Is it possible to rewrite $$\frac{1}{x\pm y}$$ as multiplication of an expression containing only $x$ and another containing only $y$ ($x$ and $y$ are real independent variables), i.e. $$\frac{1}{x\pm y}\stackrel{?}{=}f(x)g(y)$$ If it is possible, what are $f(x)$ and $g(y)$?
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I don't think so, for special cases I think we can but we can't generalise it. – Vyom Yadav Mar 18 '21 at 07:14
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If $a$ and $b$ are (explicit) constants (and $a+b\neq0$), then $\frac1{a+b}$ is just a constant, and there is no point in seeking a "decomposition" of it. On the other hand if they are in some way variable (e.g., parameters or unknowns) then as a function of them the first expression has a pole when $a+b=0$, while the second has poles at $a=0$ and at $b=0$ (unless cancelled by the numerators, which would defeat the purpose of introducing the denominators) and not at $a+b=0$ (again unless introduced in the numerators). So this is a strange thing to want to do. – Marc van Leeuwen Mar 18 '21 at 18:06
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Sure it's possible. Take $A=\frac{a}{a+b}$ and take $B=0$. Then it should be trivial to show that $$\frac Aa + \frac Bb = \frac{\frac{a}{a+b}}{a} + \frac0b = \frac{1}{a+b}.$$
In fact, taking any value of $B$, you can set $A=a\cdot\left(\frac{1}{a+b} - \frac Bb\right)$ and get the equality $$\frac{A}{a} + \frac Bb = \frac1{a+b}.$$
However, if you demand that $a,b, A, B$ are all integers, then the answer is no. A simple counterexample can be found by taking $a=b=1$. Then, $\frac{1}{a+b}=\frac12$, while $\frac Aa + \frac Bb = \frac A1 + \frac B1 = A + B$, and since $A$ and $B$ are integers, this means $\frac Aa + \frac Bb$ must also be an integer.
Since $\frac12$ is not an integer, it's clear that $\frac Aa + \frac Bb$ cannot be equal to $\frac1{a+b}$.
5xum
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$a$, $b$, $A$, and $B$ are real. I want to get rid of $a+b$ in the denominator, but in your solution $A$ is a fraction with denominator $a+b$. – Masa Mar 18 '21 at 07:36
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@Masoud Your comment is contradictory. On one hand you say that $A$ is real, indicating that there is no restriction on what $A$ could be. On the other hand, you say $A$ should not be a fraction, indicating that it should be an integer. You can't have it both ways. Like I said, if you allow $A$ to be any number, then a solution to $\frac Aa+\frac Bb=\frac1{a+b}$ clearly exists, as I have shown. If you restrict $A$ and $B$ to integers, then a solution does not always exist, as I have shown. – 5xum Mar 18 '21 at 07:42
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@Masoud Is anything in your answer left unclear? If so, please comment and tell me what is still missing. If everything is clear, you should accept the answer – 5xum Mar 18 '21 at 11:09
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converting fraction ${1}/{(x+y)}$ to something like $f(x)g(y)$ ($f$ and $g$ are arbitrary functions) that doesn't contain factors $(x\pm y)$. – Masa Mar 20 '21 at 08:24
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@Masoud That is nowhere near what you originally asked. This is like going to an ice cream shop, asking for a loaf of bread, and then being dissapointed because you didn't get steak. – 5xum Mar 20 '21 at 19:33
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I accept that my question was asked ambiguously, so I edited it to better explain my meaning. – Masa Mar 21 '21 at 07:24
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@Masoud Completely rewriting your question like you just did is not the best idea, for two reasons. One, your edited question will not appear high up on the list of new questions and will much more likely be missed by users who regularly answer new question (also, those users may notice the answer already has an answer, and move on). Two, it is not fair to the users who already (correctly) answered your original question. For this case, I will add a new answer to answer your new question, but in future, I sugges that in cases like this, you close the first question (either by [cont] – 5xum Mar 22 '21 at 09:48
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deleting it or by accepting an answer) and then post a new question. Note that there's nothing wrong with refining a question, but full on rewrites like the one you did are best served in a new question. – 5xum Mar 22 '21 at 09:48
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No, this is not possible.
For one, you have a problem when taking $x,y$ such that $x+y=0$, since in that case, $\frac{1}{x+y}$ is not defined, while $f(x)\cdot g(y)$ is a well defined number.
But let's avoid the region around $0$. Say, for example, that $x$ and $y$ are both positive. That means $x+y>0$ and you are in the clear.
Fix $y=1$ and observe your expression. It becomes $$\frac1{x+1}=f(x)g(1)$$ which means that $f(x)=\frac{1}{g(1)(x+1)}$ for all values of $x$. Similarly, you can show that $g(y)=\frac{1}{f(1)(y+1)}$ for all values of $y$. Put the two expressions together, and you get $$f(x)g(y)=\frac{1}{(x+1)(y+1)f(1)g(1)} = \frac{1}{x+y}$$ and this rewrites to $$f(1)g(1)=\frac{x+y}{(x+1)(y+1)}.$$ The right hand side of the equation is a non-constant function (just take $(1,1)$ and $(2,2)$ as examples of input values that do not return the same value), while the left hand side is a constant, which leads to a contradiction.
5xum
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