TWO THINGS! 1. Be sure to watch 3:25 for the derivative of x^(1/x) the superman way! #KickingCalculusInItsHead #CalculusFinisher 2. Please try the problem at 11:21 2.5^3 vs. 3^2.5 I do not know if that is even possible. So, any thoughts will be greatly appreciated!

“I’m not doing 2019” proceeds to do 2020 instead poor guy never saw it coming.

For the challenge question of "bases on opposite sides of the maximum" (a

Poor calculus :( why did you kick him ?

Nice video. I'm thinking of starting my own math channel in German once I'm better. Today I started chemo, and so far feeling all normal. Hope it stays this way for the rest of the cycles.

"Thank you for this cool t-shirt that I'm hiding behind my giant ball mic so you can't see it!"

Not the solution that you’re looking for in general,but since the numbers are nice... (3^2.5)^2=3^5=243 (2.5^3)^2=(2.5^2)(2.5^2)(2.5^2)=6.25^3>244 Can be done by hand if you’re up to it. So (2.5^3)^2>(3^2.5)^2 Implies 2.5^3>3^2.5

For a while, I was confused what chen lu is, but then it got clear. What made it funny is that you call it that on purpose.

Me in homeworks: "please dont make me do the second derivative" :v

*Cries in bluepen*

This is definitely a much more rigorous mathematical proof to this type of problem, and not one I would've thought of until now. The way I solved 9^10 vs 10^9 in my head almost as quickly was this: 1. For two positive integers x and y, such that the question asked is x^y vs. y^x, which is bigger? 2. Take the y-th derivative of x^y, you ultimately end up with y!x. 3. Take the x-th derivative of y^x, you ultimately end up with x!y. 4. If you compare x and y, then whichever one is bigger will have a duplicated factor in its factorial: > A. If x < y, then y!x will have an x^2 term in it, whereas x!y will merely be x!y. Therefore, x^y > y^x. > B. But if y < x, then x!y will have a y^2 term in it, whereas y!x will just be y!x. Therefore, y^x > x^y. The 10-th derivative of 9^10 is 10!x9 whereas the 9-th derivative of 10^9 is 9!x10 which is just 10!. 10!x9 is strictly greater than 10! because it has an extra factor of 9 in it. Therefore, 9^10 > 10^9. The logic of this method is this: An n-adic function (quadratic, cubic, quartic, etc.) always increases in value faster than a linear function. Therefore, taking the respective n-th and m-th derivatives of the two functions will reveal which one is n-adic and which one is linear by virtue of their coefficients. You can conclude then that one is ultimately greater than the other. However, this technique fails when either x or y are not positive integers. Although Dr. Peyam has shown how to take the a-th derivative of a functionfor some positive real number a, it's a laborious process involving integration, and doesn't mix and match with the other differentiation rules at all. And afaik it's not possible to take the a-th derivative of a function for ANY negative value of a, integer OR real. Also, when you have two numbers that are on either side of the critical value e in x^(1/x) (or the x-th root of x), something to look for is whether a or b is less than or equal to 1. If that is the case, then the function with that (a|b) e. Therefore, 1 is another special value of the function x^(1/x). I haven't done the second derivative of it yet (and I plan to after I finish posting this comment, just for the fun of it), but I would wager that x=1 will be a point of inflection for the function because it has that special property of also being the point where the left tail of the function is permanently less than the right tail, by virtue of y=1 being the horizontal asymptote and the right tail therefore always being greater than 1. But I'm curious to see where the point of inflection is for the right tail where x > e.

Oh my God You just solved one of my greatest doubts in mathematics I knew it had some sensible method using calculus But I never tried to venture and solve it Thank you very much

here's my method: 9^10 [ ] 10^9 {ln both sides} note: the [ ] is where an equal or inequality sign goes ln (9^10) [ ] ln (10^9) {by taking the powers out of the ln} 10 ln(9) [ ] 9 ln(10) {dividing both sides by 9 & dividing both sides by ln(9)} 10/9 [ ] ln(10)/ln(9) now it's obvious the left side is bigger, because the difference between ln 10 and ln 9 is very little compared to 10 and 9, the slope for any log function, including ln, is very small after you pass the base number, which is e for ln well my method is more about logic than algebra I guess

please do the second derivative

bro that video made me so excited. like, it is kind of an art and magic! thank you for those awesome videos. I have learned a lot from your videos and always kept my math passion alive.

