Very nice video, following the exploratory style that we all love. There's another nice way to distinguish between the m = 3 and m = 4 candidates for the sum of squares. So in the sum S(n) = 1^2 + 2^2 + 3^2 + . . . + n^2, each term is less than or equal to n^2. Hence since there are n terms in the sum, S(n) n^3, because n^4 grows faster than n^3. Therefore we discard the m = 4 case and proceed with m = 3. Overall very well done, hope you keep making videos. :)
@@laweaphysics4289 You're very welcome :) Yeah, that's the wonderful thing; there's always more than one way to go about it, and there's a degree of personal taste involved sometimes.
Lol nice joke/tease. Just to make it clear for other viewers this is just a joke the argument is not BS (circular) cause its just using the heights of the objects instead of the volumes. Also yes true to the label we do avoid integrals but that does not necessary mean we avoid calculus. In this case looking at limits, indeterminate forms, and a lot of perseverance to go through lots of computational algebra is all that's needed. This proof is INTEGRAL FREE!
This formula actually generalizes straightforwardly to any suspension: Any height h suspension of a 2D shape with area A, i.e., any 3D shape that has a flat 2D base with area A and goes linearly to a point at height h has volume Ah/3. In fact, any height h suspension of an (n-1)D shape with (n-1)D volume A, i.e., any nD shape that has an (n-1) base with (n-1)D volume A and goes linearly to a point at height h has nD volume Ah/n. Proof: Layer at height h-t has (n-1)D volume L_t = A(t/h)^(n-1), so the nD volume of the entire shape is int_0^h L_t dt = int_0^h A(t/h)^(n-1) dt = A/h^(n-1) int_0^h t^(n-1) dt = A/h^(n-1) h^n/n = Ah/n.
The cone is simply a stack of incrementally widening discs of radius r = Rh/H at any height h going from 0 to H with radius r going from 0 to R. Area of the disc = π r^2 = π (Rh/H)^2 and height = dh leading to the volume of each infinitesimally thin disc, dV = π (Rh/H)^2 dh = π R^2/H^2 (h^2dh). Integration can be performed between any 2 heights to get the volume of any frustum, but to get the volume of the entire cone go from h=0 to h=H. V = π R^2/H^2 (h^3/3) from h=0 to h=H = 1/3 π R^2 H
What I like is a visual proof for a piramide exists. That 3 piramids fit in a cube ( heigh* base). Thus 1/3 of base *h and then using a function that maps every slice to a cone/elliptic cone or weird complex piramide is by observation just a constant that maps very slice exactly to A (piramid) to A (any complex cone/pirmide) should logically lead to : Volume = A_piramid*h/3 to Volume = A_complex_cone/piramid*h/3, because for every slice we define the function to simply map the areas, this is a continous constant function.
I like this approach, in fact is interesting how that relates an abstract mapping with the visual example of slicing and modifying the cone/pyramid shape. And I also think is a good example to illustrate what a function does!😉😉
I think the volume of a cone is easily proved by rotating the centroid of the area of the triangle about the the y-axis. Volume = Area x 2π(Centroid) = (hr/2)(2π)(r/3).
Consider the region R delimited by the curve y = C− x^2 and the x axis. Use integration to find the value of C > 0 so that the volume of the solid obtained by rotating R about the x axis is 64√2π/15 Note: Plane sections known to the x axis are circles. Would you help me?
You could try by using cylindrical coordinates. First, note that the region could be defined in some plane xy that we’ll rotate. This is equivalent to define the region in terms of the distance from the origin (x axis when x>0) and a vertical coordinate (y) and the region is defined by the domain of integration 0
@@laweaphysics4289 Thank you so much for believing in me! Now I am preparing to my UK GCSE Maths and honestly, I have never been more motivated! These videos are pulling out a lot of the stops and they make it easier to reach the understanding needed to get things done and don't even thinking about giving up the whole mess... So... Thank you for giving your support and knowledge to the world! Peace!✌️
Form a traingle on the x & y coordinates so that the centerline of cone is on x axis and the slanting side of the cone is line passing through (0,0). Now revolve the slanting side round x axis which will generate a volume Pie .x .[y (square) dy Now integrate it Pie.x.[y (cube)/3] applying limits Volume of cone= 1/3. pie. x.[y (cube)] =1/3.pie.x .[y( cube)__0] *1/3 pie. x.y ( cube)*
@@laweaphysics4289 I have committed a mistake in it but the approach is correct. *Mistakenly in formula there is cube instead of square.*. Through integration you can also derive the formulae for area of circle, Volume of sphere and cicumfrence of circle etc
Me gustó demasiado el vídeo, las animaciones y la explicación, te felicito. Una pregunta, tú hiciste las animaciones de los cilindros infinitos en el cono? si es así, en dónde los realizaste? quedaron geniales, saludos
Hola Joel, muchas gracias por tu comentario😉😉😉. La animación de los cilindros la hice en Power Point, haciendo que aparezcan y desaparezcan los distintos grupos de cilindros (estos son relativamente transparentes para que se viese el cono de fondo)
Thanks! I mainly use Power Point to make the animations and any recording program is ok for the voice. You could also utilize audio programs as Audacity for fixing tones and noise.
