Thanks!!

Thank you.

@@bprpcalculusbasics Donyou think thisbis too unintuitive though and needlessly complicated? Hope you can respond when you can. Thanks for sharing.

Same to me, I just love this guy 😭

​@@rodrigosilva893you made me watch this again, thanks 🫂

Glad to help! Thank you!!!

same😁

Sup

That's so much simpler! The substitution thing he did in the video is too unintuitive and overcomplicates the problem in my opinion.

But WAIT can't you do the partial.fractuon decomposition WITHOUT the B term..I think you can so why doesn't he show this??

​@@leif1075 You might have one that coincidentally turns out to be zero, but in general, if you are working with a quadratic, you have to put a linear term on top of it, because fractions using a quadratic denominator have two degrees of freedom. You can either set it up as (B*x + C)/(x + 2)^2, or B/(x + 2)^2 + C/(x + 2). Either way, you'll get a partial fraction decomposition that is valid. Usually, the form of B/(x + 2)^2 + C/(x + 2) is preferable, for an application where you'd use this concept. I recommend putting terms you can get by Heaviside coverup first. So if I were given (2*x + 1)/((x + 1)*(x + 2)^2), I'd assign A to be the coefficient over (x + 1), and B to be the coefficient over (x + 2)^2. Both of these, can be found with Heaviside coverup. You get A = -1, and B = 3. There is a shortcut that can work for getting C. (2*x + 1)/((x + 1)*(x + 2)^2) = -1/(x + 1) + 3/(x + 2)^2 + C/(x + 2) Multiply through by just one instance of (x + 2): (2*x + 1)/((x + 1)*(x + 2)) = -(x + 2)/(x + 1) - 3/(x + 2) + C Let x go to infinity by taking the limit. Terms with a higher x-degree in the denominator than the numerator become zero, terms with equal degrees of x in both become a finite number. 0 = -1 - 0 + C And we can directly see that C = +1.

@@carultch indont known what heaviside cover up is nut also what you say doesn't make sense..you already have one linear term.and one quadratic term so you still.dont need the B term..see what I mean??

@@leif1075 Heaviside cover-up is a shortcut for partial fractions. The idea is that you can more directly find the coefficient over the linear terms, by plugging in the input that makes that particular factor equal zero. Then, cover up that term in the original fraction, and evaluate the rest at the same input value. If you try only having one constant coefficient over either a quadratic denominator, or a repeated linear denominator without the second term, you'll end up with an over-constrained system of equations, that is either redundant or contradictory. Try it for: (2*x + 1)/((x + 1)*(x + 2)^2) Assume it is equal to: A/(x + 1) + B/(x + 2)^2 Cross-multiply: (2*x + 1) = A*(x + 2)^2 + B*(x + 1) Expand: 2*x + 1 = A*x^2 + 4*A*x + 4*A + B*x + B This constructs the equations: A = 0 4*A + B = 2 B = 1 And you'll see a contradiction, that B needs to both equal 1 and 2, which it can't do. This is why we need a third term, of C/(x + 2), so we have a third unknown for a 3-equation system.

The easiest maths book you could ever do 😂

Glad to hear that

arctg(ln(3))+arctg(ln(2))? does it simplify somehow or smth?

@@NoNameAtAll2 Setting a := arctan( ln 2 ), b := arctan( ln 3 ) we have tan a = ln 2, tan b = ln 3, tan( a + b ) = ( tan a + tan b )/( 1 - ( tan a )( tan b ) ) = ( ln 2 + ln 3 )/( 1 - ( ln 2 )( ln 3 ) ) = ( ln 6 )/( 1 - ( ln 2 )( ln 3 ) ). The calculator shows that ( ln 2 )( ln 3 ) = 0.7615000… So tan( a + b ) > 0. Noting that 0 < a, b < π/2, 0 < a + b < π, we have a + b = arctan( ( ln 6 )/( 1 - ( ln 2 )( ln 3 ) ) ). Consequently, we obtain arctan( ln 2 ) + arctan( ln 3 ) = arctan( ( ln 6 )/( 1 - ( ln 2 )( ln 3 ) ) ).

