagittarius5466]

Ah, did not expect that!

[@gheffz]

Wow, what a great question! Both sides are extremely huge.

[@tngdwn8350]

I saw the thumbnail and wanted to give it a try, I'm really happy I've got the right solution with exactly the same path you took :)

[@river559]

I haven't tried math like this since graduating highschool in 2019, but surprisingly enough I got it right. Spent 3 minutes looking at the thumbnail doing rough calculations in my head, and my only justification was that doing exponents and then the factorial (right hand side) would be way smaller than factoring the exponent (left hand side). If the exponent is just 2x2x2x2..., then making that sequence longer just made more sense to me. Slapping the factorial in on the exponent on the left side means the exponent would be exponentially larger (pun intended lol) than doing it on the sum of 2^100. Fun mind exercise. Hadn't used math like this in years

[@e271828r]

what is the writing software being used? and what is the hardware.. the handwriting is very good

[@shivamkumbhar1266]

Taking Log of both numbers makes it easier to compare

[@jazzabighits4473]

I'm glad I got this one before the video started. It reminded me of the video that explains which is larger between TREE(g64) and g(TREE3).

[@3141minecraft]

I am pretty sure that TREE(4)>g(TREE(3))

[@jimmyh2137]

@@3141minecraft it is

[@eliteteamkiller319]

I love that I have an undefeated streak with these KZbin videos just on intuition. No idea how I know, but I've always gotten it right. Inb4 someone posts one impossible to guess by the knowledge that factorials grow faster than exponential functions. EDIT: why in this case. It felt like to me that 2^100 was a micro-machine compared to 100!. And it seemed the factorial of a relatively tiny number was going to be dwarfed by the exponent of an absolutely astonishingly huge number.

[@TheOmegaOof]

for the record, the TI-nspire CX II can compute 2^100. for the former number it returns infinity due to overflow, but for the latter it returns 2^100 (which is 1,267,650,600,228,229,401,496,703,205,376) and adds the factorial symbol after it "1,267,650,600,228,229,401,496,703,205,376!"

[@JPiMaths]

Ah I didn't know this, that's pretty cool! I think most calculators can't go beyond 69! As thereafter n!>10^100

[@sourabhsinghal253]

Bro I also thought same way even I cal that 29!>2^100 but in this case 100! Is 9.3326215443E157 and 2^100 is 1.267…E30 it's a humongous difference between power and N! N number so I done some research and i found that 2^100! Is much greater than (2^100)! I share my research on the comment section in this video if u have time then check out.

[@mrseriousv1]

yeeeee I got it right! I had significantly different logic but a win is a win

[@seaajayy]

which has a bigger impact, multiplying a number by three or putting it ot the power of three

[@rayy3923]

its obviously the second option since it has the brackets and the first one doesn't.

[@jmas3967]

Induction is also good approach

[@boi_howdy]

great video! as others already said at smaller values the RHS is bigger, i wonder at what value 2^(x!)=(2^x)!

[@Padtedesco]

A good challenge on that topic. What is the highest number possible to write using only a single symbol, 2 and 3. Tetration and pentation allowed.

[@endlessduck1642]

instead of using "this" and "that" to refer to the terms, just name the terms instead. its easier to follow which is bigger than what. otherwise good video thanks for this insight:D

[@leekyonion]

2^100 is a constant, which is the smallest order of magnitude. Since 2^(100!) is working with exponents - they ramp up much more quicky than a constant value. It stands to reason that an exponent being multiplied has a greater value compared to if a constant value is being multiplied.

[@Idk_imagine_a_cool_name]

I was searching for this. I had the exact same way of thinking. Everyone else seemed to think about this in a strange way

[@rel2tx]

For me, easiest way to look at that, is that the left side is two to the power of a multiplication of all the number between 1 to 100, and the right side is two to the power of the sum of 1 to 100

[@malikil]

My initial 5-second thought was that it would be the other way around. I know factorial grows *significantly* faster than exponents, so when I observe I see 100 terms in the factorial on the left, and way more than that in the factorial on the right. I will then ignore the power with base 2 because it grows so much slower than factorial. Coming to the conclusion that because there are more terms in the right factorial the right side must be larger.

[@rssl5500]

Really nice problem !

[@JPiMaths]

Why thank you!

[@DanielRisberg]

I tried punching it in on my TI82 and it caught fire.

[@ric6611]

My intuition told me it was down to whether 100! was bigger than 2^100 (similar to how you do it vs 99! vs 2^100), and it obviously was. But then I thought it wasn't that simple, like many others I thought "well but the one on the right is a really big factorial" and changed my mind...

[@jeffreykalb9752]

Or you can use Stirling's approximation...

[@DanBurgaud]

I love how you worded it.. HAHAHAHA! NICE!

[@jatinnarde2131]

lol I invoked the sterling black magic twice

[@benrex7775]

My reasoning was this: The exponent scales faster than the base. So the factorial in the exponent causes the larger number than the factorial in the base.

