This is what I have for 1.
We have to prove that:
Therefore is a linear map.
This problem provides a linear algebra method for computing sums of the form with positive integer.
1. Let the real vector space of all polynomials of degree . Consider the map defined by . Prove that is a linear map.
2. Prove that is constant.
3. Find the image by of the elements of the canonical basis of and determine .
4. Using 3. , find an expression for .
2. suppose T(p(x)) = 0, with . then
since this is by supposition the 0-polynomial, we have immediately that:
hence , a constant polynomial. the reverse implication is trivial.
3. (see above).
(just take my word for it,ok?).
(think: truncated pascal's triangle).
so what is the pre-image of ? this amounts to solving:
as we have seen T annihilates constants, so:
(short break for some etoufee for dinner)
ok, back. dinner was tasty!
then, as we have seen, .
this means that for all real x. so, in particular:
which is the desired formula for
(i'd like to add that this was a very entertaining problem, and that (in theory) Sn could be calculated from this method without much trouble, as the difficulty involved grows linearly with n. we still wind up with a upper triangular matrix for T, of rank n-1, so the resulting system of linear equations we get are easy to solve, although for large n, the repetition could get mind-numbing).
(Off-topic) , there is a way to solve the general case using derivatives
From , we have .
with independent term because .
Apply the -th derivative on the result is
we have then
We gonna find the other coefficients . We take the -th derivative on , with , so
, taking we have
Using and applying again the -th derivative
and this should not be surprising, because:
1) the derivative is just a difference quotient. in fact, you don't need derivatives, you could do an induction argument on successive differences.
2) the rref of the matrix for T, and the differential operator D, are the same, so this is essentially the "same process".
3)pascal's triangle rulez (binomial coefficients are great with a little cream and sugar. oh wait, that's biscotti. my bad).