Phasors: How and Why They Work
12:54
I Made a 2D Circle in 3D Space
6:47
The Bernoulli Distribution
3:37
4 ай бұрын
The Hypergeometric Distribution
3:21
The Multinomial Theorem
5:19
4 ай бұрын
The Binomial Theorem
6:47
4 ай бұрын
Permutations Vs. Combinations
4:30
Proof of Natural Logarithm Base e
6:30
Product Rule Derivation
2:17
4 ай бұрын
The SCARY 3D Rotation Formula
31:49
Пікірлер
@jtgannon4224
@jtgannon4224 Ай бұрын
10111000111
@MRT-co1sd
@MRT-co1sd Ай бұрын
Do you have part2, how this relates to Noether’s theorem?
@pioneeringproofs
@pioneeringproofs Ай бұрын
No, I don’t know about Noether’s Theorem… yet! But yeah, if I end up learning it I can do a part 2!
@MRT-co1sd
@MRT-co1sd Ай бұрын
What do you mean when you say the Force Field is conservative?
@Dddddddddddddddddddddddd388
@Dddddddddddddddddddddddd388 2 ай бұрын
Finally I have been looking for this Kind of Video!
@derivoid
@derivoid 3 ай бұрын
AMAZING!! THANK YOU!!!
@jtgannon4224
@jtgannon4224 3 ай бұрын
Nice circle ⭕️
@ilikegeorgiabutiveonlybeen6705
@ilikegeorgiabutiveonlybeen6705 3 ай бұрын
instant sub
@Khashayarissi-ob4yj
@Khashayarissi-ob4yj 3 ай бұрын
With luck and With regards
@pioneeringproofs
@pioneeringproofs 4 ай бұрын
Sorry about the weird glitches in the presentation. But I think I still got the message across, so I’ll keep the video posted
@pioneeringproofs
@pioneeringproofs 4 ай бұрын
Sorry folks, it looks like I made a small mistake. When tracing a path to a string of variables (like xzzy), the second to last line should consist of powers raised to 1, not 2. Also, to clarify, when we expand, we apply Rule 2 iteratively until we reach the non-expandable terms, as illustrated in Rule 3
@chop-plays
@chop-plays 4 ай бұрын
Isn’t ODE45 a variable time solver? So why would the simulation need step intervals? Or is it that some other component requires step intervals.
@pioneeringproofs
@pioneeringproofs 4 ай бұрын
I actually forget already, lol. But I can check
@chop-plays
@chop-plays 4 ай бұрын
@@pioneeringproofs hahahah no worries. I just finished simulating and creating a control /observer system for a magnetic levitation device so I knew at the back of my mind what ODE45 was and was only slightly confused.
@Cdictator
@Cdictator Жыл бұрын
Math genius??
@Cdictator
@Cdictator Жыл бұрын
By the way, you helped me understand this question which has bothered me for a long time.
@noodle7788
@noodle7788 6 ай бұрын
@@Cdictator dude has a massive brain
@pioneeringproofs
@pioneeringproofs 4 ай бұрын
Thanks, I’m happy to help!!
@keyamoni1602
@keyamoni1602 Жыл бұрын
It is really rare to see such a rigorous proof of a mathematical concept on internet now a days. Please keep providing us videos like this.😊
@pioneeringproofs
@pioneeringproofs 4 ай бұрын
Thank you!
@chuckadams842
@chuckadams842 Жыл бұрын
www.emis.de/proceedings/Varna/vol1/GEOM09.pdf for the article @hanniffydinn6019 refers to. FYI.
@hanniffydinn6019
@hanniffydinn6019 Жыл бұрын
Nonsense really, as practically you use quaternions/geometric algebra to do this! 🤯🤯🤯
@pioneeringproofs
@pioneeringproofs Жыл бұрын
So far, this video is my personal favorite. I feel like with this video, I've raised the bar in terms of quality, and I'm hoping that I continue to do this with future videos. Also, on the very last slide, you may hear me rambling on about how "the cow jumped over the moon." I actually did this same thing prior to presenting each slide; I just cut it out of the presentation, but I intentionally left it in the last slide for entertainment purposes. This may sound strange, but there is actually a perfectly logical reason for why I did this: I noticed that if I began speaking right away, the volume of my voice would spike and there would be a ton of background noise; but then as I continued to speak, the amplitude of my voice would reach a steady state, and the background noise would be reduced to a satisfactory level. I think that this might be because of a combination of the noise reduction algorithm that is used, as well as the fact that when a person begins speaking after a minute, the amplitude of their voice actually begins at a high level and then actually decreases a bit as they continue to speak. So, I decided to give initial audio input. But in order to do so, I had to say something; so, my mind just chose to talk about cows jumping over the moon! P.S. I used proper grammar in this post to avoid being pulled over by the grammar police
@penguinpenguin-zm2mr
@penguinpenguin-zm2mr Жыл бұрын
At 10:20 you questioned if it is possibly to find the analytical solution. If you turn 1 into (r^2 -x^2)/ (r^2 -x^2) , the x^2 from numerator will vanish. You can change x into Rt, so the dependenc on R disappears and you left with arcsin.
