It's wild that cover-up method is not taught in schools, it's one of the most awesome shortcuts ever. At least better than those "shortcuts" tiktok keeps showing me to multiply two digit numbers lol
It is taught, it’s just not very generalizable to more complicated functions as shown in the later examples. It’s more formally called the Heaviside method, which you can look up. In AP Calculus BC, all of the partial fraction problems are non-repeating linear like the first problem in the video, and some version of the coverup method is used pretty widely.
Guy does a little stretch at the beginning, then I see a whole shelf of expo markers in the background... Oh man, here we go, this guy is about to wreck my mind 🤯😅 Awesome video, thanks for the content 🙏
Other tips for examples 2 thru 4: For example 2, plug in strategic values of x, such as x=0 and x=1, that aren't already spoken-for by the coverup method. (2*x + 1)/((x + 1)*(x^2 + 2)) = -1/3/(x + 1) + (B*x + C)/(x^2 + 2) Let x = 0, to eliminate B and find C: (1)/((1)*(2)) = -1/3/(1) + (C)/(2) 1/2 + 1/3 = C/2 C = 5/3 (2*1 + 1)/((1 + 1)*(1 + 2)) = -1/3/(1 + 1) + (B + 5/3)/(1 + 2) 1/2 = -1/6 + B/3 + 5/9 B = 1/3 For examples 3 & 4: Once Cover-up finds A and C, you can find B, by multiplying by only 1 copy of the repeated term. Take the limit as x goes to infinity. Some terms cancel to zero, some terms reduce to a finite non-zero number, and B will stand on its own as a constant. This is a way to more directly find B, that allows you to ignore terms that don't affect it. It has trouble if there is an irreducible quadratic involved, but works well when only dealing with linear and linear-repeated factors. As applied to example 3: (2*x + 1)/((x + 1)*x^2) = -1/(x + 1) + B/x + 1/x^2 Multiply by just one copy of x: (2*x + 1)/((x + 1)*x) = -x/(x + 1) + B + 1/x Send x to inf, and take the limit: 0 = -1 + B + 0 B = 1
At this point: th-cam.com/video/WrGXIjXRSys/w-d-xo.html, my students follow this instruction: "It's only one more value to calculate. Then with the same strategy you used in other videos, we take a new value (like x=0, or another one, but not used), and then we get the last coefficient". I'm thinking is the fastest method. Am I wrong? Thank you for your awesome work, @bprp !!!
Do anyone know why the degree of the numerator must be only be 1 degree less than the denominator and not by less than more than 1 degree ? (I am not talking explicitly on this example ) in general i mean
In the original expression, it's OK if the numerator degree is less than the denominator degree by more than 1. It can be up to and including, a numerator of a degree that is 1 less than the denominator. Any other given fraction can use polynomial division, or alternatively: adding zero in a fancy way to generate a term that reduces to a constant. For each term you build, there are always n-1 degrees of freedom for the coefficient above any given term. To capture this possibility, we have an n-1 degree polynomial in the top, and an n degree polynomial down below. A linear factor's coefficients have just one degree of freedom, while a quadratic denominator has two degrees of freedom. A linear repeated factor is just like a quadratic factor, and has two degrees of freedom. As a result, it ultimately needs 2 coefficients to cover all possibilities.
10q but No 94, when we solve by shortway like for A we did x=-1 and found A=-1 and why you didn't use for B,(x=-2) and like this # u did that for C and by long way for B , then why ?
