Identified as superstraight

You didn't atone for your 'B' privilege at x=1. Poor A was discriminated against on account of his pole.

B should listen to Keonte Coleman's talk at the conference on Saturday, and next time "stop and think" beforehand.

Lol good one! We’re living in an era of cancel culture anyways, so B you gotta live with it!

x-1 came around, and everybody else was like "you're canceled" and B was like "sure, you can hang out with me" and then everyone saw who B was willing to associate with.

I read this when he said it

Dr. Penn has a video where he laboriously proves that a+b must be equal to b+a

I voted Heaviside for the scientist to be portrayed on the back of the new English £50 notes... Fwiw Turing won. Maybe the maxwell fans should have ganged up with the Heaviside fans and asked about a double header.

I agree, the weird notation only complicates what is actually a very straightforward approach.

Well, evaluatin trivial limits at those points technically, since the original is undefined at those values

thanks, its much clearer way to "reveal" the cover up

Exactly.

Yes, that is much easier and this is one of the most beautiful ways to solve problem in math. All teachers learn kids to solve via first method which is so boring in larger problems with exponents on x, but when you learn this way all those problems become pretty easy.

oh. That's similar to the check we have to do when checking the limit of a fraction. Good to note/know

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I think about it in terms of the fundamental theorem of algebra + theorem of complex conjugates. Every polynomial can be factored into linear roots (perhaps complex) or at worst quadratic roots. Of course in the universe of all real polynomials, most of them don’t factor nicely into rational factors but the ones we do when learning about decomp will generally play nice.

The fact it’s a basis is one way of saying why the coefficients are unique and justifying that particular step of working, not really related to why the fraction can be written as a sum in the first place. I’m not sure why that doesn’t seem obvious to you - when you add fractions together you get a fraction with some factors in the denominator, and when you start with a fraction with some factors in the denominator it’s easy to find out what sum of fractions can get you there.

@@Lucaazade uniqueness I guess is more of a conceptual stumbling point for me than existence for partial fraction decomposition

@@jkid1134 Fair enough:)

Think of it going in the other direction. What happens when you find a common denominator for two fractions and add them? Partial fractions is just the inverse operation of finding a common denominator and summing.

Does someone know how to use it, when you have something squared in the denominator?

@@trewq398 You could just iterate it. For example: 1/((x-2)^2*(x+1))=1/(x-2) * [1/((x-2)*(x+1))]. Use the method on what's inside the square brackets. You will get a sum 1/(x-2) * [A/(x-2) + B/(x+1)]. Now use the method again on the second summand and you get: A/(x-2)^2 + AB/(x-2) + B^2/(x+1).

I tend not to use the cover-up trick, since that only works when you have linear factors. I just equate coefficients.

@@herbcruz4697 That seems unnecessarily painful. You know immediately if this trick is going to work and then you can use it to reduce the number of simultaneous equations you need to work with by that many. Any additional constants you need to find can be condensed to fewer terms.

@@SmallSpoonBrigade Again, it works when you only have linear factors in your denominator. That's why it's more intuitive to simply equate coefficients. If you want to use the cover-up method, then go for it.

@Kumar Senpai ncert book have this method

@@vidhu417 can u tell me where that method is mentioned coz i didnt saw...

@@Mean_men in integral calculus chapter

@@siimplicity1459 yeah but i didnt see

@@Mean_men In integrql calculus there is a bunch of equations clunged together in page no.317. textbook no2. (12th)

_Impossible_ I thought all that engineers do was pi = e = 3 and 3^3i = -1 I have been lied to!!!

