User:Egm6322.s14.team2/report2

Report 2 PEA 2 Spring 2014 Team 2

Problems are given in lecture notes Sec66a

Problem 1

Given Information

Here is a sample equation and a sample in-text reference to the equation.

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

(Eq 1-1)

Here is a link to (Eq 1-1)

Problem Statement

The atomic-resolution TEM figure on p.66-16 is not obvious (not easy to see and to understand); search the literature (web, pdf papers, pdf reports, etc.) for better (good quality) figure(s) that show the dislocation at atomic level with a burger vector. for example, here is an image web search for “dislocation atomic TEM burger vector”.

R2.2 Roots of Associated Laguerre Polynomials

Problem Statement

Plot the associated Laguerre polynomials for $n=3$ and $\alpha =0,-1,-2,-3,-4$. Use MATLAB to find the roots of $L_{n}^{(-1)}(x)$ for $n=7,8$, and verify the numbers in the table on p.66a-7

Note: If you already have the plots made in report R1, you simply copy and paste these plots in report R2; the task then is only to find the roots of these polynomials.

Find

(a): Associated Laguerre polynomial plots

(b): Roots of Laguerre polynomials

(c): Verify numbers in the table on p.66-7

Solution

Part (a): Associated Laguerre Polynomial Plots

The plot was created using the Python scripting language.

Figure 1: Associated Laguerre Polynomial $L_{3}^{(\alpha )}$ ([1])

Part (b): Roots of Associated Laguerre Polynomials

The roots of the associated Laguerre polynomial were computed using Wolfram Alpha. The syntax for creating associated Laguerre polynomials in Wolfram Alpha: LaguerreL[n,k,x] where n is the degree, k is the family of Laguerre polynomials, and x is the independent variable.

The Laguerre polynomial given by,

$L_{3}^{(0)}={\frac {1}{6}}(6-18x+9x^{2}-x^{3})$

(Eq 2-1)

Has three roots. The approximate roots of the polynomial are:
$x\approx 0.41577$
$x\approx 2.2943$
$x\approx 6.2899$

The Laguerre polynomial given by,

$L_{3}^{(-1)}={\frac {1}{6}}(-6x+6x^{2}-x^{3})$

(Eq 2-2)

Has three roots. The roots of the polynomial are:
$x=0$
$x=3-{\sqrt {(}}3)$
$x=3+{\sqrt {(}}3)$

The Laguerre polynomial given by,

$L_{3}^{(-2)}={\frac {1}{6}}(3x^{2}-x^{3})$

(Eq 2-3)

Has three roots. The roots of the polynomial are:
$x=0$
$x=0$
$x=3$

The Laguerre polynomial given by,

$L_{3}^{(-3)}=-{\frac {x^{3}}{6}}$

(Eq 2-4)

Has three roots. The roots of the polynomial are:
$x=0$
$x=0$
$x=0$

The Laguerre polynomial given by,

$L_{3}^{(-4)}={\frac {1}{6}}(-6-6x-3x^{2}-x^{3})$

(Eq 2-5)

Has three roots. The roots of the polynomial are:
$x\approx -1.5961$
$x\approx -0.7020-1.8073i$
$x\approx -0.7020+1.8073i$

Part (c): Two Sets of Roots of Associated Laguerre Polynomials For $n=7,8$

In this section we are asked to compute the roots of the associated Laguerre polynomials with $\alpha =-1{\text{ and }}n=7,8$ . We are then asked to compare these results to previously tabulated results.

