A trick I have ignored for long enough...

[@domantasbieliunas9679]

Michael: "the B term will be canceled" B: "what the fuck did I do"

[@alexisren365]

Identified as superstraight

[@pimpomresolution5202]

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

[@balthazarbeutelwolf9097]

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

[@cornucopiahouse4204]

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

[@iabervon]

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.

[@jkid1134]

"Sketch of overkill" is some of the most self-aware I've seen this channel ever be

[@maxwellsequation4887]

I read this when he said it

[@otonanoC]

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

[@sphakamisozondi]

Heaviside is one of the underrated mathematician/engineer/physicist in my opinion.

[@joinedupjon]

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.

[@adandap]

As an army officer once explained to me, "overkill is still kill".

[@schmetterlingsjaeger]

The so-called Heavisde "cover-up method" could be much easier explained as evaluating 7x+2=A(x+2)+B(x-1) at x=1 and x=-2, respectively.

[@arsaces6936]

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

[@jkid1134]

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

[@peterlohnes1]

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

[@whiskeysprings]

Exactly.

[@riske440]

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.

[@laszloliptak611]

The second method should come with an explicit WARNING: it only works if the degree of the numerator is smaller than the degree of the denominator. The theorem on partial fraction decomposition says that every rational expression (quotient of two polynomials) can be written as a sum of a polynomial and partial fractions. The polynomial is obtained by long division, and then you can find the remaining partial fractions with the second method. Even then, in my opinion it is better to explain it by clearing the denominators and then argue that since there is a solution that will make the remaining equation an identity, we can plug in any number for the variable to get equality. Plug in the roots of the denominator to easily get the values of A and B. This will also work for multiple roots, in which case we can plug in additional numbers (such as 0) to get more equations on the unknown coefficients.

[@minktanker9705]

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

[@jkid1134]

I have wondered seriously for a long time why the existence of a partial fraction decomposition should exist, and that bit about both sides spanning a 2d vector space is the closest anything's come to helping me intuit it, so thank you for that.

[@stephenbeck7222]

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.

[@Lucaazade]

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.

[@jkid1134]

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

[@Lucaazade]

@@jkid1134 Fair enough:)

[@major__kong]

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.

[@sanjursan]

This guy produces some of the best lectures on the tube. He manages to move very quickly, but still be easy to understand and follow. Really great stuff.

[@soranuareane]

Also, this channel has by far my favorite jump cuts in any media. They're so obvious but set up so perfectly. Thank you for the entertainment!

[@BillyViBritannia]

Me: "The last one is going to be the trick" Last one: "Uno reverse card! I'm the most complex!"

[@michaelslack8900]

That cover-up trick is really cool. I always just did the 'add 0 to the numerator' thing - getting good at guessing how to do it by eye to get the right partial fractions... This was is just so much better

[@trewq398]

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

[@arsaces6936]

@@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).

[@herbcruz4697]

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

[@SmallSpoonBrigade]

@@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.

[@herbcruz4697]

@@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.

[@alejrandom6592]

Moving a disconituity into a zero is the best way someone has ever described the cover up method. Now I will remember it for ever :)

[@LucasDimoveo]

It is such a good explanation!

[@pratyachowdhury9298]

The NCERT must be feeling quite proud after watching this video.

[@lost-vidhu]

@Kumar Senpai ncert book have this method

[@Mean_men]

@@lost-vidhu can u tell me where that method is mentioned coz i didnt saw...

[@siimplicity1459]

@@Mean_men in integral calculus chapter

[@Mean_men]

@@siimplicity1459 yeah but i didnt see

[@generalginger7804]

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

[@ayanbhowmick5427]

This video just blew my mind! That last one with Complex Analysis was just too good

[@mrmanning6098]

as someone going through professor Borcherd's Complex analysis course, it was so cool to see Cauchy's integral formula being applied. I think quite a few of your viewers might be hungry for some more ways to apply Complex Analysis to problem solving.

[@isaackay5887]

What are some of your suggestions, might I ask?

[@aa1ww]

@@isaackay5887 Inverse Z-transform

[@melchiortod29]

The cover up method is so amazing. It blew my mind. Such a simple logical thought, but holy sht soo amazing and beautiful!

[@cotf-s1n]

we learn this trick in Sweden in fact there was a problem just like this one in my exam and i ended up not knowing how to use it hahaha

[@TheGlassgubben]

Partialbråksuppdelning❤️

[@TheStillWalkin]

Partialbruchzerlegung

[@mikaelvalter-lithander1247]

Handpåläggning

[@Connor-qb2yu]

Köööng!

