This is how we partial fraction, repeated linear roots, "cover-up method"
Вставка
- Опубліковано 19 січ 2018
- Here's the explanation on why and how the cover-up method for partial fractions works! Really helpful and quick!
Previous part: the setup: • This is how we partial...
blackpenredpen,
math for fun,
blackpenredpen@gmail.com
/ blackpenredpen
Please subscribe and recommend my channel
You know, I got through a physics/math double major and a physics PhD, did integrals all the time, and I confess this is the one standard integration technique from AP Calculus that I never could entirely understand or, consequently, remember. You just explained it.
The cover-up method only works when you are dealing with linear factors, though. Otherwise, you have to equate coefficients.
It was touched on again in Complex analysis, when finding residues...
I love the cover-up method, and enjoyed your explanation of WHY it works.
Sir, you have saved my math grades for countless years! Thank you once again!
This and the video about why you did B/x+2 just blew my mind. You're amazing!!!
This, sir, is why you are such a stud! Nice method and nice explanation.
mjones207 thank you!!
your the best dude....you have no idea how much time you just saved me
this is the best video for partial fractions that didnt drone on for 30 mins
That was magical 🌻
Thank you !
This is amazing! I wish I had seen this during my calc classes
Youu are a very admirable teacher!! I love your calculus tutorials
Awesome explanation!
That's a real neat trick! Thank you c:
I'm impressed !
Thanks a lot. Thanks a lot. Thank you. Thank you very much. You dont know how much u helped me now. I will always remember u for helping me out. (Tomorrow is my exam). And i really couldnt work the thing out myself.
Thank you good sir. You have saved me soo much.
Thank you i understand vividly well
the legend is back
Ayman Algeria killuaaa
You are an amazing math enthusiast...
This is amaze af! :D
Thank you bprp
I know this trick but didn't knew about proof
Thank you!!
Your videos are amazing. Is there a way I can ask you questions?
Thank you !!!!
Youre a LEGEND
HOLY MOTHER THIS IS AMAZING
Thank you very much sir
I helps me alot💕
I always did complicated systems for these... Now I will do it 10 times faster for the linear ones ! :D
Wow it saves time alot
Superb👍
this is great :)
i'm impressed😆
The young gentleman is absolutely brilliant in his teaching.
Thank you!!!
I have particularly enjoyed all the integration problems that you have solved. My favorite ones are the "Integral battles"
Flexin' with the supreme jacket lmao
Really Thank You Sir Love From India ♥️
how would this method work with a cube root?
Thank you 😊❤️
You’re welcome!
Bravo!!!
How perfectly you switch between black pen and red pen...I mean just wow... :D
Is this the reason behind the name blackpenredpen ?
Thanks
👍 super teacher
I'm amazed at how he changes the colours
Thnx sir
What if in the numerator no 1. In my case, I have s^2+ 2s + 5 would this method still work
It doesn't matter what the numerator is. The method still works.
I'll do an example with your irreducible quadratic as one of the roots.
1/((s + 1)*(s^2 + 2*s + 5)) = A/(s + 1) + (B*s + C)/(s^2 + 2*s + 5)
Since this is an irreducible quadratic, we need a linear term on top of it, hence B*s + C, instead of just B. I like to put terms I can get with H's cover-up method first.
Use H's cover-up to get A:
A = 1/((-1)^2 + 2*(-1) + 5) = 1/4
Replace A with 1/4, and multiply to clear the fraction:
1/((s + 1)*(s^2 + 2*s + 5)) = (1/4)/(s + 1) + (B*s + C)/(s^2 + 2*s + 5)
1 = (1/4)*(s^2 + 2*s + 5) + (B*s + C)*(s + 1)
Let s = 0, to find C:
1 = (1/4)*(0^2 + 2*0 + 5) + (B*0 + C)*(0 + 1)
1 = 5/4 + C
C = -1/4
Let s = 1 to find B:
1 = (1/4)*(1^2 + 2*1 + 5) + (B*1 - 1/4)*(1 + 1)
1 = 2 + 2*B - 1/2
B = -1/4
Result:
1/((s + 1)*(s^2 + 2*s + 5)) = (1/4)/(s + 1) + (-1/4*s -1/4)/(s^2 + 2*s + 5)
Will this work?
x/[(x+1)(x+2)] Will it work with cover up method?
Yes it will.
3:18 "Pick an easy number"
Could I use 0, ±1 (assuming they are not used yet) and ±∞? They are the easiest numbers for me to deal with
hold on ±∞?
