9:38 Yeah, but that worked only because the repeating factor was just "x", how can you prove the formula to be valid if the repeating factor was anything else?
Thanks for the reply! I tried the t=x+2 substitution and I ended up with the same four partial fractions of the video, except in the third I got C-4B in the numerator instead of just C and in the fourth I got 4B-2C+D instead of just D, but since they're all some sums of constant terms that do not involve x, I'm assuming I can easily replace those two numerators with, say, C' and D' respectively.
I am an engineering student, in a course, we should those partial fraction decomposition to solve Laplace questions in transform calculus for advanced math engineering, and I couldn't remember how to decompose them, and you SAVED my time and helped to remember all. Thanks!
2:01 Thank you for the shot showing the difference between the options for the various forms the rational forms should be in! Also, I really liked how the different segments describe the individual steps/possible setups in order (with a couple of examples regarding a particular step) instead of merely solving different problems like most videos! Awesome explanations & video!
Thank you! Currently taking Differential Equations and I forgot everything I learned from Calc 2 about partial fractions. Your videos always save the day
Perfect format for this kind of overview video - where you are laying out many cases and examples so we can compare side-by-side. Thank you so much! I also love that you gave us a few practice questions. 🙂
I love how you learn something like partial fractions in highschool or longdivision in elementary school, and then you dont use it for 4+ years and then get into calc 3 or diff eq at college and you need to look it up again because its expected to be memorized. I think it's less the fault of the teacher on either end but the curriculum in the middle not reinforcing some of these things that aren't just esoteric one offs for non math majors.
You have a test already by September 15? My kids just started school today. They'll learn how to do basic fractions this year, not partial fraction decomposition. They're only in 3rd grade.
@@Eltodofull Diff Eq was the hardest class I took. I studied philosophy, not math, but I really wanted to learn about chaos properly to apply it in philosophy. So I struggled through the class. Really interesting stuff, but hard.
I use patrial fractions to calculate following things 1. Integrate rational functions 2. Inverse Laplace transform I like solving recurrence relations with generating functions (ordinary and exponential) After you get generating function you have to express it as a sum of geometric series and their derivatives which looks similar to partial fraction decomposition but it is not quite the same thing In integration in repeated complex root case i prefer to use reduction formula which can be derived with integration by parts and linearity In inverse Laplace transform in repeated complex root case i use convolution theorem
There is a way to work with repeated complex roots, without using convolution. You can construct a linear combination of the transforms of sin(t), cos(t), t*sin(t), and t*cos(t). Properties of even and odd Laplace transform functions can often eliminate two of these choices. You then use the s-derivative theorem to find L{t*sin(t)} and L{t*cos(t)}, and solve for the unknown coefficients.
My first cal 2 exam I failed but i started watching your videos and doing the practice sheets you post and have became very fluent in my trig identities, trig sub, partial fractions, u sub, and improper integrals✊🏻
@@oralia3340 awesome sauce girl!! I ended up with a A in Cal 2. You got this! Currently in differential equations and its so much easier compared to cal 2.
@@12degreesnowman11 depends on the university. Im a mechanical engineering student so we need dif q for fluids. Didn't use much of cal 3 in dif q though
Sometimes the degree on the numerator may appear to be higher than the degree on the denominator. Which we can reduce by long division. I expect to gonna explain that too 💪💪
Hey! Great video! I'm taking Differential Equations and I needed to refresh on PFD! Thank you very much! You included all the details! Have a great week ❤
For 2:00, I worked it through on my own before to generalize it, and I prefer thinking of it for n > 1, since you don't really see negative exponents. So, it's just -1/ [ (a*(n-1)) * (ax+b)^n-1].
The technique of partial fraction decomposition (not really the motivation or the final steps of integrating the resulting fractions) is one you can easily do in algebra 2 or precalculus. In some books it is part of the systems of equations chapter.
