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title: User talk:Sablyth


text: File:PEW07 28-10-2009.pdf

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:PEW07 28-10-2009.pdf


text:
User:Egm6321.f09.Team1.sallstrom/HW6

= Problem 1: Laplacian in Curvilinear Coordinates = From lecture note slide 31-1, 32-1.

Given
Circular cylinder coordinates are given as:
 * {| style="width:100%" border="0" align="left"

\left. \begin{matrix} x &=& r \cos \theta &=& \xi_1 &\cos \xi_2 \\ y &=& r \sin \theta &=& \xi_1 &\sin \xi_2 \\ z &&&=& \xi_3& \end{matrix} \right. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find
 * {| style="width:100%" border="0" align="left"

\{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ $$ in terms of
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\{{\rm d} \xi_k\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

2. Find
 * {| style="width:100%" border="0" align="left"

{\rm d}s^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$ $$ Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

3. Find the Laplacian in these curvilinear coordinates.

4. Obtain the separated equations for Laplace equation on circular cylinder coordinates, and identify the Bessel differential equation
 * {| style="width:100%" border="0" align="left"

x^2 y'' + x y' - (x^2 - \nu^2) y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Solution
1. We find Eq. 2 by differentiating Eqs. 1 w.r.t. $$\displaystyle \xi_i$$
 * {| style="width:100%" border="0" align="left"

{\rm d}x_1 = \frac{\partial x_1}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_1}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_1}{\partial \xi_3} \, {\rm d}\xi_3 = \cos \xi_2 \, {\rm d}\xi_1 - \xi_1 \sin \xi_2 \, {\rm d} \xi_2 + 0 \, {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{\rm d}x_2 = \frac{\partial x_2}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_2}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_2}{\partial \xi_3} \, {\rm d}\xi_3 = \sin \xi_2 \, {\rm d}\xi_1 + \xi_1 \cos \xi_2 \, {\rm d} \xi_2 + 0 \, {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

{\rm d}x_3 = \frac{\partial x_3}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_3}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_3}{\partial \xi_3} \, {\rm d}\xi_3 = 0 \, {\rm d}\xi_1 + 0 \, {\rm d} \xi_2 + 1 \, {\rm d} \xi_3 $$ $$ So
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{matrix} {\rm d}x_1 & = & \cos \xi_2 \, {\rm d}\xi_1 - \xi_1 \sin \xi_2 \, {\rm d} \xi_2 \\ {\rm d}x_2 & = & \sin \xi_2 \, {\rm d}\xi_1 + \xi_1 \cos \xi_2 \, {\rm d} \xi_2 \\ {\rm d}x_3 & = & {\rm d} \xi_3 \end{matrix} $$ $$ 2. Use Eqs. 10 in Eq. 5
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{matrix} {\rm d}s^2 &= & \cos^2 \xi_2 \left({\rm d}\xi_1 \right)^2 - 2 \xi_1 \sin \xi_2 \cos \xi_2 \, {\rm d}\xi_1 {\rm d} \xi_2 + \xi_1^2 \sin^2 \xi_2 \left({\rm d} \xi_2\right)^2 \\ &+& \sin^2 \xi_2 \left({\rm d}\xi_1 \right)^2 + 2 \xi_1 \sin \xi_2 \cos \xi_2 \, {\rm d}\xi_1 {\rm d} \xi_2 + \xi_1^2 \cos^2 \xi_2 \left({\rm d} \xi_2 \right)^2 \\ &+& \left( {\rm d} \xi_3 \right)^2 \end{matrix} $$ $$ which simplifies to
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle {\rm d}s^2 = \left({\rm d}\xi_1 \right)^2 + \xi_1^2 \left({\rm d} \xi_2\right)^2 + \left( {\rm d} \xi_3 \right)^2 $$ $$ So
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle h_1 = 1, \quad h_2 = \xi_1, \quad h_3 = 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 13)
 * }
 * }

3. Eq. 1 on lecture slide 29-3 shows that the Laplacian is given by
 * {| style="width:100%" border="0" align="left"

\Delta = \nabla^2 = \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial}{\partial \xi_i} \right] $$ $$ We can insert the results from part 2, term by term, so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\Delta_1 = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left[ \xi_1 \frac{\partial}{\partial \xi_1} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Delta_2 = \frac{1}{\xi_1^2} \frac{\partial^2}{\partial \xi_2^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Delta_3 = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_3} \left[ \xi_1 \frac{\partial}{\partial \xi_3} \right] = \frac{\partial^2}{\partial \xi_3^2} $$ $$ Combining these yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2}{\partial \xi_2^2} + \frac{\partial^2}{\partial \xi_3^2} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 18)
 * }
 * }

