The correct answer is Г(1 + 1/p). Setting y = x^p, we have dx = y^(1/p - 1)/p dy, thus the integral becomes exp(-y) y^(1/p - 1)/p dy that is Г(1/p)/p or Г(1 + 1/p).
Well, obviously π₂ ≈ 3.141592 = 0.785398×4. And π₁ is ½ the "circumference" of a square defined by points (1,0) (0,1)(-1,0) (0,-1) exactly equal to 2√2 ≈ 2.828427.
12:37 This list looks like the area of the shape in the first quadrant with radius r=1, not \pi_p! Examples: p=2: A = \pi/4. p=1: A = 1/2. p=\inf: A = 1.
In my Uni course, this was just a general result, the desired integral is Γ(1+1/p) with a simple u = x^p substitution. Another similar result was that of ∫{0 to ∞} exp[+_ ix^(p)] dx, which you may like to make a separate video on.
I bought Squigonometry because of your suggestion, and reading it really occupies headspace. the moment I saw this problem i thought of those weird crcles.
A few issues stand out here. 1- Defining the family of curves as the parameterisation of the circle with respect to the p-norm on R^2 does not guarantee that they are well-behaved. For example: the norm given by |x| + |y| produces piece-wise parameterisations that are not continuously-differentiable. 2- The definition of pi_p as the total arc length of a single revolution is cumbersome to work with analytically, even though it has a simple geometric interpretation. 3- The integral itself can be quickly expressed in terms of the gamma function via a change of variables, rendering this roundabout journey quite moot. It is still an interesting premise, but a better - and arguably more useful - way to frame it would be expressing pi_p in terms of the gamma function.
It's really just an aesthetic preference, but I prefer to define the equation of the p-th circle as: (x^2)^(p/2) + (y^2)^(p/2) = 1 that way, we avoid the ugliness of the absolute values ;-)
Hi, Good result and proof. Amazing generalized trigs functions. I had proved this result otherwise some time ago, and I think it works even if p is not integer.
12:27. To me, 2pi_p should be defined as the value of t for which one full loop around the p-circumference is made. It may not necessarily coincide with the arc length of that curve. In fact, from the table at 12:35 I see the value of pi_1=0.5, which by no means is the arclength. (Actually, the value of pi_2 suggests that it should be pi_p/4, but then we should have pi_1 = sqrt(2)/2, not 0.5). Or maybe I have missed something?
Nice approach, great video, but just a small issue, shouldn't Pi2 in the table be equals to: 3.14156.... (or the normal pi?) why in 12:37 it shows like 0.785398?
You didn't demonstrate that the two integrals are equal. This is necessary for the given integral to equal the square root of their product. I expect that showing that the 'squig' functions are simply shifted by a quarter period would suffice -- since both integrals compose 1/4 period -- but the correct symmetry might also be needed. I begin to suspect that these functions might form an orthogonal basis of a generalized Fourier transform. I also wonder whether there is an appropriate Euler's equation...
x and y in the integrals have nothing to do with the x and y coordinates of the plane Michael chooses to draw the circles in. It is just a variable of integration like any other, you just can't use the same one twice, so he used y for the one after x. You can replace x and y with s and t, if you want.
@@byronwatkins2565 so the first equals sign is suddenly invalid because Michael later chooses to apply geometric reasoning? this integral is clearly a positive real number for any p>0, so it is equal to the square root of its square. end of story.
It just occured to me, what if you define f(x): (0, inf) -> [0, 1] as f(x) = cos_x(a) for some fixed real number a and then take the derivative with respect to x? And what would the graph of such function f(x) be?
geometrically, what you're doing is drawing a normal 2-circle/Euclidean circle around the center _c_ = (1,0), and with radius _a_. And then your function value is the x-coordinate of the singular point in the first quadrant/with positive y coordinate where the p-circle around the origin intersects your circle around _c_. It would be a function which is always strictly increasing in _x_, and for values of _a_ smaller than 1, it would always start at 1-a and climb to 1 in some (presumably smooth) manner, whereas for values of _a_ larger than 1, its limit at x=0 is 0, and its limit at x=infinity is 1-sqrt(a^2-1). sin_x(a) would behave similarly, but would start out going from 0 at x=0 to _a_ at x=infinity, and then for values of _a_ larger than 1, would switch to going from sqrt(a^2-1) to 1.
I have a summation problem, hope someone can solve it It's a double summation of 1/(mn)^2, where m goes from 1 to infinity and n goes from m to infinity, hope you can solve it
I found a general formula for these sorts of problems: The double sum m=1 to m=infinity n=m to n=infinity of (n×m)^-s is given in a closed form as ((-1)^s×zeta(s)^2)/(2×gamma(s)) |+zeta(2s)/2
I think I'm going to call these type a function called the Di-sum function (because of its relation to the Di-gamma function) . It would look like a capital sigma that takes parameter s. Have a nice day with your answer gang 💪💪
Are we using the definition of the sinp and cosp to define the dynamical system, or are we using the dynamical system to define sinp and cosp? If the latter I would say that we didn’t really prove that x^p + y^p = 1, and if the former, we didn’t really prove that the functions sinp and cosp satisfy the system, so I’m confused, it seems like circular logic.