The coverup method is just a shorthand of what I call the ‘convenient value of x’ method. It is just some algebraic techniques of solving a rational function by turning it into a polynomial function and solving from there. The solutions of the polynomial will be the same as the solutions of the rational function except where the rational function does not exist. So we aren’t really solving the original function at x =-1, but we’re just kind of going ‘if we could solve at x= -1, what would we get…’
It seems that way, but really this method is just a shortcut to the 'traditional' way of solving partial fractions. The 'traditional' way is like what he did for the middle term in the later examples, where you just multiply everything by the common denominator and then you group the terms according to their degree. Except that instead of only having 1 variable to solve for, allowing you to ignore many of the terms and solve a simple single variable equation, you will instead end up with a system of equations that you need to solve to get A, B, C, etc. And methods to solve such a system of equations generally involve finding ways to scale and combine the equations so that other terms become 0 to let you solve one variable and then substituting that value in to solve for others, etc. Well, this method of picking an x-value so that you end up multiplying most of the terms by 0 is just a shortcut to solving this system of equations. It doesn't matter that the original function isn't defined at that value, we're not evaluating the function at that value, we're solving a system of equations. Take the first example from the video, if you just fully multiply by the common denominator you get: 2x+1=A(x+2)+B(x+1) 2x+1=Ax+2A+Bx+B 2x+1=(A+B)x+(2A+B) Matching the coefficients by their degrees, we get the following system of equations: 2=A+B 1=2A+B One way you might go about solving this system is to multiply the first equation by -1 and then adding the equations together so that the Bs cancel out. But this is exactly the same as substituting x=-1 into the last form of the equation before we split it into our system of equations. You could also solve it by multiplying the first equation by -2 and adding them together to make the As cancel out, which again is no different than substituting x=-2 in earlier. This is what we're really doing, we're just able to skip a lot of the steps due to the way the equations are built.
Hes comparing coefficients, on the right side, the coefficients of x² are -1 and B (because theres a -x² and a Bx²), meanwhile on the left there are no x² terms, so the coeffient for x² on the left is 0 (for 0x²) This means you can say 0 = -1 + B as these are all of the coefficients for the x² terms, you can do the same for the x terms too, or x³ terms, or x⁴ terms, etc
Thank you so much from Croatia. Great help and excellent explanation of why we can do this. May our God Jesus Christ of Nazareth bless you and keep you.
Was needing to catch up on some specialist mathematics (Calculus 2 I guess) and I had no idea how to rationalise the 3rd/4th example using that form of partial fraction. This demonstration was so clear and now I have a simple way to rationalise and complete these problems, thanks so much!
There you just call A and B as residuals and the possibility of repeated linear factors being differentiable for the said limits (especially for Laplace and fourier type functions) but still its kinda the same
@@azzteke 1:24 Setting x = -1 after multiplying both sides by (x+1) is not legit. Whenever you multiply both side of an equation by an expression, that chain of logic is only valid when that expression is not equal to 0. But somehow it works to solve partial fractions, which is handy but very strange.
I have the same worry, although I can see the coverup method is the same method with less writing. I have for a while been thinking about finding a formal treatment of algebra's "multiplication of both sides" when the factor may be zero for some values of a variable. I wonder if it's because the meaning of = is such that the equation holds wherever neither of its sides is undefined, and maybe multiplying by (X + 1) changes so the solutions to the new equation solve the original where either of its sides are strictly no more defined than the sides of the new equation.
We're not evaluating the original expression at x=-1 though. That's why it works. For a breakdown of what's happening and when/how these undefined values become allowed, it goes something like this (using the first example from the video below): In the original expression, you are most definitely correct, x=-1 and x=-2 are not allowed, as neither side of the equation are defined at those values. So these are a restriction on our domain. We multiply both sides of the equation by the common denominator on the left in order to cancel out all of the denominators. Note: no multiplication by 0 has happened here, because the restriction on our domain ensures that these denominators cannot be 0. So we now have a new expression that contains no fractions: 2x+1=A(x+2)+B(x+1). Note: x=-1 and x=-2 are still not in our domain, so we cannot actually substitute them in. Instead, the 'traditional' way of solving this for A and B involves using some algebra to move things around and group terms into coefficients. In this case, after some distribution and regrouping we get 2x+1=(A+B)x+(2A+B). Since the coefficients on left must match the corresponding coefficients on the right, this let's us express the relationship between A and B as a system of equations: the x coefficients gives us 2=A+B, and the constant terms gives us 1=2A+B. Note: there are no x's in this system of equations. There are many methods that can be used to solve a system of equations such as this, but here are a couple of rather simple and obvious methods: first, we could multiply the first equation, 2=A+B by -1 and add it to the second so that the B's cancel out giving us -1=A; second we could multiple the first equation by -2 and add it to the first so that the A's cancel out, giving us -3=-B, or 3=B. These are perfectly valid methods for solving this system of equations. We haven't violated any of our restrictions, we haven't allowed any denominators to equal 0. However, if you look closely, you'll notice that the first equation comes from the x-coefficients in our expression. So multiplying it by some value gives the same effect letting x equal that value. Now, keep in mind that we did NOT set x equal to -1 or -2, there were no x's in our system of equations. But because of the way the system of equations is built, this does reveal a shortcut for solving some of the variables in the system of equations without needing to explicitly write out the system. It "looks" like we're multiplying by 0, but that's just a shorthand for what we're really doing. It's like when you take a limit as x->inf and you "just plug in infinity"; you can't actually plug infinity into a function. Instead, what you're doing is evaluating the function as x grows unbounded, but it's faster and simpler to just write infinity into the function as if you could do that, even though it's not really what you're doing. Same thing here, we can't really set x equal to these values, they're not in the domain, but it's a shortcut to the solution to the system of equations that give us the numerator coefficients of the partial fractions.