4. Assume
 * {| style="width:100%" border="0" align="left"

f(\xi_1, \xi_2, \xi_3) = X_1(\xi_1) X_2(\xi_2) X_3(\xi_3) $$ $$ Using Laplace operator from Eq. 18 on $$\displaystyle f$$, we can write Laplace equation
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\Delta f = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial f}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 f}{\partial \xi_2^2} + \frac{\partial^2 f}{\partial \xi_3^2} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

X_2 X_3 \frac{1}{\xi_1} \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + X_1 X_3 \frac{1}{\xi_1^2} \frac{\partial^2 X_2}{\partial \xi_2^2} + X_1 X_2 \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$ $$ Divide by $$\displaystyle X_1 X_2 X_3$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} + \frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$ $$ The last term is only dependent on $$\displaystyle \xi_3$$, and is the only term dependent on $$\displaystyle \xi_3$$, hence it must be a constant. So we substitute
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 22)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \beta^2 = 0 $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 23)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\beta^2 = -\frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} $$ $$ Now, multiply by $$\displaystyle \xi_1^2$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 24)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \xi_1^2 \beta^2 = 0 $$ $$ Now, the second term is only dependent on $$\displaystyle \xi_2$$, and it is the only term dependent on $$\displaystyle \xi_2$$. It must also be a constant. We can substitute
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\nu^2 = \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} $$ $$ So that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \nu^2 - \xi_1^2 \beta^2 = 0 $$ $$ Expand the first term
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 27)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{\xi_1^2}{ X_1 } \frac{\partial^2 X_1}{\partial \xi_1^2} + \frac{\xi_1}{ X_1 } \frac{\partial X_1}{\partial \xi_1} + \nu^2 - \xi_1^2 \beta^2 = 0 $$ $$ Multiply by $$\displaystyle X_1$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 28)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\xi_1^2 \frac{\partial^2 X_1}{\partial \xi_1^2} + \xi_1 \frac{\partial X_1}{\partial \xi_1} - \left( \xi_1^2 \beta^2 - \nu^2 \right) X_1 = 0 $$ $$ Now, substitute $$\displaystyle x = \xi_1 \beta$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 29)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

x^2 \frac{\partial^2 X_1}{\partial x^2} + x \frac{\partial X_1}{\partial x} - \left( x^2 - \nu^2 \right) X_1 = 0 $$ $$ Then, with $$\displaystyle X_1 = y(x)$$, we get
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 30)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

x^2 \frac{\partial^2 y}{\partial x^2} + x \frac{\partial y}{\partial x} - \left( x^2 - \nu^2 \right) y = 0 $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 31)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle x^2 y'' + x y' - \left( x^2 - \nu^2 \right) y = 0 $$ $$ which is the same as Eq. 6, i.e. the Bessel differential equation.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 32)
 * }
 * }

= Problem 2: Laplacian in Spherical Coordinates = From lecture note slide 31-1.

Given
Spherical coordinates are given as:
 * {| style="width:100%" border="0" align="left"

\left. \begin{matrix} x &=& r & \cos \theta & \cos \varphi &=& \xi_1 & \cos \xi_3 & \cos \xi_2 \\ y &=& r & \cos \theta & \sin \varphi &=& \xi_1 & \cos \xi_3 & \sin \xi_2 \\ z &=& r & \sin \theta &             &=& \xi_1 & \sin \xi_3 & \end{matrix} \right. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find
 * {| style="width:100%" border="0" align="left"

\{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ $$ in terms of
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\{{\rm d} \xi_k\} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

2. Find
 * {| style="width:100%" border="0" align="left"

ds^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$ $$ Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

3. Find the Laplacian in these spherical coordinates

Solution
1. We find Eq. 2 by differentiating Eqs. 1 w.r.t. $$\displaystyle \xi_i$$
 * {| style="width:100%" border="0" align="left"

{\rm d}x_1 = \frac{\partial x_1}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_1}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_1}{\partial \xi_3} \, {\rm d}\xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

{\rm d}x_1 = \cos \xi_3 \cos \xi_2 \, {\rm d}\xi_1 - \xi_1 \cos \xi_3 \sin \xi_2 \, {\rm d} \xi_2 - \xi_1 \sin \xi_3 \cos \xi_2 \, {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{\rm d}x_2 = \frac{\partial x_2}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_2}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_2}{\partial \xi_3} \, {\rm d}\xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

