Eigenfunction Do I just plug in the f(x) equation into T(f)?

[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 k²e^kx-2*ke^kx-3e^kx
I'm just not sure what I'm supposed to be doing.

 

[Muphrid]

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?

 

[Mute]

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 k²e^kx-2*ke^kx-3e^kx
I'm just not sure what I'm supposed to be doing.

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 $\mbox{(some constant that depends on}~k) \times f(x)$.

 

[klawlor419]

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.

 

[pyroknife]

Thanks so for k²e^kx-2*ke^kx-3e^kx
this would be e^kx(k^2-2k-3)=e^kx(k-3)(k+1)
But hwo does that "show for each value of k, the function f(x) is an eigenfunction for T?"

 

[klawlor419]

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.

 

[pyroknife]

For any value of k this is a well-defined function for the eigenvalue. Thus e^kx is an eigenfunction for all k.

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.

 

[klawlor419]

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.

 

[pyroknife]

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

 

[pyroknife]

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?

 

[vela]

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

 

[pyroknife]

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?

 

[vela]

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

 

[pyroknife]

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

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?

 

[vela]

Remember k represents a specific value. For each possible value of k, e^kx is an eigenfunction of T. For example, if k=2, then you have f(x)=e^2x, 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?

 

[pyroknife]

Remember k represents a specific value. For each possible value of k, e^kx is an eigenfunction of T. For example, if k=2, then you have f(x)=e^2x, and the corresponding eigenvalue is $\lambda = 3$.

So what are the allowed values of k and what values of $\lambda$ do these correspond to? Hint: If you plotted $\lambda$ vs. k, what kind of curve would you get?

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.

 

[vela]

Did you mean lambda=-3?

Oops, yes.

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.

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

 

[pyroknife]

Oops, yes.


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

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.

 

[vela]

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?