Every Math Student Should Know This

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Unlock the secrets of advanced mathematics comparing 1,000^999 to 999^1,000. Ideal for math Olympiad aspirants and enthusiasts, this tutorial delves into the intriguing world of exponential comparisons. Join us as we explore these colossal numbers, offering insightful analysis perfect for Olympiad training and mathematical challenges. Whether you're preparing for math competitions or simply love exploring complex mathematical concepts, this video is your gateway to understanding the nuances of exponentiation.
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4 дек 2023

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Комментарии : 192

@BriTheMathGuy 6 месяцев назад

To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/BriTheMathGuy . The first 200 of you will get 20% off Brilliant’s annual premium subscription.

@helpingfist1492 6 месяцев назад

Wait, how is this comment posted 3 days ago..... 🤔🤔😲😲

@India_Pakistan1 6 месяцев назад

Small^Big > big^small (except if they are both very small numbers)

@wyattstevens8574 6 месяцев назад

@@India_Pakistan1By "very small," it's just that at least one < e. (Blackpenredpen has a video about this trick!)

@DonatoMaster10 6 месяцев назад

I HATE SPONSORS

@KDestroyer9 6 месяцев назад

Another thing to note is since (1+1/x)^x is a strictly increasing function, we can say that (1+1/999)^999 is strictly less than e, rather than just approximately e.

@isaacsalzman 6 месяцев назад

Thank you for mentioning this, that's super cool!

@Trix_ooo 6 месяцев назад

Sure, but it only matters in low numbers

@HopefullyUnknown 6 месяцев назад

wow, yall are actual nerds

@not.the.real_braden 6 месяцев назад

doesnt matter, itll be either 2.718 or less, 999^1000 is greater than 1000^999 either way.

@nirorit 6 месяцев назад

@@HopefullyUnknownyup we are 😊

@derinwithaq5811 6 месяцев назад

Not only is 999^1000 larger, it’s much larger. I kind of intuitively look at the problem and say that since 999 is very close to 1000, 999^999 and 1000^999 will be relatively close. Obviously, multiplying that many times will make them less and less close, but they just can’t be different enough that multiplying by a whole extra 999 at the end won’t make a massive difference.

@singularity._. 6 месяцев назад

Yeah, that's exactly how I look at it too

@Jdogrey1 6 месяцев назад

That is what I was thinking...

@justinbrentwood1299 6 месяцев назад

That's how he solved it. He just proved that multiplying that many times won't be as great as ×999 with e.

@AlmondPeach 5 месяцев назад

To simplify, versatile, and somewhat “prove” your thinking you could also split the number 1000 into (999+1), resulting in us getting (999+1)^999 = 1 + 999^999 > 999^999

@TotalTimoTime 5 месяцев назад

⁠⁠⁠⁠@@AlmondPeachYou are missing 997 terms in your expansion. (x+y)^n =/= x^n + y^n For any n other than 1 or 0. Instead you have to determine the terms with binomial expansion formula giving an additional 997 terms.

@ColeDaNerd 6 месяцев назад

my intuition told me 999^1000 was larger but it was nice to see the proof. Very cool

@happygood18 6 месяцев назад

That was cool.

@_miobrot_603 6 месяцев назад

I instantly thought about logarithms with this problem. If you take the logarithm of both numbers, you get: 999 log(1000) log(1000) = 3 3 * 999 = 2997 1000 log(999) Because the slope of a logarithm (derivative of a logarithm with base b = log_b(e)*x^-1) tends to 0 as you use sufficiently larger numbers, this means that log(1000) ~ log(999). 1000 log(1000) = 3000 3000 > 2997 Therefore, 999^1000 is greater. This is mainly just a quick method to do it in your head; it's not very rigorous, but it works out. REAL VALUES 1000 log(999) = 2999.57

@mike1024. 6 месяцев назад

I thought of logarithms first too. It seemed reasonable. I'm still playing around with inequalities to see if I can make it rigorous though. It seems like calculus or something like it would need to get involved regardless.

@josephmellor7641 6 месяцев назад

@@mike1024.You can use the first order Taylor series approximation. log 999 is about log 1000 + (999 - 1000) / 1000 which is about 2.999, which is off by 0.00057.

@radupopescu9977 5 месяцев назад

I agree logarithm or in case of very very large number, double logarithm. If you really calculate the log (not by hand of course), the method is very rigorous.

