This is one of my favorite videos!!! Definitely please do more videos explaining WHY certain processes work it leads to a really deep understanding!! Love your channel, keep it up!!:)
I appreciate that some of these videos are devoted purely to theory. Seriously, this guy is a great teacher! It's one thing to know the material, but it takes more to be able to explain it with clarity. And it takes more still to anticipate the follow-up questions that students have before they even fully formulate them. This LITERALLY happened for me at 1:54! He answered my question as I was formulating it in my mind, almost as if he was asking it for me!)
I just started learning this and noone ever told me the reason for it, it really sucks being taught for exams and not life or for the fact that i appreciate the subject
Oh my god, I have a semester exam in pre-calc, and I recently discovered you from your imaginary number videos, and you managed to upload this today! What timing, and thank you so much!
I'm currently graduating mathematics and soon I'll get to the 2nd year of graduation.All I have to say is that this person is my savior.Hope I can meet you one day.Cheers from Greece.
So basically, you can start by treating the repeated linear factor like an irreducible. For example (Ax + B)/(x + 1)^2. But now you can split up the fraction further: in this case into A/(x + 1) + (B - A)/(x + 1)^2. So the standard method for repeated linear factors isn't really different from the method for irreducible factors, it just saves a step.
True. Repeated linear factors can be treated just like irreducible quadratics. And in your example, if you set it up with a linear term on top, you will ultimately get the same answer. Heaviside cover-up helps us when we have repeated linear factors, because the cover-up method will get us the coefficient on the highest power term. We still need to use first principles to get the remaining coefficients.
years ago you helped me with my calculus homework. Now you are helping me when Laplace transforms have reared their ugly head in my process controls class.
Awesome explanation. I tried searching for before and got way overcomplicated explications involving the Euclidean algorithm for polynomials, but this is way more understandable.
Because when combining fractions, you multiply the denominator, i just think about putting every single fraction down which has one of the factors in the denominator. There is no A/((x+1)(x+2)) term, because that can be further split into a/(x+1) + b/(x+2) which is already included. The c/((x+2)^2) = d/(x+2) + e / ((x+2)^2) which just gives back what is already written. Those two fractions are surely the only combination that when added together give the desired denominator of (x+2)^2.
Bx+C is just a general expression so that we won't miss out any possible coefficient of x. If it happen that the numerator is just a constant, then the term "B" for (Bx+C) will be zero
I wanted to be more formal about it than the answers so far. In general we might have up to a second order polynomial on the original expression instead of just one, after any long division were there a higher order numerator. That means three free parameters in the original expression, so you need three free parameters in the partial fraction expansion to match. The only rule is that there needs to result in three linearly independent equations. There are infinitely many ways to do that but the easiest is Bt+C. See Imgur link. Though it is even easier to use the building up the powers method! But of course our goal here is to show why we build up the powers. Addendum to blackpenredpen comment imgur.com/a/08C0Qk7
You've shown that building up the powers is equivalent to having a linear term over the quadratic in a change of variables. However, that begs the question, "Why must the degree of the numerator be one less than that of the denominator?" and "Why is it that such a decomposition is always possible?"
The setup being possible is a good question but irrelevant most of the time because the proof for an individual question can be done by just working backwards and adding the partial fractions using common denominators. The degree of the numerator might be more than one less than the denominator (I.e. a quadratic denominator may have a constant numerator) but we do not know that when we first do the set-up. We know the numerator cannot have equal or greater degree, because if it did than the original fraction would also have equal or greater degree on the top and we would just do long division to make the numerator smaller.