{\rm d}x_2 = \cos \xi_3 \sin \xi_2 \, {\rm d}\xi_1 + \xi_1 \cos \xi_3 \cos \xi_2 \, {\rm d} \xi_2 - \xi_1 \sin \xi_3 \sin \xi_2 \, {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{\rm d}x_3 = \frac{\partial x_3}{\partial \xi_1} \, {\rm d}\xi_1 + \frac{\partial x_3}{\partial \xi_2} \, {\rm d}\xi_2 + \frac{\partial x_3}{\partial \xi_3} \, {\rm d}\xi_3 = \sin \xi_3 \, {\rm d}\xi_1 + 0 \, {\rm d} \xi_2 + \xi_1 \cos \xi_3 \, {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }

So
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{matrix} {\rm d}x_1 & = & \cos \xi_3 \cos \xi_2 &{\rm d}\xi_1& -& \xi_1 \cos \xi_3 \sin \xi_2 &{\rm d} \xi_2& -& \xi_1 \sin \xi_3 \cos \xi_2 &{\rm d} \xi_3 \\ {\rm d}x_2 & = & \cos \xi_3 \sin \xi_2 &{\rm d}\xi_1& +& \xi_1 \cos \xi_3 \cos \xi_2 &{\rm d} \xi_2& -& \xi_1 \sin \xi_3 \sin \xi_2 &{\rm d} \xi_3 \\ {\rm d}x_3 & = & \sin \xi_3 \, &{\rm d}\xi_1& &&& +&\xi_1 \cos \xi_3 &{\rm d} \xi_3 \end{matrix} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

2. Use Eqs. 11 in Eq. 5
 * {| style="width:100%" border="0" align="left"

\begin{matrix} {\rm d}s^2 & = & &\cos^2 \xi_3 & \cos^2 \xi_2 &{\rm d}\xi_1^2 \\ &-& 2 \xi_1 & \cos^2 \xi_3 & \sin \xi_2 \cos \xi_2 & {\rm d}\xi_1 {\rm d} \xi_2 \\ &-& 2 \xi_1 & \sin \xi_3 \cos \xi_3 & \cos^2 \xi_2 & {\rm d}\xi_1 {\rm d} \xi_3 \\ &+& \xi_1^2 &\cos^2 \xi_3 & \sin^2 \xi_2 & {\rm d}\xi_2^2 \\ &+& 2 \xi_1^2 &\sin \xi_3 \cos \xi_3 & \sin \xi_2 \cos \xi_2 & {\rm d} \xi_2 {\rm d} \xi_3 \\ &+& \xi_1^2 & \sin^2 \xi_3 & \cos^2 \xi_2 & {\rm d} \xi_3^2 \\ &+& & \cos^2 \xi_3 & \sin^2 \xi_2 &{\rm d}\xi_1^2 \\ &+& 2 \xi_1 & \cos^2 \xi_3 & \sin \xi_2 \cos \xi_2 & {\rm d}\xi_1 {\rm d} \xi_2 \\ &-& 2 \xi_1 & \sin \xi_3 \cos \xi_3 & \sin^2 \xi_2 & {\rm d}\xi_1 {\rm d} \xi_3 \\ &+& \xi_1^2 & \cos^2 \xi_3 & \cos^2 \xi_2 & {\rm d}\xi_2^2 \\ &-& 2 \xi_1^2 & \sin \xi_3 \cos \xi_3 & \sin \xi_2 \cos \xi_2 & {\rm d} \xi_2 {\rm d} \xi_3 \\ &+& \xi_1^2 & \sin^2 \xi_3 &\sin^2 \xi_2 & {\rm d} \xi_3^2 \\ &+& & \sin^2 \xi_3 && {\rm d} \xi_1^2 \\ &+& 2 \xi_1 &\sin \xi_3 \cos \xi_3 && {\rm d} \xi_1 {\rm d} \xi_3\\ &+& \xi_1^2 &\cos^2 \xi_3 && {\rm d} \xi_3^2 \end{matrix} $$ $$ this simplifies to
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle {\rm d}s^2 = {\rm d}\xi_1^2 + \xi_1^2 \cos^2 \xi_3 {\rm d}\xi_2^2 + \xi_1^2 {\rm d} \xi_3^2 $$ $$ Hence
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle h_1 = 1,\quad h_2 = \xi_1 \cos \xi_3, \quad h_3 = \xi_1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 14)
 * }
 * }

3. Eq. 1 on lecture slide 29-3 shows that the Laplacian is given by
 * {| style="width:100%" border="0" align="left"

\Delta = \nabla^2 = \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial}{\partial \xi_i} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * }

We can insert the results from part 2, term by term, so that
 * {| style="width:100%" border="0" align="left"