[@nooneatall5612]

My intuition was very confident that the second one is bigger, once again, a testament as to why we shouldn't trust it

[@SequinBrain]

my calculator answered the inequality. it doesn't need to have the actual listed values.

[@user-co9zx8ur9h]

Taking ln(ln(y)) for both expressions, and using the Stirling approximation, for the one on the left we get approx. 360, but for the one on the right, we get approx. 74 Interestingly enough, the sequence 2^(n!) initially grows more slowly than (2^n)!

[@laeman7774]

Two to the one hundred! its stuck in my brain now...

[@sparshsharma5270]

Can you tell by which point or by what value of n does 2^(n!) becomes bigger than (2^n)! cause for smaller values of n (greater than 1), the latter is clearly bigger number? At n=0, 2^(n!) is bigger but both are equal at n=1 and the latter becomes bigger as value of n increases. But by what point does 2^(n!) number tend go become bigger than (2^n)!.

[@JPiMaths]

I think you can just test, and it works for n=5

[@gametimewitharyan6665]

This stumpted me too, I thought the RHS would be larger

[@arericarnau4773]

very educational tbh

[@glauberk9018]

Another way: log base 2 in both sides, 100! > sum from 1 to 100 = 5050

[@JPiMaths]

Nice idea but this doesn't quite work - on the right hand side you've forgotten about the terms (2^100-1)(2^100-2) etc meaning the RHS is much bigger than 5050

[@glauberk9018]

Oh yes, you are absolutely right! My mistake

[@ratpackenterprises1607]

You could still take a log of both sides, which gives you a large summation for ln((2^100)!), and then think of that summation as a Reimann sum of lnx. You can show that ln((2^100)!) < integral of lnx from 1 to 2^100 + 1 < ln(2^100!), which is what we want.

[@spencer1880]

My calculator for anything approaching complex has been wolfram alpha for a very long time now.

[@the_peacemaker002]

This question really messed up my intuition. If we have the composition of two functions, and you want to maximize the growth rate, usually you apply the slower function, and then the fast growing function. For example, if you have 2x and x^2, then doing (2x)^2 is better than 2(x^2) In this scenario, we have 2^x and x!, but (2^x)! is apparently worse than 2^(x!) So, what weirdness is going on here!? Any insight would be helpful.

[@leonardkarsten5005]

I had the exact same thought process. Maybe this doesn't hold in general. But then it would be interesting if a similar statement is true in general for function composition

[@whamer100]

my immediate thought was left side because factorials grow faster than exponentials, but its a factorial as an exponential looks like my hunch was right

[@jrbowler84]

Ah this is bigger than that, because this is bigger than this, and this is this that that than that. Helpful

[@IceCream_Racing]

what if its 15, replace the original question with 15 instead of 2, would it be any different?

[@JosefZeethuven]

instinctually taking the factorial of an exponent vs the sum, the exponent should be bigger

[@djsmeguk]

Just make the first term 8. 99 > 8. 98 > 2. Rinse repeat. Add a x1 at the end of the 2s so you have 1=1. Every term in the top is bigger or equal.

[@JPiMaths]

Nice!!

[@HobinderSinghA]

could also just take log

[@den_kayator]

You also can do it by taking logarithm

[@Lorenzo_der_Ritter]

When you substitute 100 by other numbers, you will always get the same result - unless it's 4 or lower. That's because 4! (=24) < 2^5 (=32) but 5! (=120) > 2^6 (=64). I got fooled by calculating the result substituting the 100 with 3 and then obviously the right side is bigger. So I wrongly concluded that the same would go for 100 and tried to prove a wrong statement.

[@user-qd5fn5ch2g]

Having watched your wonderful video, I can confidently say that if anyone ever asks me which one of these is bigger, I will not be able to prove it.

[@Sci-Marvels]

Wow 😮... Very nice 💯

[@SidereusNox]

Great video in general, but i got very confused around 2:10 when you just said "this" 50 times in 5 seconds, and I couldn't even really follow the cursor due to the speed, small size and bad contrast.

[@mattvalcarc]

I didn't understand that at all! (I'm not into maths) But thanks for sharing 😊

[@CtrlAltSpray]

my logic was: for values that we care about, if x! > y! x^x > y^y always so i could "replace" 2^(100!) with 2^(100^100), and i could replace (2^100)! with (2^100)^(2^100) which would equal 2^(100 * 2^100) in comparing 2^(100^100) and 2^(100 * 2^100), you can simplify it to 100^100 vs 100 * (2^100) and (though not rigorous) i just decided that this would be adequate. 100^100 is bigger, which means 2^(100^100) is bigger, which means 2^(100!) is bigger

[@joramarentved]

That's true, Bunny!

[@jaapfrietzak3129]

You can also take out 8, instead of 4, so that the pesky 1 doesn't matter

[@T-99K_GMD]

Very easy ;) 2^100! bigger than (2^100)!