Just to further elaborate on this, I think you can grasp an even better understanding of this if you do it the incorrect way to see where it goes wrong. Suppose we decided to try: 1/[(x+1)(x+2)^2]=A/(x+1)+B/(x+2)^2 Note that I’m only using 2 variables, A and B. Now, we don’t even need to solve for A and B to see where this goes wrong. Instead, let’s try recombining the two fractions. I’m only going to show the unsimplified result of the numerator: A(x+2)^2+B(x+1) If we slightly expand the expression above, we can get: A(x^2+4x+4)+B(x+1) The thing that is important to realize here is that there is no way for the expression above to equal a constant (specifically the value 1) no matter what value of A and B we use. It’s impossible to obtain our original expression using this approach. The Ax^2 term can’t get canceled out. Hence, we need to use: A/(x+1)+(Bx+C)/(x+2)^2 You might still be wondering how we pulled Bx+C out of our ass but it’s the middle of the night and I have spent more time than any person should writing a RUclips comment. If someone actually reads all this and wants me to explain, maybe I’ll do it another day. Thanks. Good night
I want you to explain, I am struggling. :'/ I don't understand why your expanded numerator can't result in a constant (specifically 1) for any values of A and B. Why can't: A(x^2+4x+4)+B(x+1) = 1
@@erwinsmith3312 Thanks for asking! I failed the exam at the time lol But I did the course again and succeeded :) Now I'm (almost) a math teacher, so that's awesome.
You could see that they are equivant in this way: beta*x+gamma = beta(x+2)+delta = beta*x + beta*2 + delta where gamma = beta*2+delta. Here beta = B and delta = C
To me, if you're comfortable working with complex numbers and you're okay doing more algebra to simplify the intermediate integrals, then factoring non-reducible polynomials (eg x^2 +2x +2) into complex linear factors and setting up the partial fractions as if they're linear factors would be better than trying to evaluate an integral with a linear function over a non reducible quadratic. Am I wrong for thinking this way?
is it correct that @ 5:00; if you have a denominator (x+2)^1 instead of (x+2)^2 that you would have to do long division to find the constant C; and then do partial fraction expansion to find A and B?
Excellent video again. I think it might help understand this issue even better, if you proved the equivalence of the beta-gamma approach in the end with the original setup. Just add zero in the beta-gamma-numerator like this: beta*x +gamma = beta*x +2*beta -2*beta+gamma=beta*(x+2)+gamma-2*beta if you plug this into the numerator, the whole term falls apart into beta/(x+2) + (gamma -2*beta)/(x+2)^2, which concludes to B=beta and C=gamma-2*beta.
at 2:30 why do we have to have 1 degree less at the top than the bottom ? is it just because it makes it easy to integrate ? or is there some other reasons ?
If you have a degree on the top that is higher than the degree on the bottom, then you won't be able to set up the partial fraction expansion to cover the highest degree on the top. For instance: (x^3 - x^2 - 2*x)/((x - 1)*(x-2)) If you attempt to set this up with a partial fraction expansion, you'll see that it isn't possible to generate the x^3 and x^2 coefficients. A/(x - 1) + B/(x - 2) Cross-multiply denominators: A*(x - 2) + B*(x-1) =?= x^3 - x^2 - 2*x Expand: A*x - 2*A + B*x - 2*B =?= x^3 - x^2 - 2*x You can't do it. There are no x^2 terms or x^3 terms to use. You'd have to do polynomial long division first, and end up with a polynomial as your result, with a rational term remaining. The polynomial term can be integrated via the power rule, and the rational term would then be integrated by partial fraction expansion.
Pretty interesting and well done video which shows how partial fraction is done bit I didn't see the reason why. I mean, there is no difference between (Bt+C)/(t+2)^2 and B/(t+2)+C/(t+2)^2 so we can use both of them. Because I want to know (OK, I do personally know but there are many people who don't) WHY we use second form over first, I expected to hear something like: "We use second fraction over first one while integrating such a function because the second form is just a sum of 2 basic integrals and we can instantly write down the answer. On the other side, there is no such instant answer for the first one." And one more little thing: in description you wrote "why do we HAVE TO build up the powers". If I'm right, "HAVE TO" is almost "MUST", but video clearly shows that we don't do it as must (just misunderstanding between video and it's description)
If we have a quadratic binomial factor in the denominator of the integrand such as x squared + 1 then we write Ax + B/x squared + 1. If we have (x squared + 1) squared then we write Ax + B/x squared + 1 + Cx + D/(x squared + 1) squared. Furthermore, if we have a cubic binomial factor such as x cubed + 1 then we write Ax squared + Bx + C/x cubed + 1. If we have (x cubed + 1) squared then we write Ax squared + Bx + C/ (x cubed + 1) + Dx squared + Ex + F/ (x cubed + 1) squared.
