User:Egm6322.s12.team2/HW14

= R 14.1 Analysis of Coefficient Bn in 3-Term Recurrence Relation =

Given
$$p_(n+1)(x) = (A_nx + B_n)p_n(x) - C_np_{n-1}(x)$$

If the highest coefficient of $$p_n(x)$$ is $$k_n > 0$$, then $$A_n = \frac{k_{n+1}}{k_n}$$ and $$C_n = \frac{A_n}{A_{n-1}} \frac{h_n}{h_{n-1}}$$

Find
Find the epxression for $$B_n$$ similar to the epressions for $$A_n$$ and $$C_n$$ in (4) - (5) of p.48j-39, respectively.

From lecture notes [[media:Pea2.s12.sec48.k.djvu|Mtg 48k]]

Solution
The general proof to derive An can be applied to Bn. Recall the following equation from lecture 48j.

$$p_n\in \mathcal P_n \Rightarrow p_n(x) = \sum_{i=0}^n k_i^n x^i$$ $$(14.1.1)$$ $$$$

$$s(x) := p_{n+1}(x) - B_nxp_n(x) \in \mathcal P_{n+1}$$ $$(14.1.2)$$ $$$$

$$=\sum_{i=0}^{n+1} k_i^{n+1}x^i - B_n \sum_{i=0}^{n} k_i^nx^{i+1}$$ $$=\sum_{j=-1}^{n} k_{j+1}^{n+1}x^{j+1} - B_n \sum{i=0}^{n} k_i^n x^{i+1}$$ The key is to select the coefficient that will cancel the highest order of s(x). The solution is below. $$B_n = \frac{k_{n+1}}{k_n}$$

Manuel Steele Signed Egm6322.s12.team2.steele.m2 (talk) 19:59, 25 April 2012 (UTC)

= R 14.2 - The relation of coefficient $$ h_n$$ of the polynomial $$ p_n$$=

Given
Using the following fomula

$$\displaystyle C_{n}=\frac{A_{n}}{A_{n-1}}\frac{h_{n}}{h_{n-1}}$$ (14.2.1) Where $$\displaystyle {\color{red}C_{n}} $$ is the coefficent of the following orthogonal polynomials $$\displaystyle {P_{n}(x)}$$  relations:

$$\displaystyle p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-{\color{red}C_{n}}p_{n-1}(x)$$

Find
Prove that

$$\displaystyle h_{n}=h_{0}\frac{A_{0}}{A_{n}}C_{1}C_{2}\ldots C_{n}$$ (14.2.2) where $$ h_n$$ represents the square of the magnitude of the polynomial $$ p_n$$.

Solution
From the 14.2.1, we have

$$\displaystyle \begin{array}{l} C_{1}=\frac{A_{1}}{A_{0}}\frac{h_{1}}{h_{0}}\\ \\ C_{2}=\frac{A_{2}}{A_{1}}\frac{h_{2}}{h_{1}}\\ \\ C_{3}=\frac{A_{3}}{A_{2}}\frac{h_{3}}{h_{2}}\\ \vdots\\ C_{n-1}=\frac{A_{n-1}}{A_{n-2}}\frac{h_{n-1}}{h_{n-2}}\\ \\ C_{n}=\frac{A_{n}}{A_{n-1}}\frac{h_{n}}{h_{n-1}} \end{array} $$ (14.2.3) then multiply left hand side by left hand side and right hand side by right hand side.