The cool part is that the core of the proof applies to more than cones but to pyramids as well - the base doesn't even need to be regular so far as I can tell.
Yeah, that’s true. As long as you have a “generalized cone” with section S that increases linearly with the height of the “cone” you could use the same approach even if the shape of the section changes from the base to the top. Good point this, you made me feel interesting about a new topic to study. Thanks!!!
what I appreciate about this proof is that it answers why 1/3 shows up in both pyramid and cone volume formulae, the limit of (1^2 + 2^2 + 3^2 + ... n^2) / n^3 = 1/3 . I find that much more satisfying than rotating a line :) Indeed when I first saw that pop up, I thought of the relationship between sums of cubes and squares... but when I finally remembered it (sum of cubes = square of sums) it doesn't quite fit this case... I
I combine multiple free software (cause I can’t pay expensive programs 😅😅) -I mainly use Power Point to create the geometrical objects and equations, so I can animate them. -I also employ Audacity if I want to precisely modify audio clip. -And iMovie, Movie maker or your device’s incorporated video editor for joining the video clips with audio.
Muy buen video. Una pregunta,¿ Cómo reordenas las sumas parciales para que te quede con la ecuación de un polinomio? Entendí el razonamiento para determinar el grado del polinomio y demás pero me falta justamente el asegurar que es un polinomio. Muchas gracias!!
Gracias Ángel. Si no me equivoco preguntas por cómo llego a escribir la forma general de la suma hasta n^2 en forma de polinomio. En este caso he supuesto que la suma, al ser una función de n “que se comporta normalmente”, se puede desarrollar en potencias. Si el desarrollo no es infinito te queda un polinomio, obtienes el comportamiento lineal que muestro en el vídeo, y a partir de ahí se puede seguir. La demostración más rigurosa, tal y como yo lo veo, sería en los siguientes pasos: -Planteamos todo como en el vídeo y con lo que te he explicado. Por tanto, la fórmula del polinomio de grado 3 es nuestra “candidata” a fórmula general. - Se supone que la fórmula es la correcta para la suma de los cuadrados y se demuestra formalmente por inducción matemática, con lo que se establece que realmente hemos encontrado la fórmula buena. Espero que la aclaración te ayude, y si necesitas cualquier cosa no dudes en preguntar. Un saludo!!😉😉😉
the summation of squared terms can be generalized in a easier way with a little geometry... you can find it here... ua-cam.com/video/aXbT37IlyZQ/v-deo.html
Nice try. Would be much easier to use Cavalieri's principle, pyramid with its reflection. If you are really interested in this task, ask yourself "what exactly three here"? You will be surprised how far this question goes and how many interesting things opens. Here is a clue for you for the first step: 3 here is the number of dimensions. For flat triangle you will gave 2 and for 4 dimensional cone you will have 4. But the way you find it will lead you to many discoveries in math.
Of course, the point here is to: -Show a method for the people who don’t know how to integrate. -Use a large variety of mathematical techniques. Anyway, I hope you liked the video. 😉😉
Exactly, the key is in the techniques used to solve it (graphing functions, linear equations and finding the grade of the polynomial), so I try to present them in a way you could apply them in other mathematical problems.
