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

In this video, we present two animated visual proofs demonstrating that if e is less than or equal to A and A is less than B then A^B is greater than B^A . Moreover, we include a bonus proof that is sometimes referred to as the standard way to get this result.
Correction : at 0:57 I had misspoken, so I cut 1 second, which is why the tangent line appears so fast and the audio has a break.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualPr.... Thanks!
This animation is based on two visual proofs, one by Charles D. Gallant from the February 1991 issue of Mathematics Magazine (doi.org/10.2307/2690451 ) and the other by Nazrul Haque from the January 2021 issue of Resonance (doi.org/10.1007/s12045-020-11... ). Notice that plugging in A = e and B = Pi, we get Pi^e is less than e^Pi.
If you are interested in the general question about A^B vs. B^A, check out this article from Bikash Chakraborty and Nazrul Haque: link.springer.com/article/10....
If you like this video, check out my playlists on inequalities and calculus:
• Inequalities
• Calculus
#mathshorts​ #mathvideo​ #math​ #inequality #mtbos​ #manim​ #animation​ #theorem​ #pww​ #proofwithoutwords​ #visualproof​ #proof​ #iteachmath #calculus #inequality #logarithm #naturallog #exponential
To learn more about animating with manim, check out:
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[@Kounomura]

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]

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

[@lyrimetacurl0]

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

[@hasanjakir360]

Where can I get the proof?

[@STEAMerBear]

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]

Yes. GeoGebra would be great for this.

[@asparkdeity8717]

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.

[@xjuhox]

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 👀

[@GreenMeansGOF]

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

[@plamenpenchev262]

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

[@sonicwaveinfinitymiddwelle8555]

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.

[@Achill101]

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=

[@nazrulhaque260]

Thank you sir!!!❤

[@deananderson7714]

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]

Damn I really really like that first proof

[@MathVisualProofs]

😀😎

[@kennnnn]

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

[@StevenTorrey]

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

[@yusufdenli9363]

Amazing... thanks.

[@MathVisualProofs]

👍😀

[@WhattheHectogon]

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

[@cupostuff9929]

That's what i was thinking

[@MathVisualProofs]

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]

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]

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

[@oida10000]

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

[@pauselab5569]

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?

[@DOTvCROSS]

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

[@economicist2011]

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!

[@thenew3dworldfan]

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

[@zsoltnagy5654]

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]

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]

@@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.

[@sundareshvenugopal6575]

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.

[@oniondesu9633]

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

[@MathVisualProofs]

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.

[@guessundheit6494]

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.

[@SHIVAPRASAD-hz4tj]

Incredible that this was discovered 3 years ago!

[@MathVisualProofs]

The second proof only :)

[@beaumatthews6411]

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

[@sundareshvenugopal6575]

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.

[@Naej7]

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

[@Simpson17866]

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

[@raulgalets]

ABBA is the greatest of all time

[@Dalroc]

The third is the least complicated.

[@MrEliseoD]

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

[@MathVisualProofs]

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

[@peterkron3861]

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”.

[@oinvestigard]

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

[@fabiansolisvalencia2158]

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

[@oinvestigard]

ONGLY, if a is bigger than b oppocite oppocites that

[@bilaalahmed8923]

which font do you use?

[@MathVisualProofs]

For what?

[@bilaalahmed8923]

@@MathVisualProofs math. Your animations in your videos.

[@MathVisualProofs]

@@bilaalahmed8923 That is computer modern - the standard in LaTeX.

[@rafiahamed2978]

🥰

[@BeaDSM]

2^4 = 4^2

[@_rd_5043]

2

[@BeaDSM]

@@_rd_5043 yeah exactly.

[@dtikvxcdgjbv7975]

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

[@GenericInternetter]

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

[@Andy_Hendrix_9842]

ABBA 😁

[@mikhailnikolaev4992]

ABBB💀

[@Rippertear]

@@mikhailnikolaev4992ABBC 😲

[@Naej7]

ACDC

[@mater5930]

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]

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]

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]

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

[@shivpatel3311]

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]

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) 👾 👽 ???!!!

[@lukatolstov5598]

3⁴>4³, BUT 2³

[@MathVisualProofs]

Yes. At the end I leave the most general question for you to consider. What can you say about A

[@aldojerish1328]

Bro,if e

[@kaaristotelancien3005]

why ma = 1/A * ln(A) ?

[@MathVisualProofs]

Compute the slope of a line passing through origin : it is the y-coordinate of another point divided by the x-coordinate (rise over run)

[@agustinfranco0]

theres isnt that much "visual" here tbh

[@user-lj4yg7gw8r]

So 2³ > 3² 😂

[@vangrails]

Yes, because 2 > e.

[@Sans________________________96]

Fuck

[@vidal9747]

I just passed by to say that A and B are matrices. Use x and y like a reasonable human being.