Which is bigger???
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- เผยแพร่เมื่อ 11 ก.ย. 2024
- 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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2^100! be like:
Nah I'd win
which one 2^(100!) or (2^100)!
@@GrifGrey that's probably the joke
😂😂😂
"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)
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"
If anyone is curious, (2^100)! has about 10^31 digits, whereas 2^(100!) has about 10^157 digits.
how did you get these numbers? they seem far too small.
@@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).
@@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
Actually it is 10^1571 digits and 10^311 digits, so the numbers are even larger@@shehannanayakkara4162
@shehannanayakkara4162 so in other words the real number is n×10^(10^157)?
Ah, too bad, I got it wrong because I figured the one with the bigger exclamation point would be the bigger number.
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
@@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)
@@ChepparaBabai You dont get the same result tho. 2^(3!) is less than (2^3)!
Big number calculator?
I hereby retract that question
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!
I thought the same too!!
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...
I suppose its like having your slowest for loop inside of another for loop, In a sense
Factorials have a rate of growth bigger than exponents. But these are not quite factorials and exponents, instead a combination of both.
@@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
"the numbers are far too big for calculators"
Wolfram alpha: nah, I'd compute
I remember using that for discord
I think it was "hold my beer" that it had said
Lol
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.
Nice!!
My gut feeling told me the one with the bigger exponent is bigger and I was right.
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.
2^((100^100)!) Is over 2^^5 digits
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.
Exact same
It seems from the comments that most had the same idea!
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!
Factorial grows slower than exponentials though?
Except 100! is far greater than 2^100. Factorial grows faster doesn't apply.
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
Thanks for the feedback! This is a good shout - will see if I can make my cursor more visible!
Perhaps just mark one "a" and other "b" and don't use "this" at all..
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
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. :)
3:46 Stop, he's already dead
😅😅😅😅😅.
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
Two to the one hundred factorial.
lol nice 😂
What you wrote is ambiguous. You need to write it differently.
@@robertveith6383 that's the joke
@@yeager593exactly
Your correctness is in superposition
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!
Thank you - I think it's also the nicest way of comparing them!
Me the whole time: *Struggling to understand what "this term" and "that term" meant* 💀
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
how about just look at the video and see what he's underlying u lazy fuck
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
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.
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.
Nice, although I don't see how this is different to what I did?
That's really smart. I initially thought the opposite but your reasoning was very clear and easy to follow👍
The first one is bigger. I just simplified the expression a lot. 2^(2*2) or (2^2)*2
Another way: log base 2 in both sides, 100! > sum from 1 to 100 = 5050
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
Oh yes, you are absolutely right! My mistake
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.
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.
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.
Which is bigger
2^(50^100) or (2^99)^(2^100)
Spoiler: the left is larger, they are approximations of the question posed in the video. Hopefully this is an easier way to visulize it
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
I don't really have a background in higher math to make an educated guess, so I made an uneducated one. I took out a small calculator and tried a few values for 2^n! and (2^n)! as a test. I can intuit that there are two possibilities for a series of such: either one will always be larger for any value of n, or one will start off as larger, the series will converge, and then the other will be larger.
My test calculations showed values for 2^n! of 2, 4, 64, 1.6e7, 1.3e36, >1.0e99 [overflow error]. For (2^n)! they were 2, 24, 4.0e4, 2.1e13, 2.6e35, 1.3e89. This shows a convergence at an n value between 4 and 5, after which the former series became increasingly larger.
There is no way this series could ever re-converge, therefore for an n value of 100 (or any value greater than 5) the former must be the larger.
I went with real basic: just know that in the long run, exponential is bigger than factorial, so i'd rather have 2^[big number] over [big number]!
Factorial actually grows faster. I believe it is the operator that grows fastest, even faster than exponentials
@@j-rey- yep, in general n! will eventually be greater than r^n no matter how big r is.
But also if we have n^n vs n! the first one will always be bigger. And iirc even for (n-c)^n vs n! for any constant c, no matter how big, the first function will eventually overtake the second in value, as n becomes bigger.
@@Aphanvahriusexponential function we mean n^x for some real n, not x^x which is a different kind of thing
By common sense/intuition:
2^(100!) > (2^100)!