That it looks interesting is useful data about yourself! That’s awesome. Just keep learning new math skills, and this will eventually become easy for you. 🙂
Love these examples! They’re similar to what my professor showed us! And then to prove how amazing my professor is, he then proceeded to quiz us on PFD integration of 1/(u^8+u^6+u^2+1) :’’’’)
That can be simplified to 1/(u^6+1) (u^2+1) But still, that's just evil if you expant it and try to do it by partial fraction. Yaa, it's just too long 😂😂
Thank you for this video! At 10:17, in the 6th example, how to find the values of A,B and C? Whatever we substitute to make the B and the C terms zero would also make the A term zero.
For the case of repeated roots, you can use Heaviside cover-up to find the coefficient on the highest power of the repeated term. I prefer to assign the first letters for terms I can get with Heaviside cover-up, and put those first. Then, I descend the power for all other terms of the repeated root. In his case, that's coefficient B. Coefficient C can also be found with Heaviside cover-up. Once you use x=0 for Heaviside cover-up, that value is spoken-for, and can't directly be used again. You can use other unrelated values of x, and you just need as many choices for x as you have unknowns. There is a general method for Heaviside cover-up, that involves taking derivatives and using the same value again, until you find all coefficients of the repeated power. However, I find it is more trouble than it is worth. There is a trick with repeated roots, where x=infinity can be one of your values. It involves partially clearing the fraction, by multiplying by the repeated root just once. Then take the limit as x goes to infinity. Here's how I'd solve his example: Given: (2*x - 5)/[x^2*(x + 1)] Set up partial fractions: A/(x + 1) + B/x^2 + C/x Use Heaviside coverup for A & B: At x = -1, A = (2*(-1) - 5)/[(-1)^2] = -7 At x = 0, B = (2*0 - 5)/(0 + 1) = -5 You could plug in x=1 as a strategic value to find C. Alternatively, you can partially multiply by one copy of x: (2*x - 5)/[x^2*(x + 1)] = -7/(x + 1) - 5/x^2 + C/x Multiply by 1 copy of x: (2*x - 5)/[x*(x + 1)] = -7*x/(x + 1) - 5/x + C Take the limit as x goes to infinity: 0 = -7 - 0 + C C = +7 Result: -7/(x + 1) - 5/x^2 + 7/x
Logical Proofs I will reply if I want to. You presented a problem, and I presented you a fact about the problem. It is not my problem if you do not like it. How about you present the solution yourself, since you are acting so high and mighty? I would love to see you solve this equation step by step and provide all 3 complex solutions in exact form. If you are not able to do so, then maybe you should stay silent and stop acting all arrogant for no reason.
Wow! Such a wonderful video, Blackpenredpen, you are my favorite youtuber, always try to be the first one to watch your videos, it would be nice if you could respond to this comment.
That's because I think his wrong starting @5:30 case 2: irreducible quadratic factors. (I commented this to him) "I think you're wrong on Quadratic partial fraction form, the first Constant (in this case B) should be multiplied to the derivative of the denominator + the 2nd Constant (in this case C). The partial fraction form @6:18 should be (B(2x)+C)/(X^2+4). And the rest of this video has a wrong Quadratic partial fraction form."
@@janconfusio7178 no, what he’s done is correct for case 3, as I’ve done multiple times since I watched this and it’s always like this, as that’s what your meant to do for a brackets/symbol to a power. Case 2 I’m not so certain on as I’ve always done it a different way, but case 3 he’s done correctly
Will you upload old worksheets back to your website? (e.g. work, hydrostatic force, some other stuff) Thank you for keeping my feed interesting these past years!
