User:Egm6322.s12.TEAM1.HW/HW7

Problem 1: Find a coefficient of the 3-term reccurence relation
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Find
Find an expression for $$ B_n \ $$ similar to expressions given in Eq1.1 and Eq1.2.

Solution
Knowing the three term recurrence relation is of the form: Also given that the following holds: Allowing $$ n=0 \ $$ and plugging the values of Eq1.4 thru Eq1.6 into Eq1.3, one gets the following: Plugging in for $$ p_1(x) \ $$ and with $$ A_n = \frac{leading\ coef\!ficient\ of\ p_{n+1}}{leading\ coef\!ficient\ of\ p_n } = 1 \ $$ Moving the values to opposite sides of the equivalency and combining like terms results in the following: or more generalized:

Problem 2: Proof of the magnitude of a polynomial
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Find
Prove Eq2.1 can be obtained using Eq 2.2.

Solution:
From equation (2.2) we can obtain: $$\begin{align} & {{C}_{n}}=\frac\frac \\ & {{h}_{n}}={{h}_{n-1}}{{C}_{n}}\frac \\ & {{C}_{n-1}}=\frac\frac \\ & {{h}_{n-1}}={{h}_{n-2}}{{C}_{n-1}}\frac \\ \end{align}$$ Then, we continue our derivation:

$$\begin{align} & {{h}_{n}}={{h}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}\frac{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}\frac{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{n-3}}{{C}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{0}}{{C}_{1}}...{{C}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \end{align}$$ So, eqation(2.1) has been obtained.

Problem 3: Re-framing the the 2-term recurrence relation into the 3-term recurrence relation
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Find
Put the recurrence relation RR2, Eq3.1, for the Legendre Polynomials in the framework of the recurrence realtion in Eq3.2; i.e. identify the coefficients $$ A_n,B_n,C_n $$

Solution:
According to RR2 equation (3.1): $$\displaystyle(n+1){{P}_{n+1}}-(2n+1)x{{P}_{n}}+n{{P}_{n-1}}=0$$ we have: $$\displaystyle{{P}_{n+1}}=\frac{(2n+1)}{(n+1)}x{{P}_{n}}-\frac{n}{(n+1)}{{P}_{n-1}}$$ Compared with the equation (3.2): $$\begin{align} p_{n+1}(x)=(A_nx+B_n)p_n(x)+C_np_{n-1}(x) \end{align}$$ we can obtain the corresponding coefficients:

$$\displaystyle\begin{align} & {{A}_{n}}=\frac{(2n+1)}{(n+1)} \\ & {{B}_{n}}=0 \\ & {{C}_{n}}=\frac{-n}{(n+1)} \end{align}$$

Problem 4: Obtain an expression for recurrence relation relative ot the magnitude
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Find
Reconcile the reccurence relation, Eq4.2, with the recurrence relation in Eq4.1 and obtain the expression for $$ h_n \ $$ in Eq4.3 usng Eq4.2.

Solution
$$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1} $$ $$\displaystyle h_n=h_0\frac{A_0}{A_n}C_1 C_2 \cdots C_n $$ From the legendre polynomials,  $$\displaystyle (n+1)p_{n+1}(x) - (2n+1)xp_n(x)-np_{n-1}(x)$$ $$\displaystyle{{P}_{n+1}}=\frac{(2n+1)}{(n+1)}x{{P}_{n}}+\frac{n}{(n+1)}{{P}_{n-1}}$$ According to the above relationships, we can get $$\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)+C_np_{n-1}(x) $$ Hence, we can define as below:  $$\displaystyle\begin{align} & {{A}_{n}}=\frac{(2n+1)}{(n+1)} \\ & {{B}_{n}}=0 \\ & {{C}_{n}}=\frac{n}{(n+1)} \end{align}$$

Let's look at an example for the relationship between eq(4.1) and eq(4.2). $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}=h_0\frac{A_0}{A_n}C_1 C_2 \cdots C_n $$ $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}=h_0\frac{(n+1)}{(2n+1)}C_1 C_2 \cdots C_n $$ When n = 1,  $$\displaystyle h_1=\frac{1}{3}h_{0}=h_0\frac{(2)}{(3)}\frac{1}{2}=h_0\frac{(1)}{(3)} $$ When n = 2, $$\displaystyle h_2=\frac{3}{5}h_{1}=\frac{3}{5}\frac{1}{3}h_{0}=h_0\frac{(3)}{(5)}C_1C_2=h_0\frac{(3)}{(5)}\frac{(1)}{(2)}\frac{(2)}{(3)}=\frac{1}{5}h_0 $$ From the above results, hn can be written as below: $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}$$ And from eq 4.2  $$\displaystyle h_{n-1}=h_0\frac{A_0}{A_{n-1}}C_1 C_2 \cdots C_n=h_0\frac{n}{(2n-1)}\frac{1}{n}$$ Then plugging the above equation into eq (4.1) Finally we can get  $$\displaystyle h_n=\frac{2n-1}{2n+1}h_0\frac{n}{(2n+1)}\frac{1}{n}=\frac{2n-1}{2n+1}h_0\frac{1}{(2n-1)}=\frac{1}{2n+1}h_0$$ By using inner product of the 0th order legendre polynomials, we can get the value of h0. $$\displaystyle h_0 = \int_{-1}^{1} P_0(x)^2dx = \int_{-1}^{1}1dx = 2$$ Plugging the above value, we can express as below: $$\displaystyle h_n=\frac{1}{2n+1}h_0=\frac{2}{2n+1}$$

Problem 5: Derive the Christoffel-Darboux identity for Legendre polynomials
 Solved without assistance 

Given:
The 3-term recurrence relation for any sequence of orthogonal polynomials;
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\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$ for n=0,1,2,..., and $$\,\,p_{-1}(x)=0\,\,$$
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\displaystyle A_n\,,B_n\,,C_n \in \mathcal R\,,with\, A_{n-1}A_nC_n > 0 $$ If the highest coefficient of $$\,\,p_n(x)\,\,$$ is kn, then
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\displaystyle A_n=\frac{k_{n+1}}{k_n} $$
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\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
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Find:

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\displaystyle \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k}=\frac{p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)}{A_nh_n(x-y)} $$
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Solution:
From the three term recurrence relation
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\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$ multiply both sides by pn(y):
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\displaystyle p_{n+1}(x)p_n(y)=(A_nx+B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x) $$ $$ And for the symmetry between the factors
 * <p style="text-align:right;">$$\displaystyle (Equation\;5.1 )
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\displaystyle p_{n+1}(y)p_n(x)=(A_ny+B_n)p_n(x)p_n(y)-C_np_n(x)p_{n-1}(y) $$ $$ Substract equation 5.2 from equation 5.1:
 * <p style="text-align:right;">$$\displaystyle (Equation\;5.2 )
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\displaystyle p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)=A_n(x-y)p_n(x)p_n(y)-C_n[p_n(y)p_{n-1}(x)-p_n(x)p_{n-1}(y)] $$ For
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\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$ Then
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\displaystyle \begin{align} & \frac{p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)}{A_nh_n(x-y)} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_n(x)p_{n-1}(y)-p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)h_{n-1}} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_{n-1}(x)p_{n-1}(y)}{h_{n-1}}+\frac{p_{n-1}(x)p_{n-2}(y)-p_{n-1}(y)p_{n-2}(x)}{A_{n-2}(x-y)h_{n-2}} \\ & \vdots \\ =& \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k} \end{align} $$
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