[@julianlidberg7590]

PBU

[@johnunverzagt9387]

Thank you for using the Cauchy Integral formula! When I did a Complex Variables course almost a decade ago, Cauchy’s integral was my favorite part of the course. When you set it up in terms of C, I recognized where you were going with your explanation, and it brought a big smile to my face.

[@mtbassini]

Electrical engineers, specially in systems engineering, use this heavily. I remember even using the Cauchy integral sometimes.

[@mastershooter64]

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

[@aa1ww]

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

[@ViperrKsa]

save your time and use MATLAB

[@siimplicity1459]

@@mastershooter64 also pi^2=g=10

[@MT-in3tp]

Heaviside was awarded an engineering degree alas post mortem

[@Stormskip]

First time I've heard the vector space argument as a reason for why we can use this method, it's very good to know, particularly since partial fraction decomposition is one of the most important things I learned in the process of going through my engineering classes.

[@peterklenner2563]

A follow-up on the deeper connection between partial fractions and Cauchy's contour integral would be much appreciated!

[@theash4361]

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.

[@davidherrera4837]

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: kzbin.info/www/bejne/iYrbqpijg7iGo5o 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.

[@OuroborosVengeance]

@@davidherrera4837 thank you! Good samaritan

[@davidherrera4837]

@@OuroborosVengeance The KZbin 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.

[@MrCreeper20k]

@@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?

[@yulu1248]

I can very easily recognise Heaviside due to his unique haircut that loooks like his own step function

[@sl0wsn0w]

I understood everything until last part. Last part overkilled me hardly! But still I love your videos. Learned more from your videos than in college. Thank you.

[@dmitrystarostin2814]

That's a critical trick in algebraic expressions. There was a Russian math test book from 1980s that contained thousands of problems similar to this one, but I don't remember the name of the author. In the 7th grade the math teacher would just make us randomly go through them and as a result, a good student would remember [without deduction] up to a hundred of basic algebraic transforms.

[@nathanisbored]

i think i learned this trick as the "annihilation method", and i was taught to do it during the 2nd step of the algebraic way, which is actually a little less efficient because you still have to solve for A, you just dont have to collect and compare coefficients

[@SmallSpoonBrigade]

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.

[@johncrwarner]

Oliver Heaviside is an oft-overlooked mathematician / physicist who developed notations and methods used even today that make solving particular equations and problems easier for the everyday mathematician / physicist.

[@maxwellsequation4887]

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

[@davidbrisbane7206]

@8:26 It is not really legitimate to set X = 1, as that would mean X - 1 = 0 in the LHS denominator, which means the LHS is not defined at X = 1. However, if we take the limit of X - 1, as X approaches 1, then B (X - 1) / (X + 2) approaches zero, and we have recovered the situation without encountering a discontinuity.

[@cornucopiahouse4204]

Interesting point, but I guess it would be an overkill if anyone is going to do it this way while trying just to figure out the A and the B. Somehow we are dealing with what you’re describing all the time, aren’t we, e.g. when we write x^2-1/(x-1)=x+1 when technically the LHS is undefined when x=1?

[@davidbrisbane7206]

@@cornucopiahouse4204 Yes. It is overkill. Michael speaks the words, but doesn't write the limits.

[@RichardJohnson_dydx]

You know watching this, my instructor blended in the Heaviside method. In the algebraic method, he told us to plug in the roots so those terms would cancel out. Then we could solve the system that way.

[@SmallSpoonBrigade]

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.

[@alexatg1820]

For partial fraction in the form of (ax+b)/[(cx+d)(ex+f)], there is a neat trick you can do it may seen complicated but it's actually very simple and fast if u try it yourself if abcdef are integers (or some simple fraction). Uses (7x+2)/[(x-1)(x+2)] as an example, we can think it as [7x+2+n(x-1)-n(x-1)]/[(x-1)(x+2)], where 7x+2+n(x-1) is a multiple of x+2, Noticing 7>2 while 1

[@Izerion]

Oliver Heaviside is one of my inspirations. His achievements are very often overlooked! The Maxwell's equations that we are familiar with today were actually all formulated by Heaviside.

[@carultch]

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?

[@Izerion]

@@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.

