Thank you!

i think the exponents always win unless the base numbers are close to e^(1/e) or something

@kennethgee2004

2 жыл бұрын

correct as seen in both BPRP and Dr. Peyam's videos about x^(1/x) as there is a critical number at e and it is happens to be a maximum. The number who base is closest to e in any a^b vs b^a is the winner as the power wins. Which interestingly answers the Numberphile video about which is larger g64^tree(3) or tree(3)^g64, it is easy to see that g64 is closer to e than tree(3), so g64^tree(3) wins as it is the larger power.

@SimsHacks

2 жыл бұрын

That's a^b vs b^a, not a^c vs b^d.

@nathanisbored

2 жыл бұрын

@@SimsHacks yeah i realized after i said it that it didnt make sense. i suppose when all the numbers are different, you can have some cases which are really close calls and keep it interesting

This is a clever and convincing proof, of course, but strategically speaking, it only makes sense to replace 57/62 with something smaller if you already knew that the original comparison would create a ratio greater than 1. And that's a good guess because so many of these problems have the answer "larger exponent wins!" But if you didn't have that intuition, would you just have to try to work through a few possibilities of replacements both larger and smaller until you found an argument that worked? Or was there a different strategic reason to create the inequality in the direction you did?

@drpeyam

2 жыл бұрын

Yes the ratio thing is standard

@jacemandt

2 жыл бұрын

Sure...setting it up as a ratio is standard, that makes sense. But replacing it with something smaller is only standard if you expect the ratio to be bigger than 1. Why would someone trying to solve this problem try that instead of trying to replace it with something bigger. If the answer is: "There's no reason...I just tried both, and only one worked" then I think viewers would be really helped by seeing that first possibly incorrect attempt, to get an insight into your problem-solving process. Said another way: figuring out the strategy for this proof is harder than the proof itself, and as much as I like your channel (which I do!), I'd love to see more attention paid to that decision-making process.

@MichaelRothwell1

2 жыл бұрын

Please take a look at the solution in my comment, where I think the motivation for believing which number is larger is clear.

Another idea is to use that (1-1/n)^n is approx e^{-1} to approximate (9/10)^{10}. Also, great videos, I especially like the songs, congratulations!

I, in fact, did believe it after your argument

@drpeyam

2 жыл бұрын

Good

The ratio 62^7/57^8 is less than 0.032, very far from 1.

62^7/57^8=(62/57)^7/57=A then use ln we get B=7ln(1+5/57)-ln57 less than 7*5/57-ln57 as e less than 3; e^3 less than 3^3=27 wich is less than 57 so B is less than 35/57-lne^3=35/57-3 less than zero so A is less than 1

Dr Peyam, 57^8 / 62^7 = 111429157112001 / 3521614606208 thanks to PARI/GP, but doing it by hand is REAL mathematics.

Hello Dr Pi-M, please can I convolute transform of Fredholm Integral? I know that of Volterra I can take the convolution of the integral because the limit of integration is from 0 to x

@willthecat3861

2 жыл бұрын

No need to bother the prof: you have my permission. Go ahead!

@dahiru_umar

2 жыл бұрын

Thanks for your input, but still do I need to change the limit of integration?

I felt 57⁸ would be larger than 62⁷, as (dividing both by 57⁷) the comparison is equivalent to 57 vs (62/57)⁷, and as 62/57

@drpeyam

2 жыл бұрын

Is the (1+1/n)^n < e result true though?

@MichaelRothwell1

2 жыл бұрын

@@drpeyam Yes. The most intuitive way to see this is to think of (1+1/n)ⁿ as the result of investing €1 (or $1 or whatever) at a 100% interest rate, compounded n times a year. E.g. if n=1, you receive 100% simple interest after 1 year, and end up with 2 (graph the result against time as a line segment joining (0,1) to (1,2)). If n=2, you receive 50% interest every 6 months, and end up with (1 1/2)²=2 1/4 (graph the result against time as two line segments, joining (0,1) to (1/2, 1 1/2) and from here to (2, 2 1/4)). If n=10, you receive 10% interest every 1.2 months, and end up with (1 1/10)¹⁰≈2.59, and so on (graph the result against time as ten line segments joining (0,1) to (1,(1 1/10)¹⁰)). Obviously, the more often you compound, the more interest you receive. In the limit, with continuously compounded interest at 100%, you get e (graph the result as eˣ from (0,1) to (1,e), and understand why (eˣ)'=eˣ). So (1+1/n)ⁿ is monotone increasing with limit e (this limit is of course the definition of e). Using the binomial theorem: (1+1/n)ⁿ=1 + n(1/n) + n(n-1)/2!×(1/n)² + n(n-1)(n-2)/3!×(1/n)³+... (sum of n+1 terms) =1 + 1 + 1(1-1/n)/2!× + 1(1-1/n)(1-2/n)/3!+... (sum of n+1 terms)

@LeviATallaksen

2 жыл бұрын

Yes, it's well known that (1+1/n)^n is an increasing sequence converging to e. So it never exceeds e.

@MichaelRothwell1

2 жыл бұрын

@@LeviATallaksen and in any case, I gave two proofs.

@drpeyam

2 жыл бұрын

Two proofs?

What’s the hardest version of this comparison of powers problem?

@drpeyam

2 жыл бұрын

You tell me :)

@kumardigvijaymishra5945

2 жыл бұрын

If I build a tower of these powers problem, they can become quite hard. For example, 99^81^87^96^113^67 versus 81^67^113^99^87^96. Luckily the verson Peyam solved can be easily solved using calculator.

@dugong369

Жыл бұрын

@@drpeyam 2.5^3 vs 3^2.5 ?

Quick version: the exponent always wins. Lol

57*8 > 62*7 so the first one uwu

isnt this equivalent to the problem for lim(x-5)⁸/x⁷ as x tends to 62 which in that case the problem becomes much more obvious if this is a valid approuch

Again :-)

We can easily see (9/10)^7 > 0.3

@drpeyam

2 жыл бұрын

How?

@deinauge7894

2 жыл бұрын

@@drpeyam (9/10)^n > 1-n/10 for n>1. seen easily from 0.9^2=0.81>0.8 and so on

@drpeyam

2 жыл бұрын

How do you know the first inequality?

@pauldifolco5736

2 жыл бұрын

@@drpeyam (9/10)^n = (1-1/10)^n = 1 - n/10 + … (binomial expansion) > 1 - n/10 -> (9/10)^7 > 1 - 7/10 = 0.3. 57*0.3 >1 so 57^8 wins.

@adityadwivedi4412

2 жыл бұрын

@@drpeyam maybe he is using binomial theorem