6:01 Hmmm... I didn’t expect that ending 😂 PSA one more time : I’m taking a break from homeworks in order to take build a solid collection of exercises and be able to post one every day for a long time. But if you have an exercise you want to share for today, just post it below.
Eight celebrities meet at a party. It so happens that each celebrity shakes hands with exactly two others. A fan makes a list of all unordered pairs of celebrities who shook hands with each other. If order does not matter, how many different lists are possible?
@@manijakus I believe I have the correct answer so I'll try to make the read more option appear as a spoiler warning/cover for others. There are 7+6+5+4+3+2+1=28 possible handshakes. However, we want every combination of handshakes and not the number. Since there are 2 options for each possible handshake (to shake or not to shake), there will be 2^28=268435456. Alternatively, you can sum up the combinations and find that the sum of combinations n choose k for all k from 0 to n is 2^n as a collary of the binomial theorem. Finally, to generalize for n people, the number of edges on a Kn graph is E=n(n-1)/2 and the number of lists/combinations will be 2^E.
Love the proof without words! It would be interesting to draw that then ask the question, 'What identity does this represent?' to get students thinking about the graphical representation.
@@Tiessie Most likely the video was uploaded 2 weeks ago but set to private. So this comment just had time to squeeze in before the video went private.
Nice identity, I came up with the graphical solution myself. I love these problems which seem very complex or impossible at first but can be reduced to something trivial (difference of areas of two rectangles).
The determinant of a 2x2 Matrix is the Area defined by its row-/columnvectors. Maybe u can start from there. I don't really see a connection here at first glance though
Picture at 6:01 should've come first I guess, to first get an intuitive picture of what is supposed to happen. It's funny how different physicists' and mathematicians' intuition works. A physicist would look at an expression like that, draw a picture and be like "ah, so this should be equal to...(difference of areas)", whereas a mathematician starts with something seemingly random, shuffles it around and magically shakes out the result. When I took calculus classes with proofs I always had to remember these weird starting points of all proofs - I knew the result would eventually follow after a few steps are worked out, but the beginning is totally unintuitive. I still appreciate the rigour in your videos though! Very refreshing to see assumptions and properly proven implications, rather than "everything is a "nicely behaved" function in physics" (a.k.a. it has "all properties you could ever want or need" (even if it doesn't really)) lol.
Very nice video! Again in theme of function- inverse function, a great video could be made on the following nice Problem: find out the number of solutions of the equation a^x=log_a(x) as a>1.
That's a handy result. Let's apply it! What is the integral of the cube root of x from 1 to 8? What is the integral of the inverse sine of x from 1/2 to sqrt(3)/2? Look, ma! No integral tables!
I've a question if the function is differentiable it's monotonous. But we assumpt that f(a) = c and f(b) = d, and that means that we assumpted the function is monotonus crescent, don't we have to consider also the case in which the function is monotonous decrescent? Thank you for the answer Sorry for my bad english but I'm foreign😅
Michael, when you wrote down the first step, I thought "I bet he's going to integrate by parts." But then I found myself wondering if anyone has ever used an analogous integration technique based on the quotient rule. I'd love to hear your thoughts on this.
I'm just wondering, if the result is only true for an increasing function? I did not check, but the substituion with the integral bouds f(a) = c and f(b) = d is only correct if its a monotone increasing function, if its decreasing it would be f(a) = d and f(b) ) c instead, would it not? (it would just change the sign of the one integrand, but still... )?
Using the graphical solution presented at the end, it is easy to see that the result does not hold for stricly decreasing functions. Here is a counterexample: Take f(x) = 1 - sqrt(1 - (x - 1)²) (decreasing quarter circle with radius 1) with a = 0,b = 1, c = 0, d = 1. The sum of the two integrals is just 2*(1 - 0,25*pi) = 2 - 0,5pi = ~0.4 =/= bd - ac = 1.
@@tomtheultimatepro The graphical solution is ok, but we need to be careful with the values of c and d. In this video we have c=f(a), d=f(b), and we add some areas. In the case of a decreasing function we have c=f(b), d=f(a), and we subtract some areas. We get (integral of f(x)dx from a to b) - (b-a)*f(b) = (integral of f^-1(y)dy from f(b) to f(a)) - (f(a)-f(b))*a, and then (integral of f(x)dx from a to b) + (integral of f^-1(y)dy from f(a) to f(b)) = b*f(b) - a*f(a). This is the same equation as in the video (if we write f(a) and f(b) instead of c and d, respectively).