Very nice video, following the exploratory style that we all love. There's another nice way to distinguish between the m = 3 and m = 4 candidates for the sum of squares. So in the sum S(n) = 1^2 + 2^2 + 3^2 + . . . + n^2, each term is less than or equal to n^2. Hence since there are n terms in the sum, S(n) n^3, because n^4 grows faster than n^3. Therefore we discard the m = 4 case and proceed with m = 3. Overall very well done, hope you keep making videos. :)
Wow, I love your appreciation. I like to discover new ways to solve this kind of mathematical problems. Thanks for the support! 😉😉
@@laweaphysics4289 You're very welcome :) Yeah, that's the wonderful thing; there's always more than one way to go about it, and there's a degree of personal taste involved sometimes.
Totally agree, that’s what makes people uplift!
you: let's avoid integrals
also you: let's aproximate with cylinders
😂😂😂😂 Too much accurate!
This is literally also integration. We are just starting from the definition, instead of using known integrals.
Lol nice joke/tease. Just to make it clear for other viewers this is just a joke the argument is not BS (circular) cause its just using the heights of the objects instead of the volumes. Also yes true to the label we do avoid integrals but that does not necessary mean we avoid calculus. In this case looking at limits, indeterminate forms, and a lot of perseverance to go through lots of computational algebra is all that's needed. This proof is INTEGRAL FREE!
I love these sort of archimedean proofs
I’m glad you like it!
I like how he performed integration without mentioning it through its notation once.
This formula actually generalizes straightforwardly to any suspension:
Any height h suspension of a 2D shape with area A, i.e., any 3D shape that has a flat 2D base with area A and goes linearly to a point at height h has volume Ah/3.
In fact, any height h suspension of an (n-1)D shape with (n-1)D volume A, i.e., any nD shape that has an (n-1) base with (n-1)D volume A and goes linearly to a point at height h has nD volume Ah/n.
Proof:
Layer at height h-t has (n-1)D volume
L_t = A(t/h)^(n-1),
so the nD volume of the entire shape is
int_0^h L_t dt
= int_0^h A(t/h)^(n-1) dt
= A/h^(n-1) int_0^h t^(n-1) dt
= A/h^(n-1) h^n/n
= Ah/n.
This is the best explanation that I could found for that generalization. Really helpful this insight. Thank you so much!! 😉😉😊😊
I love mathematics
The cone is simply a stack of incrementally widening discs of radius r = Rh/H at any height h going from 0 to H with radius r going from 0 to R.
Area of the disc = π r^2 = π (Rh/H)^2 and height = dh leading to the volume of each infinitesimally thin disc, dV = π (Rh/H)^2 dh = π R^2/H^2 (h^2dh).
Integration can be performed between any 2 heights to get the volume of any frustum, but to get the volume of the entire cone go from h=0 to h=H.
V = π R^2/H^2 (h^3/3) from h=0 to h=H = 1/3 π R^2 H
What I like is a visual proof for a piramide exists. That 3 piramids fit in a cube ( heigh* base).
Thus 1/3 of base *h and then using a function that maps every slice to a cone/elliptic cone or weird complex piramide is by observation just a constant that maps very slice exactly to
A (piramid) to A (any complex cone/pirmide) should logically lead to :
Volume = A_piramid*h/3 to Volume = A_complex_cone/piramid*h/3, because for every slice we define the function to simply map the areas, this is a continous constant function.
I like this approach, in fact is interesting how that relates an abstract mapping with the visual example of slicing and modifying the cone/pyramid shape.
And I also think is a good example to illustrate what a function does!😉😉
Very nice proof and excellent graphics! Thank you.
You’re welcome!!!
I think the volume of a cone is easily proved by rotating the centroid of the area of the triangle about the the y-axis. Volume = Area x 2π(Centroid) = (hr/2)(2π)(r/3).
Great👍
LIMIT is a strong & mutual bridge between algebra to calculus.....you have proven this concept with excellence.....keep it up👍
Thank you so much! I’m glad you’ve liked the approach to the problem
Spotted an error at 7:11 , the 3rd row should be : the same, but the a_1*3 instead of a_1*2. Since n=3 there.
That’s true, thank you so much. I’m adding that in the description.
Maths is consistent because, nevermind what human language we speak and where we start out from, we understand each other perfectly.
Totally agree!!
This is so elegant! Love the visual explanation for it.
Thanks! Glad to see that you liked the animation support 😁😁😉😉.
S can also be written as n(n+1)(2n+1)/6
Very nice video. Nice choice of topic and very nice presentation.