@KyberNuII 6 месяцев назад

Another interesting way to calculate it would be to consider the f(x) = x^(1/x) verify its monotonicity Consider the inequality 1000>999 Operate f on both sides Swap the powers with laws of exponents, this method is useful when handling irrational power irrational

@RealNaine 5 месяцев назад

Interesting way to solve it, and a good way to prove it. What I did was just think about how much smaller each of them should be relative to 1000^1000. 1000^999 is clearly a thousandth the size, by the definition of exponentiation. 999^1000 I can intuit is going to be **very** roughly a single-digit fraction the size, based on my experience evaluating expressions like .999^1000 to calculate probabilities.

@alexbarac 6 месяцев назад

Would this way of thinking work? Assume ab^a. Apply ln on both sides, ln(a^b)>ln(b^a) or b*ln(a)>a*ln(b) or b/a>ln(b)/ln(a). Since proportion of actual numbers is greater than proportion of their ln functions, assumption was right. Then just replace a=999 and b=1000 to solve this problem.

@pa28cfi 5 месяцев назад

After 2 and 3, smaller integer to the larger integer is always bigger than the other way around. The gap between the two grows at an insane rate. It's never going to suddenly switch.

@TheRotundRodent 5 месяцев назад

I basically thought that increasing the exponent is more substantial than increasing the base, since the gap between n^3 and (n+1)^2 increases as n gets larger, and n^3 is ALWAYS larger for positive values of n. Since this is true, the gap would only increase if the exponents were to grow by the same amount/ always be 1 apart such that n has an exponent 1 larger than (n + 1)’s exponent.

@Ninja20704 6 месяцев назад

Another way I’ve see to deal with these problems is a calculus approach. Consider the function f(x)=x^1/x You can show using differentiation that the function is increasing for 0b^a. You can apply a similar approach if a and b were both less than e. In general, the number whose base is closer to e will be the winner. The limitation of this approach is that it only works if a and b are both on the same side of e. If they were on different sides like 2.5^3 vs 3^2.5 then we need another appoach

@OffBrandChicken 5 месяцев назад

I eyeballed it and thought the higher exponent would be higher. And that is my exact proof I would give the professor.

@nitya26an92.. 6 месяцев назад

It can also be solve by Using Some Basic Application Derivatives Let a function F(x) = lnx/x F(x) is decreasing if x>e Now F(1000)

@ShomnaphasOS 5 месяцев назад

1000 won oops

@mangler241 6 месяцев назад

y = (1 + 1/x)^x is strictly monotone increasing for x > 0 with a LUB of e, so a strict inequality could be used.

@James2210 6 месяцев назад

take the natural log. As n->inf, ln(n+1) ~= ln(n). So nln(n+1) < (n+1)ln(n), which means (n+1)^n < n^(n+1)

@Alphabetatralala 6 месяцев назад

More generally, for any x,y such that x>y>e, you get : x^y < y^x and for any x,y such that e>x>y>0, x^y > y^x. You only really need to think about the cases for which x>e>y

@rodbenson5879 6 месяцев назад

That was really cool, thanks.

@sehr.geheim 5 месяцев назад

Whoever made the sponsor block segment for this one: You are awesome

@aguyontheinternet8436 6 месяцев назад

2:08 one miniscule correction, both e and the approximation at n=999 are closer to 2.72 than 2.71 :P

@tschantz 5 месяцев назад

I assumed 3 & 4 would act the same so I compared 3 to the 4th with 4 cubed. 3 to the 4th is larger, therefore 999 to the 1000th is larger.

@victorystar8586 6 месяцев назад

I just thought that the number with the greater power has more value regardless of the base number. E.g. 3 to the power of 4 (81) is greater than 4 to the power of 3 (64). So, I conclude that 999 to the power of 1000 is greater than 1000 to the power of 999. It's not the perfect plan (2 to the power of 3 and 3 to the 2) but it works for most numbers.

@bipolarminddroppings 5 месяцев назад

it works for consecutive numbers in every case except 1 and 2, and 2 and 3, if memory serves.

