Proving that a set is a set of generators

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SUMMARY

The discussion focuses on proving that the set <1, x, x^2, ..., x^n> forms a basis for the polynomial space P_n, which includes all polynomial functions up to degree n. The user has already established that this set is linearly independent and is now demonstrating that it serves as a set of generators for P_n. The conclusion is that any polynomial function f in P_n can be expressed as a linear combination of the basis elements, specifically in the form α_0 + α_1x + ... + α_nx^n, confirming that the set generates the space.

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Homework Statement


I want to show that the set
$$
<1,x,x^2,\cdots ,x^n>
$$
forms a basis of the space
$$
P_{n}
$$
where
$$ P_{n} $$ contains all polynomial functions up to fixed degree n.

The Attempt at a Solution


I have already shown that the set
$$
<1,x,x^2,\cdots ,x^n>
$$
is linearly independent and now I want to show that this set is a set of generators for $$ P_{n}.$$

Take any
$$
f\in P_{n}.
$$ Let
$$
<\alpha_{0},...,\alpha_{n}>$$
represent the coefficients of $$ f.$$ Then since
$$
\alpha_{0}\cdot 1=\alpha_{0},...,\alpha_{n}\cdot x^{n}=\alpha_{n}x^{n}
$$
adding these up gives us
$$
\alpha_{0}+\cdots+\alpha_{n}x^{n}=f.
$$

Is that correct or am I missing something? Thanks!
 
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Hi DeadOriginal! :smile:

(use # instead of $ and it won't start a new line every time! :wink:)

Yes, that looks fine, except I think you can shorten it a little:

you can say that by definition, any f in Pn is of the form ##
\alpha_{0}+\cdots+\alpha_{n}x^{n}## :wink:
 
LOL. Thanks for the advice! I will remember it.

Thank you for looking over my work too!
 

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