@thomasolson7447
@thomasolson7447 Жыл бұрын
My first time seeing this. I don't think I used it in high school. How do you cancel the units? R has to be unitless, right?
@pioneeringproofs
@pioneeringproofs Жыл бұрын
m is in Kg, v is in m/s, R is in m, and F is in kg*(m/s^2). So the units in the equation F = mv^2/R are kg*m/s^2 = kg*(m/s)^2/m, which simplifies to kg*m/s^2 = kg*m^2/s^2/m, which simplifies to kg*m/s^2 = kg*m/s^2. Both sides of the equation match, so the correct units were used
@thomasolson7447
@thomasolson7447 Жыл бұрын
@@pioneeringproofs that's cool, thanks for the video.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Whoops, sorry about the mistakes on that “Raw Form” slide at 23:03. I provided the corrections below, as well as 2 things that I didn't explain in the video that I should clarify: Correcting Mistakes at 23:03 (1): I messed up pretty much the ENTIRE procedure for substitution of "n" for "i" in that one "Raw Form" slide at 23:03 - It's a wild mess!! I still got the right answer, but I provided the wrong explanation. Let's start with the definition of the variable "n". According to that one slide, the correct equation for n SHOULD BE: n = i + 2 (and also N = I + 2). Thus, the first term in the sequence should ACTUALLY correspond to n = 0, NOT n = 1. And the second mistake I made was my explanation for WHY we can simplify to 1/n! . It SHOULD BE because of the fact that the "repeated multiplication of 2 + j from j = 0 to j = i" is equivalent to "repeated multiplication of 2 + j from j = 0 to n -2", and this is equal to "n!" . And we can see this if we say that k = j + 2: We have n! = "repeated multiplication of k from k = 2 to k = n". And this is equal to n*(n - 1)*(n - 2)*(n - 3)*...*2 . And, if we want, we can multiply by 1 also, so this is indeed equal to n! Clarification 1: I don’t think I explained well enough WHY the two forms of what I called the "expandable integral terms" (you know, for q = + or - 1 AND m = even or odd) are indeed the ONLY 2 forms of the expandable terms you will encounter in the Taylor Series. This is highlighted by starting with your general form, expanding it out to 2 more terms. You will then notice that if you DON’T simplify using the distributive property, then you obtain an expandable term at the end that is of the form of the ORIGINAL expandable term from 2 terms back in the sequence. And I show this in the presentation, but I do not HIGHLIGHT this important fact. Clarification 2: I saw online that people express the "error term" as a term that does not involve an integral. So, that surprised me, and maybe I'll try to figure out how to simplify that integral and make a video about it in the future.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
My own opinion of this video: the content and script were fantastic, but the audio can be improved: I think that the headphones made my voice a bit too muffled, and I got a little too “breathy” and “tongue-clicky”. And I unnecessarily put emphasis on words too many times. I’ll try to fix these problems with audio and narration in future videos
@pioneeringproofs
@pioneeringproofs Жыл бұрын
There's one thing I forgot to mention that's pretty important: In the diagram representing the path of the particle that is bathed in the vector field, the path isn't some arbitrary path - It is the RESPONSE to the force field. This means that the conservation of energy really is just Newton's Second Law expressed in a very useful way (it's useful because it takes into account distance!!)
@MRT-co1sd
@MRT-co1sd Ай бұрын
Do you have part2, how this relates to Noether’s theorem?