\Delta_1 = \frac{1}{\xi_1^2 \cos \xi_3} \frac{\partial}{\partial \xi_1} \left[ \xi_1^2 \cos \xi_3 \frac{\partial}{\partial \xi_1} \right] = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left[ \xi_1^2 \frac{\partial}{\partial \xi_1} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Delta_2 = \frac{1}{\xi_1^2 \cos \xi_3} \frac{\partial}{\partial \xi_2} \left[ \frac{1}{\cos \xi_3} \frac{\partial}{\partial \xi_2} \right] = \frac{1}{\xi_1^2 \cos^2 \xi_3} \frac{\partial^2}{\partial \xi_2^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Delta_3 = \frac{1}{\xi_1^2 \cos \xi_3} \frac{\partial}{\partial \xi_3} \left[ \cos \xi_3 \frac{\partial}{\partial \xi_3} \right] $$ $$ Combining these yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \frac{\partial}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \cos^2 \xi_3} \frac{\partial^2}{\partial \xi_2^2} + \frac{1}{\xi_1^2 \cos \xi_3} \frac{\partial}{\partial \xi_3} \left( \cos \xi_3 \frac{\partial}{\partial \xi_3} \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

= Problem 3: Laplacian in Spherical Coordinates (Math/Physics Convention) = From lecture note slide 31-2.

Given
Spherical coordinates are given as:
 * {| style="width:100%" border="0" align="left"

\left. \begin{matrix} x &=& r & \sin \bar \theta & \cos \varphi &=& \xi_1 & \sin \bar \xi_3 & \cos \xi_2 \\ y &=& r & \sin \bar \theta & \sin \varphi &=& \xi_1 & \sin \bar \xi_3 & \sin \xi_2 \\ z &=& r & \cos \bar \theta &             &=& \xi_1 & \cos \bar \xi_3 & \end{matrix} \right. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the Laplacian in these spherical coordinates

Solution
We have that
 * {| style="width:100%" border="0" align="left"

\bar \theta = \frac{\pi}{2} - \theta $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\bar \xi_3 = \frac{\pi}{2} - \xi_3 $$ $$ so that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\cos \xi_3 = \sin \bar \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\sin \xi_3 = \cos \bar \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\partial \xi_3 = - \partial \bar \xi_3 $$ $$ Substituting these into Eq. 19 from problem 3 yields
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \frac{\partial}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \sin^2 \bar \xi_3} \frac{\partial^2}{\partial \xi_2^2} - \frac{1}{\xi_1^2 \sin \bar \xi_3} \frac{\partial}{\partial \bar \xi_3} \left[ \sin \bar \xi_3 \left( - \frac{\partial}{\partial \bar \xi_3} \right) \right] $$ $$ or simply
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \frac{\partial}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \sin^2 \bar \xi_3} \frac{\partial^2}{\partial \xi_2^2} + \frac{1}{\xi_1^2 \sin \bar \xi_3} \frac{\partial}{\partial \bar \xi_3} \left( \sin \bar \xi_3 \frac{\partial}{\partial \bar \xi_3} \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

= Problem 4: = From lecture note slide 31-2.

Given
The following equation is given
 * {| style="width:100%" border="0" align="left"

\lambda ( \lambda + 1 ) = n ( n + 1 ) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the two solutions

 * {| style="width:100%" border="0" align="left"

\lambda = n $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\lambda = -(n + 1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Solution
Rewrite Eq. 1
 * {| style="width:100%" border="0" align="left"

\lambda^2 + \lambda = n^2 + n $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\left(\lambda + \frac{1}{2} \right)^2 - \frac{1}{4} = \left(n + \frac{1}{2} \right) - \frac{1}{4} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\left(\lambda + \frac{1}{2} \right)^2 = \left(n + \frac{1}{2} \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\lambda + \frac{1}{2} = \pm \left( n + \frac{1}{2}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \lambda_1 = n $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\lambda_2 + \frac{1}{2} = - n - \frac{1}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \lambda_2 = -\left( n +1 \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }

= Problem 5: Legendre Polynominal 1 = From lecture note slide 31-3.

Given
The general form of the Legendre polynomials can be written
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$ where $$\displaystyle [n/2]$$ means the integer part of $$\displaystyle n/2$$. Another way to write it is
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Find
Show that Eqs. 1 and 2 are equivalent

Solution
Rearrange Eq. 1
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i  x^{n-2i}}{2^i i!(n-2i)!} \frac{(2 n - 2 i)!}{2^{n-i} (n-i)! } $$ $$ Rearrange Eq. 2
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i x^{n-2i}}{2^i i!(n-2i)!} 1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right) $$ $$ Now we can rewrite Eq. 3
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i  x^{n-2i}}{2^i i!(n-2i)!} \frac{\prod_{j=1}^{2n - 2i} j}{\prod_{j=1}^{n - i} 2 \prod_{j=1}^{n - i} j } $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i  x^{n-2i}}{2^i i!(n-2i)!} \frac{\prod_{j=1}^{2n - 2i} j}{\prod_{j=1}^{n - i} 2 j } $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i  x^{n-2i}}{2^i i!(n-2i)!} \frac{1 \cdot \cancel{2} \cdot 3 \cdot \cancel{4} \cdot ... \cdot (2n - 2i - 1) \cdot \cancel{2(n-i)}}{\cancel{2} \cdot \cancel{4} \cdot ... \cdot \cancel{2(n-i)}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i  x^{n-2i}}{2^i i!(n-2i)!} 1 \cdot 3 \cdot ... \cdot (2n - 2i - 1) $$ $$ Eqs. 8 and 4 are the same, hence Eqs. 1 and 2 are equivalent.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

= Problem 6: Legendre Polynominal 2 = From lecture note slide 31-3.

