Proving Trigonometric E-Values for a Symmetric Tridiagonal Matrix

[Nusc]

Actually I finished the problem using a different method and it's a lot more elegant that the crap i just went through.

Thanks for the help tim.

you can close this thread

 

[Nusc]

Hi Nusc!

Hint: if the characteristic equation of the n x n matrix is Pn = 0, find a recurrence relation expressing the polynomial Pn in terms of Pn-1 and Pn-2.

Then either solve that recurrence relation by normal methods, or (since the question gives us the solution) define ∆ by: lambda = b + 2c.cos∆.

You should find that Pn = cos(n∆).

And then … ?

What did you mean by normal methods?

 

[tiny-tim]

Hi Nusc!

Goodness, that was a long time ago!

I meant that if a sequence {An} has the recurrence relation

C0An + C1An+1 + … + CkAn+k = 0,

then you pretend that it's a polynomial

C0 + C1x + … + Ckx^k = 0,

factor it into (x - P)(x - Q)…(x - Z),

and then the solutions are linear combinations of {An = P^n} and {An = Q^n} … and {An = Z^n},

except that for repeated roots (x - P)^2, there's an extra {An = n*P^n},

(x - P)^3, there's an extra {An = n^2*P^n}, and so on.

(In other words, recurrence relations are a lot like polynomial differential equations.)

 

[Nusc]

Let's say we had an nxn matrix: ( I didn't draw the l dots but assume its nxn)

$$\left[\begin{array}{cccc}0 & -c & 0 & 0 \\ -c & 0 & -c & 0 \\ 0 & -c & 0 & 0\end{array}\right]$$

We know previously that
$$P_n = -\lambda P_{n-1} - c^2 P_{n-2}$$

How do I solve for lambda?

 

[tiny-tim]

We know previously that
$$P_n = -\lambda P_{n-1} - c^2 P_{n-2}$$

How do I solve for lambda?

Hi Nusc!

We rewrite it $$P_n\,+\,\lambda P_{n-1}\\,+\,c^2 P_{n-2}\,=\,0$$

so we see the roots are (-λ ±√(λ² - 4c²))/2;

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]

Well in this case we're not given $$\lambda$$ so what now?

 

[tiny-tim]

Well in this case we're not given $$\lambda$$ 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.

 

[Nusc]

Hi Nusc!

We rewrite it $$P_n\,+\,\lambda P_{n-1}\\,+\,c^2 P_{n-2}\,=\,0$$

so we see the roots are (-λ ±√(λ² - 4c²))/2;

ARe you suggesting here that $$P_n= P_{n-1}= P_{n-2}$$ ? Why is that true?

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∆).

We don't know ∆

 

[tiny-tim]

ARe you suggesting here that $$P_n= P_{n-1}= P_{n-2}$$ ? Why is that true?

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

$$P_n= P_{n-1}= P_{n-2}$$ 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.

We don't know ∆

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

Then we find λ from ∆.

(It is easier to find ∆ first, then λ.)

 

[Nusc]

Sorry, what does ∆ supposed to represent in this case?

 

[tiny-tim]

...

we defined ∆ by λ = 2c cos∆

 

[Nusc]

But we don't know λ !

λ does not equal 2c cos∆

 

[tiny-tim]

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?

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∆.