DIFFERENT ENDING?! Nice, great proof without words btw (:
Morning without Michael Penn Math is like a day without sunshine.
It's like existing without existing
Loved that proof without words... could be great to see more of it
6:01 Hmmm... I didn’t expect that ending 😂
PSA one more time : I’m taking a break from homeworks in order to take build a solid collection of exercises and be able to post one every day for a long time.
But if you have an exercise you want to share for today, just post it below.
Eight celebrities meet at a party. It so happens that each celebrity shakes hands with exactly two others. A fan makes a list of all unordered pairs of celebrities who shook hands with each other. If order does not matter, how many different lists are possible?
@@manijakus is it 2520
@@Thaplayer1209 No :)
@@manijakus I believe I have the correct answer so I'll try to make the read more option appear as a spoiler warning/cover for others.
There are 7+6+5+4+3+2+1=28 possible handshakes. However, we want every combination of handshakes and not the number. Since there are 2 options for each possible handshake (to shake or not to shake), there will be 2^28=268435456. Alternatively, you can sum up the combinations and find that the sum of combinations n choose k for all k from 0 to n is 2^n as a collary of the binomial theorem.
Finally, to generalize for n people, the number of edges on a Kn graph is E=n(n-1)/2 and the number of lists/combinations will be 2^E.
@@manijakus 3507
... And the video didn't stop to this day.
Yes, there was never a good place.
Love the proof without words! It would be interesting to draw that then ask the question, 'What identity does this represent?' to get students thinking about the graphical representation.
The proof without words was simply beautiful.
I love the geometric visual proof at the end!
Excellent proof accompanied by an action movie soundtrack!
Loved the proof-without-words, but sorely missed the final comment.
dice oculto, solo yo puedo verlo? I was surprised by the graphic proof
Time traveller?!
@@Tiessie Most likely the video was uploaded 2 weeks ago but set to private.
So this comment just had time to squeeze in before the video went private.
Wolo bulo pesudo fucko bulo shito
@@goodplacetostop2973 Tienes razón, este video me indico que era privado cuando lo vi.
Wow, that visual proof was lovely. So simple, and great music, I think that’s something you could introduce more If you like it
Simple and beautiful. wow !!
I’ve been up all night trying to figure out if that’s a good place to stop
Nice identity, I came up with the graphical solution myself. I love these problems which seem very complex or impossible at first but can be reduced to something trivial (difference of areas of two rectangles).
That representation, that little graphical representation is called happiness...
That ending shrug mocks us to the end of time, as there will never be a good place to stop
This is how you integrate all inverse trig functions. Transforms them from finding areas with the x axis, to finding areas with the y axis.
bd-ac = det[[b,a],[c,d]]. I wonder if it means something, if there is any connection
The determinant of a 2x2 Matrix is the Area defined by its row-/columnvectors. Maybe u can start from there. I don't really see a connection here at first glance though
Yeah someone really needs to do the work for that. The outcome is gonna be rewarding and illuminating af.
@@uncreatedlogos it may also merely be a coincidence.
Picture at 6:01 should've come first I guess, to first get an intuitive picture of what is supposed to happen. It's funny how different physicists' and mathematicians' intuition works. A physicist would look at an expression like that, draw a picture and be like "ah, so this should be equal to...(difference of areas)", whereas a mathematician starts with something seemingly random, shuffles it around and magically shakes out the result. When I took calculus classes with proofs I always had to remember these weird starting points of all proofs - I knew the result would eventually follow after a few steps are worked out, but the beginning is totally unintuitive. I still appreciate the rigour in your videos though! Very refreshing to see assumptions and properly proven implications, rather than "everything is a "nicely behaved" function in physics" (a.k.a. it has "all properties you could ever want or need" (even if it doesn't really)) lol.
thank you. sir.
Very nice video! Again in theme of function- inverse function, a great video could be made on the following nice Problem: find out the number of solutions of the equation a^x=log_a(x) as a>1.
That's a handy result. Let's apply it!
What is the integral of the cube root of x from 1 to 8?
What is the integral of the inverse sine of x from 1/2 to sqrt(3)/2?
Look, ma! No integral tables!
please take a look at question 5 of the 1997 international math Olympiad
I remember I did one of the STEP papers and I was required to prove this identity using graph lol or at least something very similar to this one
i like the proof without words. But we still need a good place to stop :(
Hehe for the very first time, I found the answer right away when I read that f was differentiable 😅
Can't wait to see your next video
Geometrical proof is more illuminating than analytic.
