Not exactly, it is known that the series \sum \frac{1}{n} is divergent, while \sum \frac{(-1)^n}{n} is convergent.
Both have terms which converges to zero, with the same "rate".
However, the last one is, of course, not absolutely convergent.
OK, I guess I should add some hypothesis on f and change the conclusion in order to have something that could be true:
Here is the new problem :
Let \epsilon>0.
Suppose that f is in C^\infty(U, [0,1]) where U is an open of R^n and suppose that for any x_0 in \partial U (the boundary...
My question is simple :
Suppose that f is in C^\infty(U, [0,1]) where U is an open of R^n .
Is there g in C^\infty(R^n,[0,1]) such that f=g on U ?
I would say yes, but I don't know how to prove it.
Thanks
Indeed, we even have H(x,y,1) singular for any x,y !
Thank you !
I wonder if the theorem is true in dimension 2...
(it is trivially true in dimension 1).
Ok I agree. The spectrum indeed contains zero.
About terminology, Kato in its standard book (Pertubation theory of linear operators) uses the term "invertible" for injective and always precises whether or not the inverse operator is bounded.
Here is the problem:
Suppose that g is a diffeomorphism on R^n. Then we know that its jacobian matrix is everywhere invertible.
Let us define the following matrix valued function on R^n
H_{i,j} (x) = \int_0^1 \partial_i g^j(tx) dt
where g^j are the components of g.
Question : Is...
Yes, but the claim is false : There ARE invertible compact operators in infinite dimensional Banach spaces, but the inverse has to be unbounded.
In other words, the spectrum of any compact operator in an infinite dimensional space always accumulate at 0 but it does not contain 0 in general.
For example, you have :
f_1^{'}(0)=f_0(-1)=1
f_0^{'}(0)=2
so this is not smooth (not even C1) in 0.
And therefore you have no smoothness in 1, in 2 etc..
Indeed, you have :
f_2^{''}(1)=f_1^{'}(0)=f_0(-1)=1
f_1^{''}(1)=f_0^{'}(0)=2
But If you find a smooth function f_0 on [-1,0)...
Thanks saltydog, but I am not sure that
f would be smooth on R. (or on [0, +infinity) ) .
And my question is : is there a real function which is not everywhere zero,
smooth on R such that :
f '(x) = f(x-1) on R
I've been searching for an answer on the web but the only things I've found...
I didn't even know that these equations had a name. With that I think I'll find some information, thanks.
I tried to use a power serie expansion ot see what kind of coefficients one would have, but the relations are not quite simple, and I was unable to prove any existence.
As for drawing...
I think there is an misunderstanding. I know what f(x+a) means...
I know that if a>0 then x+a > x ... (Was it a joke? :bugeye: )
Maybe was I not clear : I'd like to know the solutions of this differential equation :
df/dx(x) = f (x-1) (for example) .
And if there are no other solution but...
Hi,
What do you think about the possible (nontrivial) solutions of this equation ?
(1) f '(x) = f(x-1) (f is a smooth function R -> R)
More generally, what do we know about these ?
(2) f '(x) = f(x+a) (a fixed in R)
(3) f '(x) = f(g(x)) (g smooth diffeomorphism R -> R )