@holymeto9981 6 месяцев назад

Simplify with 3^4 and 4^3 and see that 3^4 is bigger. Then, try 4^5 and 5^4 and see that 4^5 is bigger. Using this logic, we can determine that 999^1000 is bigger than 1000^999

@reinhardtristaneugen9113 6 месяцев назад

I think the right one is larger under the short-cut explanation, that you get 998001 if you square 999 and you can clearly state that if cubed the number is larger than 1000 squared... ...now this paradigm holds throughout so the number of the right is larger... ...by the way applies this for all numbers to be the base larger than 9, for up to nine my explanation does not work out... Le p'tit Daniel

@khondokarsindeed8994 6 месяцев назад

Another way to solve is by knowing this property(Idk if this property has a name or not, I just randomly found it while playing with my calculator) Where x+1=y and x>=3, x^y>y^x.

@mesplin3 6 месяцев назад

I know e

@wyattstevens8574 6 месяцев назад

Exactly how I remembered which was bigger in "almost all" (both > e) cases- I used 3 and 4 too!

@mangler241 6 месяцев назад

Not always! x^2 > 2^x when 2 < x < 4.

@mesplin3 6 месяцев назад

@@mangler241 That's why I need to pick numbers bigger than e=2.718...

@Termenz1 6 месяцев назад

@@mesplin3yet you explained that really poorly

@glowstonelovepad9294 6 месяцев назад

With a^b and b^a, whichever one where the base is closer to e is larger.

@PowdergameFan18 5 месяцев назад

I just figured exponential numbers were so overpowered at that number size that 1 singular extra multiplication would vastly overtake the other with such similar starting values. Thanks equation!

@paulchapman8023 6 месяцев назад

It seems like if a > b > 4, then b^a > a^b. Is there a hard and fast rule to that effect, or are there exceptions?

@Buphido 5 месяцев назад

Let’s assume a, b are integers and, for our initial case, a = 6, b = 5. Then 5^6 > 6^5 15625 > 7776 true. Now assume we have any true case of b^a > a^b. Inspect case a+1: b^(a+1) > (a+1)^b b*b^a > a^b + b*a^(b-1) + bC(b-2)*a^(b-2) + … + bC2*a^2 + b*a + 1 < b*a^b b*b^a > b*a^b follows from assumption and thus case true follows from b*a^b > (a+1)^b Now assume we have any true case of b^a > a^b for which a > b+1. Inspect case b+1: (b+1)^a > a^(b+1) …and I dunno how to continue here. If you could prove that second part, you‘d have a proof by induction.

@bipolarminddroppings 5 месяцев назад

with consecutive numbers higher than 3, yes, it is always true that n^n-1

@jaye1967 5 месяцев назад

Another way of looking at this problem is that 1000 and 999 are approximately equal and the exponents are multiplication, so ^1000 would be 1 more operation than ^999.

@bipolarminddroppings 5 месяцев назад

You see these kinds of problems on youtube all the time and the answer is always the same. Whichever has the higher exponent is the larger number. The 1000^999 will be "ahead" all the way until that final operation and then suddenly 999^1000 is vastly larger.

@rahikulalamsakib 6 месяцев назад

Imagine, getting this kind of math teachers in High School level. ❤

@ALLAHFOLLOWER315 6 месяцев назад

You do but in the end you get bored, make "depressing" edits and then say what is the purpose of maths, some excuses

@hellfireofdooom8876 6 месяцев назад

@@ALLAHFOLLOWER315 ?????

@ALLAHFOLLOWER315 6 месяцев назад

@@hellfireofdooom8876 this is why school is important, so you can understand English

@nicolaasvanvliet-hg3iu 6 месяцев назад

@@ALLAHFOLLOWER315they teach you that kind of stuff in like kindergarten to first grade so this man skipped out on something.

@ALLAHFOLLOWER315 6 месяцев назад

@@nicolaasvanvliet-hg3iu facts

@livedandletdie 5 месяцев назад

Unless it's 2^3 and 3^2 which are the only integers that have the lower exponent and higher base be the bigger number. Otherwise it's always base is less important than exponent. especially when it's k^(k-n) < (k-n)^k.

@iaroslav3249 3 месяца назад

Here's a theorem I (And probably lots of other people) came up with by looking at the graph xʸ ≥ yˣ ∀x,y ≥ e : If y ≥ x : xʸ ≥ yˣ There is more, however the rest of the curve is a non trivial piece wise function which involves the Lambert W Function, so it isn't simple.