You could treat (x - 2)^2 as a quadratic, and go through the same exercise to set it up as if it were an irreducible quadratic, and you'd get an equally valid result. It's generally easier if you use the fact that it is a repeated linear factor, rather than a quadratic, to your advantage, since Heaviside coverup will help you get one of the coefficients on it. Using a repeated linear factor helps get you closer to where you ultimately want to be, for an application of this process, such as integration or Laplace transforms. As an example, consider: 1/((x - 3)*(x - 2)^2) Treating (x - 2)^2 as if it were an irreducible quadratic, we get: 1/((x - 3)*(x - 2)^2) = A/(x - 3) + (B*x + C)/(x - 2)^2 H-cover-up for A, at x=3: A = 1/((3 - 2)^2) = 1 Thus: 1/((x - 3)*(x - 2)^2) = 1/(x - 3) + (B*x + C)/(x - 2)^2 Let x=0, to solve for C: 1/((0 - 3)*(0 - 2)^2) = 1/(0 - 3) + C/(0 - 2)^2 -1/12 = -1/3 + C/4 C = 1 Let x=1 to solve for B: 1/((1 - 3)*(1 - 2)^2) = 1/(1 - 3) + (B*1 + 1)/(1 - 2)^2 -1/2 = -1/2 + B + 1 B = -1 Result: 1/((x - 3)*(x - 2)^2) = 1/(x- 3) + (-x + 1)/(x - 2)^2 And this is equivalent to what you'd get, if you did use the repeated linear factor to your advantage, which is: 1/(x - 3) - 1/(x - 2)^2 - 1/(x - 2)
tan(2x) = 2·tan(x)/[1 - tan(x)^2] = 2·tan(x)/([1 - tan(x)]·[1 + tan(x)]) = ([1 + tan(x)] - [1 - tan(x)])/([1 - tan(x)]·[1 + tan(x)]) = 1/[1 - tan(x)] - 1/[1 + tan(x)]. This may be useful in a situation where you have an integrand with several trigonometric functions with an input of x but with the exception of the tangent, which has input 2x. Otherwise, I cannot imagine this being particularly useful.
6:30 Why not to use cover up method and plug complex numbers eg x=2i? 7:40 i think cover up can be generalized by using derivatives to eliminate the guested coeffs. i think is is based on residual theorem from complex analysis. am i correct? in my opinion if cover up method is used by plugging poles, then plugging other distinct x values should be used, for example, the roots of the numerator . what do u think?
6:30, you can do that with complex factors, since there is technically no such thing as an irreducible quadratic. It often ends up not helping you, since it is simpler to use the standard method, but you can do it nevertheless. 7:40, yes, you can generalize the cover-up method, using derivatives, to more directly get at the remaining solutions when you have the repeated factors in the denominator.
Given: (3*x^2 - 3*x + 8)/(x^3 - 3*x^2 + 4*x - 12) Factor the bottom: 3 sign swaps = 3 or 1 positive roots are possible 12 as the final term = 1, 2, 3, 4, 6, and 12 are possible roots, as are their negatives x = +3 is a root Use polynomial division to reduce to a quadratic and linear term: (x - 3)*(x^2 + 4) Thus, our fraction becomes: (3*x^2 - 3*x + 8)/[(x - 3)*(x^2 + 4)] Set up partial fractions: A/(x - 3) + (B*x + C)/(x^2 + 4) Heaviside coverup for A at x = 3: A = (3*3^2 - 3*3 + 8)/(3^2 + 4) = 2 Thus: (3*x^2 - 3*x + 8)/[(x - 3)*(x^2 + 4)] = 2/(x - 3) + (B*x + C)/(x^2 + 4) Let x=0, to solve for C: 8/[(-3)*(4)] = 2/(0 - 3) + C/(4) -2/3 = -2/3 + C C = 0 Let x = 1, to solve for B: (3*1^2 - 3*1 + 8)/[(1 - 3)*(1^2 + 4)] = 2/(1 - 3) + B/(1^2 + 4) -8/10 = -1 - B/5 B = -4 + 5 = 1 Result: 2/(x - 3) + x/(x^2 + 4)
4:13 What's the method you said in that timestamp? Cafra? Taphra? I'm really sorry but I've searched for this for 30 minutes and still can't find what it is
can u use your old technique of teaching with red and black pens? I loved the old version where you had a ball shaped mic in your one hand and another hand with two pens busy on the board..
Descartes set a precedent to use the beginning of the alphabet for constants, and the end of the alphabet for variables. x, y, z, and t are the most common variables you'll likely use as either inputs or outputs of functions. u is simply the alphabet neighbor of t, so it is used as a de-facto letter for substitutions. If u is spoken-for, I choose w.
9:38 Yeah, but that worked only because the repeating factor was just "x", how can you prove the formula to be valid if the repeating factor was anything else?