[@marleykristianto2836]

Thank You so much sir I just studied Laplace Transform using partial fraction today Your video helped me so much

[@gastonsolaril.237]

Great video!! I used that trick plenty of times in Control Theory classes, while dealing with the inverse Z transform, to find recurrence relations for designing digital PID systems in robot microcontrollers. Nice to know these things have real world applications!

[@cicciocareri85]

I fondly remember when I studied residuals in complex analysis... the connection with real integrals blew my mind

[@hydra147147]

We can also derive the second method by multiplying both sides by the denominator of the LHS and then substituting. For example at z=1 B(z-1)=0 and we are left with (7z+2) = A(z+2) for z=1 so A=(7z+2)/(z+2) for z=1 and similarly for B.

[@hydra147147]

@@magnusPurblind 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).

[@robertdriscoll9414]

I loved learning and using the 2nd method here (I didn't know it was the Heaviside method). It was so quick and simple. We learnt how to apply it to higher order poles and higher order denominator terms (not just linear) by simple tricks. Then we started using it in Laplace transforms and the hard work started, but we had a great teacher. Two years later I went to Uni.

[@masonholcombe3327]

I would love to see how to do that integral at the end / go in more depth about residue

[@JoshuaHernandez8a]

The only thing that I can remember for the entire control system course though

[@neilduran2335]

Step function?

[@33Euler]

You might want to add that the Heaviside cover up method only works if you have distinct linear factors in the original denominator.

[@carstenmeyer7786]

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 ;)

[@dork8656]

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

[@carstenmeyer7786]

@@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.

[@dork8656]

@@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.

[@carstenmeyer7786]

@@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"

[@גלעדקוסטין]

At 08:29, why are you allowed to set x to be equal to 1. Isn't it out of the definition area of x? (Based on the first line, x can't be 1 nor -2 becauae it resets the denominator).

[@josephcoon5809]

0:00 Beginning 1:05 Algebraic Method 4:40 Plug for the ironically named “Civil” “Discourse” Conference 5:25 Heaviside Coverup Method (HCM) 7:30 Why HCM works 9:50 Sketch of Overkill Method

[@DrDailbo]

I also avoided this trick and waded through the "algebraic" method. I'll take this as the wake up call I needed.

[@ScienceRob]

Thanks for this. I studied math in university but later ended up working in banking so I haven't had a chance to really use any math i've learned in awhile. It really took me back to my complex analysis class with the overkill method. I definitely miss university math.

[@douglasstrother6584]

"The Forgotten Genius of Oliver Heaviside: A Maverick of Electrical Science" ~ Basil Mahon

[@jefflee146]

so lucky to find this channel which provides me with novel ideas to solve math problems!

[@homerthompson416]

My favorite trick is to open up Mathematica and use the Apart function

[@AbsentMindedMathematician]

Another fancy method is to use the extended Euclidean algorithm (over the ring of polynomials) to find a, b such that a*(x-1) + b*(x+2) = n, where n is any polynomial multiple of gcd(x-1, x+2) = 1. (Choose n = 7x + 2 in this case.)

[@christophseeliger5551]

Getting the residue is possible by just looking at the coefficient of the only term of the Power Series evaluated at x=1. but that’s equivalent to the algebraic methods. The contour integral is bogus though since you need the same partial decompositionn you are looking for to solve it.

[@soranuareane]

Thank you for explaining why we can set coefficients equal to each other for solving these kinds of equations. I always loved using that trick but never took a moment to figure out why we can rely on it.

[@Jim-be8sj]

Bonus points for tying this to residue calculus. I never thought about partial fractions in that way even though I have often thought about partial fractions with complex numbers.

[@danielslilaty]

Hi Michael. Great videos! There is one other place that some undergraduate students would encounter partial-fraction decompositions. In combinatorics, solving recurrence relations with power series very often uses partial fractions.

[@NoorquackerInd]

I absolutely love how I've memorized what Heaviside looks like I just remember his hairline looks like the step function and then I'm able to instantly recognize him

[@gemmaweber4115]

This trick is fascinating!! I love it! Thank you! It's gorgeous!

[@horenyi3083]

I have absolutely no idea what is happening but youtube keeps recomending this vids so ill watch them. (high school student here)

[@l.h.308]

I have not seen the whole video but this is how I have been doing this for 5 decades: A(x + 2) + B (x - 1) = 7x + 2 (really the identity with 3 horisontal lines rather than =) This shall be equal for ALL x (an identity), so: Put x = 1 to get 3A = 9, A =3, then put x = -2 t to get -3B = -12, B = 4 immediately. A funny fact is that we use just those two forbidden x values (in the original rational function). They are no longer forbidden in the end.

