Which is larger: A^ B or B^A? (two visual proofs)

Рет қаралды 37,871

Күн бұрын

Пікірлер: 84

@Kounomura 4 ай бұрын

The magical properties of "e" are endless. There is another surprising problem similar to the one above: Let N=n1+n2+n3+ .. + nk, where we are talking about the decomposition of N into arbitrary (integer) sums. The question is, how can N be decomposed into summables so that their product is maximal? The answer: the items to be added should be as close to "e" as possible, i.e. if possible, only 3's or maybe 2's should occur among the items to be added. For example, in the case of 10=3+3+3+1, the product= 27. The 1 does not make sense here, it is not an optimal solution. The optimum: 10= 3+3+2+2 The product=3*3*2*2 = 36 (In the range of real numbers, using exponentiation instead of multiplication, the maximum would be e^(10/e) = 39.5986..)

@s4br3 4 ай бұрын

I had this exact question in an oxford interview lol. I should have thought of the naturally exponential nature of e

@lyrimetacurl0 4 ай бұрын

(1 + (1/n))^n tends to e as n increases (was another one)

@hasanjakir360 3 ай бұрын

Where can I get the proof?

@STEAMerBear 4 ай бұрын

The slopes proof is much easier to teach. I recommend first teaching it with chalk or dry erase marker, then transitioning to GeoGebra where you can move A & B around so students can see how it works.

@MathVisualProofs 4 ай бұрын

Yes. GeoGebra would be great for this.

@asparkdeity8717 4 ай бұрын

Wow, this is very similar/nearly identical to the comment I posted a while ago on e^pi vs pi^e; I recalled fixing the tangent line to log and rotating it clockwise/anticlockwise and considering the sign of the intercept, where you fixed the intercept and considered lines of different gradient (which is much easier haha); thanks for listening and making a great video that's so simple to follow! I also love how u kept the other cases open for viewers to try out, really engaging too. I copied and pasted my longer comment from before with the slightly different method for the sake of interest: with c = 0, m = 1/e gives one tangential intersection at x = e. Now, in the case c = 0, if we reduce m such that 0 < m < 1/e (visually “pivoting” the line clockwise about the origin), we will get two intersections points a,b where a < e < b. This means at a,b: the gradients of the straight line through ln(x) are equal: ==> (ln(a) - 0)/(a-0) = (ln(b) - 0)/(b-0) ==> a^b = b^a Now, same principle, but consider the case c > 0. Suppose two intersections with the straight line and ln(x) are labelled p and q, with q > p WLOG. Suppose p = e. Another lovely visual argument (pivoting the line clockwise about x = e by decreasing m) shows that we get two intersections for c > 0 with p = e and q > p, since we know there is only one intersection at e with c = 0. Again, we’ll run the same argument that the gradients at p,q must be equal: ==> (ln(p) - c)/(p-0) = (ln(q) - c)/(q-0) ==> qln(p) - pln(q) = qc - pc Now note in this case, c>0 and q>p ==> RHS > 0 ==> p^q > q^p -> Using another pivot argument and slightly adapting the one I did, try and show that p^q > q^p for any q > p >= e -> Now by considering the cases in which c < 0 yields two intersections, try and show that p^q < q^p for any 0 < p < q Notice there is a deadzone of values where it is impossible to determine which is bigger, in the case q >= e and p < e. A classic example of this 2^3 < 3^2 , 2^4 = 4^2 and 2^5 > 5^2.

@WhattheHectogon 4 ай бұрын

0:57 doesn't lnx have a slope of 1/e at x=e?

@cupostuff9929 4 ай бұрын

That's what i was thinking

@MathVisualProofs 4 ай бұрын

Yes. I misspoke. I meant the slope is 1/e and that’s why it passes through the origin. Thanks! I’ll note the mistake in the description.

@Kounomura 4 ай бұрын

The function curve is magnified by a factor of e in the y direction so that the proportions are more visible. The function shown is actually y= e*ln(x). But this does not change the explained principles

@MathVisualProofs 4 ай бұрын

I was able to kind of edit it on youtube and correct. Thank you for catching it!

@plamenpenchev262 4 ай бұрын

Very nice video. Without doubt, visual presentation facilitates understanding but using only formulas also works.

@Achill101 4 ай бұрын

Sorry, but I found the third proof the most instructive. Sure, it's more difficult to show x^(1/x) has only one maximum, but once we've got that, the cases for e=

@raulgalets 3 ай бұрын

ABBA is the greatest of all time

@xjuhox 3 ай бұрын

Let B = A+h, where h > 0. By the linear approximation ln(B) ~ ln(A) + h/A, we have B/A > ln(B)/ln(A) if A > e. That is, A^B > B^A if B > A > e 👀

@oida10000 4 ай бұрын

The first proof is the best but I prefer the 3rd over the second mathematically because integral.

@oniondesu9633 3 ай бұрын

don't think the second proof counts as a "visual proof" by any reasonable standard. first one is quite elegant and intuitive though

@MathVisualProofs 3 ай бұрын

It is a published proof without words, so by some reasonable standard it is. I do agree it requires some decent background knowledge to make the jump without writing down the formulas, but it’s doable.