$$\displaystyle C_{1}\cdot C_{2}\cdot C_{3}\cdot\ldots C_{n-1}\cdot C_{n}=\frac{\bcancel{A_{1}}}{A_{0}}\frac{\bcancel{h_{1}}}{h_{0}}\cdot\frac{\cancel{A_{2}}}{\bcancel{A_{1}}}\frac{\cancel{h_{2}}}{\bcancel{h_{1}}}\cdot\frac{\bcancel{A_{3}}}{\cancel{A_{2}}}\frac{\bcancel{h_{3}}}{\cancel{h_{2}}}\cdot\cdots\cdot\frac{\bcancel{A_{n-1}}}{\cancel{A_{n-2}}}\frac{\bcancel{h_{n-1}}}{\cancel{h_{n-2}}}\cdot\frac{A_{n}}{\bcancel{A_{n-1}}}\frac{h_{n}}{\bcancel{h_{n-1}}}$$ (14.2.4) thus

$$\displaystyle C_{1}\cdot C_{2}\cdot C_{3}\cdot\ldots C_{n-1}\cdot C_{n}=\frac{1}{A_{0}h_{0}}\frac{A_{n}h_{n}}{1} $$ (14.2.5) and the equation 14.2.2 exists.

= R 14.3 - Coefficients of Recurrence Relation =

Given
$$\displaystyle (n+1)P_{n+1}-(2n-1)\mu P_n + nP{n-1}=0$$ (14.3.1)

$$\displaystyle P_{n+1}= (A_{n}(x)+B_{n})P_{n}(x)-C_{n} P_{n-1}(x)$$ (14.3.2)

Find
Finding Coefficients $$\ A_n, \ B_n, $$ and $$ C_n$$  in Legendre  Polynomials.

Solution
Solving Eq (14.3.1) for $$ P_{n+1}$$  and  comparing  with  Eq(14.3.2) $$(n+1)P_{n+1}=(2n+1){\mu}P_n - nP_{n-1}$$

$$\displaystyle P_{n+1}=\frac{2n+1}{n+1}{\mu}P_n-\frac{n}{n+1}P_{n-1}$$ <p style="text-align:right">(14.3.3) Comparing Eq(14.3.3)  with  Eq(14.3.2) Therefore, We  get $$A_n=\frac{2n+1}{n+1}$$ $$B_n=0$$ $$C_n=\frac{n}{n+1}$$ Egm6322.s12.sungsik (talk) 02:07, 25 April 2012 (UTC)

= R 14.4 - Re-derivation of the recurrence relation of the norm of Legendre polynomials =

Given
Recurrence relation:
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_{n-1} $$     (14.4.1) Corollary for $$h_n$$:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = h_0\frac{A_0}{A_n}C_1C_2....C_n $$     (14.4.2) Definition of C:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle C_n = \frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$     (14.4.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Reconcile Eq. 14.4.2 with Eq. 14.4.1 and obtain the following expression for $$h_n$$:
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = \frac{2}{2n+1} $$     (14.4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Eq. 14.4.2 can be re-expressed for $$h_{n-1}$$ and substituted directly into Eq. 14.4.1 to easily reconcile the two equations:
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_0\frac{A_0}{A_n}C_1C_2....C_{n-1} $$     (14.4.5) The C terms in the equation can be expanded out and, using the definition of C similar to problem 14.2, it is seen that:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_0\frac{A_0}{A_n}\frac{h_{n-1}A_{n-1}}{h_0A_0} $$     (14.4.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Cancelling terms yields the same result as Eq. 14.4.1, indicating that both recurrence relations are valid for Legendre polynomials:
 * {| style="width:100%" border="0"

$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_{n-1} $$     (14.4.7) The $$h_{n-1}$$ term needs to be reduced down to the known value of $$h_0$$ term, which can easily be found by direct integration and used as a basis for all h terms. Eq. 14.4.7 can be rewritten:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} h_n = \frac{2n-1}{2n+1}h_{n-1} = \frac{2(n-1)+1}{2(n-1)+3}h_{n-1} \end{align} $$     (14.4.8) Likewise,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} h_{n-1} = \frac{2(n-1)-1}{2(n-1)+1}h_{n-2} = \frac{2(n-2)+1}{2(n-2)+3}h_{n-2} \end{align} $$     (14.4.9) Substituting Eq. 14.4.9 into Eq. 14.4.8 will result in the cancellation of the numerator of 14.4.8 and the denominator of 14.4.9; extending this method all the way down to the $$h_0$$ term yields the final equation for $$h_n$$ in terms of $$h_0$$:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} h_{n} = \frac{2(n-n)+1}{2(n-1)+3}h_{0} \end{align} $$     (14.4.10) $$h_0$$ is obtained by taking the inner product of the 0th Legendre polynomial, $$P_0=1$$:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h_{0} = \int_{-1}^1 P_0(x)^2dx = x|^1_{-1}=2 $$     (14.4.11) Substituting Eq 14.4.11 into Eq 14.4.10 yields the desired expression for $$h_n$$:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} h_{n} = \frac{1}{2(n-1)+3}(2) = \frac{2}{2n-2+3} = \frac{2}{2n+1} \end{align} $$     (14.4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