Factorials are REALLY large numbers - so if you put a REALLY large number in the exponent, it will be REALLY large... whereas, if you have a number without an ma factorial in the exponent - just a regular whole number in the exponent, it won't be as large - and even taking the factorial of that number won't produce as large a number compared to putting the factorial in the exponent itself.
Before even watching this: Replacing 100 by 6 gives 2^6! = 2^720 and (2^6)! = 2^64. Since 2^720 > 2^64 that means 2^6! > and (2^6)! Replace the 6 by any number, x (including 100) will result in the same relationship where 2^x! > (2^x)!
Before watching the video, here's my idea:
Make an educated guess as to which one is smaller.
Find an upper bound for that supposedly smaller value which is easier to compare to the supposedly larger one.
Hopefully, we can show that the larger one is larger than the upper bound, which will also prove that it is larger than the smaller value.
This method worked out well for me; I chose (2^100)^(2^100) = 2^(100 × 2^100) as an upper bound for (2^100)!. You can get rid of the bases and compare 100! to 100 × 2^100, for which it is relatively easy to show that the factorial is (quite a lot) larger, and hence, 2^(100!) > (2^100)!. To be honest, the initial guess for which was smaller was less "educated" and more about which was easier to find an upper bound for. A factorial n! has an easily remembered upper bound of n^n. To see this, notice that you are comparing *n × (n-1) × (n-2) × ... × 2 × 1* to *n × n × n × ... × n × n.* See how each factor of the first is less than or equal to the corresponding factor in the second.
After finishing the video, it turns out I did it in pretty much the same way as presented in the video (yay!). For completeness, I compared 100! to 100 × 2^100 by first dividing both sides by 100, leaving 99! vs 2^100. To do this, I compared the fraction (2^100)/(99!) to 1. This fraction is less than one if and only if 2^100 < 99!.
Now, (2^100)/(99!) can be written as (2^4 × 2 × 2 × ... × 2 × 1 × 1)/(99 × 98 × 97 × ... × 3 × 2 × 1).
We can split this into a product of many fractions: 16/99 × 2/98 × 2/97 × ... × 2/3 × 1/2 × 1/1.
We can get rid of the 1/1 at the end since it is equal to 1 and we have that a × 1 = a for all integers a.
This leaves us with a product where every factor is strictly less than 1, so the product itself must also be strictly less than 1.
Hence, 99! > 2^100, so 100! > 100 × 2^100, which means that 2^(100!) > 2^(100 × 2^100), and then by using our initial upper bound condition that 2^(100 × 2^100) > (2^100)! , we can conclude that 2^(100!) > (2^100)!.
P.S. When I found ratio of the two numbers, I put the smaller one on top. You could equivalently put the larger one on top and make sure the fraction is *bigger* than 1 (if they're equal, the fraction will come out to exactly 1), but for whatever reason I put the smaller one on top.
I don't understand why you did that. The RHS can be simplified from a factorial function into an exponential:
2^(100x99x...x3x2) [?] 2^(100+99+...+2+1)
100! > Sum(1, 100)
Because n * (n-1) > n + (n-1) if n >= 0
Then 2^(100!) > 2^sum(1, 100)
Just ignore this. I got drunk, did a typo, and I can't understand what i wrote. But your method seems a little convoluted
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.
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
I'll explain (please stick with the explanation till the end):
Now, in general, the strength of the operations is factorials >> exponentials >> multiplication >> logs. This is why doing factorial last would usually create the biggest numbers (because it means that the factorial acts on a larger base so you're making maximum usage of the factorial). However, now let's look at the question. The second option (2^100)! is a factorial of a large basic number, but that's all it is: a factorial. The operations of exponent and factorial are being done one after each other.
The option 2^(100!) is actually combining the exponent and factorial operation into a single mega-powerful operation (let's name it z where z >> factorial >> exponent >> multiply >> log).
Hope this made sense
When you take the binary logarithm on both sides, you get 100! on the left side and the sum of the binary logarithm of all numbers of 1 to 2^100 on the right side, which is strictly less than 2^100 times the largest term 100. If you take the logarithm again, you get the sum of logarithms of 1 to 100 on the left side and 100*log(2) + log(100) for the upper bound of the right side. You can do similar splitting of terms to rearrange the left side into 100 terms that are all equal or larger than log(2).
I did some quick maths: (2^100)! would be 2^(100+99+98+….) = 2^5050. So I quickly reached the conclusion the left one is larger.
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.
