How to decompose and resolve respect to x

  • Thread starter Thread starter Ignition
  • Start date Start date
AI Thread Summary
To decompose and resolve the expression \frac{a}{b\pm x}, it's important to note that it represents two separate fractions: \frac{a}{b+x} and \frac{a}{b-x}. A space is required between "\pm" and "x" to ensure proper LaTeX formatting, as the absence of this space leads to errors. The term "decompose with respect to x" needs clarification, as neither fraction can be further simplified. Understanding these points is crucial for accurate mathematical representation and manipulation.
Ignition
Messages
54
Reaction score
0
How to decompose and resolve respect to x\frac{a}{b\pmx}?
 
Physics news on Phys.org


I would say a/b+x.
 


Ignition said:
How to decompose and resolve respect to x\frac{a}{b\pm x}?
You need a space between "\pm" and "x". The way you have it, LaTex tries to find a command "\pmx" and since there isn't one, it shows nothing.
\frac{a}{b\pm x} means, of course, two values: \frac{a}{b+ x} and \frac{a}{b- x}. Now, what do you mean by "decompose with respect to x"? Neither of those fractions can be simplified more.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top