Check out my 100 algebra problems to get you ready for calculus th-cam.com/video/XtQIsKoHc8Q/w-d-xo.html
It's wild that cover-up method is not taught in schools, it's one of the most awesome shortcuts ever. At least better than those "shortcuts" tiktok keeps showing me to multiply two digit numbers lol
It is taught, it’s just not very generalizable to more complicated functions as shown in the later examples. It’s more formally called the Heaviside method, which you can look up. In AP Calculus BC, all of the partial fraction problems are non-repeating linear like the first problem in the video, and some version of the coverup method is used pretty widely.
I was taught it 30 years ago
you are contributing to make the world a better place by spreading such knowldge to everyone for free
Happy to help!
I don't know how I can thank you. You really deserve to be a professor
Just a “like” on this video is more than enough. Thanks!
Btw I am a teacher in real life 😃
@@bprpcalculusbasics I don't mind at all ♥
@@bprpcalculusbasics Where do you teach and what grades?
@@yashbrawlstars3343 Pierce College in LA
Thank you so much! I’m preparing for a competition and I knew something I this existed for these type of questions but I actually forgot the method
Guy does a little stretch at the beginning, then I see a whole shelf of expo markers in the background... Oh man, here we go, this guy is about to wreck my mind 🤯😅 Awesome video, thanks for the content 🙏
Thanks!
he is the best maths teachers out there, like seriously maths can never be this easy but his teaching makes it sound so simple.
Other tips for examples 2 thru 4:
For example 2, plug in strategic values of x, such as x=0 and x=1, that aren't already spoken-for by the coverup method.
(2*x + 1)/((x + 1)*(x^2 + 2)) = -1/3/(x + 1) + (B*x + C)/(x^2 + 2)
Let x = 0, to eliminate B and find C:
(1)/((1)*(2)) = -1/3/(1) + (C)/(2)
1/2 + 1/3 = C/2
C = 5/3
(2*1 + 1)/((1 + 1)*(1 + 2)) = -1/3/(1 + 1) + (B + 5/3)/(1 + 2)
1/2 = -1/6 + B/3 + 5/9
B = 1/3
For examples 3 & 4:
Once Cover-up finds A and C, you can find B, by multiplying by only 1 copy of the repeated term. Take the limit as x goes to infinity. Some terms cancel to zero, some terms reduce to a finite non-zero number, and B will stand on its own as a constant. This is a way to more directly find B, that allows you to ignore terms that don't affect it. It has trouble if there is an irreducible quadratic involved, but works well when only dealing with linear and linear-repeated factors.
As applied to example 3:
(2*x + 1)/((x + 1)*x^2) = -1/(x + 1) + B/x + 1/x^2
Multiply by just one copy of x:
(2*x + 1)/((x + 1)*x) = -x/(x + 1) + B + 1/x
Send x to inf, and take the limit:
0 = -1 + B + 0
B = 1
Thank you so much
You know a teacher is GOATED when he uses two markers with one hand 🐐
Thanks for this but can you help me with this x/(x^2+a^2)(x^2+b^2).😊😊
do the one where the upper term also has x squared
Dear Blackpenredpen,
Can you explain why lim x->inf [xtanh(x)-ln(cosh(x))] = ln(2)?
There is something exponentially weird with this limit.
You are a goat proffffffff!❤️🙏🏻
i love you
U just save my exam, love u ❤
At this point: th-cam.com/video/WrGXIjXRSys/w-d-xo.html, my students follow this instruction: "It's only one more value to calculate. Then with the same strategy you used in other videos, we take a new value (like x=0, or another one, but not used), and then we get the last coefficient".
