User:Csstewart10

Cameron Stewart

EML6322 Group: 2

Email: csstewart10@ufl.edu

Problem 1

Given Information

The following polynomial is given as a solution to an associated Laguerre differential equation:

$P_{N}(u)=\sum _{n=2}^{N}{(-1)^{n-1}{\frac {(N-1)(N-2)...(N-n+1)}{n!\,(n-1)!}}u^{n}},N\geq 2$

(1.1)

A general associated Laguerre polynomial has the form:

$xy''+(\alpha +1-x)y'+Ny=0$

(1.2)

$P_{N}$ is a solution to the associated Laguerre differential equation where $\alpha =-1$.

$uP_{N}''(u)-P_{N}'(u)+NP_{N}(u)=0$

(1.3)

Problem Statement

(a) Verify that the polynomial in equation 1.1 is a Laguerre polynomial except for a minus sign.

(b) Find the expressions of $P_{N}(u)$ for various values of N, and compare them to a table of Laguerre polynomials.

(c) Plot the associated Laguerre polynomials for $N=3$, and $\alpha =0,-1,-2,-3,-4$.

Solution

Part a

The falling factorial term in Equation 1.1 can be rewritten with a Pochhammer symbol.

$P_{N}(u)=\sum _{n=2}^{N}{(-1)^{n-1}{\frac {(N)_{n}}{N\,n!\,(n-1)!}}u^{n}},N\geq 2$

(1.4)

Where,

$(N)_{n}=N(N-1)(N-2)...(N-n+1)$

(1.5)

The Pochhammer symbol can be related to a binomial coefficient by the equation:

${\frac {(N)_{n}}{n!}}={\binom {N}{n}}$

(1.6)

Plugging Equation 1.6 into Equation 1.4 results in:

$P_{N}(u)=u+\sum _{n=2}^{N}{\binom {N}{n}}{\frac {(-1)^{n-1}}{N(n-1)!}}u^{n}$

(1.7)

The binomial coefficient has the following closed form for integer arguments:

${\binom {N}{n}}=\left\{{\begin{matrix}{\frac {N!}{n!\,(N-n)!}}&0\leq n\leq N\\0&{\text{otherwise}}\end{matrix}}\right.$

(1.8)

Substituting the closed form in Equation 1.8 into Equation 1.7 gives:

$P_{N}(u)=u+\sum _{n=2}^{N}{\frac {(N-1)!\,(-1)^{n-1}}{n!\,(n-1)!}}u^{n}$

(1.9)

The associated Laguerre polynomial has the closed form:

$L_{N}^{\alpha }(u)=\sum _{n=0}^{N}{\binom {\alpha +N}{N-n}}{\frac {(-1)^{n}}{n!}}u^{n}$

(1.10)

For the the case where $\alpha =-1$, Equation 1.10 has the form:

$L_{N}^{-1}(u)=\sum _{n=0}^{N}{\binom {N-1}{N-n}}{\frac {(-1)^{n}}{n!}}u^{n}$

(1.11)

Using the expression Equation 1.6 for the binomial term in Equation 1.11 yields:

${\binom {N-1}{N-n}}=\left\{{\begin{matrix}{\frac {(N-1)!}{(N-n)!\,(n-1)!}}&0\leq N-n\leq N-1\\0&{\text{otherwise}}\end{matrix}}\right.$

(1.12)

Plugging Equation 1.12 into Equation 1.11 gives:

$L_{N}^{-1}(u)=\sum _{n=1}^{N}{\frac {(N-1)!\,(-1)^{n}}{(N-n)!\,(n-1)!\,n!}}u^{n}$

(1.13)

It should be noted that the summation index in Equation 1.13 changes from $n=0$ to $n=1$. This is because when $n=0$ the binomial coefficient is zero and cannot be represented by the factorial expression. This term must be brought outside of the sum, but since the term is zero, all that results is an index change.

The first term of Equation 1.13 is brought outside the summation.

