Is y'' = y solvable by using a homomorphism and understanding its kernel?

[DanielThrice]

Let G denote the set
G = {f : R → R | f is infinitely differentiable at every point x ∈ R}.

Prove that G is a group under addition. Is G a group under multiplication? Why or why
not?

I have proved this after much trouble, using the axioms of group theory, and I think I understand the function quite well, but I'm confused on the definition of homomorphism and more so with kernel. Assuming ϕ : G → G defined by ϕ(f) = f ' , how can I prove that ϕ is a homomorphism with respect to the group's operation of addition.

And in this case what can I say about it's kernel?
Thank you I know this is the basics

 

[jbunniii]

Well, the question is not entirely group theoretic. You need to know some basic theorems from calculus as well.

In particular, you need this result: If f and g are differentiable, then (f+g)' = f' + g'. This is pretty easy to prove directly using the derivative definition.

The kernel of \phi will be the set of all functions in G whose derivatives are zero. Again, you need some basic calculus for this. Certainly all constant functions have zero derivative. Can there be any others?

 

[DanielThrice]

Of course not, the constants are the only functions with no change in how they change over time, that's why they are constants.

But how do I know to look for the zero derivatives?

 

[DanielThrice]

Nevermind, lol, that's what a homomorphism is.

I said the kernel is just all of the constants:
f ' (x) = 0
(Integrate)
f (x) = C

How about if we have a different function, % : G to G defined by % (f) = f '' - f ?
Homomorphism?
(f + g) '' + (f + g) = f '' + g '' + f + g. Well any functions in the reals added to other functions of the reals gets you another function of the real, so % is a homomorphism.
Kernel:
f '' (x) + f(x) = 0
f (x) = -f '' (x) so the kernel is the set of all negative second derivatives? Since functions are infinite can we just say that every function is the second derivative of something, thus the kernel is just f (x)?

 

[jbunniii]

How about if we have a different function, % : G to G defined by % (f) = f '' - f ?
Homomorphism?
(f + g) '' + (f + g) = f '' + g '' + f + g. Well any functions in the reals added to other functions of the reals gets you another function of the real, so % is a homomorphism.
Kernel:
f '' (x) + f(x) = 0
f (x) = -f '' (x) so the kernel is the set of all negative second derivatives? Since functions are infinite can we just say that every function is the second derivative of something, thus the kernel is just f (x)?

You are looking for specific functions that satisfy the differential equation f''(x) = -f(x). Observe that this equation is true for f(x) = cos(x), and for f(x) = sin(x). Are there others? The kernel consists of ALL functions in G that satisfy this equation.

 

[DanielThrice]

So the kernel would just be cos, sin, and e^x correct?

 

[DanielThrice]

Is there an easy way to prove that these are the only three?

 

[Fredrik]

The exponential function isn't a solution to f''+f=0. And any linear combination of two solutions is a solution.

 

[DanielThrice]

Nevermind, lol, that's what a homomorphism is.

I said the kernel is just all of the constants:
f ' (x) = 0
(Integrate)
f (x) = C

How about if we have a different function, % : G to G defined by % (f) = f '' - f ?
Homomorphism?
(f + g) '' + (f + g) = f '' + g '' + f + g. Well any functions in the reals added to other functions of the reals gets you another function of the real, so % is a homomorphism.
Kernel:
f '' (x) + f(x) = 0
f (x) = -f '' (x) so the kernel is the set of all negative second derivatives? Since functions are infinite can we just say that every function is the second derivative of something, thus the kernel is just f (x)?

Wait did I mess myself up...the original equation is % (f) = f '' - f so the homomorphism doesn't change that much, it just becomes (f + g) '' - (f + g) = f '' + g '' - f - g. Well any functions in the reals added to other functions of the reals gets you another function of the real, so % is a homomorphism.
But the kernel would be where f '' (x) = f (x), so we have the identity element and e^x right?

 

[Fredrik]

You defined % by %(f)=f''-f. This implies %(f+g)=(f+g)''-(f+g)=(f''-f)+(g''-g)=%(f)+%(g). That's the most straightforward way to verify that it's a homomorphism. The kernel is the set of all f such that 0=%(f)=f''-f. So yes, f=0 and f=exp are solutions, but they're not the only ones.

 

[snipez90]

Yes this is a straight up analysis problem, and the OP seems to have sorted out the algebraic details in an older thread. Figuring out what type of math you're really dealing with will keep you from getting lost.

If you're interested in finishing the problem, i.e., finding the solutions to y'' = y, my advice is to assume this and differentiate y^2 - (y')^2 and go from there.