@karlkastor 6 месяцев назад

I just argued that 999^1000 is the higher order polynomial of the two and should be larger for sufficiently large bases-

@GreyZeint 5 месяцев назад

The simplest way to see this is just to take log and rearrange. log(1000)/log(999) < 1000/999 because log is concave, ergo 999^1000 is larger. For a more detailed version of this argument, it follows from the definition of concavity that since log(0)>0, for t in (0,1) and x>0 we have log(tx) = log( tx+(1-t)*0 ) > tlog(x) + (1-t)log(0) > tlog(x), implying that 1/t > log(x)/log(tx). Letting t=999/1000 and x=1000 we have 1000/999 > log(1000)/log(999) which again implies that 999^1000 >1000^999

@cuzsleepisthecousinofdeath 5 месяцев назад

Explain to me how what we're presented with at 0:50 is not enough for a sufficient answer to this question.

@scottleung9587 6 месяцев назад

Cool!

@rishabhjain9721 5 месяцев назад

The alteernate way to solve this issue is To see the function As it is strictly increasing till e And then strictly decreasing

@3dplanet100 6 месяцев назад

I solved it using a different approach. 10⁹=1,000,000,000 (1 followed by 9 digits) 9¹⁰= 3 followed by 9 digits 3 is larger than 1. 9¹⁰ is larger than 10⁹ Therefore, 9¹⁰⁰⁰ is larger than 10⁹⁹⁹

@autrsurf2974 6 месяцев назад

How does this logic follow? This is equivalent to saying 3^2>2^3, so 3^200 > 2^333, which is false.

@autrsurf2974 6 месяцев назад

Specifically, the issue is that you raised 10^9 to the 111th expknent for 10^999 against 9^10 only being raised to the 100th exponent for 9^1000. Further, how would you reasonably transition this result if true to the case of 999 and 1000? 999 is 111 times 9, whilst 1000 is only 100 times 10.

@3dplanet100 6 месяцев назад

@@autrsurf2974 Remember, is Not 2³³³ but 2³⁰⁰ 2³ < 3² Same as 2³⁰< 3²⁰ Same as 2³⁰⁰ < 3²⁰⁰

@autrsurf2974 6 месяцев назад

@@3dplanet100 yes, thats why i said it is false in my first comment, and explained why in the second (100 vs 111)

@thomasrad5202 5 месяцев назад

obvious to me. increasing the base from 999 to 1000 is like 0.1% increase, while increasing the exponent is quite literally an exponential difference. with large numbers the exponent will always cause it to grow faster so the answer should be obvious

@miniraptor3211 5 месяцев назад

The video that made me realize that e is actually a limit to infinity and isn't just a random number my teacher told us to use.

@GilTheDino 5 месяцев назад

Make a video about |x|=-1

@athomist 6 месяцев назад

The Swedish SAT has a subtest that is only quantitative comparison. Similar to this. Only much simpler questions.

@Itskingmonty 5 месяцев назад

This video + Digithomelearning's AI tutoring = Math success. What a great time to be a student!

@BGFire_YT 6 месяцев назад

I just watched your 0 over 0 video and i think you started well but you assumed (like anyone else) that it is a constant. You said 0 over 0 is x and 0 over is 0 over 0 + 0 over 0 and you got it wrong. I think that 0 over 0 is not a constant but variable because it can be every number (0 over x and x over x give us 2 numbers already). And every time you use 0 over 0 in the same expression it is a different number. Unless you indexate it.

@pumpalin8661 5 месяцев назад

A number with a larger exponent is almost always is larger then a larger number with a smaller exponent, right? According to my grade 9 logic, idk any fancy stuff yet

@Jdogrey1 6 месяцев назад

You could pretty much tell this particular one intuitively, though.

@rahikulalamsakib 6 месяцев назад

Power is more important than base.

@Ludwig-MariaAKern-yz2vs 6 месяцев назад

Not always. 2^3 and 3^2, (3 bigger power) so u would say that 2*2*2>3*3 ? (That's wrong! 8

@reinhardtristaneugen9113 5 месяцев назад

well I guess that the right one is larger under the usage of the argumentum a minore ad maius... ...I took 99tothepowerof4 as a premise and squared 99 then just to have it multiplicated with 90 two times and it turns out already higher than 100tothepowerof3 in that the ladder is 1 million the first one is 79 388 100 and since this approximation is sufficient to be higher the real numbers are also... ...the only thing is I do not know if I made a mistake in the premise since this pattern has exceptions I already learned and I'd like to scrutinize the paradigms later on, on how they behave exactly... Le p'tit Daniel, just leave me a note if I am wrong....

