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$$\frac{1}{(1-x^a)(1-x^b)}=\frac{A}{(1-x)^2}+\frac{B}{(1-x)}+\sum_{r^a=1}^{ ‎ }\frac{C_r}{(1-x/r)}+\sum_{t^b=1}^{ ‎ }\frac{D_t}{(1-x/t)}$$ $(t,r\neq1; A, B, C, D$ are real numbers)

How did the author know that he need to separate the case r,t=1?

I would appreciate it really much if someone can guide me to resources to study the general rules for partial fraction expansion like this.

ChemistryLearner
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  • Are you sure about your formula? You are talking about $\frac{1}{1-x^a}$ and $\frac{1}{1-x^b}$ while I remembering fraction expansion being about $\frac{1}{(1-x)^a}$ and $\frac{1}{(1-x)^b}$. – Dominique Mar 07 '24 at 10:04
  • @Dominique, yes I'm sure about my formula. Otherwise, it wouldn't make sense to have $/frac{c}{1-/frac{x}{r}}$ with r being root of unity excluding 1. (Sorry for the mess, mathjax doesn't work in comment for some reason, but I think you still get the idea) – ChemistryLearner Mar 07 '24 at 10:08
  • @ChemistryLearner Try with braces and point the slashes the other way. \frac{c}{1} - \frac{x}{r} which renders as $\frac{c}{1} - \frac{x}{r}$ not $/frac{c}{1} - /frac{x}{r}$ LOL. – nickalh Mar 07 '24 at 11:05
  • @nickalh Thank you for your suggestion, after a few minutes I reread my message and found the error but it's too late to fix. I'm grateful that you spend time to stop by and take a look at my question. – ChemistryLearner Mar 07 '24 at 11:37
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    For now of maybe future reference, it's possible to delete and repost a comment, "enabling" edits after 5 minutes. – nickalh Mar 07 '24 at 17:38

1 Answers1

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This expansion is only valid if $a$ and $b$ are coprime. In that case, $t=r=1$ is the only multiple root of the denominator. For a root $x_0$ of the denominator with multiplicity $m$, a partial fraction expansion contains terms proportional to $(x-x_0)^{-k}$ for all $k$ up to $m$. If $t=r=1$ hadn’t been treated separately, both sums would have included a term proportional to $(1-x)^{-1}$, and the term proportional to $(1-x)^{-2}$ would have been missing. If $a$ and $b$ aren’t coprime, there are further multiple roots that need to be treated analogously.

joriki
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  • I still don't understand about this part, is there any chance you could explain a bit more? If the author include a term proportional to $(1-x)^{-2}$ but didn't treat $t=r=1$ seperately, $(1-x)^{-1}$ will be repeated but they can split the term for both fractions. The way they treat it currently implies that A will not be a function with respect to r or t, like $C_r$ and $D_t$. – ChemistryLearner Mar 07 '24 at 11:33
  • @ChemistryLearner: Unfortunately I don’t understand what you mean by “split the term for both fractions”. There shouldn’t be two identical terms; that would make the expansion indeterminate, since you could choose any combination of coefficients that add up to the right sum. Of course there’s no harm done in splitting a term up into two identical terms, but if you want to determine the coefficients from a system of equations, it’s a big unnecessary hassle to have an indeterminate system with a whole space of solutions when you could easily avoid it by only including each term once. – joriki Mar 07 '24 at 11:47
  • @ChemistryLearner: I also don’t understand what you mean about $A$ not being a function of $r$ or $t$. The coefficients $C_r$ and $D_t$ have the indices $r$ and $t$ because there’s one of them for each $r$ and each $t$. You could write $B$ as $B_1$ if you prefer, since it’s the coefficient for the root $1$, and you could use some similar notation for $A$, but that doesn’t make a difference. – joriki Mar 07 '24 at 11:49
  • Sorry for the ambiguity of my comment. I appreciate that you shed some light on my question. I mean that in order to solve for $C_r$, it will either be a constant or a function with respect to r; otherwise I will need to solve for a different values. So $A_r$ can take a value with respect to r and $A_t$ can take a value with respect to t, this could solve the indeterminate problem. (Here I mean $A_r$ can be $C_r$ where r=1 and same goes for $A_t$). – ChemistryLearner Mar 07 '24 at 12:01
  • @ChemistryLearner: It seems to me that you’re overthinking this. Maybe just look at a simple example. For $a=2$ and $b=3$ we have, with $\omega=\exp\left(\frac{2\pi\mathrm i}3\right)$,

    $$ \frac1{(1-x^2)(1-x^3)}=\frac A{(1-x)^2}+\frac B{1-x}+\frac{C_{-1}}{1-x/(-1)}+\frac{C_\omega}{1-x/\omega}+\frac{C_{\omega^2}}{1-x/\omega^2};. $$