This video inspired me a lot! I like your enthusiasm and smiley attitude!

For your follow up question, we know that the function you describe increases first then decreases. If we pick 1 < a < e < b (if a < 1, there are no solution and b^1/b will always be larger than a^1/a), we first pick a and want to find the value x for which x^(1/x) = a^(1/a) with x>e. If we have this, then we know that for b > x, b^(1/b) will be smaller than a^(1/a), and larger if e < b < x. The problem is to solve the general equation x^(1/x) = k. If we take the log, we get to ln(x) = h.x (with h = ln(k)) which can be solved using the Lambert W function: x = e^(hx) => x.e^(-hx) = 1 => -hx.e^(-hx) = -h, we apply W on both sides to get -hx = W(-h), thus x = -W(-h)/h. In our problem, we probably need to take the other branch of the W function, since we'd get x = a otherwise. If we for example plug in a = 2, this gives us x = -2 * W(-1, -ln(2)/2) / ln(2), but since W(-1, ln(2)/2) is -2ln(2) according to Wolfram Alfa, we get to the expected value x = 4. If you plug in a = 2.5, there's no shortcut, we get x = -2.5 / ln (2.5) * W(-1,-ln(2.5)/2.5), Wolfram Alpha gives and approximate value of 2.97029. You can verify that 2.5 ^ (2.5) is approximately the same as 2.97029 ^ (2.97029), within 6 decimal places.

Woah that's some real superman way to do the derivative of x^1/x I always wished if I could do that someway without taking "ln" both sides...

I love this kind of questions, because I have watched e^pi vs pi^e video. Your videos are great man 😍

One of your coolest videos imo, really nice

Even without checking the comments my first thought was to take logarithms of both sides. It is then usually trivial to demonstrate the comparative values. This works on all positive values. Maybe negative ones too. (No idea about complex numbers and quaternions.) And you don't have to take logs using base 10 or base e. Still, it was nice to be shown the general proof.

Let's just use a calculator. Let a < e < b Is it easy to find the unique c > e such that c^(1/c) = a^(1/a)? Because then we could just use our first theorem.

You can use the shape of the function and the equality 2^4 = 4^2 to split the problem in 4 intervals 0 < w < 2 < x < e < y < 4 < z. You get some insight for some values across e. Then w^y < y^w and x^z > z^x

I may be tired of this kind of questions but i never will be of this kind of explanations. Great video! Keep it up

Or use e^ln(9^(10)) and then solve Ad you did for the other video for e^π and π^e

Very nice! Wish I had known this trick when my Calc 2 prof gave us a similar problem for homework

(5/2)^3 vs 3^(5/2) is equivalent to 5^6 vs 2^6 * 3^5. You could mentally multiply this out, but it's easier to note that 3^5 = 243 < 244 = 122 * 2, and then note that 128 * 122 = 5^6 - 3^2 from the difference of squares identity. So 5^6 > 2^6 * 3^5, and thus (5/2)^3 > 3^(5/2). Of course, this solution isn't general, but I suspect there isn't one for the case a < e < b.

2.5^3 vs 3^2.5 (squaring) 2.5^6 vs 3^5 (multyplying by 2^6) 5^6 vs 2^6*3^5 5^6 vs 2*6^5 Clearly the left hand side is greater. My issue with this is that this technique likely will not work in general.

Cool! For the case in the question, I just noticed that the first two terms of the binomial expansion of (10-1)^10 cancel out and the third (dominating) term 90/2*10^8 is big enough to easily beat both the fourth term -720/6 * 10^7 and the other side of the equation, 10^9.

The second derivative yields slope on each side of e. This could be used to compare rate of change on those sides and give a criterion for the a and b on opposite sides though the formula will be more complicated than whichever is nearer to e.

Excellent video. I saw that you did a proof involving linear algebra and the transpose of The product of two matrices. You’re awesome. That was an excellent video. You are very very clear

Thanks again for lots of fun, and intriguing math! You keep my ancient brain chugging over on at least one of its many cylinders. Two, on a good day - like this!

Hey Steve, i have to say that i just saw your channel and i already love it. You really do a great job and i love the way you are fascinated by math. Just gained another fan :)

However, if 0

I enjoy your maths. You made me understand well trigonometric identities and proofs.

