How to decompose this fraction

Question

$$\begin{align}\frac{1-r(e^{ix}+e^{-ix})/2}{(r-e^{ix})(r-e^{-ix})}&=\frac{A}{r-e^{ix}}+\frac{B}{r-e^{-ix}}\\\\\end{align}$$

$$=1-r(\dfrac{e^{ix}+e^{-ix}}{2})=A(r-e^{-ix})+B(r-e^{ix})$$

I have received the instruction here (see J.G.'s reply) that I should consider the "$r^0$ and $r^1$" terms. I don't understand this, so he wanted me to set $r^0=1$ and $r^1=r$?

The end result is two equations: $1=-e^{-ix}A-e^{ix}B$ and $-\dfrac{(e^{ix}+e^{-ix})}{2}=A+B$

How do you obtain this system of equations?

Please don't use concepts in complex analysis, I have only studied partially Calculus 2.

@ Donald Splutterwit: I still don't understand, could you post a fuller answer? — James Warthington, Jan 28 '21 at 01:47
If we had \begin{eqnarray} a+br+cr^2+ \cdots + z r^{25}=1+2r+3r^2+ \cdots + 26 r^{25} \end{eqnarray} for some indeterminate $r$. ... Then we would say ... Equating coefficients of $r$, we have $a=1,b=2,c=3, \cdots z=26$. — Donald Splutterwit, Jan 28 '21 at 01:52

Sam Freedman · Answer 1 · 2021-01-28T02:25:28.067

1

Expanding out, we get $$1 - r\left(\frac{e^{ix} + e^{-ix}}{2}\right) = (A + B)r + (-Ae^{-ix} - B e^{ix}).$$

J.G.'s suggestion is to compare the LHS constant term with the RHS constant term, as well as the LHS linear (or $r^1$) term with the RHS linear term.

Edit: Here's one way to solve the system (although the algebra might get a bit messy). First, one of the equations can be rewritten as $$Be^{ix} = -1 - Ae^{-ix} $$ Or, after multiplying both sides by $e^{-ix}$, $$B = -e^{-ix} - Ae^{-2ix}.$$ After substituting in to the other equation, $$A + (-e^{-ix} - Ae^{-2ix}) = \left(\frac{e^{ix} + e^{-ix}}{2}\right).$$ Now get all the $A$s on one side, everything else on the other, factor out the $A$ and divide through. Hopefully this is a good start.

edited Jan 28 '21 at 02:25

answered Jan 28 '21 at 01:47

Sam Freedman

3,989

I plug in $r=0$ and this indeed obtain $1=A(-e^{-ix}+B(-e^{ix})$, but when $r=1$, I am lost. – James Warthington Jan 28 '21 at 01:53
I apologize for the confusion, see the reworked version – Sam Freedman Jan 28 '21 at 01:53
I still don't understand anything. Could you post a fuller post please. I thought the question here is simply to find $A$ and $B$. so $\frac{e^{ix}+e^{-ix}}{2}=(A+B)$ – James Warthington Jan 28 '21 at 01:58
I am happy to explain more, but it would help me if you could clarify your thinking. Can you tell me which part of my answer isn't making sense? – Sam Freedman Jan 28 '21 at 02:00
so I equate $-\frac{e^{ix}+e^{-ix}}{2}=(A+B)$ and $1=(-Ae^{-ix}-Be^{ix})$? – James Warthington Jan 28 '21 at 02:04
Yes, it turns out you understand everything :) Just watch that minus sign: it's $- [(e^{ix} + e^{-ix}) / 2 ] = (A + B)$ – Sam Freedman Jan 28 '21 at 02:06
Could you help me (by editing) to solve the system of these 2 equations, I have been using the one variable elimination method but I couldn't succeed. A and B is $A=-\frac12e^{ix},,B=-\frac12e^{-ix}$ – James Warthington Jan 28 '21 at 02:08
See the edit for the first steps of the algebra. – Sam Freedman Jan 28 '21 at 02:25
I don't know why but I obtain $A=\dfrac{-e^{ix}-3e^{-ix}}{2(e^{-ix}-e^{-2ix)}}$ This result is not correct, I think. – James Warthington Jan 28 '21 at 03:00

How to decompose this fraction

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