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About basis of vector space of polynomial degree less than n

Mathematics Asked on December 25, 2021

Let $r in mathbb{R}$ be an arbitrary real number. Show that $beta=left{1, x-r,(x-r)^{2},(x-r)^{3}, ldots,(x-r)^{n}right}$ is a basis of $P_{n}(mathbb{R})$.

I know that if I can show this vectors are linearly independent then this constitute basis of $P_{n}(mathbb{R})$ since dimension is $n+1$ for this space. So only need to show this are linearly independent.

I set linear combination to be 0. Then try to differntiate linear combination to get value of coeeficient .But I am not getting my result.

2 Answers

For any $p in P_n(Bbb{R})$ by induction we get $$p(t) = sum_{k=0}^n frac{p^{(k)}(r)}{k!}(x-r)^k$$ so the set $beta$ spans $P_n(Bbb{R})$. Since $|beta| = n+1$, we conclude that it is a basis.

Answered by mechanodroid on December 25, 2021

If $sumlimits_{i=0}^{n} a_i(x-r)^{i}=0$ for all $x$ then $sumlimits_{i=0}^{n} a_ix^{i}=0$ for all $x$ and when a polynomial vanishes all its coefficients also vanish.

Answered by Kavi Rama Murthy on December 25, 2021

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