[@sempre-a-la-contra-i-avant]

It is also important to solve telescopic series like Σ( 1/n(n^2 - 1) ) for n going from 2 to infinity

[@goodplacetostart9099]

0:01 Good place to start

[@maxwellsequation4887]

Nice

[@thephysicistcuber175]

Showdown against good place to stop when?

[@goodplacetostart9099]

@@thephysicistcuber175 since today

[@DJCray8472]

When we done the Residue Theorem the first time, I was really impressed. As you do the way back. You try to find the integral showed in this video at the end. You need only to find the residue, add it (deping on their "order") and multiply by 2 Pi i. And you are done.

[@barutjeh]

Another neat way to overkill it a bit is by looking at it is as linear algebra. You can find vector (A,B) by changing the vector (2,7) over base {1,x} to base {x+2,x-1}. It gets a lot crazier when there are more factors.

[@ogglieostrich2574]

In england for our a levels we use a similiar method to the cover up method but instead we multiply the whole thing out and start plugging in x values. Its clunky and long. This method is only a slight simplification but it is much more bearable and fluid to use. Thank you very much.

[@genghiskwon8388]

damn just saw the video today. would've been interested in going to the conference

[@Tentin.Quarantino]

That heaviside method is beautifully elegant in its simplicity when we look under the bonnet and see how it works.

[@Aditya_196]

Heavyside cover up is crazy ... While solving some sequence and series it required me to do partial fractions which I hated a lot thank God there's this crazy thing and actually makes sense

[@gennarobullo89]

I hope you'll cut that hair for the event! 😂 Keep up the good work!

[@matematicacommarcospaulo]

After almost 10 years teaching ODE, only this year I discovered Heaviside method via one of the my students who used it in a text. In all my videos about linear ODE, I solve equation using algebric way and more recently when I made videos about Laplace Transform and include Heaviside method.

[@topazioverde6464]

Hi, you should say something about cases like 7x+2 / (x-1)^2 (x+2)^3. Multiplicities can be dealt with in a "similar" way, but more needs to be done. No need to solve systems of linear equations though.

[@nawafspov1]

omg!! how didn't stumble on this channel before! great explanation, and great quality!! loved it! subscribed!

[@Tclack]

This makes so much sense. Great explanation! I've been doing the "algebraic way" for far too long

[@schrodingerbracat2927]

the residue can be calculated by multiplying with (z-1) and taking the limit as z->1 ... which is the same as the cover up method!

[@andreweberlein1509]

Is there some variant of the Heaviside method that works for roots of order >1? The best I've got so far is 1) multiply by the highest order to clear singularities, 2) find the coefficient for the highest order, 3) subtract the term we found from the original function to decrease the order by 1, 4) repeat until you are done

[@davidherrera4837]

Yes. See the comment I posted on this video today with an example. The idea is justified by partial fraction decompositions always exist and are unique.

[@xl000]

I want to reveal one of his editing tricks: before the tricks at 4:48, he looks at a mark on the ground at 4:36, and make sure he steps on those marks, so that his body doesn't move after editing the middle section. Obviously we don't see the second time he's doing it because it's edited out - but this is the trick.....

[@briann10]

Finally! I was looking for heaviside cover up approach since last week.

[@blankblank103]

Wow first time I've seen and understood why that cover up method works, thanks!

[@salehdanbous6783]

I have studied the Heaviside method in college in Signal Processing. My first impression was "Why did not they teach that in Math?". Made my life so much easier!

[@douglasmagowan2709]

I was thinking about the "overkill" method from the beginning of this video. This points you to a solution where the "cover-up" method fails when you have a multiplicity in the denominator. However, I would suggest the contour |z-1| = epsilon, giving the flexibility to set epsilon small enough such that only one pole is inside the contour, though not really a problem in this example.

[@SmallSpoonBrigade]

It fails when you've got more than one prime factor in the bottom. So (7+x)/((x-2)(x^2+3)) wouldn't be much of an issue, but you would have to use some other method to finish it off. I wouldn't personally consider that to be a failure as you'd very quickly get one of the coefficients and just have to figure out what the B and C are.