In general, for every x, y > 0 we have ( i ) 0 < xy < 1　⇒　arctan x + arctan y = arctan( ( x + y )/( 1 - xy ) ) ( ii ) xy = 1　⇒　arctan x + arctan y = π/2 ( iii ) 1 < xy　⇒　arctan x + arctan y = π - arctan( ( x + y )/( xy - 1 ) )

@@user-dl8nk5bf8v idk man having _fraction_ with logarithms doesn't seem simpler than 2 separate ones and arctg layer is still there, so it's cubersome either way

@@NoNameAtAll2 Oh, sorry. I don't know what expression is better for some calculation.

Glad you liked it!

Yes, that is super hard. Best wishes to you! : )

"essentially they are undoing finding the common denominator, and when you have a repeated term, you could have had every power of the repeated term up to the power of the term and the denominator would have looked the same, so you assume they could all be there, unless you show a numerator is zero." okay but why is this the case?

@@bprpcalculusbasics they’ve been real quiet after this.😂

@@bprpcalculusbasics I understood all of the video except the rule you used; the degree on the top must be 1 less than the degree on the bottom. Why is this the case?

@@dVPulse that requires some proof which can't be written down in the comments without making its too tiresome to read. But yeah, there IS a proof and I'll post a link whenever i find it.

@@chessematics where?pls post it

Thank you!

Yes indeed. If you include complex numbers, then all polynomial denominators up to the quartic, would technically be reducible. Quintics and beyond, have no closed-form solution for the roots, in elementary functions. It usually won't help you very much to do this. It's a lot easier to set up the algebraic system to solve for unknown coefficients, than to detour to the complex numbers. Even though Heaviside coverup works for linear factors of complex roots as well.

Best of luck!!!

By the similar way we have 1 = ( x + 2 )^2 - ( x + 3 )( x + 1 ), 2x + 1 = ( 2x + 1 )( x + 2 )^2 - ( 2x + 1 )( x + 3 )( x + 1 ) = ( 2( x + 1 ) - 1 )( x + 2 )^2 - ( 2( x + 2 )^2 - ( x + 5 ) )( x + 1 ) = ( x + 5 )( x + 1 ) - ( x + 2 )^2, ( 2x + 1 )/( ( x + 1 )( x + 2 )^2 ) = ( x + 5 )/( x + 2 )^2 - 1/( x + 1 ). And we also have ( x + 5 )/( x + 2 )^2 = 1/( x + 2 ) + 3/( x + 2 )^2. Hence we obatin ( 2x + 1 )/( ( x + 1 )( x + 2 )^2 ) = 1/( x + 2 ) + 3/( x + 2 )^2 - 1/( x + 1 ).

Bc if not, then we can do polynomial long division.

YT read your mind! Cheers : )

Bc if not, then we can do long division to break it down

If the degree on top, is equal to, or greater than, the degree on bottom, then you can simplify the rational expression, into a separate polynomial added to a fraction. Either by adding zero in a fancy way, to form a term you can cancel, or by using polynomial division. The polynomial terms on their own, once extracted from the fraction, can be handled with the power rule. As for why it is one less, rather than two less, the reason is that you'll get an indeterminate system of equations when attempting to solve for coefficients, if you don't have a big enough polynomial on top of any given term. You could "get lucky" and end up with a highest degree term of the numerator with a coefficient of zero, but in general, you need at least a polynomial on top of a degree one less, than the degree of the polynomial on bottom.

As an example, given: x^3/[(x + 1)*(x + 2)] I can simplify this to: x - 3 + (7*x + 6)/[(x + 1)*(x + 2)] Once in this form, the x - 3 part, is all set for calculus applications. It's just the power rule to either differentiate or integrate it. All that remains for partial fractions is: (7*x + 6)/[(x + 1)*(x + 2)] This one is a proper fraction, with a linear term on top, and a factored quadratic on bottom. The simplified expression after partial fractions is complete, is: x - 3 - 1/(x + 1) + 8/(x + 2)

😂😂😂

Perhaps because you could use long division instead.

