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!
Thank you! Currently taking Differential Equations and I forgot everything I learned from Calc 2 about partial fractions. Your videos always save the day
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 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.
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 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
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 ❤
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.
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
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. 🙂
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 💪💪
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].
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!
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.
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.
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
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
Teacher in my opinion the quadratic that appears in the denominator has one solution is it valid for a quadratic that have a solution to put linear factor over it
If a quadratic has exclusively one solution, then you have a repeated root. This means that you need two terms based on that repeated root. One term with a squared denominator, and another term with the denominator that isn't squared. I prefer to start with the highest power, and then descend the power until it's 1, since it allows me to put Heaviside coverup terms first. Example. Given: (x + 2)/[x*(x^2 - 2*x + 1)] Factor the quadratic with a repeated root: x^2 - 2*x + 1 = (x - 1)^2 Thus, our original fraction can be written as: (x + 2)/[x*(x - 1)^2] Set up partial fractions: A/x + B/(x - 1)^2 + C/(x - 1) Heaviside coverup finds A & B: At x = 0, A = (0 + 2)/(0 - 1)^2 = 2 At x = -1, B = (-1 + 2)/[-1] = -1 Thus: (x + 2)/[x*(x - 1)^2] = 2/x - 1/(x - 1)^2 + C/(x - 1) To find C, multiply by just one copy of the repeated root, to partially clear the fractions. Then take the limit as x goes to infinity: (x + 2)/[x*(x - 1)] = 2*(x-1)/x - 1/(x - 1) + C 0 = 2 - 0 + C C = -2 Result: 2/x - 1/(x - 1)^2 - 2/(x - 1)
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..
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)