[@vanshthakkar722]

This is taught us to do integration of the question easily... After breaking the question into 2,or 3 you can directly integrate each one of them and add the end result

[@DrR0BERT]

I teach the cover-up method as one of seven techniques to address partial fractions. I tell my students that it is the go to to get as many of the unknowns as possible. Still students will do the brute force method. On one test, I remember giving a PF problem with a factored sextic polynomial in the denominator: four linear factors of order 1 and one linear factor of order 2. The cover up method would crank out five of the six unknowns rather quickly. A quick computation would yield the sixth. Still there were a number of students who tried to solve a system of six equations in six unknowns. Note: The cover up method works with complex numbers too, but arithmetic in C is a bit of a struggle.

[@fabio19h]

We actually learned this method when we had integral calculus

[@Dekross]

Same

[@FreemonSandlewould]

Heaviside was no light weight.

[@okaro6595]

True, what we call Maxwell equations were actually written by Oliver Heaviside. Maxwell had 12 equations.

[@mastershooter64]

@@okaro6595 I'm pretty sure it was 20 equations

[@SimpMaker]

Thank you. Also the blue balls at the end is killer.

[@danielsgrunge]

I like to do the algebraic way because it's the same every time and it looks nice to see those long ones written down

[@billtimmons7071]

Where was this guy when I went to school? Very cool tricks. Thank you

[@griffisme4833]

I learned this about 2 days before my calc bc exam and was very happy

[@ChaosPod]

The way I learnt it in school was a cross between the algebraic way and the Heaviside cover-up method, that is, multiply both sides by the denominator, then set a value of x such that the B co-efficient is 0 and solve for A, then set a value of x such that the A co-efficient is 0 and solve for B. So we would evaluate 7x+2=A(x+2)+B(x-1) at x=1 and x=-2, respectively.

[@oliverhees4076]

I've encountered the coverup method in blackpenredpen's videos, before I took a formal calculus course. I did the more traditional way on tests too so I'm familiar with both methods.

[@MarcoMate87]

The Heaviside "cover-up" method is fantastic, but maybe it's worth pointing out that the limit is preferable to the direct substitution. Also, I can't find a way to extend the Heaviside method when the denominator has a multiple root, and so a factor of the form (ax+b)^n. Take for example the fraction x/(x-1)², whose partial fraction decomposition is A/(x-1) + B/(x-1)², with A=1 and B=1. If you cover up the (x-1)² factor and substitute x=1 (or, more rigorously, you compute the limit for x-->1) you correctly find B=1, but covering up (x-1) brings to nothing. How to find A?

[@whsheu]

I get stuck on the correctness of heavyside cover-up method. Can anyone help? In 8:10, the video says multiplying (x - 1) on both side can be seen as "take the discontinuity at x=1 from the first two terms and move it to a zero of the last term". Is there a formal way to state this idea? or any article regarding this kind of operations which eliminate zero terms from the denominator? I am confused here because I think the original equation only holds (or is only meaningful) for x not equal to 1 and -2. Therefore, the equation derived by multiplying (x - 1) on both sides only holds for x not equal to 1 or -2. By plugging x=1 into that equation, the yielded value should be meaningless (or at least we cannot be sure if the value really specifies a correct decomposition). Can we be sure that this method always produces a correct decomposition?

[@evanprinsloo6]

Brilliant presentation.

[@TheStillWalkin]

Mind blown: wonderful explanation

[@MaKaNufilms]

I used the heavy side cover up all the time after multiplication the LH denominators that way it makes the the vanish of B visual. I attempted to change because the comparing the exponents parts in the "algebraic" way seems to easier to handle with higher order exponents. I need to test your approach of cover up on higher order polynomials.

[@davidherrera4837]

See the comment I posted on this video just today. I give an example of how to use the heaviside cover up method for higher powers. The short of it is that the cover up method gives you the coefficient of the highest power A / (x-a)^n. then you take your function, subtract A / (x-a)^n from it, write it as a rational function under a single denominator and simplify. This reduces the power of the (x-a)^n term by 1. Then iterate.

[@arimermelstein9167]

The method I learned is neither of these. I learned to plug in 1 to eliminate the A term and then plug in -2 to plug in the B term.

[@MarcoMate87]

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

[@makkusaiko]

We actually used a derivative of the heavyside cover-up in our extramural maths class at school. We'd times out the equality as like in the algebraic method and then set x equal to smth that will cancel a term out. Only in desperate situations would we purely use the algebraic way (ex where there is A÷[x+1] + B[x+1]^2)

[@nefyt278]

In the hide up method's proof, when we multiplied both sides by (x-1), shouldn't we place the condition that x-1 is not 0 or x is not 1? Why did we later plug in 1 into the equation? Or did we take the limit of both sides as x goes to 1?