Great Sir
Reminds me of determinant somehow...
Maybe there is a clearer connection in multiple variables.
I remember this problem in Purcell's book
what a delightful result
I've a question if the function is differentiable it's monotonous. But we assumpt that f(a) = c and f(b) = d, and that means that we assumpted the function is monotonus crescent, don't we have to consider also the case in which the function is monotonous decrescent? Thank you for the answer
Sorry for my bad english but I'm foreign😅
Yes, that case hadn't been handled.
In that case, the identity has to have a plus sign instead of a minus sign... Or c and d reversed
@@sundeep0207 Thank you!
Wow that was awesome!!
Michael, when you wrote down the first step, I thought "I bet he's going to integrate by parts." But then I found myself wondering if anyone has ever used an analogous integration technique based on the quotient rule. I'd love to hear your thoughts on this.
Above all, Michael Penn knows how to communicate Maths.
That ending was nice
Beautiful
Awesome no words proof. What about the divergence of sum_{n>0}\dfrac1{n^{2+\cosθ}}
Without "and this is a good place to stop" I thought the video cut out on me.
err, does anyone know what the music at the end was?
It was some clip included in Final Cut Pro X
@Anant Mishra Thank you.
Pretty good ending proof...
I like the punchline.
here's a proof [blink] ... what proof (I like the geometric proof of pyth. theorem ... the last bit was great, and I did almost miss it)
Bravo!
04:51 dx'es flying through this room!! 🤣
What if f’(x) is 0 at some x in the domain on the first step?
Actually... isn't this a proof for integration by parts? Is there any way you could avoid using ibp during this derivation?
I guess that there's something different today, but what?
i prefer the graphical method. draw an arbitrary line and shade the respective area the integrals represent
How can we know that is a good place to stop?
I see it again and again, because I don't see the good place to stop 🤣
stylish
NOOOOOOO where (or when) is the good place to stop?!?!???
good
Why does f(a) necessarily equal c?
Because it is biyective in [a,b]->[c,d]
Since he assumes the function is monotonically increasing, the smallest number in the domain must map to the smallest number in the range.
thank you :))
¡ BEAUTIFUL !
Can graphing this alone prove the equation?
I'm just wondering, if the result is only true for an increasing function? I did not check, but the substituion with the integral bouds f(a) = c and f(b) = d is only correct if its a monotone increasing function, if its decreasing it would be f(a) = d and f(b) ) c instead, would it not? (it would just change the sign of the one integrand, but still... )?
replacing f by -f gives you a decreasing function as he said in the video. and then you can simply rename the variables -c and -d to c and d.
If f(x) is decreasing, then -f(x) is increasing. and you can swap sign of and integral by swapping the boundaries.
Using the graphical solution presented at the end, it is easy to see that the result does not hold for stricly decreasing functions. Here is a counterexample: Take f(x) = 1 - sqrt(1 - (x - 1)²) (decreasing quarter circle with radius 1) with a = 0,b = 1, c = 0, d = 1. The sum of the two integrals is just 2*(1 - 0,25*pi) = 2 - 0,5pi = ~0.4 =/= bd - ac = 1.
@@tomtheultimatepro The graphical solution is ok, but we need to be careful with the values of c and d. In this video we have c=f(a), d=f(b), and we add some areas. In the case of a decreasing function we have c=f(b), d=f(a), and we subtract some areas. We get (integral of f(x)dx from a to b) - (b-a)*f(b) = (integral of f^-1(y)dy from f(b) to f(a)) - (f(a)-f(b))*a, and then (integral of f(x)dx from a to b) + (integral of f^-1(y)dy from f(a) to f(b)) = b*f(b) - a*f(a). This is the same equation as in the video (if we write f(a) and f(b) instead of c and d, respectively).
Hi,
And without any good place to stop.
surprising ending 👍
Wow that is so cool
Nice sir
So this wasn't a good place to stop 🤔
Hahahhahaha very fun video! Great Job
but... but... where's the good place to stop?
I am here 🙋
Yo that's cool
Hehe, when Michael gets tired of math he can work as a pantomime actor.
Great 💖💖💖
Yes!!! Thank you sexy man!
___ ______ _ ____ _____ __ ____
😍😍😍
Ur accent is too sophisticated
But where’s a good place to stop?