The Laguerre polynomial given by,

$L_{7}^{(-1)}={\frac {-5040x+15120x^{2}-12600x^{3}+4200x^{4}-630x^{5}+42x^{6}-x^{7}}{5040}}$

(Eq 2-6)

The following roots:

Associated Laguerre Polynomial Roots of $L_{7}^{(-1)}$
Root #	Root
$x_{1}$	0
$x_{2}$	0.5277
$x_{3}$	1.796
$x_{4}$	3.877
$x_{5}$	6.919
$x_{6}$	11.235
$x_{7}$	17.65

The Laguerre polynomial given by,

$L_{8}^{(-1)}={\frac {-40320x+141120x^{2}-141120x^{3}+58800x^{4}-11760x^{5}+1176x^{6}-56x^{7}+x^{8}}{40320}}$

(Eq 2-7)

The following roots:

Associated Laguerre Polynomial Roots of $L_{8}^{(-1)}$
Root #	Root
$x_{1}$	0
$x_{2}$	0.4610
$x_{3}$	1.564
$x_{4}$	3.352
$x_{5}$	5.916
$x_{6}$	9.421
$x_{7}$	14.19
$x_{8}$	21.09

Problem 3

Given Information

(Eq 3-1) was given for the value of P_N(u) and (Eq 3-2) through (Eq 3-4) were given for the Laguerre polynomials written with the same variables as (Eq 3-1) for comparisons.

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

(Eq 3-1)

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

(Eq 3-2)

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

(Eq 3-3)

$L_{N}^{(\alpha )}(u)=\sum _{n=0}^{N}(-u)^{n}{N \choose n}\prod _{k=n+1}^{N}(\alpha +k)$

(Eq 3-4)

Problem Statement

Compare (Eq 3-1) to the various finite power series expressions for the associated Laguerre polynomials in (Eq 3-2) through (Eq 3-4) as specified in R2.3: sec.66a, Pb-66-6.

Solution

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

The falling factorial term in (Eq 3-1) can be rewritten with a Pochhammer symbol.

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

(Eq 3-5)

Where,

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

(Eq 3-6)

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

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

(Eq 3-7)

Plugging (Eq 3-7) into (Eq 3-5) results in:

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

(Eq 3-8)

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

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

(Eq 3-9)

Substituting the closed form in (Eq 3-9) into (Eq 3-8) gives:

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

(Eq 3-10)

For the the case where $\alpha =-1$, (Eq 3-3) has the form:

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

(Eq 3-11)

Using the expression (Eq 3-7) for the binomial term in (Eq 3-11) yields:

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

(Eq 3-12)

Plugging (Eq 3-12) into (Eq 3-11) gives:

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

(Eq 3-13)

It should be noted that the summation index in (Eq 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 (Eq 3-13) is brought outside the summation. To reproduce (Eq 3-8)

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

(Eq 3-14)

Problem 4

Given Information

The following equation is given for ds^2<\sup>:

$ds^{2}=\sum _{i=1}^{3}(dx_{i})^{2}$

(Eq 4-1)

x_i is written in spherical coordinates in (Eq 4-2) through (Eq 4-4).

$x_{1}=rcos{\theta }cos{\varphi }$

(Eq 4-2)

$x_{2}=rcos{\theta }sin{\varphi }$

(Eq 4-3)

$x_{3}=rsin{\theta }$

(Eq 4-4)

Problem Statement

Show that the infinitesimal length ds² in (Eq 4-1) can be written in spherical coordinates, (Eq 4-5), as given in Pb.R*7.3 on pp.39-[1,3] of sec.39 of the notes.

$ds^{2}=dr^{2}+r^{2}d{\theta }^{2}+r^{2}cos^{2}{\theta }d{\varphi }^{2}$

(Eq 4-5)

Solution

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

The derivatives of x_i are given in (Eq 4-6) through (Eq 4-8).

$dx_{1}=cos{\theta }cos{\varphi }dr-rsin{\theta }cos{\varphi }d{\theta }-rcos{\theta }sin{\varphi }d{\varphi }$

(Eq 4-6)

$dx_{2}=cos{\theta }sin{\varphi }dr-rsin{\theta }sin{\varphi }d{\theta }+rcos{\theta }cos{\varphi }d{\varphi }$

(Eq 4-7)

$dx_{3}=sin{\theta }dr+rcos{\theta }d{\theta }$

(Eq 4-8)

(Eq 4-9) was developed by substituting the squared derivatives in the summation of (Eq 4-1) and combining like terms.

