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I've gotten this question in a book I am reading.

My plan is consistent with the typical methods for uniqueness. I want to assume that there exist some other orthogonal basis of polynomials such that $<R_n, R_m> =0$ when $n \neq m$, and $R_n$ is an nth degree polynomial. However, I'm having trouble getting to the end where I conclude that $R_n = c_nP_n$. Any help is appreciated.

Rellek
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1 Answers1

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$n=0$ is clear. So inductively assume $R_n=c_nP_n$ for $n<N$. Then $P_0,P_1,\ldots,P_N$ span the subspace of polynomials of degree at most $N$ since they are orthogonal. So we can write $R_N=a_0P_0+a_1P_1+\ldots+a_NP_N$. But by assumption and induction: $0=\langle R_N,R_n \rangle=\langle R_N,c_nP_n \rangle$ for $n<N$. Since $P_n$ form an orthonormal basis, $a_n=\langle R_N,P_n \rangle=0$ for $n<N$. So $R_N=a_NP_N$.

ET93
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  • Ok, let me see if I understand. It is already clear that $<R_n, c_mP_m>$ will be zero whenever m is greater than n. By the inductive step, we then know that $<R_n, c_mP_m> = 0$ whenever m is less than n. So, the only term that is left after that will be $c_mP_m$? – Rellek May 14 '15 at 01:19
  • $R_N$ is a polynomial of degree $N$ by assumption so there are no terms with $n>N$. Now, $\langle R_N, R_n \rangle=0$ when $N \neq n$ by assumption. By induction, for $n <N$, $R_n=c_nP_n$. (I use $N$ as the fixed index and $n$ as the running index.) – ET93 May 14 '15 at 01:22