Caution: you can easily get lost in "this" and "that"
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
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!"
Ah I didn't know this, that's pretty cool! I think most calculators can't go beyond 69! As thereafter n!>10^100
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.
Nice question. But I think using logarithms at start is a much cleaner way to reach the solution and avoid excessive writing.
Log both sides:
LHS: log(2^(100!)) = 100! * log(2)
RHS: log(2^100!)
Using the logarithm addition rule:
log(2^100!) = Summation of log(n) where n goes from 1 to 2^100
Since the highest n is 2^100, we can use it as an upper bound for each term:
RHS_upper bound = 2^100 * log(2^100) = 2^100 * 100 *log 2
So, we want to prove LHS > RHS_upper or LHS = k * RHS_upper where k > 1, which would mean it will be true for the actual RHS.
Cancel log(2) and 100 on each side, we have:
99 * 98 * 97 * ... * 1 = k * 2^100 = k * 8 * 2^97
Since 99 > 8 and each of the terms like 98, 97,...2 is term wise greater or equal to 2 in 2^97. We need k>1
Hence LHS > RHS
(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)! 🌫️
Obviously this is very late, but if anyone is still confused about the f(g(x)) > g(f(x)) rule for large x if f grows faster than g:
This only really works if f and g have a different growth type (logarithms, addition, multiplication, exponentation, teration etc.), if they have the same growth type it's not clear at all:
f(x) = x^3, g(x) = x^2, f(g(x)) = g(f(x)) = x^6.
Additionally, putting functions of lower growth type into f can actually make g(f(x)) bigger:
f(x) = 2x^3, g(x) = x^2, f(g(x)) = 2x^6 < 4x^6 = g(f(x)).
And finally the example which explains what's happening in the video:
f(x) = e^(x*log(x)), g(x) = e^x, f(g(x)) = e^(x+e^x)) = e^e^(x + log(x)) < e^e^(xlog(x)) = g(f(x)). (x! roughly grows like f and 2^x roughly grows like g)
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)!
I think guessing would be enough proof 💀
the intuitive explanation was really beautiful
if you try this on desmos, when "n" reaches 5 *2^n! will be higher than (2^n)!* but when "n" is lower than 5 *(2^n)! will be higher than 2^n!*
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.
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.
I was searching for this. I had the exact same way of thinking. Everyone else seemed to think about this in a strange way
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.
Too many instances of “this is bigger than this, and that is bigger than this”. It’s not as descriptive as, for example, The left is bigger than the right, and the bound is larger than the right
Before watching the video, I'm gonna assume (2^100)! is bigger than 2^(100!) because if you replace it with a much smaller number like 3, then (2^3)! = 8! = 40320, while 2^(3!) is just 2^6 which is 64
Edit: I didn't consider the fact that a factorial power would increase at a far faster rate than just a normal factorial
I love that I have an undefeated streak with these TH-cam 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.
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).
I am pretty sure that TREE(4)>g(TREE(3))
@@3141minecraft it is
0:45 It was at this moment I knew... I shouldn't have said RHS is bigger.
* picks wrong option based on vague assumptions *
* skips through the video to see the answer *
* refuses to elaborate further *
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!👍
I have a different opinion if I try it in GeoGebra. For 2^(100!), I represent it as f:y=2^(x!). Then, for (2^x)!, I represent it as g:y=(2^x)!.
When I compare their graph visually, f>g if x
Of course you are wrong. Function's graphs have the nasty habit to intersect. And they really intersect quite soon, way before 100.
@@bestopinion9257 I see, that is the bottom line.
Thank you!
@@ccmplayer87 f becomes larger than g around x = 5:
2^(5!) is 1.33e36 while (2^5)! is 2.63e35
@@the_peacemaker002 Interesting. If you search on this page, I found the switch around x=5 without any calculator.
the number where intersect happens is more like around at 4,974, i just played it in Desmos too, but i also divided the high numbers so i could see the intersect
a simple approximation for powers of 2 is that 2^(10n) is roughly equal to 1 followed by 3n zeros. 2^10 is approximately 1000 (exact quantity of 1024), 2^20 is approximately 1,000,000, etc. knowing this, 2^100 must be in the ballpark of 1 followed by 30 zeros. using the logic that exponential growth is faster than linear growth, you can reason that raising 2 to the power of a number with 30 zeros will be larger than taking the factorial of a number with only 6 zeros (2^100).
