Partial Fraction Decomposition

แชร์
ฝัง
  • เผยแพร่เมื่อ 15 ม.ค. 2025

ความคิดเห็น • 53

  • @big_smoke7t
    @big_smoke7t 10 หลายเดือนก่อน +30

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

  • @siruseusesir
    @siruseusesir 10 หลายเดือนก่อน +9

    Very genius. Sending love and support

  • @luisclementeortegasegovia8603
    @luisclementeortegasegovia8603 10 หลายเดือนก่อน +4

    Great to remember this algorithm! Thank you professor 👍

  • @BruceLee-io9by
    @BruceLee-io9by 10 หลายเดือนก่อน +3

    Fantastic, professor. A hug and thanks from Italy!

  • @vwr32jeep
    @vwr32jeep 9 หลายเดือนก่อน +2

    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
    @edmondscott7444 10 หลายเดือนก่อน +5

    Very well explained sir, thank you.

  • @murdock5537
    @murdock5537 9 หลายเดือนก่อน +2

    This is aweome, many thanks, Sir!

  • @blacklightning7227
    @blacklightning7227 10 หลายเดือนก่อน +2

    You did it already 👍🏾👍🏼🔥

  • @neilmorrone691
    @neilmorrone691 10 หลายเดือนก่อน +1

    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!

  • @JulesMoyaert_photo
    @JulesMoyaert_photo 10 หลายเดือนก่อน +3

    Thank you, Professor!

  • @danilobarcelos7307
    @danilobarcelos7307 9 หลายเดือนก่อน +1

    Very well explained 👍

  • @blacklightning7227
    @blacklightning7227 10 หลายเดือนก่อน +3

    Thank Mr H

  • @sudipnayak7210
    @sudipnayak7210 9 หลายเดือนก่อน +1

    Excellent algebraic problem. Thanks sir.

  • @AK58008
    @AK58008 10 หลายเดือนก่อน +2

    Just learned this today! What a coincidence

  • @rcnayak_58
    @rcnayak_58 10 หลายเดือนก่อน +1

    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.

  • @travisjacobson2334
    @travisjacobson2334 9 หลายเดือนก่อน +1

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

  • @vinicius-yk1cq
    @vinicius-yk1cq 10 หลายเดือนก่อน +3

    you're good man

  • @88kgs
    @88kgs 10 หลายเดือนก่อน +1

    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 🙏🙏

  • @user-tr1kt2wo1d
    @user-tr1kt2wo1d 9 หลายเดือนก่อน +1

    awesome ❤

    • @mrhtutoring
      @mrhtutoring  9 หลายเดือนก่อน

      Thanks 🤗

  • @jimyoung-gy9lx
    @jimyoung-gy9lx 10 หลายเดือนก่อน +3

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

    • @jimbobago
      @jimbobago 10 หลายเดือนก่อน +1

      They match the factorization of the denominator on the left.

  • @N7TWL
    @N7TWL 10 หลายเดือนก่อน +3

    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
      @carultch 10 หลายเดือนก่อน +2

      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
      @N7TWL 10 หลายเดือนก่อน

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

  • @SYAN23SYAN13
    @SYAN23SYAN13 9 หลายเดือนก่อน

    Sangat bermanfaat ilmunya dan jenius

  • @bhavanpa3683
    @bhavanpa3683 10 หลายเดือนก่อน +3

    thank you sir.helped a much.

  • @dnd2008yi
    @dnd2008yi 9 หลายเดือนก่อน +1

    Thanks sir

  • @NabiruBogdan
    @NabiruBogdan 10 หลายเดือนก่อน +2

    thanks

  • @srr9281
    @srr9281 8 หลายเดือนก่อน +1

    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...

  • @Mari_Selalu_Berbuat_Kebaikan
    @Mari_Selalu_Berbuat_Kebaikan 8 หลายเดือนก่อน

    Let's always do alot of good ❤️

  • @davantesevier4650
    @davantesevier4650 หลายเดือนก่อน

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

  • @GarethDaviesUK
    @GarethDaviesUK 9 หลายเดือนก่อน +2

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

    • @carultch
      @carultch 9 หลายเดือนก่อน +2

      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
      @carultch 9 หลายเดือนก่อน +1

      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.

  • @foxhound1008
    @foxhound1008 10 หลายเดือนก่อน +1

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

    • @carultch
      @carultch 9 หลายเดือนก่อน +1

      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.

  • @abijahmarshal
    @abijahmarshal 10 หลายเดือนก่อน +1

    Professer X in real life ❤❤❤

  • @paladdin1500
    @paladdin1500 10 หลายเดือนก่อน +2

    👏

  • @NGUYENHOANGLONG-2012
    @NGUYENHOANGLONG-2012 10 หลายเดือนก่อน +1

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

  • @IxsayanIxflame
    @IxsayanIxflame 9 หลายเดือนก่อน +1

    Try jee advanced questions

  • @ASHHH_Art
    @ASHHH_Art 9 หลายเดือนก่อน +2

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

  • @brando3023
    @brando3023 8 วันที่ผ่านมา

    so clear

  • @josephmak0865
    @josephmak0865 10 หลายเดือนก่อน +1

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

    • @carultch
      @carultch 9 หลายเดือนก่อน

      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
    @joetandingan6328 10 หลายเดือนก่อน +1

    Your caption covers your writings

    • @mrhtutoring
      @mrhtutoring  10 หลายเดือนก่อน

      You need to turn them off on TH-cam.

  • @aram5642
    @aram5642 8 หลายเดือนก่อน

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

  • @abbasmehdi2982
    @abbasmehdi2982 10 หลายเดือนก่อน +2

    -7 or 7

    • @88kgs
      @88kgs 10 หลายเดือนก่อน

      It should be 5x + 7, instead of 5x - 7 ....@ 2:54
      AND
      Mr. H, even after seeing and pointing (-7), equating it to 7.... @3:38

  • @kapiluriyat5712
    @kapiluriyat5712 10 หลายเดือนก่อน +5

    sir we directly solve this equation without complication put in 3rd line first x=1 second x= -2 and third x= -1 then directly find A =2 B= -1 & C= -1

  • @EyobKebede-mo3gn
    @EyobKebede-mo3gn หลายเดือนก่อน

    -7