Is there any way to prove that:
[math] \frac{1}{x+y+z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} [/math] ?
fuck me, I meant:
Is there any way to prove that:
[math] \frac{1}{x+y+z}=\frac{A}{x}+\frac{B}{y}+\frac{C}{z} [/math] ? For any A, B and C
true for x=y=z=0
>>8133841
>Don't you mean for at least one A,B and C?
I think that's what he meant too.
>>8133841
Yeah, for at least one, my mistake
Ineed the result too look like
[math] \frac{1}{x+y+z}=\frac{A}{x}+\frac{B}{y}+\frac{C}{z} [/math]
or
[math] \frac{1}{x+y+z}=\frac{A}{x+y}+\frac{B}{z}[/math]
I remember there was a method of simplifying fractions for some tougher integrals but here x,y and z are different forms of energy and this method doesnt work.
Maybe theres some math trick which I forgot/never knew ? Ah A,B and C cant be 0 of course (same goes for x,y and z).
>>8133873
well it can't work x,y or z are 0 for example
what you are thinking of are partial fractions, but they are not what you want
[math]x=-y[/math] or [math]x=-z[/math]
>>8133892
>x=−y or x=−z
I hope you're trolling and not actually this retarded.
>>8133896
Why? It's a valid solution.
>>8133873
>I remember there was a method of simplifying fractions for some tougher integrals
Won't work here if you're reffering to what I think you are.
I suggest either you fix B and C as 1 and try to prove there exists at least one A, or that you triple integrate 1/(x + y + z) and see where you can go from there
>>8133838
Look up proof for "separation by partial fractions" where the concept is used for finding integrals.
>>8133835
Are you asking for a solution that is valid for every possible triples x,y,z? (Excluding any two of them equaling zero)
If so, then there's no solution. Is one of x,y, or z fixed?
I really don't understand your question.
>>8133918
We've all taken our Calculus here, isn't "separation by partial fractions" only for a single variable?
Because if not, then I got some shit to review this summer
>>8133950
it doesn't exist.
just plug in some values for x,y and z - this would give you linear equations for A,B and C. Since you can plug more than 3 combinations of values for x,y z you will most likely get an overdetermined system of equations that will have no solutions.