a further step to be certain would be to not compare the numbers themselves, but rather, their magnitudes. we can approximate the minimum power of 10 needed for the first number, and compare that to the maximum possible power of 10 for the second number.
first, you can set the ceiling of (2^100)! as 1 million raised to the power of 1 million. this would effectively be the number 1 followed by 6 million zeros. compared to 2 raised to the power of (1 followed by 30 zeros), we can reason that every 4th power of 2 increases the magnitude of our number by at least tenfold, since 2^4 = 16 > 10. 1 followed by 30 zeros, when divided by 4, gives 25 followed by 28 zeros. to reiterate, our final number for 2^(100!) would have a floor of (25 * 10^28) zeros. this is significantly more zeros than the ceiling of (2^100)! having 1 followed by 6 million zeros - in other words, (6 * 10^6) zeros.
Well done. Please put more contrast on your on-screen pointer.
I got the wrong answer because I did the left hand side wrong. Once I figured out the mistake was, it was obvious. The right hand side is just multiplying 2^100 terms. The left hand side goes much bigger than a basic tetration.
yeeeee I got it right!
I had significantly different logic but a win is a win
The trap of this question is that only when n≥5, 2^n! will be greater than (2^n)!. So if you replace 100 with 3 or 4, you will get the wrong result.
When i first looked at the thumbnail , without making any conclusion , i opened my "keep notes" app and did this :
2^(100!) ,(2^100)!
2^(100*99!) , (2^100)!
(2^100)^99^98^97^96... >> (2^100)(2^100-1)!
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)!.
I think you can just test, and it works for n=5
I used Stirling's approximation and compared ln(ln(LHS)) ln(ln(RHS)) and got the same result. It felt kind of counterintuitive, I expected the factorial outside the exponent to be much bigger.
Can you also solve it by setting the inequality and then dividing both sies by (2^100)!? This means that the left hand side will be equal to 2^(100! - (100 + 99 + 98 + 97 ...) which using the sum of arithmetic series simplifies down to:
2^(100! - 5050) > 1 which i think is proof enough.
Not quite, as the RHS doesn't just include 2^n for n=1,2,...,100 but it also contains 2^100-1, 2^100-2 etc
@@JPiMaths I seem to have missed that, still a very nice problem
It's pretty easy to do this using Stirling's approximation.
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!
Let's try it with a smaller number
2^2! = 2^2 => 4
(2^2)! => 4! => 24
2^3! => 2^6 => 64
(2^3)! => 8! => bigger than 64
From small sample sizes, we can easily say that right side is always greater, therefore (2^100)! Is greater than 2^100!
I tried reasoning from smaller numbers, but it seems that my logic must have tripped me up somewhere. Allow me to explain my reasoning, and someone please help me figure out where I got it wrong.
I tried to deal with the abstract question of whether 2^(x!) is always bigger than (2^x)!
For x=1:
2^(x!) = 2^(1!) = 2^1 = 2, while
(2^x)! = (2^1)! = 2! = 2.
Both sides are equal.
For x=2
2^(x!) = 2^(2!) = 2^2 = 4, while
(2^x)! = (2^2)! = 4! = 24
The right hand side is bigger.
For x=3,
2^(x!) = 2^(3!) = 2^6 = 64, while
(2^x)! = (2^3)! = 8!, and we're already bigger three factors in: 8(7)(6) = 56(6), which is clearly bigger than 64.
One again, the right hand side is bigger.
Once we got to x=4, I had to whip out the calculator:
2^(x!) = 2^(4!) = 2^24 = 16,777,216 while
(2^x)! = (2^4)! = 16! = 20,922,789,888,000
Once again, the right hand side is bigger.
From there, I assumed that the right hand side would always be bigger, and thus that 2^(100!) < (2^100)!
However, the video clearly shows that I was amiss. Can someone please clarify the flaw in my reasoning?
ADDENDUM: Thinking over the proof in the video some more, I think that I found out the flaw in my reasoning.
As x gets bigger, the expression 2^(x!) collects more factors of 2 than the expression (2^x)! collects integer factors, so the left-hand expression eventually overtakes the right-hand expression, despite the slower start. Am I on the right track here?
I got it right intuitively since 2^100 to the power of 100-1, then that to the power of 100-2 and so on is always going to be bigger than the factorial of 2^100 which is a multiplication of way smaller numbers.