$ds^{2}=(sin^{2}{\theta }(cos^{2}{\varphi }+sin^{2}{\varphi })+cos^{2}{\theta })dr^{2}+(2rsin{\theta }cos{\theta }-2rsin{\theta }cos{\theta }(sin^{2}{\varphi }+cos^{2}{\varphi }))drd{\theta }+(r^{2}(sin^{2}{\theta }(sin^{2}{\varphi }+cos^{2}{\varphi })+cos^{2}{\theta })d{\theta }^{2}+(r^{2}cos^{2}{\theta }(sin^{2}{\varphi }+cos^{2}{\varphi }))d{\varphi }^{2}$

(Eq 4-9)

Using the Pythagorean Trigonometric Identity, (Eq 4-9) reduces to (Eq 4-10).

$ds^{2}=dr^{2}+r^{2}d{\theta }^{2}+r^{2}cos^{2}d{\varphi }^{2}$

(Eq 4-10)

Problem 5

Given Information

The following equation is given for P₂(x), the Legendre Polynomial of degree n=2:

$P_{2}(x)={\dfrac {1}{2}}(3x^{2}-1)$

(Eq 5-1)

Problem Statement

Using variation of parameters, show that Equation 5-2 is a second solution to the Legendre differential equation in Equation 5-3 as given in Pb.R*6.11 on pp.37-4 in sec.37 of the notes.

$Q_{2}(x)={\dfrac {1}{4}}(3x^{2}-1)log({\dfrac {1+x}{1-x}})-{\dfrac {3}{2}}x$

(Eq 5-2)

Solution

The Legendre differential equation of degree n=2 is given in equation 5-3.

$(1-x^{2})y''-2xy'+6y=0$

(Eq 5-3)

Using reduction of order, some function, v, times P₂(x) and the first two derivatives of the product are substituted into the Legendre differential equation in equation 5-7. Furthermore, since P₂(x) is a solution to the Legendre equation the coefficient of the first order term, v, reduces to zero. Since the lowest order term cancels out, the second order differential equation is reduced to a first order differential equation where w=v' and w'=v in Equation 5-8 which can be solved through separation of variables.

$R_{2}(x)=v({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})$

(Eq 5-4)

$R'_{2}(x)=v'({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})+v(3x)$

(Eq 5-5)

$R''_{2}(x)=v''({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})+6v'x+3v$

(Eq 5-6)

$(1-x^{2})({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})v''+(6x(1-x^{2})-2x({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}}))v'+(3(1-x^{2})-2x(3x)+6({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}}))v=0$

(Eq 5-7)

${\dfrac {dw}{w}}={\dfrac {(6x(1-x^{2})-2x({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}}))dx}{(1-x^{2})({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})}}$

(Eq 5-7)

Through partial fraction decomposition, the integral is

$\int {\dfrac {dw}{w}}=\int {\dfrac {-6x}{{\dfrac {3}{2}}x^{2}dx-{\dfrac {1}{2}}}}+\int {\dfrac {2x}{1-x^{2}}}dx$

(Eq 5-8)

Through integration

$log(w)=-2log({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})-log(1-x^{2})$

(Eq 5-9)

Using properties of the natural logarithm and then inverting it using the exponential function, equation 5-9 becomes

$w={\dfrac {1}{({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})(1-x^{2})}}$

(Eq 5-10)

Substituting v' back into Equation 5-10 and again using partial fraction decomposition

$\int {dv}=\int {\dfrac {dx}{4(1-x^{2})}}-{\dfrac {3}{4}}\int {\dfrac {(3x^{2}+1)}{({\dfrac {3}{2}}x^{2}-{\dfrac {1}{2}})^{2}}}$

(Eq 5-11)