You can do substitution. Like for ex7, you can let t=x+2
And it will work like ex6 : )
Good point tho I will pin it so that others will see it. Thanks
Thanks for the reply! I tried the t=x+2 substitution and I ended up with the same four partial fractions of the video, except in the third I got C-4B in the numerator instead of just C and in the fourth I got 4B-2C+D instead of just D, but since they're all some sums of constant terms that do not involve x, I'm assuming I can easily replace those two numerators with, say, C' and D' respectively.
@@nomoremathhere just subbed to your channel, extra cool content!!
@@nomoremathhere Prove it without taking substitution
As a math tutor, this video explains it perfectly and it is not taught with this level of detail in most calc courses. Thanks BPRP!
Glad to hear. Thank you!
Did he try to bait us into the video with a sneak peak of him cancelling x like a vlog youtuber?!
I’m deadddd
Altough the board is a sacred tradition, I also enjoy this format of videos. Keep it up!
Thanks
This man saved me in 11th and 12th grade. Now is coming clutch again when Im in college! He's da GOAT
My professor showed us partial fractions today, this helps a lot, thank you!
I am an engineering student, in a course, we should those partial fraction decomposition to solve Laplace questions in transform calculus for advanced math engineering, and I couldn't remember how to decompose them, and you SAVED my time and helped to remember all. Thanks!
2:01 Thank you for the shot showing the difference between the options for the various forms the rational forms should be in! Also, I really liked how the different segments describe the individual steps/possible setups in order (with a couple of examples regarding a particular step) instead of merely solving different problems like most videos! Awesome explanations & video!
Thank you! Currently taking Differential Equations and I forgot everything I learned from Calc 2 about partial fractions. Your videos always save the day
The new format of the video is amazing, thanks!
Perfect format for this kind of overview video - where you are laying out many cases and examples so we can compare side-by-side. Thank you so much! I also love that you gave us a few practice questions. 🙂
LOVED THIS! I was confused and after watching this, it clicked. Thanks!
You’re welcome! I am happy to help!
Your videos help me A LOT. It's all so well-structured and straightforward. Thank you!
Best math channel on TH-cam, you know the deal!
I love how you learn something like partial fractions in highschool or longdivision in elementary school, and then you dont use it for 4+ years and then get into calc 3 or diff eq at college and you need to look it up again because its expected to be memorized. I think it's less the fault of the teacher on either end but the curriculum in the middle not reinforcing some of these things that aren't just esoteric one offs for non math majors.
I salute to your timing
I required this video today and you uploaded it
I literally have a test on this tomorrow thanks so much for a good review!
Good Luck :))
Play 2x speed for review
You have a test already by September 15? My kids just started school today.
They'll learn how to do basic fractions this year, not partial fraction decomposition. They're only in 3rd grade.
Mine was yesterday, a differential equation, autonomous, it had 3 factors.
Guess what, the time gave negative :D
It's likely wrong
@@Eltodofull Diff Eq was the hardest class I took. I studied philosophy, not math, but I really wanted to learn about chaos properly to apply it in philosophy. So I struggled through the class. Really interesting stuff, but hard.
It took just the first 7 seconds of the video, folks, for him to answer my biggest lingering question with partial fraction decomposition. 7 SECONDS!
: )))))) thank you!
I found the first 2 minutes super helpful as strategy for what formats to aim for !
Screenshotted the board at 2:10 for review!
same bro, neither my teacher could explain it to me in 5 minutes jeje
Great video and a very clear explanation.
Thanks Dave!
This is my favorite integration techinque and you managed to perfectly explain it in less then 10 minutes i hope you do other techinques as well
Hlo all might midoriya here 😂
🙌
Wow this channel just gets better and better
I use patrial fractions to calculate following things
1. Integrate rational functions
2. Inverse Laplace transform
I like solving recurrence relations with generating functions (ordinary and exponential)
After you get generating function you have to express it as a sum of geometric series and their derivatives
which looks similar to partial fraction decomposition but it is not quite the same thing
In integration in repeated complex root case i prefer to use reduction formula which can be derived with integration by parts and linearity
In inverse Laplace transform in repeated complex root case i use convolution theorem
There is a way to work with repeated complex roots, without using convolution. You can construct a linear combination of the transforms of sin(t), cos(t), t*sin(t), and t*cos(t). Properties of even and odd Laplace transform functions can often eliminate two of these choices. You then use the s-derivative theorem to find L{t*sin(t)} and L{t*cos(t)}, and solve for the unknown coefficients.