Thanks! I’m glad you like it.
Wrapping my mind around shapes just being equivalent to a rectangular prism with side lengths that have multiplication in them and pi was such a trip.
Consider the region R delimited by the curve y = C− x^2 and the x axis. Use integration to find the value of C > 0 so that the volume of the solid obtained by rotating R about the x axis is 64√2π/15
Note: Plane sections known to the x axis are circles. Would you help me?
You could try by using cylindrical coordinates.
First, note that the region could be defined in some plane xy that we’ll rotate.
This is equivalent to define the region in terms of the distance from the origin (x axis when x>0) and a vertical coordinate (y) and the region is defined by the domain of integration 0
Btw I love that Flamengo badge
@@laweaphysics4289 Teacher, thanks for the help.
@@laweaphysics4289 The Flamengo emblem is very beautiful. It is the most stupendous emblem in Brazil.
You’re very welcome! I like Flamengo’s emblem and it’s my favorite Brazilian team.
I really hope that I will be able to understand this level of mathematics one day !
Of course you’ll do!!!
@@laweaphysics4289 Thank you so much for believing in me!
Now I am preparing to my UK GCSE Maths and honestly, I have never been more motivated!
These videos are pulling out a lot of the stops and they make it easier to reach the understanding needed to get things done and don't even thinking about giving up the whole mess...
So...
Thank you for giving your support and knowledge to the world!
Peace!✌️
Simply marvellous👌👌👌👌👌
Nicely explained
This is really high quality, I am sure that in the future you will get loads of views and subscribers!!!
Thank you so much!!!
Very nice proof.
Great video! Loved this approach
Thank you so much!
Great explanation. 👍
I loved how you tackled sum of squares
Thanks! I hope it gave you a new insight 😉😉
interesting detour for the sum squares!
Thanks!
I like the inclusion of prerequisites in the video description.
Great! I hope them helped you to follow the whole video 😉😉.
Very detailed and informative. Thanks.
It can also be proved through Calculus by integrating the equation
y = k x
Could you explain it a little bit more? I’m interested in that procedure
Form a traingle on the x & y coordinates so that the centerline of cone is on x axis and the slanting side of the cone is line passing through (0,0).
Now revolve the slanting side round x axis which will generate a volume
Pie .x .[y (square) dy
Now integrate it
Pie.x.[y (cube)/3]
applying limits
Volume of cone= 1/3. pie. x.[y (cube)]
=1/3.pie.x .[y( cube)__0]
*1/3 pie. x.y ( cube)*
Now I see your point, that’s a good approach. Thanks!
Regretted.
There are some constraints to show and express it graphically with mathatical symbols.
@@laweaphysics4289
I have committed a mistake in it
but the approach is correct.
*Mistakenly in formula there is cube instead of square.*.
Through integration you can also
derive the formulae for area of circle, Volume of sphere and cicumfrence of circle etc
Absulutely love this video
Please excuse dumb question. At 6:00, "according to the right hand side if n is larger all these terms will vanish". Why would they vanish exactly?
cool video
Thanks! 😉😉😉
Me gustó demasiado el vídeo, las animaciones y la explicación, te felicito.
Una pregunta, tú hiciste las animaciones de los cilindros infinitos en el cono? si es así, en dónde los realizaste? quedaron geniales, saludos
Hola Joel, muchas gracias por tu comentario😉😉😉. La animación de los cilindros la hice en Power Point, haciendo que aparezcan y desaparezcan los distintos grupos de cilindros (estos son relativamente transparentes para que se viese el cono de fondo)
Amazing!
Thanks!!
nice vedio :} what program u use to explain
Thanks! I mainly use Power Point to make the animations and any recording program is ok for the voice. You could also utilize audio programs as Audacity for fixing tones and noise.
The cool part is that the core of the proof applies to more than cones but to pyramids as well - the base doesn't even need to be regular so far as I can tell.
Yeah, that’s true. As long as you have a “generalized cone” with section S that increases linearly with the height of the “cone” you could use the same approach even if the shape of the section changes from the base to the top. Good point this, you made me feel interesting about a new topic to study. Thanks!!!
what I appreciate about this proof is that it answers why 1/3 shows up in both pyramid and cone volume formulae, the limit of (1^2 + 2^2 + 3^2 + ... n^2) / n^3 = 1/3 . I find that much more satisfying than rotating a line :) Indeed when I first saw that pop up, I thought of the relationship between sums of cubes and squares... but when I finally remembered it (sum of cubes = square of sums) it doesn't quite fit this case... I
just brilliant
Thanks! 😁😁😉😉
my mind is blown
Joan la tienes en español o siemprebhaces el guión en ingles?
