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Suppose that the group $p_0(x), p_1(x),... p_N(x) $ are orthonormal in the interval $[-1,1]$, which means with 2 arbitrary functions $p_i(x), > p_j(x)$, the conditions below is satisfied.

$\int_{-1}^1p_i(x), p_j(x)dx = \begin{cases} 1,& (i = j)\\ 0,& (i \not= j) \end{cases}$$ $

In this case, $p_i(x)$ is the polynomial with degree $i$.

Prove the uniqueness of $p_N(x)$ without considering sign by the procedure as below:

  1. Generally the N-degree polynomial $f_N(x)$ can be written as below: $f_N(x) = \sum_{k=0}^N c_kp_k(x)$

Demonstrate the coefficient $c_k$ $(k = 0,1,2...,N)$by $p_0(x), > p_1(x),... p_N(x) $ and $f_N(x)$ .

  1. Let $p_N(x)^*$ is an N degree polynomial other than $p_N(x)$ so that $p_0(x), p_1(x),... p_N(x)^* $ are also orthonormal. Let $f_N(x) > = p_N(x)^*$, prove that $ p_N(x) = -p_N(x)^*$

Here is what I am thinking:

First, demonstrate $f_N(x)$ in matrix form as below:

$f_N(x)$ = $\pmatrix{p_0(x)&p_1(x)&...&p_N(x)} \pmatrix{c_0\\c_1\\...\\c_N} $.

Then multiply both sides with $\pmatrix{p_0(x)\\p_1(x)\\...\\p_N(x)} $, so we have: $\pmatrix{p_0(x)f_N(x)\\p_1(x)f_N(x)\\...\\p_N(x)f_N(x)} =\pmatrix{p_0(x)p_0(x)&p_0(x)p_1(x)&...&p_0(x)p_N(x)\\p_1(x)p_0(x)&p_1(x)p_1(x)&...&p_0(x)p_1(x)\\...........&...........&...&...........\\p_N(x)p_0(x)&p_N(x)p_1(x)&...&p_N(x)p_N(x)}\pmatrix{c_0\\c_1\\...\\c_N} $

Integral both side from ${-1}$ to $1$, and use the properties of orthonormal functions, we have:

$\pmatrix{\int_{-1}^1p_0(x)f_N(x)dx\\\int_{-1}^1p_1(x)f_N(x)dx\\...\\\int_{-1}^1p_N(x)f_N(x)dx} =\pmatrix{1&0&...&0\\0&1&...&0\\...&...&...&...\\0&0&...&1}\pmatrix{c_0\\c_1\\...\\c_N} = \pmatrix{c_0\\c_1\\...\\c_N}$

So number 1 is satisfied (we already demonstrated the coefficients by $f_N(x)$ and $p_0(x),p_1(x),...p_N(x)$

Then for 2, when $f_N(x)=p_N(x)^*$, we have:

$\pmatrix{c_0\\c_1\\...\\c_N}= \pmatrix{\int_{-1}^1p_0(x)p_N(x)^*dx\\\int_{-1}^1p_1(x)p_N(x)^*dx\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx} = \pmatrix{1\\1\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx} $

Then: $f_N(x)$ =$p_N(x)^*$=$\pmatrix{p_0(x)&p_1(x)&...&p_N(x)}\pmatrix{1\\1\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx}$ =$p_N(x)*\int_{-1}^1p_N(x)p_N(x)^*dx$

So $p_N(x)^*$=$p_N(x)*\int_{-1}^1p_N(x)p_N(x)^*dx$

Then I stuck here. Can we show that $\int_{-1}^1p_N(x)p_N(x)^*dx = -1$ in this case so that $p_N(x)^* = - p_N(x)$?

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