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Let $P_{0}, P_{1}$ ... be the orthogonal polynomial for weight $w_{1}$ on $[0,1]$ and $Q_{0}, Q_{1}$ the orthogonal polynomial for weight $w_{2}$ on $[1,2]$.

The inner product is define as usual $<P_j(x),P_i(x)>=\int_{0}^{1}P_{i}(x)P_{j}(x)\omega_{1}(x)dx$. and for $Q(x)$ it is similar.

The question is: suppose that $2w_{1}=w_{2}(2-x)$, find the relation between $P_{j}$ and $Q_{j}$.

Note: $P_j$ is a polynomial with degree j.


I am asking this question for two reasons:

  1. I am really not sure if it was actually asking us for the relation between $P_{j}$ and $Q_{j}$. Here is the original handwriting of my professor (I knew I should have asked him earlier) enter image description here Normally that little subscript that looks like a reserved c means s in his handwriting. What do you guys think is the something that he would be reasonably asking according to his notation?

  2. Assuming that it is asking for the relation between $P_j$ and $Q_j$, then I thought the solution should be like this.

Since they are both orthogonal sets, we have



where $i\neq j.$

Using the relation $$2\omega_{1}(x)=\omega_{2}(2-x)$$

$$\implies\omega_{1}(x)=\frac{1}{2}\omega(2-x)$$ we have


Let $u=2-x$, then we have $x=2-u$, $dx=-du$




Since $$\int_{1}^{2}Q_{i}(u)Q_{j}(u)\omega_{2}(x)dx=0$$

we have $$Q_{j}(u)=P_{j}(2-u)$$

That last line of logic is where I felt skeptical. It seems that it requires that:

On a given domain and a fixed weight, the set of orthogonal polynomial on it is unique.

Thinking about this in detail, I realized that seems to be only true if we require that:

In this set of polynomials , the polynomial of each degree must appear and that for each degree, there is only one polynomial.

According to Wikipedia's definition of Orthogonal Polynomial, this seems to be implied, but not directly stated. Is this condition really true? Thanks.

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Have you tried expressing the inner product associated with $Q$ in terms of $w_1(x)$? – Timmy Turner Dec 2 '11 at 6:23
good thought. I will try that. Thanks. – Xiaowen Li Dec 2 '11 at 6:42
The results is that P(x)=Q(2-x). Letting u=2-x and we get the same conclusion as before, but once again this requires the "skeptical reasoning" that I mentioned. – Xiaowen Li Dec 2 '11 at 6:48
His last written line reads: "find the relation between the $P_j$'s and $Q_j$'s." – Greg Martin Dec 2 '11 at 7:11
Thank you so much Greg! That sounds very reasonable! – Xiaowen Li Dec 2 '11 at 7:14
up vote 1 down vote accepted

There is no such thing as the orthogonal polynomials for the weight $w$, rather there exists some (and many different) orthogonal bases $(P_i)$ of polynomials such that each $P_i$ has degree $i$. If one imposes furthermore that the basis is orthonormal, that is, that $\langle P_i,P_i\rangle=1$ for each $i$, this restricts the choice of the basis to $\pm P_i$ for each $i$.

Here one considers two weight functions $w_1$ and $w_2$, defined on $I_1=[0,1]$ and $I_2=[1,2]$ respectively, such that $w_2(2-x)=2w_1(x)$ for every $x$ in $I_1$. Define the transformations $T$ and $\Theta$, such that for every polynomial $P$ on $I_1$, $T(P)$ and $\Theta(P)$ are the polynomials on $I_2$ defined by $T(P)(x)=P(2-x)$ and $\Theta(P)(x)=\frac1{\sqrt2} P(2-x)$ respectively. Obviously, for every polynomials $P$ and $Q$ on $I_1$, $$ \langle T(P),T(Q)\rangle=2\langle P,Q\rangle,\qquad\qquad \langle \Theta(P),\Theta(Q)\rangle=\langle P,Q\rangle. $$ Thus, if $(P_i)_i$ is an orthogonal basis for the weight $w_1$, then the families $(T(P_i))_i$ and $(\Theta(P_i))_i$ are two orthogonal bases for the weight $w_2$, and, if $(P_i)_i$ is an orthonormal basis for the weight $w_1$, then the family $(\Theta(P_i))_i$ is an orthonormal basis for the weight $w_2$ (and the family $(T(P_i))_i$ is orthogonal but not orthonormal).

Assume from now on that $(P_i)_i$ is an orthonormal basis for the weight $w_1$. Then, $(\Theta(P_i))_i$ is not the only orthonormal basis for the weight $w_2$ such that the degree of the $i$th polynomial in the basis is $i$ for every $i$. For example, $((-1)^i\Theta(P_i))_i$ is also an orthonormal basis for the weight $w_2$. But, if one imposes that the degree of the $i$th polynomial in the basis is $i$ for every $i$, then all the orthonormal bases for the weight $w_2$ are $(\varepsilon_i\Theta(P_i))_i$ where $\varepsilon_i=\pm1$ for every $i$.

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Hello Didier, thank you for your answer. It does make more sense to say that we are only looking for basis of a set of orthogonal polynomials since linear combination of different elements form such a set remains orthgonal. But it seems that $Q_{i}=cP_{i}(2-x), c\in\mathbb{R}$ since we are not restricting the polynomials to be normalized. Does this seem right? – Xiaowen Li Dec 2 '11 at 18:23
Your comment is puzzling. A vector space cannot contain only orthogonal polynomials... A linear combination of orthogonal polynomials is a polynomial, and it does not make sense to write that this polynomial is orthogonal. One vector is not orthogonal all alone, it is (or is not) orthogonal to some other vector(s). // As I wrote, if $(P_i)_i$ is an orthonormal family, then the family $(Q_i)_i$ is orthogonal but $\langle Q_i,Q_i\rangle=2\langle P_i,P_i\rangle=2$ hence $(Q_i)_i$ is not orthonormal. – Did Dec 2 '11 at 22:33
I was talking about the case when the normality requirement is lifted.... – Xiaowen Li Dec 3 '11 at 5:33

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