took me a hot sec to cut through the accent (not too difficult tho since my professor is German), but once I did this was a fantastic video, helped me understand this subject just in time for finals!
i still don't get the reason why when the denominator is 2nd degree we must have a linear term on the numerator?? i mean at the end we ended up with a constant (c) over t^2
Hey blackpen redpen do you have any question which has involve both feynman technique and power series in one question it would be great to see you solving that question
Can you explain why it has to be exactly 1 degree less on top? Why not 2 degrees less, or more? I can purely, intuitively see why degrees 2/4 are bad, for example, but not 1/3 or 0/2. Also, when _would_ you use Cx+D instead?
There is another thread about this with more details. The basic answer is that we know the numerator can be expressed as at most 1 degree less (because if it were the same or greater, than we just divide and get a simple polynomial with no fraction). We don't know whether it will be 1 or 2 degrees less until we work it out (incidentally, it cannot be more than 2 degrees less because every polynomial in the denominator can always be divided into 1 or 2 degree factors, because of the fundamental theorem of algebra/complex conjugate pairs theorem). So for the set-up you just assume the degree is always 1 less, and if you're lucky the "x" term will be 0.
When you are using partial fraction decomposition, and you have a quadratic factor ( meaning it has a degree of 2), you want to make the numerator one degree lower than the denominator. That is why when you have a linear factor in the denominator , you only put a constant in the numerator ( A, B,C, etc.). Whereas if you have a quadratic factor, you want the numerator in linear form (ax+b, bx+c, cx+d, etc.). This pattern repeats, for example if you have a cubed denominator, you would make put numerator in quadratic form, something like ax^2+ cx+d.
Hey BPRP! A friend asked me if I could solve a limit but I couldn't, or at least I didn't try enough. I challenge you to solve it! It is lim x->π of (x-π)[cotg(3x)+cosec(2x)]. Any method is allowed!
This is one of my favorite videos!!! Definitely please do more videos explaining WHY certain processes work it leads to a really deep understanding!! Love your channel, keep it up!!:)
Brandon Klein my pleasure!
I appreciate that some of these videos are devoted purely to theory. Seriously, this guy is a great teacher! It's one thing to know the material, but it takes more to be able to explain it with clarity. And it takes more still to anticipate the follow-up questions that students have before they even fully formulate them. This LITERALLY happened for me at 1:54! He answered my question as I was formulating it in my mind, almost as if he was asking it for me!)
I had this question in my mind 15 years ago when I was in the university but now I have the answer so thank you sir!
manamimnm my pleasure!!
You remember questions on your mind from 15 years ago? I can’t remember what I ate yesterday hahaha
@@navjotsingh2251 legend ha ha
@@blackpenredpen Why didnt you also at 9:35 write B(x plus 2)
I just started learning this and noone ever told me the reason for it, it really sucks being taught for exams and not life or for the fact that i appreciate the subject
Oh my god, I have a semester exam in pre-calc, and I recently discovered you from your imaginary number videos, and you managed to upload this today! What timing, and thank you so much!
Danesh Sivakumar nice!!!! Best of luck in ur class!
Collab with 3Blue1Brown when? I guess it would be called 3bluepen1brownpen
Derek C
Please don't get mad at John T. I am not sure why he is like that.
Some people just like to see the glass as half empty.
And you get mad at Derek's joke.
John T
And what are you??
what's the matter
I literally just searched a video of you where you explain partial fraction!!!
THE TIMING MY MAN
this was so understandable. i was so confused when my teacher taught me this but now i got the hang of it. thank you !
