A classic number battle -- Did I make this too hard?

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Опубликовано:

13 дек 2021

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Комментарии : 113

@rahenson1 2 года назад

Once you get to (1 + 1/2021)^2021 I think the simplest path is to say it is less than e which is limit of (1+ 1/x) ^x as x goes to infinity. And e/2021 is most definitely less than 1.

@roberto18594 2 года назад

you would need to prove/know that (1+1/x)^x is increasing tho

@TJStellmach 2 года назад

@@roberto18594 It's certainly not going to be different from e by a factor of 2021, though I grant that fact isn't exactly rigorous.

@johannwebersberger9477 2 года назад

Lim n->inf ( 1 + 1/n)^n = e + fast convergence. Deal done

@roberto18594 2 года назад

@Oily Macaroni I don't know about that. You can do it with calculus or by induction on the discrete variant, tho.

@roberto18594 2 года назад

@@TJStellmach yeah that's for sure, but not proving that is the same as computing (1+1/2021)^2021 with a calculator, at that point :D

@montjw 2 года назад

"Too hard" is an imaginary line in the sand, but I can say there is an alternative route using only early calculus material (derivative of log) to show log(A/B) < 0. First, log'(x) = 1/x gives us the bound log(x + y) < log(x) + y (1/x) for all x,y > 0. Using y = 1, we have log(x + 1) < log(x) + 1/x. Second, log(A/B) = x log(x+1) - (x+1) log(x) < x [log(x) + 1/x] - (x+1) log(x) = 1 - log(x) < 0 for x = 2021. log(A/B) < 0 implies A/B < 1. --------------------------------------- Clarification on the bound log(x + y) < log(x) + y (1/x): LHS = height of log at x+y. RHS = height of tangent line through (x,log(x)) at x+y. Since every line tangent to log is strictly greater than log away from the point of intersection (concave down) then LHS < RHS.

@GreenMeansGOF 2 года назад

If we compare the reciprocals of A and B, we compare 2021^2022 and 2022^2021. Having done so many of these, we know 1/A is greater than 1/B because 2021 is closer to e. Thus, B is greater than A. QED

@chessematics 2 года назад

I was about to say exactly the same thing.

@OscarCunningham 2 года назад

The exponent always wins (except when it doesn't).

@chessematics 2 года назад

@@OscarCunningham i am always alive, except when I'm not

@iwersonsch5131 2 года назад

Multiply both sides by 2022^2022 Left side is (2022/2021)^2022 = (1 + 1/2021)^2021 * (2022/2021) which is smaller than e * 2. Right side is 2022. Right side is larger.

@quzpolkas 2 года назад

I see an alternate solution to this. This method is usually used for comparisons of the form (a^b ? b^a), but we can easily adapt it to our situation. 1. First, consider the function f(x) = (lnx) / x, x > 0. It's derivative f'(x) = (1 - lnx) / x^2 is positive for x between 0 and e, and negative for x bigger than e. That means that f(x) is increasing on the interval (0;e) and decreasing for x > e. Keep it in mind. 2. Go back to the original question, log both sides, use the exponent rule ("?" is an unknown comparison symbol, one that we have to determine): (1/2021)^2022 ? (1/2022)^2021 2022 ln(1/2021) ? 2021 ln(1/2022) Logarithm rules: 2022 ln(1/2021) ? 2021 ln(1/2022) -2022 ln(2021) ? -2021 ln(2022) Swap negative numbers around, divide both sides by 2021*2022: -2022 ln(2021) ? -2021 ln(2022) 2021 ln(2022) ? 2022 ln(2021) ln(2022) / 2022 ? ln(2021)/2021 This becomes the question of comparing the values of f(x) for x = 2022 and x = 2021, respectively. As we established earlier, f(x) decreases for x > e, and both 2021 and 2022 are definitely bigger than e. So that means f(2022) < f(2021), and our symbol "?" was actually a "

@fix5072 2 года назад

it iis way easier to consider the function f(x)=x^x, which is strictly decreasing for x

@quzpolkas 2 года назад

@@fix5072 I'm sorry, but I can't see how x^x help us here, base and exponent are different in our case? We're not comparing (1/2022)^2022 with (1/2021)^2021, but (1/2022)^2021 with (1/2021)^2022. Not to mention computing derivatives of x^x is kinda tedious, and definitely harder than those of lnx/x. Can you please elaborate?

@bosorot 2 года назад

Learn someting new everyday . Thank and cheer mate.