Given
The first 5 Legendre polynomials are
 * {| style="width:100%" border="0" align="left"

P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{1}{2} \left(3 x^2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{1}{2} \left(5 x^3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Find
Verify that polynomials given by Eqs. 1, 2, 3, 4, and 5 can be written as
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Solution
0. Substitute $$\displaystyle n=0$$ into Eq. 6
 * {| style="width:100%" border="0" align="left"

P_0(x) = \sum_{i=0}^{[0/2]} (-1)^i \frac{(2 \cdot 0 - 2 i)! x^{0 - 2i}}{2^0 i! (0-i)! (0-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_0(x) = \sum_{i=0}^0 (-1)^i \frac{(- 2 i)! x^{ - 2i}}{i! (-i)! (-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_0(x) = \frac{0! }{0! (0)! (-0)!} = \frac{1}{1\cdot 1\cdot 1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_0(x) = 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }

1. Substitute $$\displaystyle n=1$$ into Eq. 6
 * {| style="width:100%" border="0" align="left"

P_1(x) = \sum_{i=0}^{[1/2]} (-1)^i \frac{(2 \cdot 1 - 2 i)! x^{1 - 2i}}{2^1 i! (1-i)! (1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_1(x) = \sum_{i=0}^{0} (-1)^i \frac{(2 - 2 i)! x^{1 - 2i}}{2 (i!) (1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_1(x) = \frac{(2)! x}{2 (1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_1(x) = \frac{2 x}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_1(x) = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 15)
 * }
 * }

2. Substitute $$\displaystyle n=2$$ into Eq. 6
 * {| style="width:100%" border="0" align="left"

P_2(x) = \sum_{i=0}^{[2/2]} (-1)^i \frac{(2 \cdot 2 - 2 i)! x^{2 - 2i}}{2^2 i! (2-i)! (2-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_2(x) = \sum_{i=0}^{1} (-1)^i \frac{(4 - 2 i)! x^{2 - 2i}}{4 i! (2-i)! (2-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{(4)! x^{2}}{4 (1!)(2!) (2!)} -1 \cdot \frac{2!}{4} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{24 x^2}{16} -\frac{1}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_2(x) = \frac{1}{2} \left( 3 x^2 - 1 \right) $$ $$ 3. Substitute $$\displaystyle n=3$$ into Eq. 6
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 20)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_3(x) = \sum_{i=0}^{1} (-1)^i \frac{(6 - 2 i)! x^{3 - 2i}}{8 (i!) (3-i)! (3-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{6! x^3}{8 (3)! (3)!} - \frac{(4)! x}{8 (2)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 22)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{5 }{2 } x^3 - \frac{3 x}{ 2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 23)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_3(x) = \frac{1}{2} \left( 5 x^3 - 3x \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 24)
 * }
 * }

4. Substitute $$\displaystyle n=4$$ into Eq. 6
 * {| style="width:100%" border="0" align="left"

P_4(x) = \sum_{i=0}^{2} (-1)^i \frac{(8 - 2 i)! x^{4 - 2i}}{16 (i!) (4-i)! (4-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 25)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{(8)! x^{4}}{16 (4)! (4)!} - \frac{(6)! x^{2}}{16 (3)! (2)!} + \frac{(4)! }{16 (2)! (2)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 26)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_4(x) = \frac{35 }{8} x^{4} - \frac{15 }{4} x^{2} + \frac{3}{8 } $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 27)
 * }
 * }

= Problem 7: Legendre Polynomial 3 = From lecture note slide 32-1.

