life changing partial fractions trick

[@MihaiNicaMath]

Very nice trick! I have seen this before in a complex analysis context but for some reason never thought to apply to partial fractions. One note is that in your final example of A,B with a repeated root you can at least extract B first by doing lim (x-a)^2*(func(x)) and then once you know B subtract it off to reduce to devious case. (Annoying if you have more than double root!)

[@renesperb]

This is a very useful formula (well known). If you have multiple zeros in the denominator things get a little more complicated. But there are formulae for these cases as well. Maybe you could present them in a future video.

[@maximespinelli7929]

Good old Heaviside cover-up method. I always talk about it in class.

[@DarinBrownSJDCMath]

If you know the Heaviside coverup method, you can see why this works. Think of the derivative of (x-a_1)(x-a_2)...(x-a_n), by the general product rule, the derivative is the sum of n terms, each with a single factor missing. If you substitute in a_i, all those terms vanish except for the term with all (x-a_j), j not equal to i. But this is nothing more than what you would be evaluating and dividing by to isolate your unknown variable using Heaviside. Note this only works specifically for a denominator that is a product of distinct linear factors with linear coefficient 1.

[@UberHummus]

I teach my students this method; it seems sorta like how L’Hospital’s works. But, because high school algebra skills have been a bit lacking in my own observations the last couple of years, I often have them use the algebraic method.

[@QuinzerOO0O]

Dr. I can only tell you that your videos are special for some reason because they teach us the easiest and very understandable methods that they don't teach us in college so Thank you for the content you do!

[@drpeyam]

I'm so glad you're enjoying them! 😊

[@mertaliyigit3288]

Thank you dr peyam ! This is very helpful

[@drpeyam]

You’re welcome :)

[@damianwrobel5715]

My questions: 1) 02:27 The formula for "A" is: A = f(1)/g'(1). WHY does it worki? How does the Proof look like? 2) 07:22 If we would have B/(x+3)^2, how would the formula for "B" look like? Would it be B = 2* f'(-3)/g''(-3) ? 3) 07:22 If we would have B/(x-2)^3, how would the formula for "B" look like? Would it be B = 3* f''(2)/g'''(2) ? 4) 07:22 How about this situation: (A*x + B) / (a*x^2 + b*x + c), where ∆ = b^2 - 4*a*c is < 0 ? How does the formula for "A" look like? How does the formula for "B" look like? 5) 07:22 How about this situation: (A*x + B) / (a*x^2 + b*x + c)^2, where ∆ = b^2 - 4*a*c is < 0 ? How does the formula for "A" look like? How does the formula for "B" look like? 6) 07:22 How about this situation: (A*x + B) / (a*x^2 + b*x + c)^3, where ∆ = b^2 - 4*a*c is < 0 ? How does the formula for "A" look like? How does the formula for "B" look like?

[@DarinBrownSJDCMath]

Hint for A: think about the general product rule for differentiation.

[@Aman_iitbh]

It can proofed by idea similar to lagrange interpolation, As g(x) can be factored in 3 linear factor ,taking two at a time will form a basis( if diffrent linear factors are there), then find c1,c2,c3 (as we do in lagrange interpolation) ,you will find out c1=f(a1)/g'(a1), similarly for other.

[@Aman_iitbh]

Wrting f in term of g ,as degree of numerator is small it will always lie in vector space of denominator polynomial having basis i wrote abaove , so one can always get c1,c2,c3 for f. And at last chain rule for product of linear function will leave you with result

[@michaelz2270]

My life has changed forever.

[@raghvendrasingh1289]

❤ we also use this formula in problem of finding inverse Laplace transform and residue. actually (x-a)f(x)/g(x) , where g(a) = 0 takes the indeterminate form 0/0 as x approaches 'a' hence by L- hospital's rule result will be f(a)/g'(a)

[@jordimayorgisbert6490]

Me too !! :-)

[@jyotsanabenpanchal7271]

Very nice!