@maxsch.6555 2 года назад

@@quzpolkas (1/2021)^2022 < (1/2022)^2021 (1/2021)^(1/2021) < (1/2022)^(1/2022), so you can easily use f(x) = x^x and some calculus to prove this

@fix5072 2 года назад

@@maxsch.6555 yes, the derivative also is not that hard (if you write it as e^(lnx*x) you can even do it in your head) and since the e part is always positive it really isnt hard to solve lnx=-1

@brainlessbot3699 2 года назад

x^x is decreasing in (0, 1/e) and as 0

@ahmadawlagi6481 2 года назад

I think you can do it much easier by flipping the fraction and using the rule that says (a^b > b^a if e

@DjKryx 2 года назад

The problem is the same as showing that ln(x)/x is strictly decreasing function after some natural number smaller than 2021. Taking a derivative, you can easily show this holds true for every number greater than e, so ln(n)/n > ln(n+1)/(n+1). After that conclusion, you can get the answer which number is bigger

@MyOneFiftiethOfADollar 2 года назад

Right! , which is also equivalent to proving x^(1/x) is decreasing on interval (e, infinity). Used it recently to show e^pi > pi^e

@user-mf3mp2ko2d 2 года назад

As an engineer I said, since 1/2021 is really small a good approximation would be :=1+x*n for x=1/2021 and n=2021, problem sovled

@benjaminbrat3922 2 года назад

So basically, 1/2021 is not very different from 1/2022 (in the basis) but multiplying by a supplementary factor reduces the result significantly, because in the end one is 2000 smaller than the other.

@sharpnova2 2 года назад

isn't this just the same as comparing 2021^2022 to 2022^2021?

@roquedefrutos8667 2 года назад

(1+1/n)^n is increasing and converges to e Then (2022/2021)^2021=(1+1/2021)^2021

@heh2393 2 года назад

What was the last 25 s of blackscreen for tho?

@someuser257 2 года назад

Well it could‘ve been a little faster if we used the number e as its definition (1+1/x)^x as x->♾

@damascus21 2 года назад

Clever

@twarisko 2 года назад

The backflips 🤧😥

@manucitomx 2 года назад

@HamzaJamali I agree

@DrPhipster 2 года назад

Missing the good old days, huh?🥲

@twarisko 2 года назад

@@DrPhipster yeah"((

@Tiqerboy 2 года назад

An equivalent problem would be. If you can solve that below, you have the answer to the puzzle. Lim { x^(x+1) / (x+1)^x } x → ∞ Taking the logarithm of that expression makes it easier to handle log ( x^(x+1) / (x+1)^x ) = log (x^(x+1) ) - log ((x+1)^x) = (x+1)log(x) - x(log(x+1)) = xlog(x) + log(x) - xlog(x+1) . as x → ∞ log(x) ~ log (x+1) so the expression is ~ log(x) -0.43 . for very big x.

@kriswillems5661 2 года назад

The engineering solution : the first factor is about e, e/2021

@vince8638 2 года назад

Once you get to (1+1/n)^n with n large, just use the Taylor expansion (1+1/n)^n = 2 + O(1/n^2) and that's it

@GDPlainA 2 года назад

just pretty much learnt from bprp's video about a^b vs b^a. basically we start with 1/(2021^2022) and 1/(2022^2021) if 1/a > 1/b then b > a from bprp's video about how to solve a^b vs b^a, we can say that 2021^2022 > 2022^2021. therefore 1/(2022^2021) is greater

@beirirangu 2 года назад

I went about it in a different way: (1/a)^(a+1) vs (1/(a+1))^a and as long as a is greater than ~2.23 then (1/(a+1))^a will always be greater than (1/a)^(a+1)

@annaarkless5822 Год назад

i did it based on vibes exponentiation makes numbers less than 1 smaller quicker than adding 1 to the denominator does, so B feels bigger. QED i will accept my fields medal now

@forgetittube5882 2 года назад

Ok… (using ?

@shreyassathe1286 2 года назад

You can check with (1/x)^(1/x) being an increasing function in the domain of [2021, 2022] by getting the derivative. now just revert this nd you get this result. Infact you’ll find for all x>e, (1/x+1)^x > (1/x)^x+1 . Thus done easily using simple calculus.

@davidwright5719 2 года назад

It’s easier to take ln of both quantities and use ln(1+x) ~ x.

@Snidbert 2 года назад

I came to the same conclusion by simply taking the log of both sides and plugging it into my phone calculator, lol.

