Proving Trigonometric E-Values for a Symmetric Tridiagonal Matrix

In summary: Do not...)In summary, the matrix has e-values lambda_i = b +2cos((i * pi)/(N+1)), and e-vectors x_i = [sin ((i* pi)/(N+1), sin ((2*i*pi)/(N+1)), ..., sin((N*i*pi)/(N+1))].
  • #36
Well in this case we're not given [tex]\lambda [/tex] so what now?
 
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  • #37
Nusc said:
Well in this case we're not given [tex]\lambda [/tex] so what now?

Well, remember that the whole point was that the characteristic equation is Pn = 0.

So we just choose A and B and ∆ so that Pn = 0, and so that the the values for P1 and P2 (which are esy to find directly) are correct. :smile:
 
  • #38
tiny-tim said:
Hi Nusc! :smile:

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;

ARe you suggesting here that [tex]P_n= P_{n-1}= P_{n-2}[/tex] ? Why is that true?
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∆). :smile:
We don't know ∆
 
  • #39
Nusc said:
ARe you suggesting here that [tex]P_n= P_{n-1}= P_{n-2}[/tex] ? Why is that true?

No … P_N is the characteristic determinant of the matrix in the question.

[tex]P_n= P_{n-1}= P_{n-2}[/tex] is an equation which we use for calculating P_N. We do so because P_N is far too complicated to calculate directly.

So we build it up, using that equation, until we get to n = N.

Once we have P_N, we put P_N = 0 … because that's what the characteristic determinant is for!

But we don't put any other P_n = 0 … if we did, it would give the wrong result.

To put it another way … we have been looking for what we have been calling λ.

It would probably have made you happier if we'd called it λN.

Solving PN = 0 gives us values for λN.

Solving Pn = 0 for any other value of n gives us values for λn, not λN. :smile:
We don't know ∆

We find ∆ (or ∆N, if you prefer) from PN = 0.

Then we find λ from ∆. :smile:

(It is easier to find ∆ first, then λ.)
 
  • #40
Sorry, what does ∆ supposed to represent in this case?
 
  • #41
...
tiny-tim said:
we defined ∆ by λ = 2c cos∆
 
  • #42
But we don't know λ !

λ does not equal 2c cos∆
 
  • #43
Nusc said:
But we don't know λ !

Well of course we don't know λ …

λ is what the question asks us to find, isn't it? :confused:
λ does not equal 2c cos∆

Yes it does …

We defined the substitution b - λ = -2.c.cos∆, to make the equations easier, and you've chosen b = 0 (in your post #34), so λ = 2c cos∆. :smile:
 

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