Completing the integration

$v={\dfrac {-1}{8}}log({\dfrac {1+x}{1-x}})-{\dfrac {3}{4}}{\dfrac {x}{3x^{2}-1}}$

(Eq 5-12)

Substituting v multiplied by the arbitrary constant 8 back into Equation 5-4 then identifying the new solution as Q₂

$Q_{2}(x)={\dfrac {1}{4}}(3x^{2}-1)log({\dfrac {1+x}{1-x}})-{\dfrac {3x}{2}}$

(Eq 5-12)

Problem 6

Given Information

The following equations are given for heat conduction on a cylinder

$x=r\cos \theta =\xi _{1}\,\cos {\xi _{2}}$

(Eq 6-1)

$y=r\sin \theta =\xi _{1}\,\sin {\xi _{2}}$

(Eq 6-2)

$z=\xi _{3}$

(Eq 6-3)

Problem Statement

(a) Find

$\left\{dx_{i}\right\}=\left\{dx_{1},dx_{2},dx_{3}\right\}$

(Eq 6-4)

in terms of

$\left\{\xi _{j}\right\}=\left\{\xi _{1},\xi _{2},\xi _{3}\right\}$

(Eq 6-5)

and

$\left\{d\xi _{k}\right\}=\left\{d\xi _{1},d\xi _{2},d\xi _{3}\right\}$

(Eq 6-6)

(b) Find

$ds^{2}=\sum _{i}(dx_{i})^{2}=\sum _{k}(h_{k})^{2}(d\xi _{k})^{2}$

(Eq 6-7)

Identify $\{h_{i}\}$ in terms of $\{\xi _{i}\}$

(c) Find $\Delta u$ in terms of cylindrical coordinates

(d) Use separation of variable to find the separated equations of Laplace's equation in cylindrical coordinates and compare to the Bessel equation:

$x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+(x^{2}-\alpha ^{2})y=0$

(Eq 6-8)

Solution

Part a

The total differential of $x$ can be written as:

$dx={\frac {\partial x}{\partial \xi _{1}}}d\xi _{1}+{\frac {\partial x}{\partial \xi _{2}}}d\xi _{2}+{\frac {\partial x}{\partial \xi _{3}}}d\xi _{3}$

(Eq 6-9)

Similar expressions can be written for $y$ and $z$.

$dy={\frac {\partial y}{\partial \xi _{1}}}d\xi _{1}+{\frac {\partial y}{\partial \xi _{2}}}d\xi _{2}+{\frac {\partial y}{\partial \xi _{3}}}d\xi _{3}$

(Eq 6-10)

$dz={\frac {\partial z}{\partial \xi _{1}}}d\xi _{1}+{\frac {\partial z}{\partial \xi _{2}}}d\xi _{2}+{\frac {\partial z}{\partial \xi _{3}}}d\xi _{3}$

(Eq 6-11)

Plugging in the expressions from (Eq 6-1), (Eq 6-2) and (Eq 6-3):

$dx=\cos {\xi _{2}}\,d\xi _{1}-\xi _{1}\,\sin {\xi _{2}}\,d\xi _{2}$

(Eq 6-12)

$dy=\sin {\xi _{2}}\,d\xi _{1}+\xi _{1}\,\cos {\xi _{2}}\,d\xi _{2}$

(Eq 6-13)

$dz=d\xi _{3}$

(Eq 6-14)

Part b

Plugging in (Eq 6-12), (Eq 6-13), and (Eq 6-14) into (Eq 6-7).