@GKinWor 2 года назад

bruh

@manucitomx 2 года назад

Thank you, professor! I agree with @HamzaJamali, I miss the backflips!

@fedorlozben6344 2 года назад

You could use the limit (1+1/x)^x It is growing function and it is always less then e.

@laurendoe168 2 года назад

When comparing A^(A+1) to (A+1)^A, the former will be larger than the latter any time A > 3. As a result, when comparing the inverse of these two, the former will be less than the latter any time A > 3.

@blake121666 2 года назад

It's actually the case whenever A > e (Euler's number). Since Euler's number is the monotonically increasing limit of (1 + 1/n)^n, anything of that form, (1+1/n)^n, is always less than e. Your example is simply a rearrangement of this inequality.

@laurendoe168 2 года назад

@@blake121666 Ok, I will admit I did this in like 15 minutes on Excel and considered only integers. Now that I broaden my outlook, and consider it more carefully, e makes a lot more sense. I was just looking for some "simple" demonstration that when considering numbers which are orders of magnitude greater 3... you're just wasting your time. :D

@blake121666 2 года назад

@@laurendoe168 Looking over my post I see I didn't actually write out the simple demonstration. Here is that: (1 + 1/N)^N < e - this is Euler's well-known definition of his e as the limit as N -> infinity So ((N+1)/N))^N < e Therefore (N+1)^N < e * N^N < N^(N+1) if N > e This is all the video needed to show for the general case and show that his particular numbers satisfy this inequality - as you mentioned.

@shoryaprakash8945 2 года назад

Another way using calculus that x^x is dec from (0,1/e] and 1/2021>1/2022 is between (0, 1/e) so (1/2021)^(1/2021) (1/2021)^2022

@antormosabbir4750 2 года назад

x^1/x and calculus

@jainko8471 2 года назад

x --> x^x is non-increasing on [0 ; 1/e ] 2022>2021>e 1/2022 < 1/2021 < 1/e (1/2022)^(1/2022) > (1/2021)^(1/2021) (1/2022)^2021 > (1/2021)^2022 (both side ^ 2021*2022) and i think it's a good place to stop.

@ibrahimkachal6759 2 года назад

Let a,b and c real positives numbers satifying: abc= a²b² + b²c² +c²a² Find the greatest value of real number m where: m= a² +b² +c²

@adandap 2 года назад

Let f(x) = log(1+x) -x. Then f(0) = 0 and f'(x) = 1/(1+x) -1 < 0 for all x >0, which tells us that f(x) < 0 for all positive x. Set x = 1/2021 then we have 2021 log(1 + 1/2021) < 1 < log(2021). Putting that together we have -log(2021) + 2021 log(1 + 1/2021) < 0, which shows that (2022)^2021 < (2021)^2022. (In fact, numerically the difference is a factor ~ 743.7)

@egillandersson1780 2 года назад

It seams easier to take the (2021.2022)th root of each and then study the function xˣ.

@keinernichts3531 2 года назад

Hmm, I took the root of order 2021*2022 from both to have same basis and exponent then just studied the behavior of the function x^x, saw where its minimum is and compared based on that.

@lavneetjanagal 2 года назад

Much simpler is to write them as power of e. That is A = e^{-2022 x Log(2021)} B = e^{-2021 x Log(2022)} Now, think of the function e^{-x Log (y)} The graph of f1(x)=Log(x) is always below f2(x)=x for positive integer x and both functions are >0 . Therefore, f2(x+1) f1(x) > f2(x) f1(x+1). In our case 2022 Log(2021)> 2021 Log(2022) Or B>A.

@MyOneFiftiethOfADollar 2 года назад

Wow, the discrete version of things versus continuous, to wit x^(1/x) decreasing on (e, infinity) implies (2021^(1/2021))^(2021x2022) = 2021^2022 is the larger number

@user-zd1jb4id8p 2 года назад

In general when does (x+1) to the x become shorter than x to the (x+1)? Between 2 and 3, about 2,29316?

@sergebertrand5681 2 года назад

Ln(1+x)=x+o(x^2) is sufficient to conclude quickly

@charleyhoward4594 2 года назад

forgotten my " rules of Exponentiation" (b *c ) ^ n = (b ) ^ n * (c ) ^ n

@mackenziekelly1148 2 года назад

Take f(x)=x^(1/(2001*2002)). This is a simple nth root function therefore f is always increasing on the interval (0, infinity). As a result, if two numbers a and b are positive, then a

@tomaytimur5471 2 года назад

just compare 2021^2022 and 2022^2021, or 2022log2021 and 2021log2022. It is well known that the lhs of these are larger. Therefore -2022log2021 is smaller than -2021log2022. q.e.d.

