- #36

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- #36

- #37

- #38

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He wrote that the sum of ##c_i## for ##i## from 1 to ##j## is zero, for all values of ##j## from ##1## to ##n##. That means that ##c_1 = 0## (since that's the only term in the sum for ##j = 1##), and then, by induction, all of the other ##c_i## for ##i## from ##1## to ##n## must also be zero (because each time we increase ##j## by ##1## we include just one additional term, and all the previous terms are already known to be ##0## from the previous values of ##j##).He did ***not** write ##c_i = 0, i \in \{1, \dots, j\}##. Rather it was thesumof the ##c_i##.

That might not be what he intended to write, but that's what he wrote. If he intended to write something else, he should tell us what.

- #39

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Aa sample element of ##S## would look like:

$$

p(t) = c_0 + c_1t +c_2t^2 + \cdots + c_nt^n

$$

So if you are trying to find a basis for ##S##, you should be thinking in terms of picking ##n## sets of

The fact that each member of ##S## can be described by a set of ##n## numbers (the ##c_i##) is what suggests the "vector" terminology. (Strictly speaking, to show that these sets of numbers actually are vectors requires proving that they satisfy the vector space axioms.)

- #40

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Even if you pick the "right" constraint on the coefficients (where they don't all draw from the same coefficients), setting all of them equal to zero still satisfies the criteria, so no, the set A is not guaranteed to be a basis without additional constraints.

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