@Andy_Hendrix_9842 4 ай бұрын

ABBA 😁

@mikhailnikolaev4992 4 ай бұрын

ABBB💀

@Rippertear 4 ай бұрын

@@mikhailnikolaev4992ABBC 😲

@Naej7 4 ай бұрын

ACDC

@StevenTorrey 4 ай бұрын

I've always wondered whether there was a formula for solving that question, since it is a favored KZbin question.

@DOTvCROSS 4 ай бұрын

Let A be a matrix (2x2)'s of trace(2) and zeros, let B be a matrix with trace(3) and zeros. |A|

@GenericInternetter 3 ай бұрын

Doesn't really count as a "visual proof" when most of the proof is done using formulas.

@sonicwaveinfinitymiddwelle8555 3 ай бұрын

At first glance this question seems easy but once you look at it for a while it seems hard but when you actually try and solve it's actually pretty simple.

@deananderson7714 4 ай бұрын

To find the derivative in the third method you can also do this y=x^(1/x) => ln(y)=(1/x)ln(x) instead of taking both sides to the power of e, take the derivative of both sides (1/y)(dy/dx) = (-1/x^2)ln(x)+1/x^2 => (1/y)(dy/dx) = (1-ln(x))/x^2 => dy/dx = y(1-ln(x))/x^2 => dy/dx = (x^(1/x))(1-ln(x))/x^2 we can combine the powers of x to get dy/dx = (x^(1/x-2))(1-ln(x)) From here we can see this is only zero when 1-ln(x) is zero(as the other term is nonzero for all x) and we see that only happens when x=e (we have a point where the derivative doesn’t exist at x=0 but that point isn’t in the original function’s domain so it isn’t a critical point)

@FishSticker 4 ай бұрын

Damn I really really like that first proof

@MathVisualProofs 4 ай бұрын

😀😎

@rafiahamed2978 4 ай бұрын

🥰

@economicist2011 4 ай бұрын

As much as I love calculus, I love even more the ability to explain concepts like this one without it. Calculus is beautiful but it makes elementary topics like A^B B^A seem "gatekept" to people who haven't taken it. Good show!

@mathysvienne3032 2 ай бұрын

Can someone explain me how the fundamental theorem was used in the second visual proof ? I didn't get it

@beaumatthews6411 4 ай бұрын

2 < e! I was just teaching a kid about this yesterday

@Dalroc 4 ай бұрын

The third is the least complicated.

@oinvestigard 4 ай бұрын

ONGLY, if a is bigger than b oppocite oppocites that

@guessundheit6494 4 ай бұрын

Even with only knowledge of natural numbers, it's clear that if A>=3, there will never be a case where B^A is greater. And even if you know decimals, as long as A>=3 and B>A (e.g B=3.00000000...1), then A^B will be greater. But since 2 and 3 break that trend (2^3 < 3^2, 2^4 = 4^2), the point where it changes must be somewhere between. After fooling around with spreadsheets, it's interesting how exponentially large B becomes as A decreases (e.g. A=1.1, B=43.55774178) for A>B to remain greater.

@zsoltnagy5654 3 ай бұрын

The inverse function f^(-1) for f(x) = x^(1/x) is f^(-1)(x) = exp(-W(-ln(x))) for x∈[0, e^(1/e)]. So if 1 < A < e < B, then A^B > B^A, if and only if A^(1/A) > B^(1/B) and if and only if f^(-1)(A^(1/A)) > f^(-1)(B^(1/B)) as f^(-1) is a monotonically increasing function. So *if 1 < A < e < B, then A^B > B^A, if and only if A > exp(-W(-ln(B^(1/B)))) = exp(-W(-ln(B)/B)).* Otherwise it should be trivial, that *if A ≤ 1 < B, then A^B < B^A.* For example 2 < e < 3 and 2 < exp(-W(-ln(3)/3)) ≈ 2.48. So it's also, that 2^3 < 3^2 (which is also trivially true of course). So at the same time 2.5 < e < 3 and 2.5 > exp(-W(-ln(3)/3)) ≈ 2.48. So 2.5^3(≈15.63) > 3^2.5(≈15.59).

@sachavalette1437 3 ай бұрын

So instead of just seeing the result of a^b and b^a you must calculate 2 more complicated formulas with two receprical functions, that’s genius 💀💀💀

@zsoltnagy5654 3 ай бұрын

@@sachavalette1437 Well, alternatively you can also compare B to the following: A^B > B^A, _approximately_ if and only B > (e-1)²/exp(-W(-ln(A)/A))+1. With my previous example 3 < (e-1)²/(exp(-W(-ln(2)/2))-1)+1≈3.9525, such that 2^3 < 3^2.

@peterkron3861 4 ай бұрын

In the area proof, the critical step is applying the fundamental theorem of calculus. However, this handwaving “Applying the FToC we see that…” is totally unsatisfactory for what is offered as a “visual proof”.