I'm currently graduating mathematics and soon I'll get to the 2nd year of graduation.All I have to say is that this person is my savior.Hope I can meet you one day.Cheers from Greece.
Sir you are superb.....every teacher should be like you and school won't be hell anymore
Be very proud of yourself, the work you are doing here is outstanding. This simple demonstration makes everything very clear.
As on mid of 9th grade, learnt partial fractions few classes ago. Finally understood how those work
So basically, you can start by treating the repeated linear factor like an irreducible. For example (Ax + B)/(x + 1)^2. But now you can split up the fraction further: in this case into A/(x + 1) + (B - A)/(x + 1)^2. So the standard method for repeated linear factors isn't really different from the method for irreducible factors, it just saves a step.
True. Repeated linear factors can be treated just like irreducible quadratics. And in your example, if you set it up with a linear term on top, you will ultimately get the same answer.
Heaviside cover-up helps us when we have repeated linear factors, because the cover-up method will get us the coefficient on the highest power term. We still need to use first principles to get the remaining coefficients.
Woooow, I’ve been thinkin’ ‘bout this for sooo long🤯 Thank you sooo much🙏
years ago you helped me with my calculus homework. Now you are helping me when Laplace transforms have reared their ugly head in my process controls class.
I'd not thought earlier about this question. Thought this when saw the thumbnail and got answer after watching the video. Thanks for this.
Awesome explanation. I tried searching for before and got way overcomplicated explications involving the Euclidean algorithm for polynomials, but this is way more understandable.
Thank you
Because when combining fractions, you multiply the denominator, i just think about putting every single fraction down which has one of the factors in the denominator.
There is no A/((x+1)(x+2)) term, because that can be further split into a/(x+1) + b/(x+2) which is already included.
The c/((x+2)^2) = d/(x+2) + e / ((x+2)^2) which just gives back what is already written.
Those two fractions are surely the only combination that when added together give the desired denominator of (x+2)^2.
I love watching your channel. Your explanation makes me clear.
Worrawate Leela-apiradee thank you l!!!
7:56 Why does the degree of the polynomial above t2 have to be exactly one less, i.e. why couldn't we have had just B instead of Bx+C?
Bx+C is just a general expression so that we won't miss out any possible coefficient of x. If it happen that the numerator is just a constant, then the term "B" for (Bx+C) will be zero
@@hhtan1346 i did not understand can you pls clarify
@@DD-rl7xo The "B" in Bx+C could be 0 or another real number so you have to put a linear factor over the cuadratic one to be sure
Great explanation guys!
I wanted to be more formal about it than the answers so far. In general we might have up to a second order polynomial on the original expression instead of just one, after any long division were there a higher order numerator. That means three free parameters in the original expression, so you need three free parameters in the partial fraction expansion to match. The only rule is that there needs to result in three linearly independent equations. There are infinitely many ways to do that but the easiest is Bt+C. See Imgur link. Though it is even easier to use the building up the powers method! But of course our goal here is to show why we build up the powers.
Addendum to blackpenredpen comment imgur.com/a/08C0Qk7
Thank you so much for that explanation. It's really intuitive and easy to understand.
wow, I will write a calc. test tomorrow and just asked myself this. Then I saw this video, perfect sir!
This is amazing. Thank you for this. I was having a debate with my self for a good 15 minutes.
OMG...thank you so much
I was thinking about it for over a week😍
You've shown that building up the powers is equivalent to having a linear term over the quadratic in a change of variables. However, that begs the question, "Why must the degree of the numerator be one less than that of the denominator?" and "Why is it that such a decomposition is always possible?"
The setup being possible is a good question but irrelevant most of the time because the proof for an individual question can be done by just working backwards and adding the partial fractions using common denominators.
The degree of the numerator might be more than one less than the denominator (I.e. a quadratic denominator may have a constant numerator) but we do not know that when we first do the set-up. We know the numerator cannot have equal or greater degree, because if it did than the original fraction would also have equal or greater degree on the top and we would just do long division to make the numerator smaller.