Also the pi_p values don’t seem right, if p = 2 value is the unit circle, then wouldn’t pi_2 = pi not pi/4 like we have there? Or does it have to do with the fact that we are restricting ourselves to the first quadrant? So we are taking the angle it takes to go halfway around the first quadrant which would be (2pi/4)/2
![ค้างคืนโรงเก็บกระสวยอวกาศร้างกลางทะเลทราย [Around The World Ep. 3]](http://i.ytimg.com/vi/E5bWZQhABJE/mqdefault.jpg)
They: which kind of girl you like?
Me: someone as loyal as goodplacetostop.
Lol
Even better, someone as loyal as Michael with the floor function.
When backflip? 😮
This guy looks like he can control backflip
@@eqwerewrqwerqre In some of his videos he indeed performs a backflip as part of a transition.
The correct answer is Г(1 + 1/p).
Setting y = x^p, we have dx = y^(1/p - 1)/p dy, thus the integral becomes exp(-y) y^(1/p - 1)/p dy that is Г(1/p)/p or Г(1 + 1/p).
no, michael has not forgotten that there is an r in the above expression, so effectively the result seems to be Gamma(2/p)/p
@@thierrychenevier3508 effectively? Γ(1+1/x)≠Γ(2/x)/x identically so only at most one can be correct.
Mathematica (when given the original problem) agrees with @AlexanderKhlebushchev's answer.
Why do you call this "the" correct answer? There are several ways to state the answer, Michael's answer also was correct.
My answer also: (1/p)*gamma(1/p)
valid for p=2
12:35 Is that a chart of p and π_p, or a chart of p and 1/4 π_p? π_2=π, and the given value for p=2 looks like π/4.
For p = 1 it'd have to be 2sqrt(2) I think, no? So a factor of 4sqrt(2) somehow?
It's 1/4 of pi. On the chart it tends to 1, and will be 1 at infinity
@@plislegalineu3005For p=1, the value should be 0.707107, not 0.5.
For p=1, the value should be 0.707107, instead of 0.5, assuming the values tabled are pi_p/4.
Well, obviously π₂ ≈ 3.141592 = 0.785398×4.
And π₁ is ½ the "circumference" of a square defined by points (1,0) (0,1)(-1,0) (0,-1) exactly equal to 2√2 ≈ 2.828427.
15:22
12:37 This list looks like the area of the shape in the first quadrant with radius r=1, not \pi_p! Examples: p=2: A = \pi/4. p=1: A = 1/2. p=\inf: A = 1.
In my Uni course, this was just a general result, the desired integral is Γ(1+1/p) with a simple u = x^p substitution.
Another similar result was that of ∫{0 to ∞} exp[+_ ix^(p)] dx, which you may like to make a separate video on.
I bought Squigonometry because of your suggestion, and reading it really occupies headspace. the moment I saw this problem i thought of those weird crcles.
I like that when p=2 we get the normal answer back.
It wouldn’t be a generalization otherwise!
A few issues stand out here.
1- Defining the family of curves as the parameterisation of the circle with respect to the p-norm on R^2 does not guarantee that they are well-behaved. For example: the norm given by |x| + |y| produces piece-wise parameterisations that are not continuously-differentiable.
2- The definition of pi_p as the total arc length of a single revolution is cumbersome to work with analytically, even though it has a simple geometric interpretation.
3- The integral itself can be quickly expressed in terms of the gamma function via a change of variables, rendering this roundabout journey quite moot.
It is still an interesting premise, but a better - and arguably more useful - way to frame it would be expressing pi_p in terms of the gamma function.
The solution for the differential equation is well defined only for p > 1 due to Picard lindeloff theorem and the associated lipschitz condition
@@firemaniac100Well for the singular p=1 you don't need it and for p
14:34 “a suitably gnarly integral” haha
It's really just an aesthetic preference, but I prefer to define the equation of the p-th circle as:
(x^2)^(p/2) + (y^2)^(p/2) = 1
that way, we avoid the ugliness of the absolute values ;-)
That's not ugly! |x|ᵖ+|y|ᵖ = ||(x, y)||ₚᵖ is the p-th power of the p-norm. The equation |x|ᵖ+|y|ᵖ = 1 draws the sphere by the p-norm: ||(x, y)||ₚ = 1.
I would use Gamma function,
but first I would use substitution to get it
Hi,
Good result and proof. Amazing generalized trigs functions.
I had proved this result otherwise some time ago, and I think it works even if p is not integer.
2:21 orange is p < 1 actually. The pointy edges only go out after the rotated square at p = 1.
I wish you could explore the solution more in your video. like showing different examples for P and diagrams of the starting eq etc.
Cos_p(t) = sign(cos(t)) (|cos(t)|)^(2/p) .. so we can express them as a function of normal cos and sin.
