How to re-write one fraction as two others.

Question

I have the two following fractions.

$$ \dfrac{A}{Bx^{\alpha+1}}$$ and $$ \dfrac{C}{Dx^{\alpha+\beta}}$$ The form i want $$ \dfrac{E}{Fx^{\alpha+\beta+1}}$$

I was thinking to do partial fractions or potentially doing a Laplace or Fourier transform, then rearranging and transforming back. How could one possibly get the two fractions into one fraction of this form.\ $ A,B,C,D,E$ and $F$ can be number or possible a factorial if we go down the laplace transform route, as can $\alpha, \beta$

Answer 1

I'm assuming, as you seem to have implied, that $A,B,C,D,E,F$ are arbitrary, and, in particular, that $E$ and $F$ can be defined as some combination of $A,B,C,D$. $$ \frac{A}{Bx^{\alpha+1}}=\frac{A}{Bx^{\alpha+1}}\cdot\frac{x^\beta}{x^\beta}= \frac{Ax^\beta}{Bx^{\alpha+1}x^\beta}=\frac{Ax^\beta}{Bx^{\alpha+\beta+1}} $$ And, similarly, $$ \frac{C}{Dx^{\alpha+\beta}}=\frac{Cx}{Dx^{\alpha+\beta+1}} $$ So you can have $$ \frac{Ax^\beta}{Bx^{\alpha+\beta+1}}-\frac{Cx}{Dx^{\alpha+\beta+1}}\cdot x^{\beta-1}= \frac{Ax^\beta}{Bx^{\alpha+\beta+1}}-\frac{Cx^{\beta}}{Dx^{\alpha+\beta+1}}= \frac{AD-BC}{BDx^{\alpha+\beta+1}} $$ and thus, we can let $E=AD-BC$ and $F=BD$ to get $$ \frac{E}{Fx^{\alpha+\beta+1}}=\left(\frac{A}{Bx^{\alpha+1}}\right)- \left(\frac{C}{Dx^{\alpha+\beta}}\right)x^{\beta-1} $$

Edit:

This also shows that you cannot get appropriate values of $E,F$ from a linear combination. Suppose, for the sake of contradiction, that there are constants $M,N$ such that $$ M\left(\frac{A}{Bx^{\alpha+1}}\right)+N\left(\frac{C}{Dx^{\alpha+\beta}}\right)= \frac{E_1}{F_1x^{\alpha+\beta+1}} $$ for some $E_1,F_1$. Then we have $$ \frac{E/E_1}{F/F_1}\left(M\left(\frac{A}{Bx^{\alpha+1}}\right)+ N\left(\frac{C}{Dx^{\alpha+\beta}}\right)\right)= \frac{E/E_1}{F/F_1}\frac{E_1}{F_1x^{\alpha+\beta+1}}=\frac{E}{Fx^{\alpha+\beta+1}} $$ For simplicity, let $\epsilon=\frac{E/E_1}{F/F_1}$. Then, \begin{align} \epsilon M\left(\frac{A}{Bx^{\alpha+1}}\right)+ \epsilon N\left(\frac{C}{Dx^{\alpha+\beta}}\right)&= \left(\frac{A}{Bx^{\alpha+1}}\right)- \left(\frac{C}{Dx^{\alpha+\beta}}\right)x^{\beta-1}\\ \epsilon M\left(\frac{A}{Bx^{\alpha+1}}\right)&= \left(\frac{A}{Bx^{\alpha+1}}\right)- \left(\frac{C}{Dx^{\alpha+\beta}}\right)\left(x^{\beta-1}+\epsilon N\right)\\ \epsilon M&=1-\left(\frac{BC}{ADx^{\beta-1}}\right)\left(x^{\beta-1}+\epsilon N\right)\\ \epsilon M&=1-\frac{\epsilon BCN}{ADx^{\beta-1}}\\ M&=\frac{1}{\epsilon}-\frac{BCN}{ADx^{\beta-1}} \end{align} And this contradicts the assumption that $M$ is constant.

Answer 2

Assuming that I properly understood (which is not sure, I must confess) : let $$P=\frac{A}{Bx^{\alpha+1}}$$ $$Q=\frac{C}{Dx^{\alpha+\beta}}$$ So $$P^{\frac{1}{\alpha+1}}=\Big(\frac{A}{B}\Big)^{\frac{1}{\alpha+1}}\frac 1x$$ $$P^{\frac{1}{\alpha+1}}Q=\Big(\frac{A}{B}\Big)^{\frac{1}{\alpha+1}}\Big(\frac CD \Big)\frac{1}{x^{\alpha+\beta+1}}=\dfrac{E}{Fx^{\alpha+\beta+1}}$$ So $$\dfrac{E}{F}=\Big(\frac{A}{B}\Big)^{\frac{1}{\alpha+1}}\Big(\frac CD\Big)$$ You can choose any value you want for $E$ or $F$ and compute the other one from the above equation.