Sequences where \sum_{n=k}^{\infty} a_n=\sum_{n=k}^{\infty}a^2_n I was recently looking at the series

Shamar Padilla

Shamar Padilla

Answered question

2022-01-26

Sequences where n=kan=n=kan2
I was recently looking at the series n=1=sinnn, for which the value quite cleanly comes out to be 12π1, which is a rather cool closed form.

Answer & Explanation

trnovitom06

trnovitom06

Beginner2022-01-27Added 12 answers

Notice that a similar property holds for the related integral
0sinxxdx=0(sinxx)2dx=π2
This makes me wonder if such sequence has to do something with the property of orthogonality. Two functions f and g are said to be orthogonal with weight 1
0f(x)g(x)=0
If we impose the condition that g(x)=1−f(x) then we are looking at a special case of orthogonality where
0f(x)(1f(x))=0
which is analogous to question since n=kan(1an)=0
bemolizisqt

bemolizisqt

Beginner2022-01-28Added 16 answers

This is one way of looking at these sequences. As your starting k might not be 0, we're inside a linear subspace of l2, the space of sequences (a0,a1,) whose sum {n=0}|an|2 converges. As we also need the sum {n=0}an to converge, that places us inside a somewhat smaller linear subspace V of l2
In there, we're looking for sequences a=(a0,a1,) such that f(a)=0, where f(a)={n=k}an{n=k}an2. This is a smooth function on V, and its differential at a in the direction of b is
xdbf(a)=n=kbn2n=kanbn
If a is the zero sequence in V (that is, any sequence a with an=0 for nk), then df(a)=0. If a is not the zero sequence, then there exists a b such that dbf(a)0, take for example b=12a.
Thus f is a smooth function whose differential is nondegenerate away from the linear subspace N of V defined by the zero sequences in V, and degenerate on that subspace. This means that the set X={aVf(a)=0} of sequences that satisfy your identity is an infinite-dimensional hypersurface that is nonsingular outside of N, and has singularities on that set.
there is also a smooth projection function p:VNXN that is, a smooth function p such that p(a)=a for any sequence aX.

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