"heavily"? I see what you did there; clever.

save your time and use MATLAB

@@mastershooter64 also pi^2=g=10

Heaviside was awarded an engineering degree alas post mortem

He is more deserving of being fanboyed than an engineer who did nothing new

It's not the same as a video explanation, but why it works is as follows. Let (x-a) be a unique factor of the denominator. Then the partial fraction decomposition will contain a term c/(x-a) and some other terms d/(x-b) with a =/= b. Taking the residue at x=a we see all terms d/(x-b) vanish, leaving us only with the residue of c/(x-a), which is a first order pole and gives c. Cauchy's contour integral doesn't have a special relation here, but Cauchy's residue theorem states that that the contour integral of a closed path in the complex plain is given by the sum of its residues times their winding number times 2 pi i. By picking a small circle centered about our residue as our path we can ensure that this is simply 2 pi i res(f at x=a), so dividing by 2 pi i gives us our residue, which is the coefficient c. The formatting as a comment is a bit wonky so I recommend you look up Cauchy's residue theorem to get a better picture, but the gist is that contour integrals can be calculated using the sum of their residues, which in this case is chosen to be the only the residue giving our coefficient.

To add to what TheAsh said, a standard result in complex analysis is that is a function f(z) "blows up" (its absolute value goes to infinity) as z converges to a point a, then f(z) is approximately equal C/(z-a)^n for some constant C and some power n. We say that f(z) has a pole of order n at a in this case. This is captured in the fact that any complex function with a pole of order n at a point a can be represented as f(z) = C_n / (z-a)^n + C_{n-1} / (z-a)^{n-1} + ... + C_2 / (z-a)^2 + C_1 / (z-a) + h(z) where h(z) is analytic (and hence continuous) near a. Integration in complex analysis around a smooth curve is like integrating a line-integral of conservative vector field in the plane. As long as the region inside does not have any holes in it where the vector field is integral, then the integral is always guaranteed to be zero. In complex analysis, when you calculate the contour (another way to say a curve with an orientation, a way of telling which direction to integrate along the curve) integral of a function, then all the terms in the expression f(z) = C_n / (z-a)^n + C_{n-1} / (z-a)^{n-1} + ... + C_2 / (z-a)^2 + C_1 / (z-a) + h(z) give a zero integral except the C_1 / (z-a) term. This is related to the fact that when doing calculus you can always find the antiderivative of x^n as x^{n+1}/(n+1), except in the case that n = -1. In that case you get a logarithm. In the complex analysis case, the function is z^n and it has an antiderivative z^{n+1}/(n+1) except when n = -1. These antiderivatives that are powers are functions that you can apply the fundamental theorem of calculus to conclude have a zero integral when you integrate around a circle. However when taking the antiderivative of 1/z, you get the complex logarithm which is not a "function", but a multivalued function. See: kzfaq.info/get/bejne/ib-orMies7m3nWw.html for an introduction to the problem of how to define the complex logarithm. What you see in the video is that there is something funny going on when you rotate z around the complex plane in the circle. The complex logarithm increases by 2pi*i. See the second picture on en.wikipedia.org/wiki/Complex_logarithm This spiral-staircase represents the fact that if you go around in circles around the origin, you end up at a different place. So, all of this is to say that when you have an expression like f(z) = C_n / (z-a)^n + C_{n-1} / (z-a)^{n-1} + ... + C_2 / (z-a)^2 + C_1 / (z-a) + h(z) and you integrate around a circle (counterclockwise), you get 2*pi*i*C_1. You normalize the integral by dividing by 2*pi*i to get that the integral of the function f(z) around the contour then divided by 2*pi*i gives you C_1. This means that if you have an expression like f(z) = (z^2+z-1) / [ (z+10) (z-2*i) (z-4)] then you expect the partial fraction decomposition (its proof involves linear algebra) to give you f(z) = A / (z+10) + B / (z-2*i) + C / (z-4). To find a coefficient, say A, then you integrate f(z) around a circle that contains only the pole a = -10 and not the poles 2*i or 4. Divide this answer by 2*pi*i to obtain A.

@@davidherrera4837 thank you! Good samaritan

@@OuroborosVengeance The KZfaq Channel "Richard E. BORCHERDS" has a nice introduction to the topic. He likes to give high-level introductions, not getting bogged down in the details but to give a survey of the topic. His Complex Analysis course is directed toward undergraduates but I found his graduate level courses very informative. It is very interesting, even for me having taken a course on this years ago.