Given
The first 5 Legendre polynomials are (from lecture note slide 31-3
 * {| style="width:100%" border="0" align="left"

P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{1}{2} \left(3 x^2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{1}{2} \left(5 x^3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Find
Verify that $$\displaystyle P_0$$, $$\displaystyle P_1$$, $$\displaystyle P_2$$, $$\displaystyle P_3$$, and $$\displaystyle P_4$$ are solutions to the Legendre polynomial (from lecture slide 14-2)
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Solution
0. For $$\displaystyle n=0$$, we substitute Eq. 1 into Eq. 6
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) 0 - 2 x 0 + 0(0+1) 1 = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 7)
 * }
 * }

1. For $$\displaystyle n=1$$, we substitute Eq. 2 into Eq. 6
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) 0 - 2 x + 1(1+1) x = -2x + 2x = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

2. For $$\displaystyle n=2$$, we substitute Eq. 3 into Eq. 6
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) 3 - 6 x^2 + 2(2+1)\frac{1}{2} \left(3 x^2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = 3-3 x^2 - 6 x^2 + 3 \left(3 x^2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = \left(-3-6+9\right) x^2 + 3 - 3 = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

3. For $$\displaystyle n=3$$, we substitute Eq. 4 into Eq. 6
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) 15 x - 2 x \frac{1}{2} \left(15x^2 - 3\right)+ 3(3+1)\frac{1}{2} \left(5 x^3 - 3x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = 15 x - 15 x^3 - 15x^3 + 3x + 30 x^3 - 18x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = 18 x - 30 x^3 + 30 x^3 - 18x = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 14)
 * }
 * }

4. For $$\displaystyle n=4$$, we substitute Eq. 5 into Eq. 6
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \left(\frac{105}{2} x^2 - \frac{15}{2} \right) - 2 x \left(\frac{35}{2} x^3 - \frac{15}{2} x\right) + 4(4+1) \left(\frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = \frac{105}{2} x^2 - \frac{15}{2} - \frac{105}{2} x^4 + \frac{15}{2} x^2 - 35 x^4 + 15 x^2 +  \left(\frac{175}{2} x^4 - 75 x^2 + \frac{15}{2}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = - \frac{105}{2} x^4 + \frac{175}{2} x^4 - 35 x^4 + \frac{105}{2} x^2 + \frac{15}{2} x^2  + 15 x^2 - 75 x^2 + \frac{15}{2}- \frac{15}{2} = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 17)
 * }
 * }

= Problem 8: Sums of Even and Odd Functions = From lecture slide 33-3

Given
Let
 * {| style="width:100%" border="0" align="left"

f = \sum_i g_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that 1. if the family $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd. 2. if the family $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
1. The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$.
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 2)
 * }
 * }

2. The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. So, if every $$\displaystyle g_i$$ is even, then
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 3)
 * }
 * }

= Problem 9: Legendre Polynomials' Even and Oddness = From lecture slide 33-3

Given
The Legendre polynomials are given by
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that, with $$\displaystyle k = 0,1,2,...$$, 1. $$\displaystyle P_{2k}(x)$$ is even, and 2. $$\displaystyle P_{2k+1}(x)$$ is odd

Solution
1. The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ in Eq. 1
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$ $$ Then, since both $$\displaystyle i$$ and $$\displaystyle k$$ are integers,
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} = P_{2k}(-x) $$ $$ 2. The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ in Eq. 1
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$ But Eq. 6 at $$\displaystyle -x$$ is
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$ $$ Comparing Eqs. 7 and 8 shows that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 9)
 * }
 * }

= Problem 10: Compute Legendre Coefficients Using the Gram Matrix Method = From lecture slide 33-4

Given
The polynomial
 * {| style="width:100%" border="0" align="left"

q(x) = \sum_{i=0}^4 c_i x^i $$ $$ with the coefficients
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Find
1. Find the coefficients $$\displaystyle \{ a_i \}$$ s.t. $$\displaystyle q(x) = \sum_{i=0}^4 a_i P_i(x)$$, where $$\displaystyle P_i(x)$$ are Legendre polynomials.

2. Plot
 * {| style="width:100%" border="0" align="left"

q = \sum_i c_i x^i $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

q = \sum_i a_i P_i(x) $$ $$ to verify that they are equal.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Solution
1. The first 5 Legendre polynomials are
 * {| style="width:100%" border="0" align="left"

\underbrace{ \left\{ \begin{matrix} P_0(x) \\ P_1(x) \\ P_2(x) \\ P_3(x) \\ P_4(x) \end{matrix} \right\} }_{\mathbf{P}(x)} = \left\{ \begin{matrix} 1 \\ x \\ \frac{1}{2} \left(3 x^2 - 1\right) \\ \frac{1}{2} \left(5 x^3 - 3 x\right) \\ \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8} \end{matrix} \right\} = \underbrace{ \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \end{matrix} \right] }_{A} \underbrace{ \left\{ \begin{matrix} 1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{matrix} \right\} }_{\mathbf{x}} $$ $$ Then we have that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

q(x) = \mathbf{a} \cdot \mathbf{P}(x) = \mathbf{a}^T A \mathbf{x} = \mathbf{c}^T \mathbf{x} $$ $$ the above is true for all $$\displaystyle \mathbf{x}$$ if
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mathbf{a}^T A = \mathbf{c}^T $$ $$ or
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mathbf{a}^T = \mathbf{c}^T A^{-1} $$ $$ The inverse of $$\displaystyle A$$ is
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