That's really really cool,man!

Me: but what if ... 1 minute later - 11:02 bprp: *_I KNOW_*

I love Chan luuu! *Gives a flying kiss*

Love the way you teach math,enjoyable

This reminds me about some math a did a couple years ago with tetration! I found that when you infinitely tetrate x, you get a function of x=y^(1/y). But only for 0

He and calculus are like BFF.....who personally beats him and publicly promotes his friend. LOL!🤣

At first I forgot that they have to be gte e, thought "hey, 8 < 9", and then guessed wrong.

Thank you for the general case. For the case involving 10 and 9, considering 'decimal log' and the fact that 'log' is an increasing function:: It's known from HS log(3)=0.4771... --> 20*log(3) > 20*0.477 --> log(9^10) > 9 = log(10^9). --> 9^19 > 10^9.

If numbers are on the opposite side in some cases there is a way. 1. We have numbers a and b, that 0 < a

Great you always bring something unique :)

2.5 is going to be tough, but since most people will probably just use 2 as one of the numbers, it's useful to know that the graph has the same value at 4 allowing you to still use this method. 3^2 is "closer" to e than 2^3 because 2 is as "far" (vertically) as 4.

These videos are so fucking satisfying! I love this!!!! YES!

We can create another bound where it becomes easy again if we use the horizontal asymptote y=1. Choose any value a such that f(a) < 1 and we know it can never be greater than f(b) if b >= e

Use binomial theorem.. write 9^10 as (10-1)^10 then compare the two

I keep forgetting the order so it helps me to put in outrageous numbers, like 3^1000 vs 1000^3. Obviously the former is greater.

"I- I'm not gonna do a second derivative" I felt that

I think i get one extra year of life understanding that. Finally someone explain it clearly

he said: don't use a calculator, me: uses a calculator.

Loved how you worked with the differentiation part! Is there a proof for this?

Regarding the derivative y'=x^(1/x), there's a faster way: take y=exp{ln(x^(1/x))}. Then y'=[exp{ln(x^(1/x))}][ln(x^(1/x)]'=x^(1/x)[(1/x^2)-(1/x^2)lnx]. Nice explanation, as usual!

Hmm. The equation x^(1/x)=a has two solutions for 1

Such problems used to haunt me like anything. Thanks for this brilliant approach!!!!

I think 2.5^3 is is bigger coz 2.5 is closer to e than 3

Very nice explanation!!!

Case 1 and Case 2 - when both a and b are greater than e, then the exponent dominates regardless. Case 3 and Case 4 - when both a and b are less than e, then the base dominates regardless. Case 5 - When b

With *a < e < b* there is no simple rule, since all cases may occur. Counterexamples: *2^3 < 3^2 2^4 = 4^2 2^5 > 5^2* Concerning the challenge: Define the function *f(x) = x^(1/2)* increasing for *x > 0* and notice *3^(2.5) / (2.5)^3 = f( 3^5 / (2.5)^6 ) = f( 2 * 6^5 / 5^6 ) = f( 15552 / 15625 ) =: X* Since *f* maps *(0; 1) -> (0; 1),* the result *X* lies in *(0; 1),* leading to *3^(2.5) < (2.5)^3*

Wouldn't it be easier to take the ln on both sides of y=x^(1/x), bring the 1/x in front, and then take the derivative? Although I believe it is essentially the same thing, it would make more sense to me.

Fantastic work!👍👍👍

Watching this at 3 am and not understanding the reason behind half the stuff you did but I still understood the concept. I'll try this myself when I'm free lol

I mean come on bprp you're basically begging for us to ask you to do the second derivative!

I liked the inequality and its maxima

Wow 😮 what a trick. It definitely helped me ❤❤❤Love from India🇮

There is one case where you do know the answer of the comparison when a is below e and b above e: if 0 < a 1 for any b > e will yield that b^(1/b) > 1 >= a^(1/a). Which means a^b < b^a.

Please give lectures on real analysis

In a math exam would you have to demonstrate why you pick one or the other, or would you just pick one and say "that one is bigger" ?

I remember Doing it using the variations of the function lnx/x. Edit: this is a decreasing function for all x>e.. (in fact (e,1/e) is an absolute maximum) then for e lnb/b then blna > alnb, or ln(a^b)>ln(b^a), which finally gives a^b > b^a.