I just wanted to say thank you for everything! I just took my second cal 2 exam today and know I did awesome thanks to you!.
My first cal 2 exam I failed but i started watching your videos and doing the practice sheets you post and have became very fluent in my trig identities, trig sub, partial fractions, u sub, and improper integrals✊🏻
@@JesusMartinez-zu3xl I know this was 7 months ago but great job!! I'm tryna be like you haha
@@oralia3340 awesome sauce girl!! I ended up with a A in Cal 2. You got this! Currently in differential equations and its so much easier compared to cal 2.
@@JesusMartinez-zu3xldo you need calc 3 for differential equations or is calculus 2 enough ?
@@12degreesnowman11 depends on the university. Im a mechanical engineering student so we need dif q for fluids. Didn't use much of cal 3 in dif q though
Sometimes the degree on the numerator may appear to be higher than the degree on the denominator.
Which we can reduce by long division. I expect to gonna explain that too 💪💪
Hey! Great video! I'm taking Differential Equations and I needed to refresh on PFD! Thank you very much! You included all the details! Have a great week ❤
I swear!! you're the best ever
One of the only videos I could find that explained how to break up the fraction when it's like the last example, very useful thank you
PFD also stands for Personal Floatation Device aka a life jacket. A good analogy.
Thanks for simplifying this. This is the first time to open it and I may have an exam on it tomorrow.
Was doing complex calculus and needed a quick refresher on partial fractions when this popped up on my feed. Great video!
I mostly used in fractional integration, Thnx for enhancing in advanced level. 👍👍👍👍
Thank you! we just finished integrations in school and this helped a lot!!
Love from india man❤❤. The way of teaching was just fab😍
Awesome video !!!!! like the new format
So you have changes the format of your lesson?
Well done 💪💪
Thank you so much Sir, I have a test coming up tomorrow.
Could you show how to solve integrals of the type:
sqrt(x^2+a^2)
sqrt(x^2-a^2)
1/sqrt(x^2+a^2)
1/sqrt(x^2-a^2)
Thanks and keep up the good work!!
The first two can be solved using trigonometric substitution, the latter two require you to know the derivatives of inverse trigonometric functions
For 2:00, I worked it through on my own before to generalize it, and I prefer thinking of it for n > 1, since you don't really see negative exponents. So, it's just -1/ [ (a*(n-1)) * (ax+b)^n-1].
Me: Doesn't understand -everything- anything
whatever it looks interesting
Michael Wu first let me wooooosh myself, then I’ll say that this is an integration technique which you will learn if you learn calculus
@@ericw2391 also useful in telescoping series sometimes
王Eric You also use it in Laplace transform.
The technique of partial fraction decomposition (not really the motivation or the final steps of integrating the resulting fractions) is one you can easily do in algebra 2 or precalculus. In some books it is part of the systems of equations chapter.
That it looks interesting is useful data about yourself! That’s awesome. Just keep learning new math skills, and this will eventually become easy for you. 🙂
Love these examples! They’re similar to what my professor showed us! And then to prove how amazing my professor is, he then proceeded to quiz us on PFD integration of 1/(u^8+u^6+u^2+1) :’’’’)
That can be simplified to 1/(u^6+1) (u^2+1)
But still, that's just evil if you expant it and try to do it by partial fraction. Yaa, it's just too long 😂😂
I learnt a lot, thank you so much BPRP!!
It has really met my expectations
We truly appreciate your hard work sir 👏👏 👏👏 you have taught us to do maths with fun....😃
Honourable sir, You are genius 💙👼🙏👌🏻👌🏻😍😊
Thank you for explaining this so well!