A^{-1} = \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \end{matrix} \right] $$ $$ Hence
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mathbf{a}^T = \left\{ \begin{matrix} 3 \\ 10 \\ 15 \\ -1 \\ 5 \end{matrix} \right\}^T \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \end{matrix} \right] = \left\{ \begin{matrix} 9 \\ \frac{47}{5} \\ \frac{90}{7} \\ -\frac{2}{5} \\ \frac{8}{7} \end{matrix} \right\}^T $$ $$ So the answer is
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \mathbf{a} = \left\{ \begin{matrix} 9 \\ \frac{47}{5} \\ \frac{90}{7} \\ -\frac{2}{5} \\ \frac{8}{7} \end{matrix} \right\} \approx \left\{ \begin{matrix} 9 \\ 9.4 \\ 12.8571 \\ -0.4 \\ 1.1429 \end{matrix} \right\} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

2. The two functions are plotted

Matlab source code:

= Problem 11: = From lecture slides 22-2, 22-3. If you did this in homework 4, just link to it.

= Problem 12: Gauss-Legendre = From lecture slide 34-2

Given
Consider the boundary condition
 * {| style="width:100%" border="0" align="left"

\psi(1,\theta) = f(\theta) = T_0 \cos \theta $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(\mu) = T_0 \sqrt{1 - \mu^2} $$ $$ is even, and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mu = \sin \theta $$ $$ Eq. 5 from lecture slide 33-2
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

A_n = \frac{2n + 1}{2} <f, P_n> $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Find
1. Without calculation, find property of $$\displaystyle A_n$$, i.e.
 * {| style="width:100%" border="0" align="left"

\begin{cases} A_{2k} = 0 \\ A_{2k+1} = 0 \end{cases} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

2. Compute 3 non-zero coefficients of $$\displaystyle A_n$$

2a. Analytical, using either$$\displaystyle \theta$$ or $$\displaystyle \mu$$ as integrating variable

2b. Numerically, using Gauss-Legendre, accurately within $$\displaystyle 5\%$$

Solution
1. Eq. 3 from lecture slide 33-1
 * {| style="width:100%" border="0" align="left"

<f, P_n> = \int_{-1}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ We can split the integral up in two intervals
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_n> = \int_{-1}^0 f(\mu) P_n(\mu) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ When $$\displaystyle n$$ is odd, so is $$\displaystyle P_n$$, and $$\displaystyle f(\mu)$$ is $$\displaystyle even$$. Hence, when $$\displaystyle n$$ is odd, we have that $$\displaystyle n = 2k + 1$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = \int_{-1}^0 f(-\mu) (-P_{2k+1}(-\mu)) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$ We can substitute $$\displaystyle \xi = -\mu$$ in the first integral so that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = \int_1^0 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = - \int_0^1 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu = 0 $$ $$ Hence
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle A_{2k+1} = \frac{2k + 2}{2} <f, P_{2k+1}> = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

2.

2a.

2b.

= Problem 13: Gauss-Legendre Quadrature = From lecture slide 35-3

Find
1. Verify the expressions on Wikipedia for the Legendre polynomials $$\displaystyle P_n$$ with $$\displaystyle n = 0, 1,..., 6$$. (See homework on lecture slide 31-3)

2. Verify the table for Gauss-Legendre quadrature in Wikipedia, analytical expressions of $$\displaystyle \{x_j\}$$ and $$\displaystyle \{w_j\}$$, $$\displaystyle j=1,...,5$$, and $$\displaystyle n=1,...,5$$, where $$\displaystyle n$$ is the number of integration points.

3. Evaluate numerically $$\displaystyle \{x_j\}$$ and $$\displaystyle \{w_j\}$$ and compare results with Abramovitz & Stegum. See lecture plan.

