MHB Partial fraction decomposition

AI Thread Summary
The discussion focuses on the partial fraction decomposition of the expression 4x²y divided by the product of two quadratic polynomials. A clever algebraic manipulation is suggested to simplify the expression, leading to the identification of 4xy as the difference of two quadratic terms. The proposed decomposition involves expressing the original fraction as a sum of two simpler fractions, each with one of the quadratic polynomials in the denominator. Through coefficient comparison, the values for A and D are determined, resulting in the final decomposition of the expression. The solution demonstrates an effective approach to tackling complex algebraic fractions.
Drain Brain
Messages
143
Reaction score
0
please help decompose$\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}$

I've used the cases I know for this problem but to no avail. please help me.
 
Mathematics news on Phys.org
This takes a bit of trickery, note that :

$$\begin{align}\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} &= \frac{x \cdot 4xy}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} \\ &= x \cdot \frac{(x^2+2xy+2y^2) - (x^2-2xy+2y^2)}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}\end{align}$$

Can you proceed?
 
mathbalarka said:
This takes a bit of trickery, note that :

$$\begin{align}\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} &= \frac{x \cdot 4xy}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} \\ &= x \cdot \frac{(x^2+2xy+2y^2) - (x^2-2xy+2y^2)}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}\end{align}$$

Can you proceed?

sure, from here It seems that I can decompose it. but how come you replaced $4xy$ to $(x^2+2xy+2y^2) - (x^2-2xy+2y^2)$ ??
 
Uh, not sure if I understand your question, but it follows from basic algebra

$$(x^2+2xy+2y^2) - (x^2-2xy+2y^2) = \cancel{\color{red}{x^2}} + 2xy + \cancel{\color{green}{2y^2}} - \cancel{\color{red}{x^2}} + 2xy - \cancel{\color{green}{2y^2}} = 2xy + 2xy = \boxed{4xy}$$
 
While mathbalarka's suggestion is quite clever and makes light work of the problem, I would have assumed the decomposition would take the form:

$$\frac{4x^2y}{\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)}=\frac{Ax+By+C}{x^2-2xy+2y^2}+\frac{Dx+Ey+F}{x^2+2xy+2y^2}$$

and then plodded along with the resulting cumbersome algebra.

Hence:

$$4x^2y=(Ax+By+C)\left(x^2+2xy+2y^2\right)+(Dx+Ey+F)\left(x^2-2xy+2y^2\right)$$

$$4x^2y=(A+D)x^3+(C+F)x^2+(2A+B-2D+E)x^2y+(2A+2B+2D-2E)xy^2+(2C-2F)xy+(2C+2F)y^2+(2B+2E)y^3$$

Comparing coefficients, we obtain:

$$A+D=0$$

$$C+F=0$$

$$2A+B-2D+E=4$$

$$A+B+D-E=0$$

$$C-F=0$$

$$B+E=0$$

From the 2nd and 5th equations, we immediately find:

$$C=F=0$$

From the 1st, 4th and 6th, we find:

$$B=E=0$$

Thus, we are left with:

$$A=-D$$

$$A=D+2$$

Thus, $$A=1,\,D=-1$$ and so we find:

[box=green]$$\frac{4x^2y}{\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)}=\frac{x}{x^2-2xy+2y^2}-\frac{x}{x^2+2xy+2y^2}$$[/box]
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
11
Views
2K
Replies
3
Views
2K
Replies
8
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
3
Views
546
Replies
6
Views
200
Back
Top