$ds^{2}=(h_{1})^{2}(\cos {\xi _{2}}\,d\xi _{1}-\xi _{1}\sin {\xi _{2}}\,d\xi _{2})^{2}+(h_{2})^{2}(\sin {\xi _{2}}\,d\xi _{1}+\xi _{1}\cos {\xi _{2}}\,d\xi _{2})^{2}+(h_{3})^{2}(d\xi _{3})^{2}$

(Eq 6-15)

Expanding (Eq 6-14) out, and applying the following trig identity:

$\sin ^{2}{\xi _{2}}+\cos ^{2}{\xi _{2}}=1$

(Eq 6-16)

Results in the following expression:

$ds^{2}=(d\xi _{1})^{2}+(\xi _{1})^{2}(d\xi _{2})^{2}+(d\xi _{3})^{2}$

(Eq 6-17)

Comparing (Eq 6-17) to the left hand side of (Eq 6-7):

$h_{1}=1$

(Eq 6-18)

$h_{2}=\xi _{2}$

(Eq 6-19)

$h_{3}=1$

(Eq 6-20)

Part c

The Laplacian operator for a scalar function $u$ in general curvilinear coordinates is defined as:

$\Delta u={\frac {1}{h_{1}h_{2}h_{3}}}\left[{\frac {\partial }{\partial \xi _{1}}}\left({\frac {h_{2}h_{3}}{h_{1}}}{\frac {\partial }{\partial \xi _{1}}}\right)+{\frac {\partial }{\partial \xi _{2}}}\left({\frac {h_{1}h_{3}}{h_{2}}}{\frac {\partial }{\partial \xi _{2}}}\right)+{\frac {\partial }{\partial \xi _{3}}}\left({\frac {h_{1}h_{2}}{h_{3}}}{\frac {\partial }{\partial \xi _{3}}}\right)\right]u$

(Eq 6-21)

Plugging in (Eq 6-18), (Eq 6-19) and (Eq 6-20).

$\Delta u={\frac {1}{\xi _{1}}}\left[{\frac {\partial }{\partial \xi _{1}}}\left(\xi _{1}{\frac {\partial }{\partial \xi _{1}}}\right)+{\frac {\partial }{\partial \xi _{2}}}\left({\frac {1}{\xi _{1}}}{\frac {\partial }{\partial \xi _{2}}}\right)+{\frac {\partial }{\partial \xi _{3}}}\left(\xi _{1}{\frac {h_{1}h_{2}}{h_{3}}}{\frac {\partial }{\partial \xi _{3}}}\right)\right]u$

(Eq 6-22)

$\Delta u={\frac {1}{\xi _{1}}}{\frac {\partial u}{\partial \xi _{1}}}+{\frac {\partial ^{2}u}{\partial \xi _{1}^{2}}}+{\frac {1}{\xi _{1}^{2}}}{\frac {\partial ^{2}u}{\partial \xi _{2}^{2}}}+{\frac {\partial ^{2}u}{\partial \xi _{3}^{2}}}$

(Eq 6-23)

$\Delta u={\frac {1}{r}}{\frac {\partial u}{\partial r}}+{\frac {\partial ^{2}u}{\partial r^{2}}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}u}{\partial \theta ^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}$

(Eq 6-24)

Part d

Laplace's equation in cylindrical coordinates is:

$\Delta u=0$

(Eq 6-25)

${\frac {1}{r}}{\frac {\partial u}{\partial r}}+{\frac {\partial ^{2}u}{\partial r^{2}}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}u}{\partial \theta ^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}=0$

(Eq 6-26)

Assume a solution to (Eq 6-26) has the following form:

$u(r,\theta ,z)=R(r)\Theta (\theta )Z(z)$

(Eq 6-27)

Plugging (Eq 6-27) into (Eq 6-26):

$\Theta (\theta )Z(z){\frac {1}{r}}{\frac {dR(r)}{dr}}+\Theta (\theta )Z(z){\frac {d^{2}R(r)}{dr^{2}}}+R(r)Z(z){\frac {1}{r^{2}}}{\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}+R(r)\Theta (\theta ){\frac {d^{2}Z(z)}{dz^{2}}}=0$

(Eq 6-28)

Dividing by $u$

${\frac {1}{rR(r)}}{\frac {dR(r)}{dr}}+{\frac {1}{R(r)}}{\frac {d^{2}R(r)}{dr^{2}}}+{\frac {1}{\Theta (\theta )}}{\frac {1}{r^{2}}}{\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}+{\frac {1}{Z(z)}}(\theta ){\frac {d^{2}Z(z)}{dz^{2}}}=0$

(Eq 6-29)

The terms in z are then "seperated" and set equal to a constant: $\lambda ^{2}$.