Thank you for this video!
At 10:17, in the 6th example, how to find the values of A,B and C? Whatever we substitute to make the B and the C terms zero would also make the A term zero.
For the case of repeated roots, you can use Heaviside cover-up to find the coefficient on the highest power of the repeated term. I prefer to assign the first letters for terms I can get with Heaviside cover-up, and put those first. Then, I descend the power for all other terms of the repeated root.
In his case, that's coefficient B. Coefficient C can also be found with Heaviside cover-up. Once you use x=0 for Heaviside cover-up, that value is spoken-for, and can't directly be used again. You can use other unrelated values of x, and you just need as many choices for x as you have unknowns. There is a general method for Heaviside cover-up, that involves taking derivatives and using the same value again, until you find all coefficients of the repeated power. However, I find it is more trouble than it is worth.
There is a trick with repeated roots, where x=infinity can be one of your values. It involves partially clearing the fraction, by multiplying by the repeated root just once. Then take the limit as x goes to infinity. Here's how I'd solve his example:
Given: (2*x - 5)/[x^2*(x + 1)]
Set up partial fractions:
A/(x + 1) + B/x^2 + C/x
Use Heaviside coverup for A & B:
At x = -1, A = (2*(-1) - 5)/[(-1)^2] = -7
At x = 0, B = (2*0 - 5)/(0 + 1) = -5
You could plug in x=1 as a strategic value to find C. Alternatively, you can partially multiply by one copy of x:
(2*x - 5)/[x^2*(x + 1)] = -7/(x + 1) - 5/x^2 + C/x
Multiply by 1 copy of x:
(2*x - 5)/[x*(x + 1)] = -7*x/(x + 1) - 5/x + C
Take the limit as x goes to infinity:
0 = -7 - 0 + C
C = +7
Result:
-7/(x + 1) - 5/x^2 + 7/x
I got a completely different math test Friday. Linear Algebra! Already completed Calc I and II
thanks it was so helpful ❤
Thanks for the video,it's very helpful for integrations
How did you know? We were just started doing solving integration by partial fractions in our class
Let's modify the Tejas question
√(5-x)= 5-x³
Solve for x.
From what I understand, it is not possible to express the solution to this using analytic methods.
@@angelmendez-rivera351 if you are not able to do this question then don't reply.
Logical Proofs I will reply if I want to. You presented a problem, and I presented you a fact about the problem. It is not my problem if you do not like it. How about you present the solution yourself, since you are acting so high and mighty? I would love to see you solve this equation step by step and provide all 3 complex solutions in exact form. If you are not able to do so, then maybe you should stay silent and stop acting all arrogant for no reason.
@@angelmendez-rivera351 He is not arrogant.
this format was probably more work for you but it was just as good.
super helpful, thank you so much!
Glad to help 😃
Wow! Such a wonderful video, Blackpenredpen, you are my favorite youtuber, always try to be the first one to watch your videos, it would be nice if you could respond to this comment.
see if he responds
Better luck next time :)
Thank you so much for the video, it was well explained 👏👏👏👏
Wow i literally needed it,Thank you sir...
I see you from Honduras! 🇭🇳
9:42 earns you a like.
He's back to regular uploads, I hope. :)
this is like a pep talk:
you know in life you won't always get what you want
This video make it easy.. thanks a lote.
@ BPRP can u plz derive each and everything about integration factor of differential equations, how it came , how to use it etc
U r back!!!!!!!!!
Thank you! I loved the video
Interesting format!
Thank u sir from india
thanks I am now able to do DEs with laplace transform easily
thank you,it's really helpful.
You save me men ......
I was looking for that like that 😩😩😩,... when I get your Vedio notification my face :😲😲
pov:11:30 * BlackPenRedPen you won't ask that question in the exam but my calc professor will!!! we need more professors like u
Thank you sir,it was really helpful.
9:38 is exactly what I what I was getting confused on
That's because I think his wrong starting @5:30 case 2: irreducible quadratic factors. (I commented this to him) "I think you're wrong on Quadratic partial fraction form, the first Constant (in this case B) should be multiplied to the derivative of the denominator + the 2nd Constant (in this case C). The partial fraction form @6:18 should be (B(2x)+C)/(X^2+4). And the rest of this video has a wrong Quadratic partial fraction form."