@BaoNguyen6742 2 года назад

a^b > b^a with a,b>e so the answer is obviously 1/2022^2021

@danielparry7643 2 года назад

Prove (1+1/x)^x is monotonic as x\to \infty. Would be a lot simpler.

@jimmorgan6213 2 года назад

Too hard?? It’s actually eextremely eesie.

@pneujai 2 года назад

i solved this when i saw the thumbnail :)

@ieatgarbage8771 2 года назад

Well, both of these numbers are above e 2021^2022>2022^2021 (1/2021)^2022

@tubemaan 2 года назад

Without going through the complex math presented, to me, A is simply greater than B.

@TechnocratiK 2 года назад

For x > 0, ln(x) is monotonically increasing (since (ln(x))' = 1 / x > 0 for all x > 0). Thus, by taking the ln of both sides, (1/2021) ^ 2022 < (1/2022) ^ 2021 iff -2022 ln(2021) < -2021 ln(2022). Rearranging, the last is true iff 2022 / ln(2022) > 2021 / ln(2021). Let f(x) = x / ln(x), then f'(x) = (ln(x) - 1) / ln(x) ^ 2, and f(x) is monotonically increasing on all intervals for which f'(x) > 0---that is, for all x > e. Since 2021 > e, f(x) = x / ln(x) is monotonically increasing for 2021

@davidovic4845 2 года назад

The title of this video...? That's what she said

@kevinmartin7760 2 года назад

So (1/1)^2 > (1/2)^1 and (1/2)^3 > (1/3)^2 but (1/3)^4 < (1/4)^3 [1>1/2 and 1/8>1/9 but 1/81

@stephenhousman6975 2 года назад

Try using e. Usually when x appears in bot the base and exponent e tends to pop up.

@yehonatankane1192 2 года назад

Overkill…

@mattcarnevali 2 года назад

My intuition was that A > B since the larger base usually wins out, but I was wrong. Interesting problem.

@luislopez-tx4tl 2 года назад

what lmfao

@playgroundgames3667 2 года назад

A = B

@TrimutiusToo 2 года назад

so "good place to stop" is at 5:58, and then we have bunch of black screen of half a minute or so...

@RipleySawzen 2 года назад

A bit too handwavy over parts

@mohitkrjain9396 2 года назад

Isn't it contradictory? I mean 1/2021>1/2022 and 2022>2021. So the base and exponent are both greater for A than for B. Then how come A < B

@anushrao882 2 года назад

It isn’t contradictory. Both the bases are smaller than 1. In fact, if you were to increase the exponent of A further, its value would decrease.

@bro_vega_1412 2 года назад

Would you consider 0.2^2 to be bigger than 0.1^1 ? You won't , right? It doesn't always work for base less than 1. That's why he took such approach.

@seroujghazarian6343 2 года назад

x^(1/x) is descending for x>e

@ibrahimkachal6759 2 года назад

Can anyone solve this

@kenbrady119 2 года назад

After 2+ views I followed you up until the binomial expansion, and then it was like "I'm gonna trust you on this", until you got past that bit. Not doubt after 3+ views I would get it, but also perhaps for slower viewers you could taken a little more time to expand upon your expansions?

@ramza2779 2 года назад

The task is more difficult 5^51 or 2^118 ?

@AdamCheng 2 года назад

log(5^51*2^51)/log(2^169) 51/(169*log2) 51/169*0.301=51/50.869>1 5^51 > 2^118

@davidbrisbane7206 2 года назад

Once it is shown that 5⁵¹ > 2¹¹⁸, it can then be shown that 5⁵¹ < 2¹¹⁹ as follows. 2⁷ = 128 > 125 = 5³ ⇒ 2⁷ > 5³ ⇒ (2⁷)¹⁷ > (5³)¹⁷ ⇒ 2¹¹⁹ > 5⁵¹ So, 2¹¹⁸ < 5⁵¹ < 2¹¹⁹

@ramza2779 2 года назад

@@davidbrisbane7206 2¹¹⁹ > 5⁵¹ agree! but that doesn't mean that 2¹¹⁸ < 5⁵¹

@davidbrisbane7206 2 года назад

@@ramza2779 Yes. But I already showed that 5⁵¹ > 2¹¹⁸ in another comment.

@playgroundgames3667 2 года назад

Neither. They're both the same answer