${\frac {1}{rR(r)}}{\frac {dR(r)}{dr}}+{\frac {1}{R(r)}}{\frac {d^{2}R(r)}{dr^{2}}}+{\frac {1}{\Theta (\theta )}}{\frac {1}{r^{2}}}{\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}+\lambda ^{2}=0$

(Eq 6-30)

Multiplying by $r^{2}$:

${\frac {r}{R(r)}}{\frac {dR(r)}{dr}}+{\frac {r^{2}}{R(r)}}{\frac {d^{2}R(r)}{dr^{2}}}+{\frac {1}{\Theta (\theta )}}{\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}+(\lambda r)^{2}=0$

(Eq 6-31)

Now the $\theta$ term can be separated and set to $-\alpha ^{2}$.

${\frac {r}{R(r)}}{\frac {dR(r)}{dr}}+{\frac {r^{2}}{R(r)}}{\frac {d^{2}R(r)}{dr^{2}}}+-\alpha ^{2}+(\lambda r)^{2}=0$

(Eq 6-32)

Making a change of variable:

$\rho (r)=\lambda r$

(Eq 6-33)

${\frac {dR(r)}{dr}}=\lambda {\frac {dR(\rho )}{d\rho }}$

(Eq 6-34)

${\frac {d^{2}R(r)}{dr^{2}}}=\lambda ^{2}{\frac {d^{2}R(\rho )}{d\rho ^{2}}}$

(Eq 6-35)

Plugging the above change of variable into (Eq 6-32) and multiplying by $R(\rho )$ gives:

$\rho ^{2}{\frac {d^{2}R(\rho )}{dr^{2}}}+\rho {\frac {dR(\rho )}{dr}}+(\rho ^{2}-\alpha ^{2})R(\rho )=0$

(Eq 6-36)

Which is Bessel's equation given in (Eq 6-8).

Problem 7

Given Information

The following equations is given

$e^{-2}\int _{2}^{\infty }{\frac {1}{x\log ^{2}(x)}}\,dx$

(Eq 7-1)

and

$J_{0}(z)={\frac {1}{\pi }}\int _{0}^{\pi }\cos(-z\sin t)\,dt={\frac {2}{\pi }}\int _{0}^{\infty }\sin(z\cosh t)\,dt$

(Eq 7-2)

(Eq 7-1) is motivated by quadrature tests written by Dr. John Burkardt found here:

http://people.sc.fsu.edu/~%20jburkardt/c_src/test_int_laguerre/test_int_laguerre.html

Problem Statement

1. Find the values of the integral of the function (7-1) by other means (exact or approximate), then evaluate these integrals with the Gauss-Laguerre quadrature for comparison.

2. Find and appropriate transformation to show the equality of (7-2) and (7-3). Find the value of $J_{0}(1/2)$ in some table and use the Gauss-Laguerre quadrature to find an approximate value of the integral to compare.

Contributing Team Members

• Cameron Stewart Solved problems 1 and 4
• Elizabeth Bartlett Solved problem 2 and 3
• Kevin Frost Reviewed and formatted
• Christopher Neal Solved Problems 1 and 2

Problem Assignments
Problem #	Solved by	Reviewed by
1	Neal, Christopher	all
2	Neal, Christopher	all
3	Bartlett, Elizabeth	all
4	Stewart, Cameron	all
5	Bartlett, Elizabeth	all
6	Bartlett, Elizabeth	all
7	Bartlett, Elizabeth	all