@@janconfusio7178 no, what he’s done is correct for case 3, as I’ve done multiple times since I watched this and it’s always like this, as that’s what your meant to do for a brackets/symbol to a power. Case 2 I’m not so certain on as I’ve always done it a different way, but case 3 he’s done correctly
Please can you do a video where you will demonstrate thet we effectively can do this kind decomposition !
thank you so much for this
Very helpful!
Love from India
Great work
Hey, what does the Chinese Math curriculum look like? How are you guys so good at Mathematics?
Fantastic!
Will you upload old worksheets back to your website? (e.g. work, hydrostatic force, some other stuff)
Thank you for keeping my feed interesting these past years!
0:51-0:54 I get a image of you smiling when I first heard this lol
You're awesome.
Thank you very much senpai☺️
Amazing ❤️
thank you soo much🥰🥰
Your work is very good.
Is (x - 2)^2 not quadratic?
Why did you use a constant for its numerator instead of a linear function?
Thanks
when you can factorised it, its become linear (x=2). if can't factorised, then remain it as quadratic
You could treat (x - 2)^2 as a quadratic, and go through the same exercise to set it up as if it were an irreducible quadratic, and you'd get an equally valid result. It's generally easier if you use the fact that it is a repeated linear factor, rather than a quadratic, to your advantage, since Heaviside coverup will help you get one of the coefficients on it. Using a repeated linear factor helps get you closer to where you ultimately want to be, for an application of this process, such as integration or Laplace transforms.
As an example, consider:
1/((x - 3)*(x - 2)^2)
Treating (x - 2)^2 as if it were an irreducible quadratic, we get:
1/((x - 3)*(x - 2)^2) = A/(x - 3) + (B*x + C)/(x - 2)^2
H-cover-up for A, at x=3:
A = 1/((3 - 2)^2) = 1
Thus:
1/((x - 3)*(x - 2)^2) = 1/(x - 3) + (B*x + C)/(x - 2)^2
Let x=0, to solve for C:
1/((0 - 3)*(0 - 2)^2) = 1/(0 - 3) + C/(0 - 2)^2
-1/12 = -1/3 + C/4
C = 1
Let x=1 to solve for B:
1/((1 - 3)*(1 - 2)^2) = 1/(1 - 3) + (B*1 + 1)/(1 - 2)^2
-1/2 = -1/2 + B + 1
B = -1
Result:
1/((x - 3)*(x - 2)^2) = 1/(x- 3) + (-x + 1)/(x - 2)^2
And this is equivalent to what you'd get, if you did use the repeated linear factor to your advantage, which is:
1/(x - 3) - 1/(x - 2)^2 - 1/(x - 2)
thank you.
Thank you bro
thanks a lot!
We can do partial fraction also in tan2x =a/(1+tanx)+b/(1-tanx) is it useful in the integration or not ?
In some rare circumstances, it may be, but not generally speaking.
tan(2x) = 2·tan(x)/[1 - tan(x)^2] = 2·tan(x)/([1 - tan(x)]·[1 + tan(x)]) = ([1 + tan(x)] - [1 - tan(x)])/([1 - tan(x)]·[1 + tan(x)]) = 1/[1 - tan(x)] - 1/[1 + tan(x)]. This may be useful in a situation where you have an integrand with several trigonometric functions with an input of x but with the exception of the tangent, which has input 2x. Otherwise, I cannot imagine this being particularly useful.
First Like, then watch!!!
6:30 Why not to use cover up method and plug complex numbers eg x=2i?
7:40 i think cover up can be generalized by using derivatives to eliminate the guested coeffs. i think is is based on residual theorem from complex analysis. am i correct?
in my opinion if cover up method is used by plugging poles, then plugging other distinct x values should be used, for example, the roots of the numerator . what do u think?
6:30, you can do that with complex factors, since there is technically no such thing as an irreducible quadratic. It often ends up not helping you, since it is simpler to use the standard method, but you can do it nevertheless.