12:27. To me, 2pi_p should be defined as the value of t for which one full loop around the p-circumference is made. It may not necessarily coincide with the arc length of that curve. In fact, from the table at 12:35 I see the value of pi_1=0.5, which by no means is the arclength. (Actually, the value of pi_2 suggests that it should be pi_p/4, but then we should have pi_1 = sqrt(2)/2, not 0.5). Or maybe I have missed something?
Nice approach, great video, but just a small issue, shouldn't Pi2 in the table be equals to: 3.14156.... (or the normal pi?) why in 12:37 it shows like 0.785398?
The list is wrong, it shows the area of a quarter of the p-circle!
wow thats brilliant
You didn't demonstrate that the two integrals are equal. This is necessary for the given integral to equal the square root of their product. I expect that showing that the 'squig' functions are simply shifted by a quarter period would suffice -- since both integrals compose 1/4 period -- but the correct symmetry might also be needed.
I begin to suspect that these functions might form an orthogonal basis of a generalized Fourier transform. I also wonder whether there is an appropriate Euler's equation...
x and y in the integrals have nothing to do with the x and y coordinates of the plane Michael chooses to draw the circles in. It is just a variable of integration like any other, you just can't use the same one twice, so he used y for the one after x. You can replace x and y with s and t, if you want.
@@droid-droidsson Go to the NEXT step.
@@byronwatkins2565 so the first equals sign is suddenly invalid because Michael later chooses to apply geometric reasoning? this integral is clearly a positive real number for any p>0, so it is equal to the square root of its square. end of story.
Can we find the derivatives of cos_p and sin_p using the standard way like we do for the usual cos and sin and not the differential eq. definition?
12:34 I think I'm missing something, shouldn't pi(2) be our normal 3.14... pi?
Very interesting
Thank a lot
It just occured to me, what if you define f(x): (0, inf) -> [0, 1] as f(x) = cos_x(a) for some fixed real number a and then take the derivative with respect to x? And what would the graph of such function f(x) be?
geometrically, what you're doing is drawing a normal 2-circle/Euclidean circle around the center _c_ = (1,0), and with radius _a_. And then your function value is the x-coordinate of the singular point in the first quadrant/with positive y coordinate where the p-circle around the origin intersects your circle around _c_.
It would be a function which is always strictly increasing in _x_, and for values of _a_ smaller than 1, it would always start at 1-a and climb to 1 in some (presumably smooth) manner, whereas for values of _a_ larger than 1, its limit at x=0 is 0, and its limit at x=infinity is 1-sqrt(a^2-1).
sin_x(a) would behave similarly, but would start out going from 0 at x=0 to _a_ at x=infinity, and then for values of _a_ larger than 1, would switch to going from sqrt(a^2-1) to 1.
12:37 for p=2, shouldn't pi_p=pi?
wait
why does the exponent become r^p instead of -r^p?
can anyone answer this for me?
This is beautiful
I have a summation problem, hope someone can solve it
It's a double summation of 1/(mn)^2, where m goes from 1 to infinity
and n goes from m to infinity, hope you can solve it
7(pi) ^4/360 😁😁
Or (zeta(2)^2+zeta(4))/2
I found a general formula for these sorts of problems:
The double sum m=1 to m=infinity n=m to n=infinity of (n×m)^-s is given in a closed form as ((-1)^s×zeta(s)^2)/(2×gamma(s)) |+zeta(2s)/2
I think I'm going to call these type a function called the Di-sum function (because of its relation to the Di-gamma function) . It would look like a capital sigma that takes parameter s. Have a nice day with your answer gang 💪💪
@@OrangeFigure-wl3kp Correct thanks
11:58 In which video did you calculate the different values of Pi_p? I can't find it😭
Is there a missing negative sign in the exponent at th-cam.com/video/Nn5WadFKAew/w-d-xo.html ? (i.e the integrand r exp(-r^p)
Yes
why did you copy the link to this video???
Wow wow wow
why is pi_2 not pi?
it's a fourth of pi
The list is wrong! It shows the area of the p-circle in the first quadrant (pi/4 for p=2), not the values of pi_p!
@@rainerzufall42 yes, you are correct.
@@thefirstavenger14 U2!
Are we using the definition of the sinp and cosp to define the dynamical system, or are we using the dynamical system to define sinp and cosp? If the latter I would say that we didn’t really prove that x^p + y^p = 1, and if the former, we didn’t really prove that the functions sinp and cosp satisfy the system, so I’m confused, it seems like circular logic.
Also the pi_p values don’t seem right, if p = 2 value is the unit circle, then wouldn’t pi_2 = pi not pi/4 like we have there? Or does it have to do with the fact that we are restricting ourselves to the first quadrant? So we are taking the angle it takes to go halfway around the first quadrant which would be (2pi/4)/2
Are you shedding weight?
AsymptoticSum[(2 E^-(t x)^2)/Sqrt[\[Pi]] x/z/.t->n/z,{n,1,z},z->Infinity]