Find a polynomial P_{n} \in C[X] such as S_{n}=ker P_{n} (D) With S_{n}(n \geq 1)

sunshine022uv

sunshine022uv

Answered question

2022-01-14

Find a polynomial PnC[X] such as Sn=kerPn(D)
With Sn(n1) all of the functions such as y(n)=y (where y(n) is the nth derivative of y)
D the endomorphism sending the functions C on their derivatives D:yy
I already proved that D is an endomorphism and Sn a vector space containing the functions kex.
But my probleme is that i dont really understand the meaning of Pn(D), and so dont see what Sn=ker Pn(D) could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything.

Answer & Explanation

Paineow

Paineow

Beginner2022-01-15Added 30 answers

Let Pn=Xn1
Saying xker(Pn(D)) is precisely saying Pn(D)(x)=0.
Then, we have Dn(x)Id(x)=x(x)x=0, which exactly x(x)=x.
Also, Pn(D) is a polynomial evaluated at an endomorphism.
If f is an endomorphism of a K-vector space E and P=anXn++a1X+a0K[X] a polynomial, then P(f) is the endomorphism anfn++a1f+a0 where fi is f composed with itself i times.
Laura Worden

Laura Worden

Beginner2022-01-16Added 45 answers

Let me just try to improve the formulation of the question, which is quite confused. Before talking about endomorphisms, one should state which vector space one is considering; apparently this is the space E=C(R) of smooth functions on R (with values in the field we are working over, which I presume is C or maybe R).
Now the operation D of differentiation is an endomorphism of E (because it is a linear operation, and also because differentiating a C function results in another C function; note that the latter would not have worked for any finite differentiability class). Composition is a (bilinear) operation on endomorphisms, so DD (differentiation twice) is another endomorphism, and so is the k-fold composition Dk for any kN; the latter is written Dk:ff(k) on the level of individual vectors (C-functions) fE. (For k=0 one has f(0)=f.)
Since the set of endomorphism is also a vector space, we can form (finite!) linear combinations of these Dk for different values of k with coefficients ck to get a quite general d-th order differential operator k=0dckDk that maps fk=0dckf(k). It is natural to consider this in relation to the polynomial P=k=0dckXk, namely as the result of substituting D for X into P. I will then write k=0dckDk=P[X=D], which (since X is the only candidate to substitute for) can be shortened to P[D] (but I refuse to write this as P(D) as almost everybody does, since that would suggest that P is used as a function, which it is not). Finally, like for any endomorphism, one can form ker(P[D]) to denote the subspace {fEP[D](f)=0}={fEk=0dckf(k)=0} of vectors that are set to the zero vector by P[D].
Now the question becomes for which polynomial P this kernel gives the set of solutions to the differential equation f(n)=f. It should be immediately obvious that P=Xn1 is the unique polynomial that satisfies the requirement.

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