Man, you are the Man, the BEST like superman with powers of maths
Wow this really changed how I look at partial fractions
Gotcha.. easy to catch.. man.. loved it .. now I know why this happens..
Simple algorithm: find component for non-repeating factor, subtract, the result is (cx+d)/(x-b)^2, which can be broken up using x=t+b.
8:26 AWW YEAHH!!!
*YOU ARE GREATEST OF ALL*
Edit: *Subscribed* 👍
I've learnt a lot of things and strategies about integrals from you. Thank you, sir. You're the best mathematician. ❤
That's where I was stuck at,Thanks I got the intuition ✌️✌️
He has explained repeated factor very well
you literally saved my life, thank you
Love this video! Thank you so much!
our god has finally arrived with real answers
Nice Michael McDonald profile pic
You're an excellent teacher! I love your videos! Thank you thank you thank you!!!!
9:25 I felt that fanciness of using gamma as soon as he paused lol
I was doing this problem on a test and treated the t square as an irreducible quadratic, fair to say that was the end of my career. lol
SO THAT'S WHY.
Thanks!
Just to further elaborate on this, I think you can grasp an even better understanding of this if you do it the incorrect way to see where it goes wrong. Suppose we decided to try:
1/[(x+1)(x+2)^2]=A/(x+1)+B/(x+2)^2
Note that I’m only using 2 variables, A and B. Now, we don’t even need to solve for A and B to see where this goes wrong. Instead, let’s try recombining the two fractions. I’m only going to show the unsimplified result of the numerator:
A(x+2)^2+B(x+1)
If we slightly expand the expression above, we can get:
A(x^2+4x+4)+B(x+1)
The thing that is important to realize here is that there is no way for the expression above to equal a constant (specifically the value 1) no matter what value of A and B we use. It’s impossible to obtain our original expression using this approach. The Ax^2 term can’t get canceled out. Hence, we need to use:
A/(x+1)+(Bx+C)/(x+2)^2
You might still be wondering how we pulled Bx+C out of our ass but it’s the middle of the night and I have spent more time than any person should writing a RUclips comment. If someone actually reads all this and wants me to explain, maybe I’ll do it another day. Thanks. Good night
I want you to explain, I am struggling. :'/
I don't understand why your expanded numerator can't result in a constant (specifically 1) for any values of A and B.
Why can't: A(x^2+4x+4)+B(x+1) = 1
This has helped extremely!
You are a good tutor this real help me 🙏🙏
Nice video!
Tomorrow I have a linear algebra exam :D
haha me too and I just asked myself this question
Its been 3 years, update us on your situation!
@@erwinsmith3312 Thanks for asking! I failed the exam at the time lol
But I did the course again and succeeded :)
Now I'm (almost) a math teacher, so that's awesome.
@@weerman44 oh congrats, hope you are achieving success and happiness in your life😃😃
I'm very sentimental for maths right now!
The numerators A,B and C are constants, meaning the power of x is 0. So why cant we straghtway right. 1/(x+1)(x+2)^2 = A/(x+1) + B/(x+2)^2
The part from our lecture that was slightly confusing is between 2:15 and 4:25
Amazing explanation. You are top. Thank you and such a beautiful breakdown.
Thank you, I was not sure if A, beta and gamma from 9:29 would generally evaluate to the same values as A, B and C from 3:34
You could see that they are equivant in this way:
beta*x+gamma = beta(x+2)+delta = beta*x + beta*2 + delta
where gamma = beta*2+delta.
Here beta = B and delta = C
To me, if you're comfortable working with complex numbers and you're okay doing more algebra to simplify the intermediate integrals, then factoring non-reducible polynomials (eg x^2 +2x +2) into complex linear factors and setting up the partial fractions as if they're linear factors would be better than trying to evaluate an integral with a linear function over a non reducible quadratic.
Am I wrong for thinking this way?
Thank you, keep the good work up.
When we saw this in class it didn't make sense to me and the teacher couldn't explain me why, thanks
Thank you for the proving!!!