@pioneeringproofs
@pioneeringproofs Жыл бұрын
It looks like this is my most popular video so far! I wasn’t really expecting this, but I’m glad it seemed to help so many people!! 🎉 P.S. Sorry about the whispering at the end of the video; I had to be quiet since I was recording late at night. And I made a small mistake in displaying the brackets in the equation in that one slide toward the beginning, but the overall idea that I conveyed remains accurate despite this minor error
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Whoops, I forgot to mention something. As you may notice, I fed in a whole bunch of values for delta_x into the approximation. HOWEVER, in doing this, I sometimes made the upper bound of the summation a decimal number instead of an integer! To avoid this issue entirely, I could have used the upper bound of summation variable n as the input instead of delta_x directly, as this value would be an integer and thus force the correct values for delta_x. But little did I know, this problem was actually corrected in my code, because I forced the upper bound to be an integer by stripping away the decimal points (this operation is called taking the "floor" of a decimal number, which I technically could have included in the upper bound of the fully processed, final formula); from my perspective, I was just correcting errors that the compiler threw at me, but I just now realized that it has real meaning in the summation approximation! So, for each delta_x, the decimal points after the corresponding upper bound in the summation were simply stripped away, and for reasons similar to being able to ignore that last n value of 2R/delta_x, we can also ignore the contribution of the fraction of delta_x in the summation, since this component will also approach zero as delta_x approaches zero.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
The real proof actually starts around 7:30. Before that, I'm just giving justification for why we are changing the bounds to negative and positive infinity
@pioneeringproofs
@pioneeringproofs Жыл бұрын
One thing I didn't explain but that I probably should explain is that although each basis function has a different minimal period over which they will each repeat themselves, they all have the fundamental period T in common BECAUSE of the fact that each basis function has a period that can be multiplied by some integer to obtain T. Therefore, it is true for ALL basis functions in our set that they repeat themselves after every interval in the domain of length T; in other words, it is indeed true for each basis function f(t) in our set that f(t) = f(t+T), which also implies that f(t) = f(t+nT), where n is an integer.
@thomasolson7447
@thomasolson7447 Жыл бұрын
I noticed something the other day. If a rule exists with complex numbers, you have to find a way to apply it to the vectors. If you have a negative in a square root, that is probably going to transform your vectors in some way. <x/sqrt(y^2-x^2), y/sqrt(y^2-x^2)>, if x>y than this expression becomes <y*sqrt(y^2-x^2)/(y^2-x^2), -x*sqrt(y^2-x^2)/(y^2-x^2)> Or something like that. That might not be 100% correct. But something like that happens. So, pretend that vector is x/sqrt(y^2-x^2)+i*y/sqrt(y^2-x^2) instead. Or use quaternions if it's 3d. Could you do the area and perimeter of this thing? I honstely don't know how. <(1/2)*e^(cos(θ)-sin(θ))+(1/2)*e^(cos(θ)+sin(θ)), (1/2)*e^(cos(θ)+sin(θ))-(1/2)*e^(cos(θ)-sin(θ))>
@pioneeringproofs
@pioneeringproofs Жыл бұрын
I wish I could answer your question, but it seems a little too advanced for me. Maybe chatgpt could help?
@thomasolson7447
@thomasolson7447 Жыл бұрын
@@pioneeringproofs Chatgpt is suprising unhelpful. I did find the answer to that and I know what I'm doing now. I am done with it for now. I'm working on this atm. E^2 = mc^2((cos(arctan(i*v/c)))^2+(sin(arctan(i*v/c)))^2). E^2=m^2*c^4*((c/sqrt(c^2-v^2))^2+(i*v/sqrt(c^2-v^2))^2). E^2=m^2*c^4*((c/sqrt(c^2-v^2))^2-(v/sqrt(c^2-v^2))^2). E^2=m^2*c^4. E=m*c^2. Area= (1/2)*arctanh(v/c)*(m*c^2)^2. Area= (1/2)*ln((c+v)/sqrt(c^2-v^2))*(m*c^2)^2. Area=(1/2)*(pi/4)*(m*c^2)^2, for something moving at the speed of light. So, area is [W]*[W], so I would need to square root that. sqrt((1/2)*(pi/4)*(m*c^2)^2). I want to test this on non-relativistic speeds. So I need to construct a sentence with what I have here. I need to move c^2 into scalars and redistribute the units to say something that makes sense. sqrt((1/2)*ln((c+v)/sqrt(c^2-v^2))*(m*c^2)^2)=(1/2)*sqrt(2)*m*c^2*sqrt(ln((v+c)/sqrt(c^2-v^2))) I don't know how to make sqrt(ln((v+c)/sqrt(c^2-v^2))) prettier. Angles are unitless quatities, I think. W=F*s*cos(θ). F=m*a, I have [M][L]^2[T]^(-2). I guess I need to think about this more.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
@@thomasolson7447 Yeah, Chatgpt is my go-to solution these days, but sometimes it isn't that good at answering math/physics related problems I've noticed, and of course it's going to inevitably be even worse at answering the more advanced questions, as you've pointed out. I'm guessing you must be a physicist of some kind?