"Let me tell you, I have a secret" I hope it does not land you in difficulties with the law.

03:20 WOW! Taking the derivative of x^(1/x) using "power rule plus exponential rule". I am wondering if it works for all such functions, for example for sin(x)^ln(x).

If a is less than 1 and b is greater than e then it's pretty easy.

When considering functions like y = x^f(x) you may also take the natural logarithm of both sides and differentiate implicitly.

Your way of teaching is awesome also u help me to memorise chp. Monotonocity by this ques as I have done this type of ques during me jee prep .

Why did you add the two parts of the derivative? It's not a product rule situation is it?

What is the curvature of the graph of y=x^(1/x)? (:

Awesome as always....!!!

You can also use the tangent line to y=ln(x) (which is concave down) at x=9 to get ln(10)

Wouldn't differentiation using implicit differentiation also be easy?

can binomial theorem be used to solve this? (Can write 9^10 as (10-1)^10 and then expand it..?)

if abs(e-a) almost = abs(e-b) then and 0

There is also another way to determine which number is larger. Here if we let a=9^10 and b=10^9. Take the log of both sides using two equations, and we get log(a)=10log(9) and log(b)=9. Without looking at log(a) and log(b), we know that log(9) is somewhat a bit less than 1, which is log(10). Multiply this by 10, and the value is somewhat greater than 10. This implies that this value is greater than 9. Therefore, if you go back to the original equation and compare the numbers here, 9^10>10^9.

It’s “these kinds of questions” or “this kind of question.”

I demand a proof of that way of doing the derivative 😉

Just way too goooooood!!!! Elegant.

I think that 2.5 is closer to e than 3 is, so therefore I say 2.5^3 is bigger than 3^2.5. it's not a bullet proof way to argue, since obviously you could take 0.9^10 vs 10^0.9, and you know that 0.9^10 is strictly less than 1, whilst 10^0.9 has to be more than 1.

What kind of whiteboard is that? I don't see any marker residue or ghosting at all. How do you keep it so clean?

Interesting. I arrived to the same conclusion instinctively, because 2⁴ = 4² and exponential functions rise faster than polynomial functions, so for numbers greater than 2/4 it was the most likely answer. PS: what’s going on with the first part of your derivative? Isn’t it just chain rule?

Sir ,you have mind blowing way of teaching .....👌👌👌👌👌

When function like x^(1/x) is spotted, I think the Lambert W function can came to rescue again. The 'other' solution to x^a=a^x is (see wikipedia page for derivation) -a/log(a) W[0, -log(a)/a ], if a>e & -a/log(a) W[-1, -log(a)/a ], if a

I somehow remember the values of log3 and log5 (base 10) (crammed for an exam) and that gives me 3log2.5 > 2.5log3 Edit: upto 5 decimal places

Vier fünfmal vervierfacht macht mehr als fünf viermal verfünffacht.

Three rules of derivation one followed by the other....this is fascinating!!!!

I love your videos!!!!

I heard doraemon tune again at 0.19 although its very faint

I did the lazy estimate method. I used 10^10 to divide both. 10^10 / 10^9 = 10 10^10 / 9^10 = (10/9)^10 (10/9)^10 = (10/9)^9 * 10/9 (10/9)^10 = [(1000/729)^3 * 10/9] < [2^3 * 1.25] (which is 10) Hence 9^10 is larger.

HOLY SHIT MATE I ABSOLUTE LOVE THIS

Okay, here goes an attempt at the general form of the problem at 11:21 Start with the equation y=x^(1/x). The lim as x goes to infinity of this is 1. This means that for every value of a between 1 and e there exists a value of b >= e such that a^(1/a)=b^(1/b). If you rotate this function pi/2 radians counter clockwise and then solve for y in the range x= -(e^(1/e)) to -1 (because this would be rotated as well) you should have two solutions. To rotate any function, you can use the equation y*cos(theta) - x*sin(theta) = f(y*sin(theta) + x*cos(x)). Since our theta is pi/2 and our f(x)=x^(1/x) we get: y*0 - x*1 = (y*1 + x*0)^(1/[y*1+x*0]) which simplifies to: -x=y^(1/y) Now, solving this for y is completely beyond me. According to wolframalpha(sorry, but I just don't know this), for -(e^(1/e))