I'm thinking is the fastest method. Am I wrong? Thank you for your awesome work, @bprp !!!
Do anyone know why the degree of the numerator must be only be 1 degree less than the denominator and not by less than more than 1 degree ? (I am not talking explicitly on this example ) in general i mean
In the original expression, it's OK if the numerator degree is less than the denominator degree by more than 1. It can be up to and including, a numerator of a degree that is 1 less than the denominator. Any other given fraction can use polynomial division, or alternatively: adding zero in a fancy way to generate a term that reduces to a constant.
For each term you build, there are always n-1 degrees of freedom for the coefficient above any given term. To capture this possibility, we have an n-1 degree polynomial in the top, and an n degree polynomial down below. A linear factor's coefficients have just one degree of freedom, while a quadratic denominator has two degrees of freedom. A linear repeated factor is just like a quadratic factor, and has two degrees of freedom. As a result, it ultimately needs 2 coefficients to cover all possibilities.
10q but No 94, when we solve by shortway like for A we did x=-1 and found A=-1 and why you didn't use for B,(x=-2) and like this # u did that for C and by long way for B , then why ?
I solve No94 just like for B ,x=-2 and for C by the long way solving method but u solved that opposite using method of i solved;
W H Y why? ?? ?
I worry about this method. The RHS of the original equation is undefined at X = -1, are you relying on l'hopital's rule in some hidden way?
The coverup method is just a shorthand of what I call the ‘convenient value of x’ method. It is just some algebraic techniques of solving a rational function by turning it into a polynomial function and solving from there. The solutions of the polynomial will be the same as the solutions of the rational function except where the rational function does not exist. So we aren’t really solving the original function at x =-1, but we’re just kind of going ‘if we could solve at x= -1, what would we get…’
It seems that way, but really this method is just a shortcut to the 'traditional' way of solving partial fractions. The 'traditional' way is like what he did for the middle term in the later examples, where you just multiply everything by the common denominator and then you group the terms according to their degree. Except that instead of only having 1 variable to solve for, allowing you to ignore many of the terms and solve a simple single variable equation, you will instead end up with a system of equations that you need to solve to get A, B, C, etc. And methods to solve such a system of equations generally involve finding ways to scale and combine the equations so that other terms become 0 to let you solve one variable and then substituting that value in to solve for others, etc.
Well, this method of picking an x-value so that you end up multiplying most of the terms by 0 is just a shortcut to solving this system of equations. It doesn't matter that the original function isn't defined at that value, we're not evaluating the function at that value, we're solving a system of equations.
Take the first example from the video, if you just fully multiply by the common denominator you get:
2x+1=A(x+2)+B(x+1)
2x+1=Ax+2A+Bx+B
2x+1=(A+B)x+(2A+B)
Matching the coefficients by their degrees, we get the following system of equations:
2=A+B
1=2A+B
One way you might go about solving this system is to multiply the first equation by -1 and then adding the equations together so that the Bs cancel out. But this is exactly the same as substituting x=-1 into the last form of the equation before we split it into our system of equations. You could also solve it by multiplying the first equation by -2 and adding them together to make the As cancel out, which again is no different than substituting x=-2 in earlier.
This is what we're really doing, we're just able to skip a lot of the steps due to the way the equations are built.
In the second example there's another reason you can't use the cover-up method. x²+2 is never zero in the real world.
I wish you could give us a pdf sheet related to the concept every video 😢
I liked you and your explanation🥰
Could follow perfectly fine until 14:20 where he does something crazy with no explanation whatsoever...
Hes comparing coefficients, on the right side, the coefficients of x² are -1 and B (because theres a -x² and a Bx²), meanwhile on the left there are no x² terms, so the coeffient for x² on the left is 0 (for 0x²)
This means you can say 0 = -1 + B as these are all of the coefficients for the x² terms, you can do the same for the x terms too, or x³ terms, or x⁴ terms, etc
Nvm the guy above me said it
PLEASE HELP WHAT IS THIS PARTIAL FRACTION FOR IF WE DONT GET X IN THE END
?
?