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.
2^(100!) has roughly 100*lg(100) bits. But (2^(100))! has roughly 2^(100 + lg(100)) bits. So clearly, (2^(100))! is much much larger.
i firgured out the answer at 0:56
See,
2^100! = (2^100)^99! = a
and
(2^100)! = b
Comparing a and b
lets put x = 2^100
a = (x)^99!
b = (x)!
Now we can very easily compare a and b..
Hence a it much bigger..
i.e. 2^100! is bigger...
i found out that the powers of 2 in 99! will be 95 and it will have powers of 3 and others as well. while the other one that is (2^100) * (2^100 - 1) * ... * 2 * 1 is 2^100 items which although is having 5 more times 2s power than the first exp i.e. 2^100 * 2^100 * ... * 2^100, 99! times, is still lesser because it will be having 2^100 by far larger times.
Maybe I'm missing something here, but I did this at a smaller scale, and it doesn't check out. 2^(3!) < (2^3)! and 2^(4!) < (2^4)! and so on. I graphed it too, and it looks like the right hand side grows considerably faster than the left hand side. How can this be explained?
I actually just read a few more comments, and saw that RHS is bigger but only to a certain point, I just didn't go far enough.
2^100! - 100! factors
(2^100)! - only 2^100 factors, 2^100! > (2^100)!
More precisely, (2^100)!
I guessed the left one because of my intuition on how fast exponentiation grows.
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)!
great video! as others already said at smaller values the RHS is bigger, i wonder at what value 2^(x!)=(2^x)!
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
Without a proof it should be quite obvious which is bigger since the magnitude of the difference is so big
Feels like the math version of Gojo vs Sukuna, both too powerful/big and I can't decide who'd win at first glance.
Spoiler -- handheld calculators might not be able to deal with numbers this big, but Wolfram Alpha can.
It's trivial to see that until 2⁴, the expression 2⁴! Is bigger. From 2⁵!, it's smaller (10³⁵ vs 10³⁶). Since these are continuous functions, this remains like this. So, 2¹⁰⁰! is the smaller number by far.
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.
There's a possibly slightly cleaner way to arrange the final step, by comparing sequences of the same size:
98! = 98 * 97 * ... * 3 * 2 * 1 (product of 98 terms)
2^97 * 1 = 2 * 2 * ... * 2 * 2 * 1 (product of 98 terms)
So trivially 98! >= 2^97 * 1 since each individual term is >=, therefore 99 * 98! > 8 * 2^97 * 1 since 99 > 8, therefore 99! > 2^100.
What do you think?
Way better explanation of induction than I received through a whole CS education
I solved this using Monotonicity and the steepness of the curve
I just did this
2^100*99…=(2^100)^99…
So it’s that giant power tower compared to the factorial of a number
That's an interesting way to think about it
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
My intuition was very confident that the second one is bigger, once again, a testament as to why we shouldn't trust it
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.
Nice!!
2^(100!) wins.
Just by metro calculations (currently in the subway) I can see a ton of huge huge factors.
There is no way (2^100)! gets there when the hugest factor is 2^100.
To prove it...
(2^100)!2^(2^107)>(2^100)^(2^100)>(2^100)!
Edit: I'm home now, I can watch the video.
I like how by working on both sides so that the base is 2^100 you can get to the right inequality faster.
Ah, my "trick" of replacing 100 with 3 (getting 2^6 = 62, and (8! = 40320), and assuming it works for all sufficiently large N, works again. My intuition is strangely usually wrong when it comes to exponentials.
(2^3)! = 2^3 x 2^2 = 2^(3+2) = 2^5
2^(3!) = 2^(3x2) = 2^6
2^6 > 2^5 ; 2^(3!) > (2^3)!
2^(100!) > (2^100)!
i used a calulator, and 2^(100!) is bigger
Actually (2^x)! > 2^x! for x up to around 4.97399, then 2^x! begins to be bigger indefinitely
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.
Before going into the video, I would guess the one on the left of the thumbnail.
thoughts before watching the video:
It's just a matter of which function grows faster. Factorial grows faster so I assume it is the exponent inside the factorial.
After reading some comments I realize the flaw in my thinking. If factorial grows faster then it should be placed in the exponent to create the largest number.
Ah, did not expect that!