@bikrampurkait3786 6 месяцев назад

It's a great explanation, love it ❤.

@edwin2054 2 месяца назад

I just did square root of 999 of both. For 1000 we get 1000^1 For 999 we get 999^1.001 Now we can use a calculator to check or just see that 999 is larger Now when thinking about it, it would have probably been better to sqrt 1000 so we get 1000^0,999 and 999^1 since we’re raising by 0,(something) the number is going to go down.

@riddhamdhariwal 6 месяцев назад

The way bro emphasised at EEE

@kenkai7257 5 месяцев назад

Now this is the real question

@WilliamDye-willdye 6 месяцев назад

Problem is, I didn't know that e could be approximated that way.

@JBergmansson 6 месяцев назад

The general case, if I'm not mistaken, is that for any two numbers x and y, if y > x > e, then it follows that x^y will be bigger than y^x. I.e, large exponent beats large base for all numbers bigger than e.

@bipolarminddroppings 5 месяцев назад

if n>3 then n^n-1

@FundamSrijan 6 месяцев назад

It can be solved easily , I did it in mind It's silly to use pen and paper for this question when you know that the base 1000 is just 1 more than 999 but the 1000 in the power of 999 means multiplying the no. entirely one more time by 999 . So , it's that easy to calculate that 1000⁹⁹⁹

@chazzer5968 6 месяцев назад

This was how I did it. People see the similarity and think they're close but adding 1 to exponent does ALOT more than adding one to the base.

@whitecrow20XX 6 месяцев назад

@chazzer5968 true. it's called exponent because it is increased exponentially. 1 power matters.

@gmaasry 6 месяцев назад

Interesting though I would argue the much more logical (and rapid) path to a solution is simply to compare 2^3 with 3^2 and 3^4 with 4^3 in order to identify the pattern.

@BananaWasTaken 6 месяцев назад

But doesn’t the first pair get you 2^34^3 So the examples you gave don’t really tell you a pattern

@gmaasry 6 месяцев назад

Ha - very good point! The pattern only starts there. So, yes, fair enough.

@Ch1l1C0nCarnag3 5 месяцев назад

I just looked at it and went: "Okay, they'll have the same amount of numbers at the end. The only difference is one starts with a 1, and the other would start with 999, so the one starting with 999 would be larger."

@KillaCatzGames 5 месяцев назад

I mean, its 1000 multiplied by itself by 999. 999 on the other hand, is multiplied by itself by 1000. Thats an extra 999 to multiply, so its definitely larger.

@m.h.6470 5 месяцев назад

For any aᵇ vs bᵃ where b = a + 1 and a > 2.29317 you can say that aᵇ > bᵃ In other words: As long as the smaller base is larger than 2.29317, the larger exponent causes that number to be bigger. So for integer solutions, 1² < 2¹ (1 < 2) and 2³ < 3² (8 < 9) but everything after that is reversed. So 3⁴ > 4³ (81 > 64), 4⁵ > 5⁴ (1024 > 625) The 2.29317 comes from solving (a+1)ᵃ = aᵃ⁺¹

@amityosifovich1805 5 месяцев назад

called it

@malourano 5 месяцев назад

I am disapointed... I thought he would have said e^1 is approximately 3:(

@sodabutnofizz1294 6 месяцев назад

Im new in this, but cant we use binomial approximation to round of (1+1/999)^999 as 1+(1/999)(999) or 1+1= 2?

@mxb3ar69 6 месяцев назад

We cannot be cause according to Garfield’s rule book. You can’t. Hope this helps.

@sodabutnofizz1294 6 месяцев назад

@@mxb3ar69 so thats just bs?

@tszhinbow6087 6 месяцев назад

That's the approach I went for too but you can't just stop at the second term. The coefficent of the third term is (999×998)÷(1×2)×999^997 < 999^999/2 for example. So you end up with something like 1000^999 < 999^999 × (1+1+1/2+1/6+...) < 999^999 × e

@sodabutnofizz1294 6 месяцев назад

@@tszhinbow6087 i realised that.. thanks

@tynado7807 5 месяцев назад

And if you didn’t know that was the approximation of e?