Great but why it have to be 1 degree less than the bottom
My math teacher doesn't explain it.
Now that i know why sometimes i need to use Bx+C in Partial fraction
Thank you!!!
Baby Kwong
I am sorry. Please do not get mad at JT. I am not sure why he is like that.
I enjoy maths and it's all okay, Thank you!!!
is it correct that @ 5:00; if you have a denominator (x+2)^1 instead of (x+2)^2 that you would have to do long division to find the constant C; and then do partial fraction expansion to find A and B?
great concept thanks blackpenredpen
that was a very good explanation thank you
Wow! Amazing explanation
Excellent video again. I think it might help understand this issue even better, if you proved the equivalence of the beta-gamma approach in the end with the original setup.
Just add zero in the beta-gamma-numerator like this:
beta*x +gamma = beta*x +2*beta -2*beta+gamma=beta*(x+2)+gamma-2*beta
if you plug this into the numerator, the whole term falls apart into beta/(x+2) + (gamma -2*beta)/(x+2)^2, which concludes to B=beta and C=gamma-2*beta.
This is one of my favorite video. Thank you very much. It helps a lot for me. Please do more vedio
wow amazing!! I finally understood it, especially after u proved it using (t-1)
Thank you Sir, this is what I was looking for.
Thanks a lot sir. Very helpful videos.
Try doing this limit question:
lim n -> inf of (e^nx)/(x^n!)
0
But what's your working out?
A=1 , B=-1, C=-1 ; isn't it?
fackingcopyrights yes!
How did you get that tho
stamsarger i have part 2 coming soon
Use cover up rule for A and C
at 2:30
why do we have to have 1 degree less at the top than the bottom ? is it just because it makes it easy to integrate ? or is there some other reasons ?
If you have a degree on the top that is higher than the degree on the bottom, then you won't be able to set up the partial fraction expansion to cover the highest degree on the top.
For instance:
(x^3 - x^2 - 2*x)/((x - 1)*(x-2))
If you attempt to set this up with a partial fraction expansion, you'll see that it isn't possible to generate the x^3 and x^2 coefficients.
A/(x - 1) + B/(x - 2)
Cross-multiply denominators:
A*(x - 2) + B*(x-1) =?= x^3 - x^2 - 2*x
Expand:
A*x - 2*A + B*x - 2*B =?= x^3 - x^2 - 2*x
You can't do it. There are no x^2 terms or x^3 terms to use. You'd have to do polynomial long division first, and end up with a polynomial as your result, with a rational term remaining. The polynomial term can be integrated via the power rule, and the rational term would then be integrated by partial fraction expansion.
Mindblown.
i’ve always wondered about this
Pretty interesting and well done video which shows how partial fraction is done bit I didn't see the reason why. I mean, there is no difference between (Bt+C)/(t+2)^2 and B/(t+2)+C/(t+2)^2 so we can use both of them. Because I want to know (OK, I do personally know but there are many people who don't) WHY we use second form over first, I expected to hear something like:
"We use second fraction over first one while integrating such a function because the second form is just a sum of 2 basic integrals and we can instantly write down the answer. On the other side, there is no such instant answer for the first one."
And one more little thing: in description you wrote "why do we HAVE TO build up the powers". If I'm right, "HAVE TO" is almost "MUST", but video clearly shows that we don't do it as must (just misunderstanding between video and it's description)
Igor Zigmaker Actually, he did explain it in the video.
Igor Zigmaker From 9:30 to 9:50 he explains
If we have a quadratic binomial factor in the denominator of the integrand such as x squared + 1 then we write Ax + B/x squared + 1. If we have (x squared + 1) squared then we write Ax + B/x squared + 1 + Cx + D/(x squared + 1) squared. Furthermore, if we have a cubic binomial factor such as x cubed + 1 then we write Ax squared + Bx + C/x cubed + 1. If we have (x cubed + 1) squared then we write Ax squared + Bx + C/ (x cubed + 1) + Dx squared + Ex + F/ (x cubed + 1) squared.