    That’s the correct expansion. If you disagree, perhaps write out the expansion that you think should result.

     – joriki Mar 07 '24 at 12:12
  • Thank you for your promptness to reply, please bear with me. I, by no means, want to invalidate your answer, I just thought a sum of $C_r$ must be related in some way. In my example, I can work out $A=\frac{1}{ab}$, $B=\frac{a+b-2}{2ab}$, $C=\frac{1}{a(1-r^b)}$, $D=\frac{1}{b(1-r^a)}$. But it was all thanks to the author brilliant expansion. Had he not guide me to this step, I wouldn't had worked out an elegant solution for A, B, C, D and stuck trying to find a solution for A, C, D because B hadn't existed. – ChemistryLearner Mar 07 '24 at 12:36
  • Therefore, I want to know the thought process or general rules behide the terms B which leads to it having a different function from C, D. $\‎ $

    P.S. It hadn't occured to me until now that I made foolish mistake of typing A instead of B, that probably is the main reason why my writing was so confusing. Additionally, I might have used the word "function" incorrectly since I don't learn math in English. Please pardon my ignorance.

     – ChemistryLearner Mar 07 '24 at 12:38
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    @ChemistryLearner: I’m sorry, I’m really trying to be helpful, but I’m still not able to understand what additional “general rules” you’re looking for. The general rule is the one I gave in the answer: Terms up to $(x-x_0)^{-k}$ (or equivalently $(1-x/x_0)^{-k}$) need to be included up to $k=m$ for a root $x_0$ of multiplicity $m$. Here you have one root (namely $1$) of multiplicity $2$, and all other roots have multiplicity $1$ (if $a$ and $b$ are coprime). That’s all there is to know. Everything else is just how you write it down. – joriki Mar 07 '24 at 12:45
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    There needs to be one linear term for each $a$-th root of $1$ and one linear term for each $b$-th root of $1$, but for the double root $1$ (which is both an $a$-th root of $1$ and a $b$-th root of $1$), there musnt’t be two linear terms, and instead there should be a linear term and a quadratic term. If you want to write the solution in general form, with sums over the $a$-th roots of $1$ and the $b$-th roots of $1$, then you need to exclude their common value $1$ and treat it separately, because that’s what the general rule that I gave requires. – joriki Mar 07 '24 at 12:45
  • I apologise my disrespectful behavior, I just wanted to reassure you that I had no hostility towards you but it backfired horribly. I wish to be an enquiring and student who would follow suggestions with an open mind, not somebody who is overtly fixated on his own opinion. I didn't take notice of the negative implication that my sentence might have. I want you to know I'm indebted to the generosity of a conscientious stranger. I think the rule you gave is the correct one and it shall abundantly answer my question. – ChemistryLearner Mar 07 '24 at 13:13
  • Just one more thing, please. Did you mean coefficient term when you said linear term and linear term when you said quadratic term? By the way, my school didn't teach me about partial fraction expansion but this book treats it as a basic knowledge. Most resources I can find online only cover real roots so I figured that there's a gap in my understanding. If you know more about partial fraction expansion that wasn't included in my question and your previous answers, could you provide me more information or sources of information where I can find out more. I thank you from the bottom of my heart. – ChemistryLearner Mar 07 '24 at 13:24
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    @ChemistryLearner: I didn’t feel any hostility at all, there was no problem as far as I’m concerned – just that we didn’t seem to be understanding each other well :-). By “linear term”, I meant a term with a denominator of the form $x-x_0$, and by “quadratic term” I meant a term with a denominator of the form $(x-x_0)^2$. To be more precise I should have said “reciprocal of a linear/quadratic term”. – joriki Mar 07 '24 at 13:39