= R 14.5 Derivation of Christoffel-Darboux Identity for Legendre Polynomials =

Given
We are given the Christoffel-Darboux (CD) identity for Legendre polynomials. $$\sum_{k=0}^n \frac{P_k(x) P_k(y)}{h_k} = \frac{P_{n+1}(x) P_n(y) - P_n(x) P_{n+1}(y)}{\beta_n h_n(x-y)}$$

Find
Derive the form of the Christoffel-Darboux identity as applied to the Legendre polynomials in (1) p.45d-16e.

From lecture notes [[media:Pea2.s12.sec48.l.djvu|Mtg 48l]]

Solution
The general Christoffel-Darbou identity was derived with the 3-term recurrence relation, use of symmetry, subsequent subtraction, and rearrangement with appropriate terms to get the form given in lecture. However, this form can be modified for Legendre polynomials. As with the general derivation for polynomials, a derivation specifically applied to Legendre polynomials requires a review of Legendre equations.

Step 1: Review Legendre Equations <p style="font-size:100%" align="center">$$(n+1)P_{n+1}(x) - (2n+1) x \, P_n(x) + nP_{n-1}(x) = 0$$ <p style="text-align:right;">$$(14.5.1)$$ <p style="font-size:125%" align="center">$$$$

<p style="font-size:100%" align="center">$$\langle P_m, P_n \rangle = \int_{\mu = -1}^{\mu = 1} P_m(\mu) \, P_n(\mu) \, d\mu = \frac{2}{2n+1} \delta_{mn}$$ <p style="text-align:right;">$$(14.5.2)$$ The modified version needed for our derivation is below. <p style="font-size:100%" align="center">$$h_n := \left \langle P_n,P_n \right \rangle = \frac{2}{2n+1}$$ The following relationships are also required to understand the derivation. The leading coefficient of P_n(x) is $$\alpha_n$$ <p style="font-size:100%" align="center">$$P_n(x) = \alpha_n \pi (x)$$ The coefficient, $$\beta_n$$, is also defined below. <p style="font-size:100%" align="center">$$\beta_n := \frac{\alpha_{n+1}}{\alpha_n}$$ Step 2: Analyze Solution Pattern from Lecture 48l The key is to identify the primary coefficient in the 3-term recurrence relation for the solution. In the general case, the coefficient $$A_n$$ was chosen. So the other two coefficients were eliminated. In the 1st equation (14.5.3) a primary 3-term relationship was defined by multipliying the 3-term relationship with $$p_n(y)$$. <p style="font-size:100%" align="center">$$p_{n+1}(x)p_n(y) = (A_nx + B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x)$$ <p style="text-align:right;">$$(14.5.3)$$ The relationship was used to generate a 2nd modifed 3-term relationship, a 2nd equation (14.5.4). <p style="font-size:100%" align="center">$$p_{n+1}(y)p_n(x) = (A_ny + B_n)p_n(x)p_n(y) - C_np_n(x)p_{n-1}(y)$$ <p style="text-align:right;">$$(14.5.4)$$ Equation 17.5.4 was subtracted from 17.5.3. This resulted in a new equation where $$B_n$$ was eliminated. <p style="font-size:100%" align="center">$$p_{n+1}pn(y) - p_{n+1}(y)p_n(x) = A_n(x-y)p_n(x)p_n(y) - C_n[p_n(y)p_{n-1}(x) - p_n(x)p_{n-1}(y)]$$ Next, $$C_n$$ is eliminated by defining the relationship below. <p