- #36
Nusc
- 760
- 2
Well in this case we're not given [tex]\lambda [/tex] so what now?
Nusc said:Well in this case we're not given [tex]\lambda [/tex] so what now?
tiny-tim said:Hi Nusc!
We rewrite it [tex]P_n\,+\,\lambda P_{n-1}\\,+\,c^2 P_{n-2}\,=\,0[/tex]
so we see the roots are (-λ ±√(λ² - 4c²))/2;
We don't know ∆tiny-tim said:we defined ∆ by λ = 2c cos∆, so we can rewrite this as:
-c cos∆ ± ic sin ∆, = -c e^{±i∆}.
So the solutions are Pn = A(-c^n)e^{in∆} + B (-c^n)e^{-in∆},
or A(-c^n)cos(n∆) + B (-c^n)sin(n∆).
Nusc said:ARe you suggesting here that [tex]P_n= P_{n-1}= P_{n-2}[/tex] ? Why is that true?
We don't know ∆
tiny-tim said:we defined ∆ by λ = 2c cos∆
Nusc said:But we don't know λ !
λ does not equal 2c cos∆