@@pedrosso0 ahhh okay. thanks!
@@aronpacino8009 that wasn't an answer,I was asking a question, what do you mean by +- infinity,
I don't think you can substitute infinity.
@@pedrosso0 You aren't really substituting infinity, but rather taking the limit as x approaches infinity. This works nicely, when you have repeated linear terms, and allows you to more directly solve for the coefficient.
Example:
(x + 3)/(x^2*(x + 1))
Set up partial fraction expansion:
A/(x + 1) + B/x^2 + C/x
Heaviside coverup solves for A & B:
A = (-1 + 3)/((-1)^2) = 2
B = (0 + 3)/(0 + 2) = 3
Construct result thus far:
(x + 3)/(x^2*(x + 1)) = 2/(x + 1) + 3/x^2 + C/x
Multiply by just one instance of x, to partially clear it:
(x + 3)/(x*(x + 1)) = 2*x/(x + 1) + 3/x + C
Take the limit of each term, as x approaches infinity.
For (x + 3)/(x*(x + 1)), since there is ultimately a higher power of x in the denominator than numerator, this approaches zero.
For 2*x/(x + 1), there is an equal power of x in both numerator and denominator. As x approaches infinity, x dwarfs 1, and makes this become 2*x/x, which evaluates in the limit to equaling 2, since we have identical functions of x both approaching infinity.
For 3/x, this approaches zero, since infinity is in the denominator.
And C is just a constant.
This simplifies to:
0 = 2 + C, meaning C =-2
Thus the result is:
2/(x + 1) + 3/x^2 - 2/x
could you explain why you're able to plug in any number for B?
Genius
What college do you Dr. Peyam work at?
Pretty much the same as 1 = A(X+2)^2 + B(X+1)(X+2) + C(X+1), and zero out the terms, or pick an easy number to set up a system of equations to solve for A, B, C...
oh no, i have to desubscribe again, so I can subscribe?
AndDiracisHisProphet yes, please do so hahahaha
As you wish.
Do you get a notification anytime?
Hmm, I dont get notifications from subscriptions
I just noticed that I have 4 subscribers. WTF?
@@AndDiracisHisProphet now you have 21
An easier way to get B=-1, with complex analysis: if you integrate around a big circle of radius R, the length of the circle is 2 Pi R but the integrand is basically 1/R^3, so the integral goes to 0 as R to infinity. This means that the sum of the residues must be 0, so A+B = 0.
Bruh
How about just using logic. C=B cuz they are repeated roots.
@@iben1195 exactly bro, bros overthinking for no reason💀
When you have done all the side quests, then return to an easier boss enemy:
from where u are? sir
We dont need repeated roots case of partial fraction , For integration we have Ostrogradski isolation of rational part of integral
and for inversion of Laplace transform we have convolution
can you do calc 3 lessons?
IRedSonI sorry maybe not sometime soon. But I can keep this in the future
Keep in mind*
Thank you
Wished BPRP to teach Calc 3. Calc 3 is also fun.
@@blackpenredpen what about your promise
he really is flexin the supreme tho👀
What if the power of (x+2) is 3 or higher?
It still works. H's cover-up method works for the term with the highest exponent of each fraction denominator, when there are repeated roots.
For example:
1/((x+1)*(x + 2)^3) = A/(x + 1) + B/(x+2)^3 + C/(x+2)^2 + D/(x + 2)
A and B can be found with the cover-up method, but C and D will need another strategy to find them.
A = 1/(-1 + 2)^3 = +1
B = 1/(-2 + 1) = -1
Plug in A and B and continue:
1/((x+1)*(x + 2)^3) = 1/(x + 1) - 1/(x+2)^3 + C/(x+2)^2 + D/(x + 2)
When you plug in x=0 and x=1 as strategic values for x, you'll get the following solutions for C&D:
C = -1, D=-2
Thus the total solution is:
1/((x+1)*(x + 2)^3) = 1/(x + 1) - 1/(x+2)^3 - 1/(x+2)^2 - 1/(x + 2)
0:56 to 1:05 this is how we do discussions all the time
You have shown me de wae
Nice channel name
What if we don't have 1 in the numerator but any x linear term
It still works. Ultimately, what is required for this method to work, is that the total degree on the top is at least one less than the total degree on the bottom. If there isn't, you'd start by using polynomial division to reduce it first.