If the denominator is greater than or equal to the numerator, then there are other ways of simplifying the fraction, to prepare them for integration or inverse Laplace transforms. The numerator doesn't have to be just 1 less than the denominator, just that it is less than the denominator in general. As a couple examples, consider: (x^2 + 2*x + 4)/(x + 1) With synthetic division, it will reduce to: x + 1 + 3/(x + 1) Only the 3/(x + 1) needs to be integrated with methods of integrating algebraic fractions. The x + 1 part can simply be integrated with the power rule. As another example, that has an equal power to its denominator, consider: (x^2 + 2*x + 4)/(x^2 + 5*x + 6) For this one, we can add zero in a fancy way, to separate it down to a polynomial expression, and a rational expression. x^2 + 2*x + 4 + (3*x - 3*x) + (2 - 2) = x^2 + 5*x + 6 - 3*x - 2 This allows us to rewrite it as: (x^2 + 5*x + 6)/(x^2 + 5*x + 6) - (3*x + 2)/(x^2 + 5*x + 6) And ultimately reduce it to: 1 - 7/(x + 3) + 4/(x + 2) So the fact that there were equal numerator and denominator orders, just means that there is a polynomial term out in front, followed by a fraction where the numerator order is at least one less than the denominator.

Because x^2+2 vis not factorable over the reals and so a linear factor is allowed to be a remainder

Thank you, this helped me a lot! ^_^@@mathisnotforthefaintofheart

If you have two irreducible quadratic denominators, then you have two separate irreducible quadratic terms, each with two unknown coefficients forming linear numerators. For your example: x/[(x^2+x+1)(x^2+1)] I recommend completing the square first, when you have an x-term in the middle. Most applications of partial fractions, will eventually require completing the square. x^2 + x + 1 = (x + 1/2)^2 + 3/4 Set up the partial fractions, with each quadratic denominator, and two arbitrary linear numerators: (A*x + B)/(x^2 + 1) + (C*(x + 1/2) + D)/((x + 1/2)^2 + 3/4) Note that since we have an offset quadratic for the second term, instead of using C*x + D, I've opted for C*(x + 1/2) + D instead. You'll see why. Equate to original expression: (A*x + B)/(x^2 + 1) + (C*(x + 1/2) + D)/((x + 1/2)^2 + 3/4) = x/[(x^2+x+1)(x^2+1)] Plug in strategic values for x, to solve for the coefficients. Usually 1 and 0 are strategic values to use, since they keep your math simple. In this case, -1/2 is also a strategic value. I'll also use x=-1. Be careful with x=-1 and x=+1, since they could give redundant information. You can also use x=infinity in some cases, though it won't help us here. For x = 0, this makes the A disappear: (A*0 + B)/(0^2 + 1) + (C*(0 + 1/2) + D)/((0 + 1/2)^2 + 3/4) = 0/[(0^2+0+1)(0^2+1)] B + C/2 + D = 0 For x = -1/2, this makes C disappear: (-1/2*A + B)/((-1/2)^2 + 1) + (C*0 + D)/((-1/2 + 1/2)^2 + 3/4) = -1/2/[((-1/2)^2 - 1/2 + 1)((-1/2)^2+1)] -4/10*A + B*4/5 + 4/3*D = -8/15 -6*A + 12*B + 20*D = -8/3 For x = 1: (A*1 + B)/(1^2 + 1) + (C*(1 + 1/2) + D)/((1 + 1/2)^2 + 3/4) = 1/[(1^2+1+1)(1^2+1)] A/2 + B/2 + C/2 + D/3 = 1/6 3*A + 3*B + 3*C + 2*D = 1 For x = -1: (-A + B)/(1 + 1) + (-C/2 + D)/((-1 + 1/2)^2 + 3/4) = -1/[((-1)^2+-1+1)((-1)^2+1)] -A/2 + B/2 -C/2 + D = -1/2 -A + B - C + 2*D = -1 Solve system: A = 0; B = 1; C=0; D=-1 Solution: 1/(x^2 + 1) - 1/((x + 1/2)^2 + 3/4)