Do all the arithmetic sequences of length n form a vector space over R? How about Geometric Sequences?
Arithmetic and Geometric Sequences
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0What do you exactly mean by "arithmetic sequence of *length n*"? This is: How do you define the length of an arithmetic sequence, and then how do you define sum and product by a scalar with the set of these sequences? – 2012-10-17
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1@DonAntonio I think it would just be the standard addition and scalar multiplication. I don't think there's any ambiguity in "a sequence of length $n$" for a given $n$. – 2012-10-17
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0I'm not talking of ambiguity but of the definition of length in this case: does it mean a set with n numbers all of which are in arithmetic sequence? – 2012-10-17
2 Answers
I will leave the verification of the algebraic axioms to you. Instead let me focus on the closure axioms. If we have arithmetic sequences $(a_k)$ and $(b_k)$ then by definition we must require $$a_{k+1} - a_k = c$$ $$b_{k+1} - b_k = d$$ for constants $c$ and $d$. If we take the sum of these sequences, so $(a_k + b_k)$, we will have $$(a_{k+1} + b_{k+1}) - (a_k - b_k) = c + d$$ which is indeed an arithmetic sequence. So arithmetic sequences are closed under addition. We can similarly verify that arithmetic sequences are closed under scalar multiplication as well.
Geometric sequences on the other hand, are not closed under addition. Consider $$(1,\ 1,\ 1,\ 1 \cdots ) + (1,\ 2,\ 4,\ 8, \cdots ) = (2,\ 3,\ 5,\ 9, \cdots)$$ You can verify that there will be no $x$ such that $2x^k$ satisfies the sequence.
Edit: As suggested by Robert Mastragostino, if we define vector space addition by element-wise multiplication and scalar multiplication by element-wise exponentiation then the resulting operations are closed for positive geometric sequences. If we have geometric sequences $(a_k) = rx^k$ and $(b_k) = sy^k$ then $$(a_k) + (b_k) = rsx^ky^k = rs(xy)^k$$ which is again a geometric sequence. For scalar multiplication, we have $$c(a_k) = (rx^k)^c = r^c(x^c)^k$$
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0It might be worth noting that replacing addition with element-wise multiplication, and scalar multiplication with element-wise exponentiation *does* make a vector space of geometric series. – 2012-10-17
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0@RobertMastragostino Very nice catch. Thank you. I think the modification of addition is enough. Term-wise multiplication for scalar multiplication would work for geometric sequences as well. – 2012-10-17
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0Nevermind, I see why not now. – 2012-10-17
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2@RobertMastragostino: That would only work if you restrict to geometric series with strictly positive terms (what would be half of $((-1)^i)_{i\in\mathbf N}$?), and in that case, taking logarithms, it is just arithmetic series in disguise. – 2012-10-17
There is another way to make the geometric sequences into a vector space. Let us pick $n=4$, but it works for any $n$. There is a natural mapping between $(s,r)$ as a two component vector and $(s,sr,sr^2,sr^3)$ as a sequence. I will just regard the sequence as a funny name for the vector and define addition and scalar multiplication in terms of the two component vector. So $(1,2,4,8)$ is the name for $(1,2)$ and $(1,3,9,27)$ is the name for $(1,3)$. Then $(1,2,4,8)+(1,3,9,27)=(2,10,50,250)$ and $4(1,2,4,8)=(4,32,256,2048)$