"Show that $p(t) = p_0 + p_1t + p_2t^2 + p_3t^3$ , where $t \in [a,b]$ and the coefficients $p_k$ are real scalars, forms a vector space. What is the dimension of this vector space? Explain"
How is it closed under addition? If you have $t_1$ and $t_2$, $p(t_1) + p(t_2) = 2p_0 + (t_1 + t_2) p_1 + ({t_1}^2 + {t_2}^2) p_2 + ({t_1}^3 + {t_2}^3) p_3$.
I apologize in advance for the terrible formatting. I'm new here, and I don't know how to format mathematically
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You're not meant to add two values of the same polynomial $p(t)$ and different times $t_1$ and $t_2$. The result would be a polynomial over those two indeterminates, which is not in the original vector space, as you say.
You're meant to add two polynomials having the same indeterminate $t$, let's say $p(t)=p_0+p_1t+p_2t^2+p_3t^3$ and $q(t)=q_0+q_1t+q_2t^2+q_3t^3$.
You are not supposed to take two different values for $t$, you are supposed to take two cubic polynomials in $t$, add them, and note that you get another cubic polynomial in $t$. So you have $p(t)=p_0+p_1t+p_2t^2+p_3t^3$ and $q(t)=q_0+q_1t+q_2t^2+q_3t^3$. What is $p(t)+q(t)?$ Is it a cubic polynomial in $t$?