Rajendra Misir you can factor cubic binomials of this form though
+Angel Mendez-Rivera Angel, I agree. Whenever cubic binomials are factorable, by convention, we must factor them.
You are amazing!!! Very well.
I can't like this video enough 💯
brother you are amazing.
At last i'm going to subscribe in your chanel . you 've solved one of my mathematical problem 😉👌
Without knowing this I had completed my intermediate
took me a hot sec to cut through the accent (not too difficult tho since my professor is German), but once I did this was a fantastic video, helped me understand this subject just in time for finals!
Next up, repeated irreducible quadratic factors.;-)
You must like pain^^
Erick Cobb Isn’t (x+2)^2 already irreducible?
No, an example of a repeated irreducible quadratic would be 1/((x+1)(x^2+1)^2) = 1/(x^5+x^4+2x^3+2x^2+x+1).
Jinger McBlabbersnitch Oh. I was thinking it just meant no factors.
So what makes them being irreducible special? The normal method doesn’t work?
When you have a irreducible factor like x^2+1 (not to be confused with (x+1)^2) it alone gives you two unknowns for the fractionated (?) function.
6:39 "Waeyo?" Did i heard it right? Hahaha kyeoottt
Thankyou very much💃
love how u explain and holding the marker. hahahhahaha. tq~
Thank you so much sir
Y'all notice the drip prof is wearing 🔥💦💸
THANK YOU THANK YOU !!!!
i still don't get the reason why when the denominator is 2nd degree we must have a linear term on the numerator?? i mean at the end we ended up with a constant (c) over t^2
THANK YOU SIR
Hey blackpen redpen do you have any question which has involve both feynman technique and power series in one question it would be great to see you solving that question
Resolve1/(x-1)² (x-2) into partial fractions
Can you help me this sum
Can you explain why it has to be exactly 1 degree less on top? Why not 2 degrees less, or more?
I can purely, intuitively see why degrees 2/4 are bad, for example, but not 1/3 or 0/2.
Also, when _would_ you use Cx+D instead?
There is another thread about this with more details. The basic answer is that we know the numerator can be expressed as at most 1 degree less (because if it were the same or greater, than we just divide and get a simple polynomial with no fraction). We don't know whether it will be 1 or 2 degrees less until we work it out (incidentally, it cannot be more than 2 degrees less because every polynomial in the denominator can always be divided into 1 or 2 degree factors, because of the fundamental theorem of algebra/complex conjugate pairs theorem). So for the set-up you just assume the degree is always 1 less, and if you're lucky the "x" term will be 0.
at 9:23 isn´t there a mistake?, i think that it has to be B(X+2)+C/(X+2)^2
that supreme jacket oof
Thanks you 🤩
thank you so much
i love partial fraction
What if it was an unfavorable quadratic rather than a perfect square like (x+2)(x^2 +x+1)
7:44 sorry, but i dont get why you added C, could you please explain that to me
When you are using partial fraction decomposition, and you have a quadratic factor ( meaning it has a degree of 2), you want to make the numerator one degree lower than the denominator. That is why when you have a linear factor in the denominator , you only put a constant in the numerator ( A, B,C, etc.). Whereas if you have a quadratic factor, you want the numerator in linear form (ax+b, bx+c, cx+d, etc.). This pattern repeats, for example if you have a cubed denominator, you would make put numerator in quadratic form, something like ax^2+ cx+d.
I just let the a, b, c,... be in terms of x and solve for them the expected way .
Why does the degree on the top have to be one less on the bottom 2:32
Ziheng Liao so that it is a proper fraction
3:26 - why isn't this numerator Cx+d?
Yep you’re right, but I think he just ignored that it was a quadratic factor for the purpose of focusing on teaching repeated linear factors.
Ty sir so much
Hey BPRP! A friend asked me if I could solve a limit but I couldn't, or at least I didn't try enough. I challenge you to solve it! It is lim x->π of (x-π)[cotg(3x)+cosec(2x)]. Any method is allowed!