@kennnnn 3 ай бұрын

A^B B^A // log_A() both sides B A*log_A(B) this is simple comparison now 👍

@sundareshvenugopal6575 4 ай бұрын

If the sum of the numbers is a constant, their product is maximum only when the numbers are equal. To maximize the product N must be split into √N equal parts. This seemingly trivial fact, we choose to overlook, dismiss, or ignore, always has ginormous implications and consequences, not to mention applications, and uses. So it is with all things in life. The littlest of things always make the biggest of different. Every number always counts and all numbers always count.

@oinvestigard 4 ай бұрын

if a isnt eaqual to b, b^a is the biggest.

@fabiansolisvalencia2158 3 ай бұрын

If, let's say, a = 3, b = 4, then, a^b > b^a. 3^4 > 4^3. 81 > 64. a ≠ b. a < b. ∀ (e ≤ a) ^ (a < b) ⇒ a^b > b^a. Read out loud, "For all values where e (clarification, e is Euler's number, a constant, which is weakly approximate [=~] to 3 and e < 3) is smaller or equal than a, and a is smaller than b, then a to the power of b is greater than b to the power of a, which I've already demonstrated (3^4 > 4^3). This holds true both ways. Given a = 1, b = 2, In which case, a^b < e, a^b < b^a. 1^2 < 2^1. 1 < 2. So, in conclusion, given that b is greater than a, and a is greater than 3, a^b is always greater than b^a. Unless we use negatives, in which case it gets... Complicated

@thenew3dworldfan 4 ай бұрын

I have another question, if we change the domain from real numbers to Natural numbers, how do we solve is a^b

@agustinfranco0 3 ай бұрын

theres isnt that much "visual" here tbh

@sundareshvenugopal6575 4 ай бұрын

Most of them do not even seem close. If A is an n digit number then A^(B) is an n*B digit number, give or take a digit.

@pauselab5569 4 ай бұрын

It’s difficult to know whether something like e-1 to the e+1 is bigger than e+1 to the e-1 how do you get that instead of trying?

@Naej7 4 ай бұрын

And that’s why you need a long password, not a complex one

@Simpson17866 3 ай бұрын

There's an XKCD about that :D "correct horse battery staple"

@GreenMeansGOF 4 ай бұрын

I have done the second challenge problem in the past. If 1

@BeaDSM 4 ай бұрын

2^4 = 4^2

@_rd_5043 3 ай бұрын

@BeaDSM 3 ай бұрын

@@_rd_5043 yeah exactly.

@dtikvxcdgjbv7975 4 ай бұрын

Obe must watch video at speed of 0.25 to catch all.

@nazrulhaque260 4 ай бұрын

Thank you sir!!!❤

@yusufdenli9363 4 ай бұрын

Amazing... thanks.

@MathVisualProofs 4 ай бұрын

👍😀

@MrEliseoD 4 ай бұрын

What about in the case of 2^4 and 4^2? Kinda falls apart there…

@MathVisualProofs 4 ай бұрын

Yes. That’s the ending challenge in the video. Figure out what to do when one is larger than e and one is smaller.

@mater5930 4 ай бұрын

Speaking as someone who only know elementary mathematics. The question in the title is simple but the explanation seems verbose. If you take a base 1 and raise it to the power of 1 billion, it will always be 1. But 1 billion to the power 1 is a billion time larger. My assumption is there are an infinite number of cases where the larger base is greater than a larger exponent. Similarly, there are an infinite number of cases where the opposite is true. The question cannot be answered. Unless we can determine infinity.

@JFxPress0 4 ай бұрын

Yes indeed, likely there are infinite solutions when the modal input is approaching limits ranges where A = [ 0, ±1, ∞, f(nⁿ) ] and B = [ 0, ±1, ∞, f(nⁿ) ] etc.

@eddtlpz 4 ай бұрын

As you may have noticed, there's no way to determine the question in the title unless A and B are both less than e or greater than e, where the answer is, as shown in the video, the base closest to e is gonna be greater once raised to the power of the other number.

@beaverbuoy3011 3 ай бұрын

The question has been answered by breaking it into cases, as seen in the video

@shivpatel3311 3 ай бұрын

Just because 2 things are infinite in size does not mean there are equal amounts of both things. In math this concept is known as countable vs uncountable infinites. Fun fact there are more numbers between 0 and 1 than there are integers (even though both are infinite)

@JFxPress0 3 ай бұрын

Yes, yes indeed. Fun Fact: negative infinity is not equal to positive infinity -∞ ≠ +∞ Nor are certain continuums necessarily equal to certain "countable" ( limit approaching ) infinity groups or even unlimited infinity groups like integers... Fascinating how mathematics still has Undefined groups that are not even calculable according to modern theories. We need new terminology as addressed in statistics, stochastics, data science and AI to perhaps develop new improved concepts like 0 zero infinity or zero 0 continuums or null set or 1/0 x/0 undefined or undefined infinity or UI or AI or imaginary numbers or countable undefined or countable containable unidentified undefined function formulae ordination organization ( UFO) 👾 👽 ???!!!