# Homework Help: Eigenfunction Do I just plug in the f(x) equation into T(f)?

1. Nov 5, 2012

### pyroknife

The problem is attached.

I am not quite sure how to do the 3 parts.

for part a:

Do I just plug in the f(x) equation into T(f)?
If so that gives me k2ekx-2*kekx-3ekx
I'm just not sure what I'm supposed to be doing.

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2. Nov 5, 2012

### Muphrid

Re: Eigenfunction

What does it mean to be an eigenfunction? If you apply a linear differential operator to one of its eigenfunctions, what should the result be?

3. Nov 5, 2012

### Mute

Re: Eigenfunction

The question defines an operator, T, which acts on functions f(x), which have two derivatives. If you apply the operator T to some function g(x) and you find that $Tg(x) = \lambda g(x)$ for some number $\lambda$, then g(x) would be called an eigenfunction of the operator T, because the operator acts on g(x) and gives back g(x) times a constant.

The question tells you to apply the operator T to the function f(x) = exp(kx), and asks you to show that it is an eigenfunction. You applied T to the function f(x) and got the result that you listed in your post; now you just have to manipulate that expression into the form $\mbox{(some constant that depends on}~k) \times f(x)$.

4. Nov 5, 2012

### klawlor419

Re: Eigenfunction

All eigenvalue problems have the form T f(x)= t f(x), where t is a constant. If you have an operator and it acts on a function in this way, you have an eigenfunction and eigenvalue.

So for a.) you have to show that the eigenvalue exists for all values of k in the eigenfunction.

Part b.) kind of follows from part a.) at least I think, unless I'm missing something.

Part c.) involves some guessing. Find a function that satisfies that equation.

5. Nov 5, 2012

### pyroknife

Re: Eigenfunction

Thanks so for k2ekx-2*kekx-3ekx
this would be ekx(k^2-2k-3)=ekx(k-3)(k+1)
But hwo does that "show for each value of k, the function f(x) is an eigenfunction for T?"

6. Nov 5, 2012

### klawlor419

Re: Eigenfunction

So your eigenvalue for that function is (k-3)(k+1). For any value of k this is a well-defined function for the eigenvalue. Thus e^kx is an eigenfunction for all k.

7. Nov 5, 2012

### pyroknife

Re: Eigenfunction

I'm not quite sure if i understand why this means for all values of k, this is a well defined function for the eigenvalue.

8. Nov 5, 2012

### klawlor419

Re: Eigenfunction

So I labeled my eigenvalue in the first post as t, a number. You found t(k)=(k-3)(k+1). This function is well-defined for all k.

Lets say that you found t(k)=(k-3)/(k+1). At k=-1 you have t=-4/0 which is infinite or undefined.

The eigenvalue you found has no problems in its range like this. So it is well-defined for all values of k.

9. Nov 5, 2012

### pyroknife

Re: Eigenfunction

Oh I see I think I understand this, thanks. I'll take a look at the other parts now.

10. Nov 7, 2012

### pyroknife

Re: Eigenfunction

So for part b)
Since e^kx(k^2-2k-3)=e^kx(k-3)(k+1)

Does that mean the eigen values are just k=3 and k=-1?

11. Nov 7, 2012

### vela

Staff Emeritus
Re: Eigenfunction

No. You need to learn what an eigenvalue and an eigenfunction are instead of just guessing. What does your book say?

12. Nov 7, 2012

### pyroknife

Re: Eigenfunction

There's one section that says:
Let V be a vector space
and T:V −→ V be a linear operator. A number λ is an eigenvalue of T provided
that there is a nonzero vector v in V such that T (v) = λv. Every nonzero vector that
satisfies this equation is an eigenvector of T corresponding to the eigenvalue λ.

Nothing is said about eigenfunctions, so I guess I have to apply this definition.

T(v)=e^kx(k^2-2k-3)=e^kx(k-3)(k+1)=λe^kx
Canceling out e^kx
gives λ=(k-3)(K+1)

Doesn't that mean the eigenvalues are λ=3, -1?

13. Nov 7, 2012

### vela

Staff Emeritus
Re: Eigenfunction

No, why would it? How are you going from $\lambda = (k-3)(k+1)$ to $\lambda=3, -1$?

Last edited: Nov 7, 2012
14. Nov 7, 2012

### pyroknife

Re: Eigenfunction

hmmm, I wish setting lambda=0, which is incorrect.
Hmmm λ=(k-3)(k+1)=k^2-2k-3

Does this mean lamba is all real numbers?

So when it says Find the corresponding eigenvalues for each
eigenfunction f (x) = e^kx.

Basically lamda is all real values?

15. Nov 7, 2012

### vela

Staff Emeritus
Re: Eigenfunction

Remember k represents a specific value. For each possible value of k, ekx is an eigenfunction of T. For example, if k=2, then you have f(x)=e2x, and the corresponding eigenvalue is $\lambda = -3$. Your answer to (b) is what you wrote: λ=(k-3)(k+1).

If you want to know what the possible eigenvalues of T are, think about what the allowed values of k are and what values of $\lambda$ these correspond to. Hint: If you plotted $\lambda$ vs. k, what kind of curve would you get?

Last edited: Nov 7, 2012
16. Nov 7, 2012

### pyroknife

Re: Eigenfunction

Did you mean lambda=-3?

I would be a parabolic curve. k can be all real #s as a previous user stated. Lambda would be all real #s. But I am unsure how I would describe lambda for each eigenfunction? Lambda is a parabolic function of all real k values.

17. Nov 7, 2012

### vela

Staff Emeritus
Re: Eigenfunction

Oops, yes.

I edited my previous post, and it probably answers your question now.

18. Nov 8, 2012

### pyroknife

Re: Eigenfunction

Thanks.

I guess this λ=(k-3)(k+1)=k^2-2k-3
Pretty much answers part a and b.

Since it's saying that for all k, the function f(x)=e^(kx) is an eigenfunction because k can be any real #.

For part b, lambda=k^2-2k-3

for part c, is that just a differential equation? (First time seeing a diff eq in linear algebra)

If so then the characteristic equation for it is r^2-2r-3=0
(r-3)(r+1)=0
r=-1, 3

so the 2 functions are
y(t)=Ae^(-t)
G(t)=Be^(3t)
Where A and B are constants.

19. Nov 8, 2012

### vela

Staff Emeritus
Re: Eigenfunction

You can find the answer to (c) that way, but it's probably not what was intended. How might you answer (c) based on the other parts of the problem?