Partial Fraction Decomposition

Рет қаралды 14,454

Mr H Tutoring

Күн бұрын

Пікірлер: 53

@big_smoke7t 11 ай бұрын

The most useful 6 minutes video that i have watched today, thanks sir

@siruseusesir 11 ай бұрын

Very genius. Sending love and support

@luisclementeortegasegovia8603 11 ай бұрын

Great to remember this algorithm! Thank you professor 👍

@BruceLee-io9by 10 ай бұрын

Fantastic, professor. A hug and thanks from Italy!

@vwr32jeep 10 ай бұрын

I’m here to refresh some very old math classes and figured out I’m in completely new territory lol. Still enjoyable to try to follow along. 🍻

@edmondscott7444 10 ай бұрын

Very well explained sir, thank you.

@murdock5537 10 ай бұрын

This is aweome, many thanks, Sir!

@neilmorrone691 11 ай бұрын

Thank you mrh, again! Oh, yes brings me back to 2nd Half of The Calculus l. Lessons requiring Proper and impeccable use of Algebraic operations is first and foremost in Decomposition of polynomial fractions. Followed some months later with the treatment of The Conics, Geometrically and Algebraically, via the General Form of the Equations of Geometric plain figures. Math IS fun!

@blacklightning7227 10 ай бұрын

You did it already 👍🏾👍🏼🔥

@JulesMoyaert_photo 11 ай бұрын

Thank you, Professor!

@travisjacobson2334 10 ай бұрын

This is one thing in calculus that I never mastered. 😂

@danilobarcelos7307 10 ай бұрын

Very well explained 👍

@sudipnayak7210 10 ай бұрын

Excellent algebraic problem. Thanks sir.

@AK58008 11 ай бұрын

Just learned this today! What a coincidence

@rcnayak_58 11 ай бұрын

We can calculate the values of A, B and C from the above lines i.e., 5x + 7 = A (x+1)(x+2) + B (x-1)(x+2) + C (x-1)(x+1) in an easier way. If we put x = 1, we get 12 = A . 2. 3, that is, A = 2. Same way if put x = -2, we get -3 = C . (-3). (-1), that is C = -1 and putting x = -1, we get 2 = B. (-2). (1), that is B = -1.

@blacklightning7227 10 ай бұрын

Thank Mr H

@vinicius-yk1cq 11 ай бұрын

you're good man

@bhavanpa3683 11 ай бұрын

thank you sir.helped a much.

@88kgs 11 ай бұрын

It should be 5x + 7, instead of 5x - 7 @2:54 And, Mr. H, even after seeing and pointing to (-7) equating it as 7 @3:37 Regards 🙏🙏

@srr9281 9 ай бұрын

Mr. H should have shown first how he got the three different denominators for the right hand side of the equation. He probably factored the original polynomial using Synthetic Division? Perhaps next time make it a two-part solution showing ALL the steps. So, 3 out of 5 stars...

@N7TWL 11 ай бұрын

It was so long ago that I took Integral calculus, I don't remember if we covered partial fractions or not. I am wondering if decomposition works with rational expressions where the denominator is not factorable? Very nice presentation.

@carultch 11 ай бұрын

For a denominator factor such as (x^2 - 2), this has no rational factors, but it is factorable as (x - sqrt(2))*(x + sqrt(2)). Irrational factors are preferred when necessary over the alternative, for applications of partial fractions. You may also get a factor with no real roots, such as (x^2 + 4). It is factorable with complex factors, and you can take a complex number detour if you prefer. But it usually is easier to set up the irreducible quadratic term, with an arbitrary linear numerator. Instead of just a constant, it is (B*x + C). Example: (x - 9)/((x + 1)*(x^2 + 4)) Setup: A/x + (B*x + C)/(x^2 + 4) Solution: -2/x + (2*x + 1)/(x^2 + 4) For an irreducible cubic, there'd be a quadratic on top, with 3 unknowns. In general, the polynomial on top is of degree one less than the polynomial on bottom. This form isn't very useful for most applications, so usually you'll get at least one rational root to help you. If you had an "irreducible" cubic in an application, you'd need a cubic formula to reduce it to at least one linear factor.

@N7TWL 10 ай бұрын

@@carultch Thank you for an incredibly useful reply. I very much appreciate your time an expertise!

@ASHHH_Art 10 ай бұрын

JEE aspirant laughing in a corner 😁 but I really appreciate the way Sir solved it.

@user-tr1kt2wo1d 10 ай бұрын

awesome ❤

@mrhtutoring 10 ай бұрын

Thanks 🤗

@davantesevier4650 Ай бұрын

Man if only we actually had time to factor like this people might not struggle with this so bad

@NabiruBogdan 11 ай бұрын

thanks

@jimyoung-gy9lx 11 ай бұрын

Mr H, can you please remind me how you decide what the denominators are in the partial fractions?