[@SmallSpoonBrigade]

He's essentially removing the discontinuity by using the limit at that point as if it's the actual value. Since A does not depend upon the value of X, he's able to use an equivalent expression to calculate the A value without needing to worry about what the B value is. This is one of the reasons why the = sign should probably be an endangered species as the expressions are equivalent, except when X=1 or X=-2. But, because the A and B are constants, he's justified in using expressions that are equivalent everywhere else and extend the domain to cover the holes.

[@davidgolden1607]

Sorry I missed the conference on civil discourse, sounds interesting and a topic of great importance on campuses and elsewhere.

[@jpm3616]

I knew you were thinking residues when you described terms “trailing off to 0”. Then I realized your partial fraction decomposition was just the Laurent series at the 2 poles. Thanks for the tip - residues via partial fractions! I don’t remember them teaching me that in complex variables 😎

[@diegosantiagosantiago2632]

When I find linear denominators or easy ones to split I usually do something like this. Write denominator as 3x+4x+6-4. Then (3x+6)+(4x-4). It seems hard to explain but for me was easier sometimes.

[@krzysiekczajkowski427]

I learned that one at our uni. So cool!

[@mtaur4113]

I don't generally bother with this because it basically amounts to evaluating at x=c for convenient c, just don't distribute too much and it's good to go from there. In some context where you know for sure that you'll only get distinct linear factors, it could pay off, though. If there are repeated or irreducible quadratic factors, you don't get enough solutions with just the zeroes. You can differentiate the polynomial form of the equation and the zeroes of the repeated linear factors can be plugged in again. You still have to choose something else for the quadratics I guess. You can always plug in non-convenient numbers when you run out of the best ones.

[@pinklady7184]

I let two long ads play out to their ends, so that you get paid and make more videos. Your tutorials are great. Thank you for teaching.

[@kiro9291]

I'm curious about more complicated cases with (Ax+B)/(ax²+bx+c) and Cx/(dx-e)², would the Heaviside method still be easy?

[@paulconnor1040]

The quadratics are necessarily irreducible in this situation, there isn't a nice root to plug in to the other terms.

[@insouciantFox]

For the first case, the method works with complex roots. The second case doesn't work at all really beacuse of the fact that jt needs to be of the form A/(Dx+E) + Bx/(Dx+E)

[@David-km2ie]

@@insouciantFox You can generalize the 2nd by taking derivatives. To be specific: write down the partial fraction form, then multiply by (x-e/d)^2. Plug in e/d will give you one coefficient. Taking the derivative and then plugging in e/d gives you the other coefficient.

[@David-km2ie]

This way you can derive the general formula very easily

[@ElchiKing]

You can actually use the Heaviside method (and in my math for Physicists course we actually teach them how to do it) for any pole order. The algorithm is as follows: 1) Determine all (complex) roots of the denominator. Cancel common terms as far as possible (i.e. numerator and denominator should have no common factors). (If the degree of the numerator is bigger than the degree of the denominator you should then do a polynomial division to get the polynomial part). Determine the degree of each pole. 2) The (complex) partial fraction decomposition of the fractional part will look like A_1/(x-a)+A_2/(x-a)^2+...+A_k/(x-a)^k+B_1/(x-b)+B_2/(x-b)^2+... (with a,b... distinct roots and k the pole order of a). a) For pole a, we start with the highest order term (i.e. A_k). As in the Heaviside-trick: Multiply both sides with (x-a)^k and evaluate at x=a (cover up). We will get A_k (and this should be some nonzero value). b) now, subtract A/(x-a)^k from both sides and shorten the fraction on the LHS (at least one factor should be canceled). The pole order at a is now reduced by at least 1. (the remaining pole order determines the order of the next highest non-zero coefficient A_i) c) repeat steps a) and b) until A_1 is computed. 3) repeat with all other poles. 4) If we want a real partial fraction decomposition: Combine complex conjugate fractions to one real term. Note: If we only care about the highest order terms of the poles, we can also skip the repetitions and subtractions, i.e. don't need a full set of solutions

[@linggamusroji227]

7x+2 = A(x+2) + B(x-1) From here, just substitute x=-2 to cancel out the A, then subt. x=1 to cancel out the B