@@davidherrera4837It's been a while since my intro to complex analysis class, that f(z) you define in the second paragraph, is that called Laurent's series or related to it?

It is such a good explanation!

What are some of your suggestions, might I ask?

@@isaackay5887 Inverse Z-transform

We weren't taught the Heaviside method, or really any specific method for solving these. We were taught to factor the bottom, if applicable, and how to decide what the denominators and numerators would be, but from there we were largely left to our own devices to solve that. Heaviside cover up is really the most obvious way of solving the problems when it applies. The only reason you wouldn't do that, is if you've already decided that you want or need to use matrices to solve the expression. Pretty much whenever you can eliminate terms when solving systems of equations, you're going to want to do that.

That's what most people that are thinking about what they're doing do. The only time that it really fails is if you don't have any of those linear expressions to work with. Worst case tends to be that you immediately move to a different method, next worst case is that you're dealing with fewer terms for the system of equations.

What's the story behind how Oliver Heaviside became the namesake of the cover-up method? Like what kind of problem was he trying to solve, when he discovered it?

@@carultch It is like Michael said at the beginning of the video :) Heaviside was regularly using Laplace transforms for differential equations, and basically invented the cover-up method as a faster way to do the partial fraction decomposition. He had a knack for making mathematical shortcuts and coming up with ideas to simplify the calculations he was trying to solve. For example, he is credited as pioneering the use of complex numbers in electrical circuit analysis. Not to mention, he basically invented vector calculus. Although, I must admit that my earlier post could be mistaken. Although Heaviside is widely regarded to be the first person to write Maxwell's equations in the form we know them today, it may actually have been Lorentz who discovered this first! I still looking deeper into this :) The story of Heaviside is fascinating. I recommend the short biographical paper "Oliver Heaviside: A first-rate oddity" as an introduction if you are interested.

Nice

Showdown against good place to stop when?

@@thephysicistcuber175 since today

Step function?

If you have poles of higher multiplicity, the cover-up method still works - but only for the coefficient of highest multiplicity of each distinct pole! In most cases, that means you can directly compute all but one or two coefficients via cover-up method, greatly simplifying the problem. With the remaining terms you may use the analytic method resulting in a much smaller linear system of equations. This combination of "cover-up + analytic method" is the fastest general method I know for doing PFDs ;)

I thought of that too! Maybe if higher multiplicity, some might not work since multiple terms will be cancelled.

@@dork8656 Your instinct was right :) It's actually a great exercise to try and prove the method "cover-up + analytic method" in the general case; we did that as an exercise in our "math 1" class to improve the analytic method that was derived during the lecture. It's slightly more work to write everything down compared to the simpler case of distinct poles of first order, but the steps and ideas remain exactly the same.

@@carstenmeyer7786 nice. But this is only useful for multiplicity 1, still I'd keep it. The one trick I do when doing partial fraction decomposition on integrals is that: Suppose x(x² + x + 3) be a denominator. I'd let the fractions be A/x and [B(2x + 1) + C]/(x² + x + 3) for easier u-substitution. Handy one and always works.

@@dork8656 Of course you don't split up complex pole pairs ;) However, your example did not have poles with multiplicity > 1. In those most general cases, you only use the "cover-up" method for the highest multiplicity of each _distinct real pole:_ *Example: 1 / [ x^2 * (x+1) * (x-1) * (x+2) ]* You can calculate _all but one_ coefficients directly via "cover-up" -- only the coefficient for *1 / x* must be obtained via "analytic method"

@@attyfarbuckle The original expression no, but the expression after multiplication is. And since they are equal for all points except for two, we just have to figure out which A and B make the second expression work (the same trick is applied in the first method with equality of polynomials).

No, if you plug in 1 you eliminate the B term and find the A term.

Same