@syed3344 6 месяцев назад

we can use derivatives to find the answer too:)

@victorgoji4364 6 месяцев назад

Witch is smaller 1/1000 or 1/999

@lebollsong 5 месяцев назад

Larger denominator means it's the smaller number. Like you know 1/4 is smaller than 1/2 because 0.25

@andreifuturegm5075 6 месяцев назад

NICE

@user-xy5yg6se1k 5 месяцев назад

cool

@GurkenGamer9 6 месяцев назад

exponents are really powerful one extra at that scale is obviously larger ;/

@classymuffin4589 5 месяцев назад

Tbh I was just lazy and went "well 2^3 is larger than 3^2 and follows a similar rule to this so fuck it, 999^1000 is bigger"

@urbanoverlord2412 6 месяцев назад

Why I cant just take the 999th root of both to easily see that we get 1000 ^1 vs 999^2, and obviously 999^2 > 1000, Since both 999 and 1000 are > 1. Am I missing something??

@YT-AleX-1337 6 месяцев назад

Knew it

@whyisntfauciinprison2192 5 месяцев назад

wouldnt it be the same

@SoloGD-ql3dw 5 месяцев назад

The class: 2+2 = 4 The homework: 2^3 = ? The exam:

@user-te8vv5fv1p 6 месяцев назад

I thought everyone knew that the leading exponent always dominates

@avonire 5 месяцев назад

I'm not much of a maths person, but since 10² is 100 and 10³ is 1000 i figured 999¹⁰⁰⁰ was larger

@mint._.1078 6 месяцев назад

Seeing the thumbnail, I simply thought of 3² & 2³ instead of 1000⁹⁹⁹ & 9¹⁰⁰⁰ [3²=9 & 2³ = 8. So therefore, 3² > 2³.] In the same manner, 1000⁹⁹⁹ > 9¹⁰⁰⁰ After watching the video, I realized there's alot more stuff i should learn about.

@bipolarminddroppings 5 месяцев назад

your only mistake was to start with 2 and 3, the only case in which the reverse is true. if you had picked 3^4 and 4^3 or any consecutive numbers higher than 3, then n^n-1

@donteven4673 6 месяцев назад

if we take log on both sides, 999 log(1000) and 1000 log(999) then subtract 1000 log(999) - 999 log(1000) = 999 log(1000/999) + log(999) we can see it's > 0, yeah ur method is bettter

@saltedquackers5264 5 месяцев назад

"well 3^4 is greater than 4^3 so, 999^1000 is greater than 1000^999." That probably doesn't math but it's my inductive reasoning lol

@bipolarminddroppings 5 месяцев назад

n^n-1

@saltedquackers5264 5 месяцев назад

@@bipolarminddroppings thanks i kinda forgot about those powers

@Mrnulet 5 месяцев назад

999^1000 is larger

@DoxxTheMathGeek 6 месяцев назад

I once looked at Desmos at x^y

@wyattstevens8574 6 месяцев назад

All cases where both are bigger than e are in "that region." That's how I do it myself.

@capjus 5 месяцев назад

Wow impressed.. guessed the opposite

@Fireyyy1 6 месяцев назад

Rah

@cakedon 6 месяцев назад

im kinda trash at math but i know that 999 ≈ 1000 so 1000^999 < 1000^1000

@lenia272 5 месяцев назад

while im sure this is good for weird random numbers this example takes litterally low math skills and a few seconds of logic to figure out without any work. you simply start off with the first step of 999x1000 and that number now has the same 999 rounds to go as the 1000 starter does but is already insanely larger to begin with. your now looking at aprox 998001 to the 999 vrs 1000 to the 999

@harshitthebaby1240 6 месяцев назад

i got oh my god

@nizari116 5 месяцев назад

I knew this instantly without these weird alien calculations.

@SomeCowguy 6 месяцев назад

All the math nerds in these comments making hard equations i dont understand, all while im just going with common sense and picking the right one

@Luna5829 6 месяцев назад

for anyone wondering, it's larger by 3.67*10^2999

@The_Story_Of_Us 6 месяцев назад

if you look at this as a VS match between two functions: n to the power n-1 VS n-1 to the power n already by n = 4 does the higher exponent overtake the other function. And as you keep going along, you'll see that it is in fact growing way faster. By n = 9 it's already over 3 times bigger so imagine what the difference is at n=1000 It's not even remotely close which one is bigger than the other. This is getting dangerously close to full on googology.