$L_{N}^{-1}(u)=-u+\sum _{n=2}^{N}{\frac {(N-1)!\,(-1)^{n}}{(N-n)!\,(n-1)!\,n!}}u^{n}$

(1.14)

Comparing Equation 1.14 to Equation 1.9, it is seen that the following relation holds:

$L_{N}^{-1}(u)=-P_{N}(u)$

(1.15)

Part b

Table 1.1 Comparison of $L_{N}^{-1}(u)$ and $P_{N}(u)$

N	$L_{N}^{-1}(u)\,$	$P_{N}(u)\,$
1	$-u\,$	$u\,$
2	${\frac {1}{2}}(u^{2}-2u)\,$	${\frac {1}{2}}(-u^{2}+2u)\,$
3	${\frac {1}{6}}(-u^{3}+6u^{2}-6u)\,$	${\frac {1}{6}}(u^{3}-6u^{2}+6u)\,$
4	${\frac {1}{24}}(u^{4}-12u^{3}+36u^{2}-24u)\,$	${\frac {1}{24}}(-u^{4}+12u^{3}-36u^{2}+24u)\,$
5	${\frac {1}{120}}(-u^{5}+20u^{4}-120u^{3}+240u^{2}-120u)\,$	${\frac {1}{120}}(u^{5}-20u^{4}+120u^{3}-240u^{2}+120u)\,$

Part c

Problem 4

Given Information

The following finite power series,

$P_{N}(u)=\sum _{n=1}^{N}a_{n}u^{n}$

(4.1)

Is given as a solution to the associated Laguerre differential equation where $\alpha =-1$.

$uP''_{N}(u)-P'_{N}(u)+NP_{N}(u)=0$

(4.2)

Problem Statement

Verify that the solution given by Equation 4.1 has the following coefficients:

$a_{1}=1$

(4.3)

and,

$a_{n}=(-1)^{n-1}{\frac {(N-1)(N-2)...(N-n+1)}{n!\,(n-1)!}}$

(4.4)

Solution

Taking the derivatives of Equation 4.1 gives the following:

$P_{N}(u)=\sum _{n=1}^{N}a_{n}u^{n}$

(4.1)

$P'_{N}(u)=\sum _{n=1}^{N}n\,a_{n}u^{n-1}$

(4.4)

$P''_{N}(u)=\sum _{n=2}^{N}n(n-1)\,a_{n}u^{n-2}$

(4.5)

Note: The summation index for Equation 4.5 must start at 2; we are assuming a polynomial solution. As such, derivatives must also remain polynomials (powers of $u$ must be non-negative).

Plugging Equation 4.1, Equation 4.4 and Equation 4.5 into Equation 4.2 gives:

$\sum _{n=2}^{N}n(n-1)\,a_{n}u^{n-1}-\sum _{n=1}^{N}n\,a_{n}u^{n}+N\sum _{n=1}^{N}a_{n}u^{n}=0$

(4.6)

A change of index is needed to equate the power of $u$ in the first sum.

$\sum _{n=1}^{N}(n+1)(n)\,a_{n+1}u^{n}-\sum _{n=1}^{N}n\,a_{n}u^{n}+N\sum _{n=1}^{N}a_{n}u^{n}=0$

(4.7)

Since the summation indices are all identical, Equation 4.7 can be written under a single summation.

$\sum _{n=1}^{N}[(n+1)(n)\,a_{n+1}-n\,a_{n}+N\,a_{n}]u^{n}=0$

(4.8)

In order for Equation 4.8 to hold true, the term inside the square brackets must sum to zero giving the recurrence relation:

$(n+1)(n)\,a_{n+1}-n\,a_{n}+N\,a_{n}=0$

(4.9)

With a shift of index Equation 4.9 can be written as:

$a_{n}=a_{n-1}{\frac {(N-n+1)}{n(n-1)}}$

(4.10)

The coefficient $a_{1}$ can be found by observing the case where N=1. Equation 4.2 then becomes:

$uP''_{1}(u)-uP'_{1}(u)+P_{1}(u)=0$

(4.11)

And $P_{1}(u)$ and it's derivatives become:

$P_{1}(u)=a_{1}u$

(4.12)

$P'_{1}(u)=a_{1}$

(4.13)

$P''_{1}(u)=0$

(4.14)

Plugging Equation 4.12, Equation 4.13 and Equation 4.14 into Equation 4.11 yields:

$-a_{1}u+u=0$

(4.15)

Since this must hold everywhere:

$a_{1}=1$

(4.3)

The relation $a_{1}=1$ acts as an starting condition for the recurrence relationship in Equation 4.10. $a_{2}$ can then be found by the recursive relation in Equation 4.10 (and subsequently $a_{3}$ ,$a_{4}$ etc.). The $n^{th}$ coefficient is found to be:

$a_{n}=(-1)^{n-1}{\frac {(N-1)(N-2)...(N-n+1)}{n!\,(n-1)!}}$

(4.4)