@jimbobago 11 ай бұрын

They match the factorization of the denominator on the left.

@NGUYENHOANGLONG-2012 11 ай бұрын

Didn't watch the whole video but i know it will be useful for me later on

@IxsayanIxflame 10 ай бұрын

Try jee advanced questions

@SYAN23SYAN13 10 ай бұрын

Sangat bermanfaat ilmunya dan jenius

@GarethDaviesUK 10 ай бұрын

Why does A+B+C=0? Surely it's the x^2 that is equal to 0?

@carultch 9 ай бұрын

After setting this up, and multiplying to eliminate the fractions, you end up with: 5*x + 7 = A*(x + 1)*(x + 2) + B*(x - 1)*(x + 2) + C*(x - 1)*(x + 1) You can see that there are no x^2 terms on the left. This means, all x^2 terms on the right, have to add up to zero. Each of the three products on the right, will have an x^2 term, after they expand, each multiplied by the coefficient. When expanding, we get: 5*x + 7 = A*x^2 + 3*A*x + 2*A + B*x^2 + B*x - 2*B + C*x^2 - C Gather coefficients on like-terms of x: 5*x + 7 = (A + B + C)*x^2 + (3*A + B)*x + (2*A - 2*B - C) And we can equate coefficients, to produce a 3x3 system of equations to solve.

@carultch 9 ай бұрын

There is a shortcut, that is my preferred method, called Heaviside coverup. For simple linear factors, and the highest denominator power of repeated linear factors, Heaviside coverup allows you to more directly get at the answer. The trick works by covering up the factor in the original expression, that corresponds to each partial fraction denominator. Then, figure out what you can make the variable be, so that what is under your hand is equal to zero. Evaluate what remains, and this tells you the value of the unknown coefficient. For this example: (5*x + 7)/[(x + 1)*(x + 2)*(x - 1)] = A/(x - 1) + B/(x + 1) + C/(x + 2) A = (5*x + 7)/[(x + 1)*(x + 2)*covered], evaluated at x=+1 A = (5*1 + 7)/[(1 + 1)*(1 + 2)] = 2 B = (5*x + 7)/[covered*(x + 2)*(x - 1)], evaluate at x = - 1 B = (5*-1 + 7)/[(-1 + 2)*(-1 - 1)] = -1 C = (5*x + 7)/[(x + 1)*covered*(x - 1)], evaluate at x = -2 C = (5*-2 + 7)/[(-2 + 1)*(-2 - 1)] = -1 The underlying idea of why this works, is that near the pole of x=+1, the A/(x - 1) term will dominate the behavior of the equation, and the other terms will be insignificantly small. So you are ultimately taking the limit as x approaches the problem point of -1 and the other poles, and matching the behavior on both sides. This is related to a concept in complex analysis, called the residue of a pole.

@abijahmarshal 11 ай бұрын

Professer X in real life ❤❤❤

@dnd2008yi 10 ай бұрын

Thanks sir

@Mari_Selalu_Berbuat_Kebaikan 9 ай бұрын

Let's always do alot of good ❤️

@foxhound1008 10 ай бұрын

Can you do a video on partial fractions expansion when the denominator is not repeated linear?

@carultch 10 ай бұрын

The simple case is when you have all distinct linear factors. The setup is a simple unknown constant on top of each one of them. There is a shortcut to find that constant, called Heaviside coverup. Repeated linear terms require us to either build-the-power, or my preference, to descend the power. So if (x+1)^2 were one of your denominator factors, then A/(x+1)^2 + B/(x + 1) would be the setup associated with it. For (x + 1)^3 as a denominator factor, you'd have 3 terms, A/(x+1)^3 + B/(x+1)^2 + C/(x+1). I like descending the power instead of building the power, so I can find the first constant with Heaviside coverup first. Quadratic factors with no real roots, require an arbitrary linear term on top, such as (B*x + C)/(x^2 + 1). Cubics and beyond, would start with an arbitrary polynomial, that is one degree less than the denominator. They aren't that useful for most applications of this process.

@paladdin1500 11 ай бұрын

👏

@josephmak0865 10 ай бұрын

this will be useful for one integrating...in calculus 2. Probably the only use... and in real life. probably 0 use ever

@carultch 10 ай бұрын

This also has use in Laplace transforms, which are methods to solve differential equations. Laplace transforms are methods of converting time-domain calculus into s-domain algebra, for purposes of solving differential equations. Applications in real life of this method, are vibrations, control systems, and electric filter & amplifier circuits.

@joetandingan6328 11 ай бұрын

Your caption covers your writings

@mrhtutoring 11 ай бұрын

You need to turn them off on KZbin.

@brando3023 Ай бұрын

so clear

@aram5642 9 ай бұрын

...and all this came already shortened - because you came with the polynomial already factored out. Otherwise one needs to do synthetic division first.