Example:
x/((x + 4)*(x + 2)) = A/(x + 4) + B/(x + 2)
Heaviside cover-up for A, at x = -4:
-4/(-4 + 2) = 2
Heaviside cover-up for B, at x = -2:
-2/(-2 + 4) = -1
Result:
x/((x + 4)*(x + 2)) = 2/(x + 4) - 1/(x + 2)
How does 1+2^2 = 4?
I thought you have to use Cx + D as the numerator of a quadratic factor?
Frank Rodriguez watch previous video
to any STEM students, remember this trick. trust it saves so much time on exams in fact the professors probably expect you to know this trick anyway
Thank you
I am from Bangladesh 🇧🇩
Its crazy
@blackpenredpen sir How to do it for (1+2x)/(x+2)²(x-1)²?....I tried and I'm confused...plz some one help me
Given:
(2*x + 1)/((x+2)^2*(x - 1)^2)
Set up the partial fractions:
A/(x+2)^2 + B/(x + 2) + C/(x - 1)^2 + D/(x - 1)
Use H's cover-up method, to get A and C:
A = (2*(-2) + 1)/(-2 -1)^2 = -1/3
C = (2*(1) + 1)/(1+2)^2 = +1/3
Fill in A and C:
(-1/3)/(x+2)^2 + B/(x + 2) + (1/3)/(x - 1)^2 + D/(x - 1)
Multiply each numerator by the variants of the terms that aren't in its corresponding denominator:
(-1/3)*(x - 1)^2 + B*(x+2)*(x - 1)^2 + (1/3)*(x+2)^2 + D*(x - 1)*(x + 2)^2 = 2*x + 1
Let x=0:
(-1/3)*(0 - 1)^2 + B*(0+2)*(0 - 1)^2 + (1/3)*(0+2)^2 + D*(0 - 1)*(0 + 2)^2 = 2*0 + 1
2*B - 4*D + 1 = 1
Let x=-1 (can't use 1, because we used it earlier):
(-1/3)*(-1 - 1)^2 + B*(-1+2)*(-1 - 1)^2 + (1/3)*(-1+2)^2 + D*(-1 - 1)*(-1 + 2)^2 = 2*(-1) + 1
4*B - 2*D - 1 = -1
Solve for B and D:
B = 0, D=0
It is just a coincidence that these terms came out this way, from this particular problem. This doesn't mean they always will, for repeated linear fractions.
Result:
(-1/3)/(x+2)^2 + (1/3)/(x - 1)^2
@@carultch thank you so much
More videos
he is the ling ling of mathematics
another reason why you shouldn’t use x=-1 or x=-2 for finding B is that you’d get 1/0 on the LHS and THAT’S NOT VERY NICE, *isn’t it?*
Ironically, that's precisely what you should use, in order to make the problem as simple as possible to solve. That's the entire idea of the Heaviside coverup method.
Ultimately, what you are doing is taking the limit as x approaches the problem point, rather than directly evaluating the original function at the problem point. It pens out to be a direct evaluation, because we end up with a removable zero in the expressions we derive for each coefficient. The removeable zero cancels, and also creates numerator zeros for all the other constants, and we're left with an expression that we can evaluate directly, and not get zero over zero anymore.
Hi blackpenredpen I'm a 14 and I'm from Italy. I have a problem for you: find the integral from 0 to 1 of (-1)^x
Luca Zara convert (-1)^x into base e first. Or if you like you rewrite -1as i ^2 so that you can use Euler’s equation.
Man la ok thank you I think the solution is 2i/pi
@@lucazara9137 You'll end up with a path-dependent integral in the complex plane. When you convert to e^(ln(-1)*x), there are multiple solutions for ln(-1), and the solution to this integral will depend on which one you choose. What you've evaluated, is the Cauchy principal value.
What if non linear factors..like
1 /(1+x²)(x+1).
????? Easy method?
One way you can do it, is to force (x^2 + 1) to become two linear factors, with imaginary roots. Then you can use the cover-up method to get all three coefs in this problem.
But it is much easier to use the cover-up method only for (x-1). And to use a traditional approach for the other two.