Por el momento el idioma principal es en inglés, aunque en un futuro ampliaré el contenido.
What software you use to make this videos , please tell me
I combine multiple free software (cause I can’t pay expensive programs 😅😅)
-I mainly use Power Point to create the geometrical objects and equations, so I can animate them.
-I also employ Audacity if I want to precisely modify audio clip.
-And iMovie, Movie maker or your device’s incorporated video editor for joining the video clips with audio.
He said: "no integrals"... And then he used what is integral in disguise.
Calculus proof ftw
Muy buen video. Una pregunta,¿ Cómo reordenas las sumas parciales para que te quede con la ecuación de un polinomio? Entendí el razonamiento para determinar el grado del polinomio y demás pero me falta justamente el asegurar que es un polinomio. Muchas gracias!!
Gracias Ángel.
Si no me equivoco preguntas por cómo llego a escribir la forma general de la suma hasta n^2 en forma de polinomio. En este caso he supuesto que la suma, al ser una función de n “que se comporta normalmente”, se puede desarrollar en potencias. Si el desarrollo no es infinito te queda un polinomio, obtienes el comportamiento lineal que muestro en el vídeo, y a partir de ahí se puede seguir.
La demostración más rigurosa, tal y como yo lo veo, sería en los siguientes pasos:
-Planteamos todo como en el vídeo y con lo que te he explicado. Por tanto, la fórmula del polinomio de grado 3 es nuestra “candidata” a fórmula general.
- Se supone que la fórmula es la correcta para la suma de los cuadrados y se demuestra formalmente por inducción matemática, con lo que se establece que realmente hemos encontrado la fórmula buena.
Espero que la aclaración te ayude, y si necesitas cualquier cosa no dudes en preguntar. Un saludo!!😉😉😉
@@laweaphysics4289 muchas gracias, si respondisteis a mi duda, no había caído en lo del desarrollo en potencias.
¡Genial!
the summation of squared terms can be generalized in a easier way with a little geometry...
you can find it here...
ua-cam.com/video/aXbT37IlyZQ/v-deo.html
Nice try. Would be much easier to use Cavalieri's principle, pyramid with its reflection. If you are really interested in this task, ask yourself "what exactly three here"? You will be surprised how far this question goes and how many interesting things opens. Here is a clue for you for the first step: 3 here is the number of dimensions. For flat triangle you will gave 2 and for 4 dimensional cone you will have 4. But the way you find it will lead you to many discoveries in math.
Yaa it's really cute.
Cute pronunciation 😂😂😂
it is easier to use integration with solids of revolution
Of course, the point here is to:
-Show a method for the people who don’t know how to integrate.
-Use a large variety of mathematical techniques.
Anyway, I hope you liked the video. 😉😉
Sum of consecutive squares again! Why do figurate numbers appear everywhere!!!! And BTW I love figurate numbers - think they are cool.
Protocalculus!
An integral does the same.
5:28 bruhhhhhhhhhhhhhhhhhhhhhhhh
The word "determines" is not pronounced as you think it is pronounced.
Sorry😅😅, I’m trying to improve that.
It is hr^2 but not 1/3 pi hr^2
You guys will need to be re-educated
you literally just used the Riemann Integral.
Exactly, the key is in the techniques used to solve it (graphing functions, linear equations and finding the grade of the polynomial), so I try to present them in a way you could apply them in other mathematical problems.
That spelling of Thale's theorem is meaningless Grssk gibberish.
It would spell: TELLSS'S TESTHRSM
I just tried to use the Greek alphabet as a base font, but thanks for the correction. I hope you liked the rest of the video at least!
@@laweaphysics4289 I did.
I find Grssk (GRΣΣΚ) and Yatsssiai (Яцѕѕіаи) hilarious.
Sorry, I misunderstood you 😅😅, I agree with you in that case 😂😂. Glad to know you like it!
999th like and 100th comment somehow
Also thanks
Thanks to you for watching!! 😉😉
Hawa pani.
Astronomy.