Thank you so much from Croatia. Great help and excellent explanation of why we can do this.
May our God Jesus Christ of Nazareth bless you and keep you.
Was needing to catch up on some specialist mathematics (Calculus 2 I guess) and I had no idea how to rationalise the 3rd/4th example using that form of partial fraction. This demonstration was so clear and now I have a simple way to rationalise and complete these problems, thanks so much!
you really helped me so much. Thank you. Im not an english native speaker but i understood everything so well. Congratulations ❤️
Nice ❤
yep... good as always
1st example is also called Heaviside method
😎💙
牛逼
you are amazing at explaning, habibi
Thank you very much this video really helped
thank u sir
We, students, used complex numbers in Calculus 2, but I don’t remember this
There you just call A and B as residuals and the possibility of repeated linear factors being differentiable for the said limits (especially for Laplace and fourier type functions) but still its kinda the same
u helped me very much thank you
Thank you!!
Multiplying both sides of the equation by 0 ... somehow works? But it does.
Nonsense. Nobody multiplies LHS and RHS by Zero. Bad joke!
@@azzteke 1:24 Setting x = -1 after multiplying both sides by (x+1) is not legit. Whenever you multiply both side of an equation by an expression, that chain of logic is only valid when that expression is not equal to 0. But somehow it works to solve partial fractions, which is handy but very strange.
Original expression is not defined for x=-1. So this method can not be correct. The cover up method is also too magical to my taste.
I have the same worry, although I can see the coverup method is the same method with less writing. I have for a while been thinking about finding a formal treatment of algebra's "multiplication of both sides" when the factor may be zero for some values of a variable. I wonder if it's because the meaning of = is such that the equation holds wherever neither of its sides is undefined, and maybe multiplying by (X + 1) changes so the solutions to the new equation solve the original where either of its sides are strictly no more defined than the sides of the new equation.
We're not evaluating the original expression at x=-1 though. That's why it works. For a breakdown of what's happening and when/how these undefined values become allowed, it goes something like this (using the first example from the video below):
In the original expression, you are most definitely correct, x=-1 and x=-2 are not allowed, as neither side of the equation are defined at those values. So these are a restriction on our domain. We multiply both sides of the equation by the common denominator on the left in order to cancel out all of the denominators. Note: no multiplication by 0 has happened here, because the restriction on our domain ensures that these denominators cannot be 0.
So we now have a new expression that contains no fractions: 2x+1=A(x+2)+B(x+1). Note: x=-1 and x=-2 are still not in our domain, so we cannot actually substitute them in. Instead, the 'traditional' way of solving this for A and B involves using some algebra to move things around and group terms into coefficients. In this case, after some distribution and regrouping we get 2x+1=(A+B)x+(2A+B). Since the coefficients on left must match the corresponding coefficients on the right, this let's us express the relationship between A and B as a system of equations: the x coefficients gives us 2=A+B, and the constant terms gives us 1=2A+B. Note: there are no x's in this system of equations.
There are many methods that can be used to solve a system of equations such as this, but here are a couple of rather simple and obvious methods: first, we could multiply the first equation, 2=A+B by -1 and add it to the second so that the B's cancel out giving us -1=A; second we could multiple the first equation by -2 and add it to the first so that the A's cancel out, giving us -3=-B, or 3=B. These are perfectly valid methods for solving this system of equations. We haven't violated any of our restrictions, we haven't allowed any denominators to equal 0.
However, if you look closely, you'll notice that the first equation comes from the x-coefficients in our expression. So multiplying it by some value gives the same effect letting x equal that value. Now, keep in mind that we did NOT set x equal to -1 or -2, there were no x's in our system of equations. But because of the way the system of equations is built, this does reveal a shortcut for solving some of the variables in the system of equations without needing to explicitly write out the system.
It "looks" like we're multiplying by 0, but that's just a shorthand for what we're really doing. It's like when you take a limit as x->inf and you "just plug in infinity"; you can't actually plug infinity into a function. Instead, what you're doing is evaluating the function as x grows unbounded, but it's faster and simpler to just write infinity into the function as if you could do that, even though it's not really what you're doing. Same thing here, we can't really set x equal to these values, they're not in the domain, but it's a shortcut to the solution to the system of equations that give us the numerator coefficients of the partial fractions.