Given:
1 /((1+x²)(x+1))
Setup expansion. Constant on the first term, linear expression on the 2nd term:
A/(x + 1) + (B*x + C)/(x^2 + 1)
Use cover-up method to get A, using x=-1:
A = 1/((-1)^2 + 1) = 1/2
Reconstruct with known A:
(1/2)/(x + 1) + (B*x + C)/(x^2 + 1)
Multiply out the expression, and equate to the original numerator:
(1/2)*(x^2 + 1) + (B*x + C)*(x + 1) = 1
Let x=1, and let x=0, to develop our two equations and two unknowns to get B & C:
(1/2)*(0^2 + 1) + (B*0 + C)*(0 + 1) = 1
1/2 + C = 1
C = 1/2
Plug in known C, as we set x=1 to find B:
(1/2)*(1^2 + 1) + (B*1 + 1/2)*(1 + 1) = 1
2*B + 2 = 1
B = -1/2
Reconstruct our solution:
(1/2)/(x + 1) + (-x/2 + 1/2)/(x^2 + 1)
I was on a diplomatic mission to Alderaan then this Math Sith showed up with The Death Star in his hand and killed all my terms
ua-cam.com/video/jGF_6cCYVwI/v-deo.html
Tough integration problem
what if we have a term that can't be 0 like (x^2 + 1) am stuck help 🆘😞
That is called an irreducible quadratic. One option is to find its complex solutions, and use them to construct a pair of conjugate linear terms. Another option, is to use the partial fractions method for irreducible quadratic, and place an arbitrary linear term on top of it, which is the example I'll show.
Given:
2/((x + 1)*(x^2 + 1)), construct the partial fraction expansion of A/(x + 1) + (B*x + C)/(x^2 + 1)
Use H's cover-up method to get A:
A = 2/((-1)^2 + 1) = 1
Plug in A and cross-multiply:
2/((x + 1)*(x^2 + 1)) = 1/(x + 1) + (B*x + C)/(x^2 + 1)
2 = 1*(x^2 + 1) + (B*x + C)*(x + 1)
Let x=0 to get C:
2 = 1*(1) + C*(1)
C = 1
Let x = 1 to get B:
2 = 1*(1 + 1) + (B*1 + 1)*(1 + 1)
2 = 2 + 2*B + 2
-2 = 2*B
B = -1
Result:
2/((x + 1)*(x^2 + 1)) = 1/(x + 1) + (-x + 1)/(x^2 + 1)
Challenge: lim(x -> +0) (x^x - 1) * ln(x)
(1+n^2)/n(n+1) by cover up rule...
Answer??
Given:
(1 + n^2)/(n*(n + 1))
First, observe that the degree of the numerator is the same as the denominator. This means, we must first reduce it so the fraction part has a numerator that is at least 1 degree less than the denominator.
Expand the denominator, for reference: n^2 + n
Add zero in a fancy way, so that n^2 + n appears in the numerator:
(n^2 + n - n + 1)/(n*(n + 1))
This forms a term we can cancel, because (n^2 + n)/(n^2 + n) = 1. Thus we have:
1 + (-n + 1)/(n*(n + 1))
Now we can use partial fractions:
(-n + 1)/(n*(n + 1)) = A/n + B/(n + 1)
Heaviside coverup solves for both A & B:
at n = 0; A = (-0 + 1)/(0 + 1) = 1
at n = -1; B = (-(-1) + 1)/(-1) = -2
Thus, the result is:
1 + 1/n - 2/(n + 1)
1:09
How about 1/(s+2)^5
Hii did u find answer for it pls share
what if he have
(x+2)^3
Then you'd construct A/(the other original denominator term) + B/(x + 2)^3 + C/(x + 2)^2 + D/(x + 2)
H's cover-up method can tell you A and B, assuming the other original denominator term is a simple linear term like (x + 1). You then need alternate methods to find C and D, such as plugging in strategic values of x to construct two equations to solve for both C and D.
Wait did you say "that KILLS this turd" @ 6:12? lol
Term not turd
but what if we had more than one unknown value like B? This wouldn't work because you wouldn't be able to solve anymore with two unknown values
At that point you could multiply the entire equation out and put together like terms and perform a systems of equations and plug in the values you already know.
You killed my terms! Murderer!
Hello sir I am from uttar Pradesh
i luv u
Nice drip
U r 👌🤘
😢😊
Residue Theorem FTW
My brain hurts...... How can we possibly let X = -1? In the original fraction, that would imply we were dividing by zero???
Remember we are covering up not making something to zero. Thou shall not divide by zero
@@uchindamiphiri1381 Yeah, but at 0:50 or so, he explicitly says "we are covering up x+1 and making it equal to 0" or something like that. And then talks about plugging in -1 for X. Say what??? Just because we covered up x+1 doesn't mean it's gone--how can we legitimately plug in -1 for X?
@@randyhelzerman Did you watch the explanation part of the video starting at about 5:15?