7:40, yes, you can generalize the cover-up method, using derivatives, to more directly get at the remaining solutions when you have the repeated factors in the denominator.
Using his example at 6:30, here's how it could work with complex factors:
Given:
(4*x^2 - 9*x + 2)/((x + 3)*(x^2 + 4))
Factor the quadratic with complex roots:
(4*x^2 - 9*x + 2)/((x + 3)*(x + 2*i)*(x - 2*i))
Partial fractions:
(4*x^2 - 9*x + 2)/((x + 3)*(x + 2*i)*(x - 2*i)) = A/(x + 3) + B/(x + 2*i) + C/(x - 2*i)
H-coverup:
at x =-3, A = (4*(-3)^2 - 9*(-3) + 2)/((-3)^2 + 4) = 5
at x = -2*i, B = (4*(-2*i)^2 - 9*(-2*i) + 2)/(((-2*i) + 3)*((-2*i) - 2*i)) = -1/2 - 3/2*i
at x = +2*i, C will equal the conjugate of B, thus, C = -1/2 + 3/2*i
Thus, the expression expands as:
5/(x + 3) + (-1/2 - 3/2*i)/(x + 2*i) + (-1/2 + 3/2*i)/(x - 2*i)
After simplifying it to remove the imaginary numbers, it will become:
5/(x + 3) - x/(x^2 + 4) - 6/(x^2 + 4)
Sir how to solve Q3 in the "You Try" section
Given:
(3*x^2 - 3*x + 8)/(x^3 - 3*x^2 + 4*x - 12)
Factor the bottom:
3 sign swaps = 3 or 1 positive roots are possible
12 as the final term = 1, 2, 3, 4, 6, and 12 are possible roots, as are their negatives
x = +3 is a root
Use polynomial division to reduce to a quadratic and linear term:
(x - 3)*(x^2 + 4)
Thus, our fraction becomes:
(3*x^2 - 3*x + 8)/[(x - 3)*(x^2 + 4)]
Set up partial fractions:
A/(x - 3) + (B*x + C)/(x^2 + 4)
Heaviside coverup for A at x = 3:
A = (3*3^2 - 3*3 + 8)/(3^2 + 4) = 2
Thus:
(3*x^2 - 3*x + 8)/[(x - 3)*(x^2 + 4)] = 2/(x - 3) + (B*x + C)/(x^2 + 4)
Let x=0, to solve for C:
8/[(-3)*(4)] = 2/(0 - 3) + C/(4)
-2/3 = -2/3 + C
C = 0
Let x = 1, to solve for B:
(3*1^2 - 3*1 + 8)/[(1 - 3)*(1^2 + 4)] = 2/(1 - 3) + B/(1^2 + 4)
-8/10 = -1 - B/5
B = -4 + 5 = 1
Result:
2/(x - 3) + x/(x^2 + 4)
Thanks
What animation are you using? This is a good way to project math
4:13 What's the method you said in that timestamp? Cafra? Taphra?
I'm really sorry but I've searched for this for 30 minutes and still can't find what it is
Figured it out now, sorry. It was cover-up method
can u use your old technique of teaching with red and black pens?
I loved the old version where you had a ball shaped mic in your one hand
and another hand with two pens busy on the board..
great!
My legend.
one thing i dont understand that doesnt effect my understanding of substitution is why is the letter "u" the most common one?
u and v were available. So many other letters have full-time jobs already.
Descartes set a precedent to use the beginning of the alphabet for constants, and the end of the alphabet for variables. x, y, z, and t are the most common variables you'll likely use as either inputs or outputs of functions. u is simply the alphabet neighbor of t, so it is used as a de-facto letter for substitutions. If u is spoken-for, I choose w.
Hi , please can you solve this integral for me
2s+2/(s²+1)(s-1)³ ds
it is necessary now ، i have a midterm tomorrow
I don't think you meant that. I think you meant this: (2s + 2)/[(s² + 1)(s - 1)³] ds.
is there any idea for the integration of 1/(x^m(1-x^n)) for m>0, n>0?