[@MarkusAldawn]

I had a slightly different approach near the end: when we got down to 2¹⁰⁰ vs 99!, it became a lot easier to work out since 99! necessarily has 49 instances of numbers which are multiples of two, and then half of those are instances with 2•2, half of _those_ are 2•2•2 and so on, which means you have 49+24+12+6+3+1=95 multiples of 2 in 99! Only by counting how many multiples of 2 there are in 99!, we can show that by using only half its numbers, 99! has reduced 2¹⁰⁰ down to 2⁵, 32. Since you haven't used 33 yet, 99! is bigger. I know it's a woollier way of thinking about factorials vs exponents, but it helps me grasp how the relative sizes of things matter. For example, since 99! is made up of 2⁹⁵ and 33 multiples of 3 (and 11 of 9, 3 of 27) and 19 of 5 (3 of 25) and 14 of 7 (2 of 49) and so on, we essentially know that 99! is made up of 2⁹⁵•3⁴⁷•5²²•7¹⁶•11⁹•... And thus by counting them it's immediately obvious to me that 99! cannot be smaller than for example 2²⁰⁰ without checking (each 3² is at least 2³, so 3⁴⁷ is at least 2⁷², each 5¹ is at least 2², so 5²² is at least 2⁴⁴, and we're already at 2²⁰⁰). Breaking the number up into prime factors means they're manipulable as blocks, rather than staring at 99! and trying mentally to peel off 99, and then 98, and then and then.

[@Mike_Rottchburns]

I guessed it wrong. This is interesting

[@Iceman-oq7ok]

take log on both side and compare

[@yonataneavri8298]

I liked the part where he said: “this is bigger than that”

[@teslagamerz]

Seems like i got it right 🚁 , i just figured second one will be 2^5050 and 1st one has 100! so obviously bigger

[@arcanine_enjoyer]

Knowing how quickly exponents rise, I was right in assuming the left. Because it's 2 multiplying itself 100! times, while the other side only goes up to 2¹⁰⁰ then it multiplies by itself with 1 less the other until 3 × 2 × 1. Maybe I am just biased because I'm left-handed though

[@vincentjiang6358]

You are very very underated

[@JPiMaths]

Very kind, thank you!

[@PoRRasturvaT]

You'd need something around your cursor, or work on improving contrast/visibility/size. Because all I hear is this bigger than this and that and that... And can never see where you're pointing, making it hard to follow. This is especially true when looking at low resolution version of your video and/or smaller screen sizes. Thanks.

[@bigfatpandalaktana2747]

Nice length

[@m-w-x8859]

Around 2:10 you start saying "this" and "that" too much and it's hard to keep up with which value you are referring to at each instance

[@ghimiraysahaj278]

all i learned is this is bigger than this because this is this

[@skyethebi]

Ima be honest I started to make this way more complicated than necessary by using the gamma function followed by a u sub and the definition of an integral as the limit of a Riemann sum and then some more manipulations to give the left hand side being equal to: *lim n->\infin [(\prod k=0,n {(k/n)^(1/n)})^ln(2)]*

[@divadbyzero2793]

The reason why 2^(100!) is larger is because in that expression the exponents multiply, whereas in (2^100)!, the bases multiply, which makes the exponents add. The multiplication in the exponents is strong enough to overcome the factorials. The reason why the factorial function doesn't win over the exponential function in this scenario, despite factorials growing faster in general, is because these functions aren't placed on equal footing. The exponential function gains factors more quickly than the factorial function. When the number in the factorial increases by 1, the number of factors in the exponential function is raised to an ever-increasing power, whereas in the factorial function the number of factors is raised to the power of 2. Example: 2^(7!) has 6! factors of 2^6. Compared to 2^(6!) this is a factor of 2^6 (64 times) more factors of 2^6 than 2^(6!) But (2^7)! has only twice as many factors in its factorial compared to (2^6)!

[@Uyhn26]

My guess is the left one, because exponentials go blurrrrr….!!!

[@mr_flor]

fifty shades of "this" and "that"😔

[@imbicyl6667]

take log of both sides and the answer is obvious

[@iankrasnow5383]

Without doing the math, I assumed that 2^(100!) was bigger because that's the power set of the 100 factorial, while the other one is the factorial of the power set of 100. The power set is a much faster growing function than factorial, so it seemed logical that it should be bigger. Turns out that was correct. This isn't a proof or logical argument, just a heuristic from knowing a little about how recursion works.

[@godowskygodowsky1155]

Before I watch: The left is larger. Take log log and apply Sterling's approximation. You get something on the order of 100 log 100 - 100 = 200 log 10 - 100 = 300 on the left and 100 log 2 = 70 on the right.

[@nicolastorres147]

(2^100)^(2^100) is way bigger than (2^100)!. 99! is way bigger than 2^100, which implies (2^100)^(99!) is way way bigger than (2^100)^(2^100), which is way bigger than (2^100)!. Therefore, we can conclude that 2^(100!) which equals (2^100)^(99!) is way way way bigger than (2^100)! 🌫️

[@hk8450]

Why don't we choose a calculable number, take the factorial of the factorial or exponent of the number and compare it?

[@JPiMaths]

Interesting - see what happens when you swap 100 with a smaller number, say 3 or 4...