@thomasolson7447
@thomasolson7447 Жыл бұрын
@@pioneeringproofs nope. You're not going to like me if I say this... I'm just unemployed and bored. I finished grade 12. I think I'm starting to figure out this madness. E^2=m_0^2*c^2*(c^4+v^2)/(c^2-v^2) v:=iv m_0^2*c^2*(c^4-v^2)/(c^2+v^2) So, I think at some point somebody put the gamma function in, and I just reversed it with i*v. So, instead of cos(arctan(i*v/c)), it's now cos(arctan(v/c)). The denominator isn't going to infinity anymore. The energy is always mc^2*cos(theta). The dot product and cross product can probably be used for the rate of change. I don't know what the reason is for gamma though. If v=c, then E=m*c^2*cos(pi/4). Which is really good. I can now start to try and find out if there is an exact value for h of the p=h*lambda.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
@@thomasolson7447 No way, I thought you were a PhD student!! And no, I respect that. It makes it even more impressive if you're just teaching yourself complex topics. Very few people actually have the passion and motivation to do something like that, so keep it up. Anyway, again, good luck with everything!
@thomasolson7447
@thomasolson7447 Жыл бұрын
Shouldn't those tangents be longer? The head part is extended to where the origin vector would conect if it had continued beyond the radius. Can I take a gradient from <x/sqrt(x^2-(y^2-z^2)),y/sqrt(x^2-(y^2-z^2)),z/sqrt(x^2-(y^2-z^2))>. I haven't made up my mind if the sqrt should be x^2-(y^2-z^2) or x^2-y^2-z^2. But I guess it doesn't really matter at this point. It's Pi time btw, 3:14 am.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Ah yes, Pi time, the optimal time to communicate about such matters. I’m also writing this at Pi time. Anyway, I think you’re right but the overall concept should still be correct (I think). And I’m curious about this vector field you’re talking about, I’ll have to graph it and see. I’m not too familiar with tensors (at least not yet), but I think you’ll get a 2nd order tensor of some kind when you take the gradient of it?
@thomasolson7447
@thomasolson7447 Жыл бұрын
@@pioneeringproofs I don't think anything can graph that. When there is a negative square root, there is a change between x and y-z axis. Half the values would be complex and the computer wouldn't know what to do with them. You would have to program it. This is cosh and sinh on the complex 2d plane nonsese converted to 3d. <x/sqrt(x^2-y^2-z^2), y/sqrt(x^2-y^2-z^2), z/sqrt(x^2-y^2-z^2)>. This is x^2-y^2-z^2=1 graph.If the square root is negative, you pretend you're on the complex plane x/sqrt(x^2-y^2-z^2)+sqrt(y^2+z^2)/sqrt(x^2-y^2-z^2). <-sqrt(y^2+z^2)/sqrt(-x^2+y^2+z^2),... Yeah, I don't know what I'm talking about. I'll have to think about this. There are also singularities that you might have to patch up. I don't think they'll be a problem here, but if you try to do some math with them.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
​@@thomasolson7447 There always seems to be these subtle details that end up making things more complicated than originally expected, especially in special cases like these. But it seems like an interesting problem to solve - Good luck!
@PhysicsRaja-ul3kc
@PhysicsRaja-ul3kc Жыл бұрын
good work, keep up man, thanks
@amritawasthi7030
@amritawasthi7030 Жыл бұрын
Impressive 👍🏼thnk u
@jiahaozhu7702
@jiahaozhu7702 Жыл бұрын
Thank you!! Love you
@jiahaozhu7702
@jiahaozhu7702 Жыл бұрын
Love you 😘
@Kiwi-zz8ri
@Kiwi-zz8ri Жыл бұрын
Thank you! I’ve was wondering why the gradient vector was orthogonal when I was finding the equation of tangent planes and this helped
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Happy to help!
@ds2718
@ds2718 Жыл бұрын
Very good and useful video, I wanted to know how it's formula derive for a long time I search everywhere but I couldn't find satisfactory answer (derivation). Now you satisfy my question. Thank you very very much for this Great Video. Edited:Make more videos like this they will help other people who want to seek knowledge like this.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Glad it was helpful!
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Response to edit: Thanks, I appreciate the support!! I'll do my best to continue to make new videos whenever possible!
@deinahrsnoidson7961
@deinahrsnoidson7961 2 жыл бұрын
Very good presentation. I like this approach more than the one using Taylor series. I did find a few mistakes though: at 7:31 y0 is missing (should be y0+L/2) and at 11:24 the 1 in g1 should be a subscript. Also, at 3:10 where you argue that the vector field on a boundary is the same because the boundary is very small would also imply that the vector field on all of the boundaries would be the same since they are infinitely close to each other, which is not depicted in your figure.
@pioneeringproofs
@pioneeringproofs Жыл бұрын
Right, thanks for pointing those out! I pinned this comment so that others can see
@pioneeringproofs
@pioneeringproofs 2 жыл бұрын
This was my first video and it was mostly a proof of concept. I will be narrating in future videos