Numerical PDE's

tiny-tim
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plug P_n = ((-c)^n).sin((n+1)∆)/sin∆ (and similarly for P_n-1 and P_n-2) into P_n = -2c.cos∆P_n-1 - c^2 P_n-2.

DOne. Doing that gives me:
sin((n+1)∆) + sin((n-1)∆) = 2.cos∆.sin(n∆),

And i got this after substituting Pn and Pn-1 and Pn-2. not through standard trig. Is this correct? So once I've shown this what does this mean? how does this verify that these are the e-values? Then I have to show that the e-vectors are as stated in the first post

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tiny-tim
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And i got this after substituting Pn and Pn-1 and Pn-2. not through standard trig. Is this correct?
Yes, you got it from substitution. Then you still have to use standard trig to check that that equation is right!

If you put any old P_n in, you'd still get some trig equation … but it would be rubbish! Only a valid P_n will give a valid trig equation!
how does this verify that these are the e-values?
I've been assuming that you've done, in class, solution of eigengvalues λ of a matrix M by "characteristic equation" … that is by the equation M - λ.I = 0.

Have you?

yeah

So recap, I found a general characteristic equation for this nxn symmetric -tridiagonal matrix.

we assumed that lamdba = b+2c.cos(theta) is an e-value but I need to prove this and I dont see how what we did proved this at all.

we stuck lambda back in P_n and it gave us a solution to the general characteristic equation i dont know what this represents.

One could have just avoided this method completely and solved
AX=Lambda*B but since I'm this far I might as well finish it off

Eventually we have

sin((n+1)∆) + sin((n-1)∆) = 2.cos∆.sin(n∆) now I need to verify this with standard trig. How do I do this?
Then I stil have to find the e-vectors.
Was there not a more concise way of doing this problem?

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

Thanks for the help tim.

you can close this thread

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
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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.)

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
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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∆).

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

tiny-tim
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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.

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
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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.)

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

tiny-tim
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....
we defined ∆ by λ = 2c cos∆

But we don't know λ !

λ does not equal 2c cos∆

tiny-tim
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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∆.