style="font-size:100%" align="center">$$C_n = \frac{A_n}{A_{n-1}}$$ This gives the new relationship defined with $$A_n$$ as a main coefficient. <p style="font-size:100%" align="center">$$\frac{p_{n+1}(x)p_n(y) - p_{n+1}(y)p_n(x)}{A_n(x-y)}= p_n(x)p_n(y) + \frac{p_n(x)p_{n-1}(y) - p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)}$$ This equation can be modified by decrementing n by 1 in an iterative approach until the Christoffel-Darboux identity is obtained. Step 3: Apply Solution Pattern to Legendre Polynomials This process can be reapplied to the Legendre equations with Legendre-based coefficients. The key is to understand those coefficients for an appropriate ansatz. <p style="font-size:100%" align="center">$$(n+1)P_{n+1}(x) - \frac{2}{h_n} x \, P_n(x) + nP_{n-1}(x) = 0$$ Rearrange the equation and multiply by $$P_n(y)$$. <p style="font-size:100%" align="center">$$(n+1)P_{n+1}(x)P_n(y) = \frac{2}{h_n} x \, P_n(x) P_n(y) + nP_{n-1}(x)P_n(y)$$ <p style="text-align:right;">$$(14.5.5)$$ Next, create a 2nd equation that flips the "x" and "y" symmetry from the former equation. <p style="font-size:100%" align="center">$$(n+1)P_{n+1}(y)P_n(x) = \frac{2}{h_n} y \, P_n(y) P_n(x) + nP_{n-1}(y)P_n(x)$$ <p style="text-align:right;">$$(14.5.6)$$ Subtract 14.5.6 from 14.5.5 to get the form below. <p style="font-size:100%" align="center">$$(n+1)P_{n+1}(x)P_n(y) = \frac{2}{h_n} x \, P_n(x) P_n(y) + nP_{n-1}(x)P_n(y)$$ The rearrangement gives the following form. <p style="font-size:100%" align="center">$$P_n(x)P_n(y)*(x-y)*(2n+1) = (n+1)(P_{n+1}(x)P_n(y) - P_{n+1}(y)P_n(x)) - nP_{n-1}(x)P_n(y) + nP_{n-1}(y)P_n(x)$$ A term-by-term rearrangement requires a quantitative description, but due to lack of time, the summarized outcome is given below. <p style="font-size:100%" align="center">$$\frac{P_{n+1}(x)P_n(y) - P_{n+1}(y)P_n(x)}{\beta_nh_n(x-y)} = P_n(x)P_n(y) + \frac{P_n(x)P_{n-1}(y) - P_n(y)P_{n-1}(x)}{\beta_{n-1} h_{n-1}(x-y)}$$ <p style="font-size:100%" align="center">$$$$ As in step 1, the equation above can be modified by decrementing n by 1 in an iterative approach until the Christoffel-Darboux identity is obtained but for Legendre form. Manuel Steele Signed Egm6322.s12.team2.steele.m2 (talk) 20:00, 25 April 2012 (UTC)

=Contributing Members=

= References for Report 14 = Bell, W.W. "Special Functions for Scientists and Engineers." Mineola, NY: Dover Publications, Inc., 1996. King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003. http://en.wikipedia.org/wiki/Legendre_polynomials Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48i) [[media:Pea2.s12.sec48.i.djvu|Mtg 48i]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48j) [[media:Pea2.s12.sec48.j.djvu|Mtg 48j]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48k) [[media:Pea2.s12.sec48.k.djvu|Mtg 48k]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48l) [[media:Pea2.s12.sec48.l.djvu|Mtg 48l]] Spring 2012.