very helpful video thanks! one thing that's hard for me to understand is at 5:48 when you're describing a chord, I'm not clear on how the 4:5:6 ratio relates to the frequencies 554, 698, 831?
HELLLLLLLOOOOO. Can I ask you how or which app did you use to apply Fourier transform and obtain those frequency domain graphs? I NEED IT FOR A MATH PROJECT. PLEASE HELP ME
Great video. Thanks for making it! My only criticism is that the background music was a bit too loud (and stylistically perhaps not everyone's cup of tea)
Hi Peter that was an excellent video. I had trouble in understanding translating the time waveform about the origin, is it polar coordinate transformation of the sine wave ? I didn't understand the part where you mentioned about the degree of freedom to change frequency about the origin. Can you please clear that by siting any reference ?
The volume problem can be much simpler than that. To find the height of the cross sectional area, take (1/2a)^2 + h^2 = a^2. You have 1/4a^2 * h^2 = a^2. Subtract 1/4a^2 from both sides, then you have h^2 = 3/4a^2. Taking the square root, you have sqrt(3)/2a. Multiply 1/2(sqrt(3)/2a * a) and you get sqrt(3)/4a^2. Now you're integrating from 0 to h(height of tetrahedron), with the cross sectional area so a =0 and a = h. To find the height of the tetrahedron, draw a line connecting the left corner tetrahedron to the centroid of the base and draw another line connecting the centroid to the tip of the tetrahedron. Project that onto an a equilateral triangle with the side length of a. You should have a line bisecting the left corner of the triangle, which is x and has a 30 degree angle from the base . So using SOHCAHTOA, 1/2a/x = cos(30), x being the hypotenuse and a being the adjacent side. (1/2a)/x = sqrt(3)/2. a = xsqrt(3)/2. So x = a/sqrt(3)/2, which x = 2/sqrt(3)1/2a. x = 1/sqrt(3)a, which is sqrt(3)/3a. To find the height of the right triangle, it is h^2 + (sqrt(3)/3a)^2= a^2. So h^2 + 3/9 = a^2. H^2 = 6/9a^2, which means h= sqrt(6)/3a. So plugging that back into the bounds we have a = h, h being sqrt(6)/3a. Solving for a, we get sqrt(6)/3. So now we have the integral of sqrt(3)/4a^2 from 0 to sqrt(6)/3. Evaluating the integral we get sqrt(3)/4(a^3/3) * sqrt(6)/3 - 0. This simplifies to sqrt(18)/36(a^3). This further simplifies to 3sqrt(2)/36 (a^3). Factoring 3 out, we get sqrt(2)/12(a^3). That is our final volume.
I enjoyed the creativity in this video and learned a few things, although I share the view of others that the piano soundtrack detracts from the experience. (Thanks nevertheless)
Your calculation is correct. Clearly my python implementation of the calculation is perfectly correct, but it's really not that far off. I believe the actual value stored in my code was like 1.999999(...). But my program just took the first 3 digits to write to the display.
Reminds me of functions like fx,y)=sin(cosh(x^2+y^2).If you plot it,it's like a normal riple in the center but further out (at a finite radius) the "frequency" becomes infinite.
That function will only ocsilate infinitely fast at infinity. It is well defined for all real x and y. It will, however, ocsilate extremely quickly when x or y is large.
What a perfect video! I wish that all of my classes on any subject would be like this, well explained and treating everything as non-trivial, thank you so much for your work Peter!
That was awesome, i knew about the fourier transform but that draw you made by wrapping the wave in a circle around the origin just shined a light in my brain about the transform i never had thought before, now the formula actually makes clear sense to me, very nice very nice thank you very much!
you have the issue of computational limitations, the amount of accuracy you can get is only proportional to the number of times itterated, and the accuracy of pi. all together a near infinitly small change in pi could result in the result being way off (because that change is compounded every itteration). there is no easy way around it, all of our computers will try to store the rusult and pi in a finite space each decimal place takes up space, and because we don't know if the size of the circle could be a non rational number we can't say for certain what that constant is. much like pi you can approximate it, but it also relies on an approximation of pi. unless you want to find every digit of pi by hand, then find every digit of this circle by hand, you can only really guess. and a guess on a guess couldn't be very accurate. (nit picking) that of course is to find a perfect solution. if you want an approximation do as the engineer do, look at it, measure it, call it 8.71. much like the engineer version of pi: 3.14, it is rounded to be simple and easy to use and be good enough for the job. for pretty much any measurement going over 4 decimal places is way too far, specially because with engineering there is tolerences and they normally remove material so if it is close youre changing it anyway might as well round it. sounds like im side stepping the problem, i most certainly am, it is a problem i have no need to contribute to, partily because i am not qualified in math to, and because 8.71 is good enough for me.
First, I loved the way you closed your comment in the last paragraph, spoken like a true engineer. Apart from that, you brought up a really good point, I didn't actually consider the compounding error from using an appropriation of pi. Meaning theoretically (I think) your answer can actually become increasingly less accurate with further iterations bc of your approximation of pi. That was a very good insight.
The pacing for this video is really unnecessarily padded. You don’t need to show us a table of contents before every single section. And we know what a tetrahedron is, so there’s no need to explain that so slowly. Cut to the chase!
Hey! Fair enough. The hole point of this project is to show how a bunch of trivial steps and build up to something non-trivial. So I don't believe you are exactly my target audience, but I hope you found some value in it nonetheless.
I actually have done a lot of work on such a project. I think the Laplace transform deserves a good video...it's just a bit of a messy creature. I'm not sure how I want to show him to the world