@@randyhelzerman The reason why we GET to make one of the denominator terms equal to zero, is that what we really are doing is taking the limit as x approaches that value. When rearranging to solve for the unknown constant, you'll see that the solution has a removeable zero at the problem point in question (call it x = p). That removable zero cancels out for solving for the constant in question to become (x - p)/(x - p), and also cancels out all the other unknown constants.
Thanks@@carultch
im the first nerd *dab*
The logic is faulty.
The original equation has no solution when x = -1 since no values for A, B, or C can make 1/0 = A/(x+1) + ...
So why does this work despite that?
anon8109 first u multiply the whole equation by (x+1)AND THEN PLUG x=-1 u would get A=1
If that's allowed then suppose 1/(x+1) = A. Then 1 = A(x+1), so 1= Ax + A. Now substitute -1 for x, giving
1 = A(-1) + A, so 1 = 0.
In general, if a value for x is not a solution, you can't later substitute it into the equation since you can get a contradiction. It's a logically invalid step.
You're not looking for solutions, you're looking to rewrite the expression
anon8109 Well you cannot assume something and then multiply by 0 on both sides and say its true BUT When u know something is true u can surely do that if it gives u some useful result as in this case!
Ps- even i got confused for a sec abt the explanation but then thought of this ;-)
The original equation had holes at x=-1 and x=-2. x That is the values on either side approach each other. The new equations each fix one of those holes. It doesn't matter that you can't reverse here, because A doesn't depend on x. Assuming x=-1 will give you the same A as assume x=3
second,yaaaah
?!
Cannot see for the bright light on board.
what is your job ?
Mihai Ciorobitca
Math teacher!!
greatest ever maths teacher (my opinion)
blackpenredpen i would like you to be my teacher
Buuuut.... what if there's an x at the top..?
It still works. Ultimately, what is required for this method to work, is that the total degree on the top is at least one less than the total degree on the bottom. If there isn't, you'd start by using polynomial division to reduce it first.
Example:
x/((x + 4)*(x + 2)) = A/(x + 4) + B/(x + 2)
Heaviside cover-up for A, at x = -4:
-4/(-4 + 2) = 2
Heaviside cover-up for B, at x = -2:
-2/(-2 + 4) = -1
Result:
x/((x + 4)*(x + 2)) = 2/(x + 4) - 1/(x + 2)
Anyone else wondering how it’s legal to divide by 0 😂
What you are really doing, is taking the limit as x approaches the problem point, and the behavior of the pole of the function will dwarf the rest of the function. You are matching the coefficient on a rational function of ONLY that particular pole, to make a function that has a similar behavior immediately at the pole.
As an example, consider (x - 1)/((x + 1)*(x + 2)), which has a partial fraction expansion of 2/(x + 2) - 3/(x + 1).
Nearby the pole x = -2, the function has a similar behavior as the component of it, which is 2/(x + 2).
Nearby the pole x = -1 the function has a similar behavior as the component of it, which is -3/(x + 1).
@@carultch does this have anything to do with what we learned way back in precalc with certain zeros either crossing or bouncing back depending on if that zero came from an odd or an even power? When it gets close to zero, you’re saying it essentially overtakes all other aspects bc 1/0+ = inf?
@@darcash1738 Here's what is happening.
Consider solving for A, for the following partial fractions:
4/((x - 3)*(x - 1)) = A/(x - 1) + B/(x - 3)
Solve for A:
A/(x - 1) = 4/((x - 3)*(x - 1)) - B/(x - 3)
A = 4*(x - 1)/((x - 3)*(x - 1)) - B*(x - 1)/(x - 3)
Take the limit as x approaches 1. I'll do this numerically, starting at x=1.1. You'll get similar results if you try this from the negative side of x=1as well, but I'll opt to show it from the positive side
At x=1.1: A = -2.10526 + B*0.05263
At x=1.05: A =-2.05128 + B*0.02564
At x = 1.01: A = -2.0101 + B*0.005025
At x=1.005: A = -2.00501 + B*0.002506
At x = 1.001: A = -2.0010 + B*0.0005003
As you can see, the coefficient on B diminishes to getting closer and closer to zero, so that B has less and less effect on the value of A. The remaining portion of the expression, keeps getting closer and closer to A = -2. You can see that the expression for A, contains a removable zero that we can cancel. This means there is a hole at x=1, and the value of the function that would fill that hole, is A = -2, with no dependence on B. This is why this method has the advantage, because it allows us to more directly get at the answer without needing to find the other coefficients. The same thing will also happen, when solving for B, near x = 3.
Preme 😤