[@士-x7e]

A=Residue at 1. Interesting.

[@AriosJentu]

Nice trick. Could you explain how it works, please ;)

[@DarinBrownSJDCMath]

Hint: think about the general product rule for differentiation.

[@MichaelHPT]

thank you so much.

[@mtheorylaboratory]

wow no way!! amazing

[@Fluxinate]

You're wonderful.

[@drpeyam]

Thank you!! 😊

[@alipourzand6499]

Great video, look forward the demonstration. BTW next time one of your students still try to solve painful algebric equations for PFs, it means that he/she hasn't subscribed to your channel! 😁

[@drpeyam]

It's true! 😉

[@ekadria-bo4962]

Interesting one: So P'(x) make all zero in P(x) not root anymore

[@DarinBrownSJDCMath]

Be careful all the linear coefficients are 1. For example, if the denominator factors as (x+1)(2x-1), you need to factor out 2 first.

[@agytjax]

Can repeated roots be expressed as A/(x-1) + 2*B/(x-2) ? I guess the coefficient would still be the same ?

[@TranceFat]

Not only is it partial fractions, but it is also partial arm! (Thumbnail for the video) 😂

[@drpeyam]

Badum tsssss 😂

[@xwyl]

it's actually residue, and some trick when verified. Just imagine when x->1 the equal sign must still hold. And the derivative part is a trick e.g. g'(1)=(1+1)(1-2)

[@UltraMaXAtAXX]

Ah yes, residue theorem lite.

[@nicogehren6566]

Very good

[@drpeyam]

Glad you liked it! 😄

[@patipateeke]

Great movie; didn't know about this formula. However, in the formula at 7:36, i believe it should be simply (2!)*(f(2)/g''(2)) and not (2!)*(f'(2)/g''(2))

[@RohanHemanth]

What if the polynomial in the numerator has the higher order, will this work? But thank you for sharing

[@parasgaur1856]

You can just divide the numerator with the denominator to get a polynomial + a rational function. Then you can apply the partial fraction part to the rational function and just simply the reverse power rule for the polynomial.

[@Andyg2g]

Why does this work?

[@omograbi]

Applying Heaviside technquie would be much easier

[@drpeyam]

Explain

[@DarinBrownSJDCMath]

​@@drpeyamMultiply through by denominator, then substitute in roots. It's equivalent to what you're doing.

[@GoodWill-s8j]

How tf he factorized the denominator? 😳

[@ardademirhan2452]

Great video. Residue calculus has great applications, I love teaching complex analysis! Am I missing something or should not the last expression be 2 f(2) / g’’(2), so f’ should not appear I think. For example, think of 1/(x-2)^2. f’(2)=0. A better example would be (3x-7)/(x-1)^2 = (3/(x-1))-(4/(x-1)^2). I applied the Residue formula for f(z)/(g(z)/(z-z_0)) where f has no zero at z_0, g has a double zero. We want the Laurent coefficient c_(-2) for f/g about z_0 which is the c_(-1)=Residue of f(z)/(g(z)/(z-z_0)). Multiplying this by z-z_0 and taking the limit of it as z goes to z_0 along with L’Hôpital’s Rule would give 2 f(z_0)/ g’’(z_0) unless I made a calculation error.

[@asmauwamahoro342]

Dear Dr Peyam, ce nouvel éclairage (depuis plusieurs vidéos) fait mal aux yeux !

[@drpeyam]

Desole 😞

[@jorgeapg]

genial :D

[@thomasheljeschrodingertama4874]

Im so interested about the proof, Would you please make a video!!!! How wonderful is this algorithm technique? and btw thx for make this kind of videos, i really appreciate you work Mr Peyam!!!!

[@DarinBrownSJDCMath]

Hint: think about the general product rule for differentiation.

[@marcelob.5300]

Life changing... will these make me rich with cryptos, doctor?

[@drpeyam]

Of course hahaha