I want to rewrite this statement:
$$f(x)=\frac{x}{-x^2-x+1}$$ into this statement $$f(x)=\frac{\frac{1}{\sqrt5}}{1-(x \cdot \phi)}+\frac{-\frac{1}{\sqrt5}}{1-(x \cdot (1-\phi))}$$
provided $\phi = \frac{1+\sqrt5}{2}$
I tried for 2 days and still can't. I know how to do partial fraction decomposition but at the end I ended up with:
$$\frac{-\frac{1}{\sqrt5}\phi}{(x+\phi)} + \frac{\frac{-1+\sqrt5}{2\sqrt5}}{(-x+\frac{-1+\sqrt5}{2})}$$
which is quite far from the goal.
can anyone solve this?
please give me very detail and step by step demonstration. I'm newbie.
$$\frac x{-x^2-x+1}=\frac A{1-x\phi}+\frac B{1-x(1-\phi)}=\frac{A-xA+x\phi A+B-x\phi B}{(1-x(1-\phi))(1-x\phi)}$$ $$(1-x(1-\phi))(1-x\phi)=-x^2-x+1$$
Hence $$\phi A-A-\phi B=1$$ $$A+B=0$$
So that $$A=\frac 1{\sqrt5}$$ and $$B=-\frac{1}{\sqrt5}$$
From where you were:
$$\frac{-\frac{1}{\sqrt5}\phi}{(x+\phi)} + \frac{\frac{-1+\sqrt5}{2\sqrt5}}{(-x+\frac{-1+\sqrt5}{2})}=\frac{-\frac{1}{\sqrt5}\phi}{(x+\phi)} + \frac{\frac{1}{\sqrt5}(\phi-1)}{-x+(\phi-1)}$$
And keeping in mind that $1/\phi=\phi-1$ we get $$\frac{-\frac{1}{\sqrt5}}{(\frac x\phi+1)} + \frac{\frac{1}{\sqrt5}}{-\frac x{\phi-1}+1}=\frac{\frac{1}{\sqrt5}}{1-x \phi}+\frac{-\frac{1}{\sqrt5}}{1-x (1-\phi)}$$