= Problem 14: Second Solution to the 0th Legendre Polynomial = From lecture slide 36-4

Given
The second solution to the 0th Legendre polynomial is
 * {| style="width:100%" border="0" align="left"

Q_0 = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that
 * {| style="width:100%" border="0" align="left"

Q_0 = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right) = \tanh^{-1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Solution
We have that
 * {| style="width:100%" border="0" align="left"

\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$ $$ If Eq. 2 holds, then $$\displaystyle \tanh(Q_0) = x$$. Hence, we substitute Eq. 1 into Eq. 3
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\tanh(Q_0) = \frac{e^{Q_0} - e^{-Q_0}}{e^{Q_0} + e^{-Q_0}} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{e^{\frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} - e^{-\frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)}}{e^{\frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)} + e^{-\frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)}} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{\sqrt{\frac{1+x}{1-x}} - \sqrt{\frac{1-x}{1+x}}} {\sqrt{\frac{1+x}{1-x}} + \sqrt{\frac{1-x}{1+x}}} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{\sqrt{\frac{(1+x)^2}{1-x^2}} - \sqrt{\frac{(1-x)^2}{1-x^2}}} {\sqrt{\frac{(1+x)^2}{1-x^2}} + \sqrt{\frac{(1-x)^2}{1-x^2}}} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{\sqrt{(1+x)^2} - \sqrt{(1-x)^2}} {\sqrt{(1+x)^2} + \sqrt{(1-x)^2}} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{1+x -1 +x} {1+x +1-x} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \frac{2x} {2} $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 4)
 * }
 * }

= Contributing Team Members =

title: User:Egm6321.f09.Team1.sallstrom/HW6


text:
User:Egm6321.f09.Team06.sada/HW5

Problem 2
p.29-4 Find equation (4)for i=3.

Solution

We know that, from p.29-3

$$\displaystyle \Delta \psi = \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial \psi}{\partial \xi_i} \right] $$

Using $$\displaystyle h_1 = 1, \quad h_2 = r \cos \theta, \quad h_3 = r $$

and

$$\displaystyle \xi_1 = r, \quad \xi_2 = \varphi, \quad \xi_3 = \theta $$

For i=1 by using the above relations, we get equ. 2 in p.29-4

$$\displaystyle \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial r} \left[ \frac{r^2 \cos \theta}{1} \frac{\partial \varphi}{\partial r} \right] $$

For i=2

$$\displaystyle \Delta \psi_2 = \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial \xi_2} \left[ \frac{h_1 h_2 h_3}{h_2^2} \frac{\partial \psi}{\partial \xi_2} \right] = \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial \varphi} \left[ \frac{r^2 \cos \theta}{r^2 \cos^2 \theta} \frac{\partial \psi}{\partial \varphi} \right] $$

$$\mathcal {4}\psi_{2}={1\over r^2cos^2\theta}\Bigg[{\partial^2 \psi \over \partial \phi^2}\Bigg]$$

For i=3

$$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Bigg[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Bigg]$$ becomes

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over (r)^2}{\partial \psi \over \partial \theta}\Bigg]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over r^2}{\partial \psi \over \partial \phi}\Bigg]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[cos\theta{\partial \phi\over \partial\theta}\Bigg]$$

$$\mathcal {4}\psi_{i=3}=\frac{1}{r^2 \cos \theta} \left[ \sin \theta \frac{\partial \psi}{\partial \theta} + \cos \theta \frac{\partial^2 \psi}{\partial \theta^2} \right] $$

title: User:Egm6321.f09.Team06.sada/HW5


text:
Fostering Classroom Norms: Using Wikispaces

 * {| cellpadding="4" cellspacing="0" width="450px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: CORE subjects and Electives
 * }

 * {| cellpadding="4" cellspacing="0" width="450px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: Character Education/Earth Charter Principles
 * }

 * {| cellpadding="4" cellspacing="0" width="450px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: 2 class periods (1 hour each)/On-going
 * }

 * {| cellpadding="4" cellspacing="0" width="450px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: Wikispaces
 * }
 * }

{| cellpadding="5" cellspacing="5" style="width: 90%; background-color: inherit; margin-left: auto; margin-right: auto"
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Background
How do we get students to take ownership of their learning? How can we make learning meaningful for students? These are just a few of the questions that this lesson will address. In my classroom, I want students to collaborate using authentic discussion, have pride in their work, and be supportive of their peers. In addition to these objectives, I strive to teach students using the Principles of The Earth Charter. Specifically, the principle that this lesson will address is Principle I.3: Build democratic societies that are just, participatory, sustainable, and peaceful. In order to instill these principles in my students, I have found that there are many environmental conditions that need to be established in the classroom for these objectives to take place. In a case study of a 5th grade classroom, it was found that several of these environmental conditions surfaced when authentic discussion took place (Hadijioannou, 2007). Those conditions included the following:

 * 1) Physical environment-areas in the classroom for students to work comfortably
 * 2) Curricular demands-using backwards design and inquiry methods
 * 3) Teacher beliefs-openness to a variety of interpretations from students
 * 4) Student beliefs-positive attitude toward learning
 * 5) Relationships between members-taking the time for students to get to know one another
 * 6) Classroom procedure-rule of respect
 * 7) Norms of class participation-opinions need to be supported

These conditions should should serve as a guide for facilitating a rigorous learning environment. Yet, in order for students to begin taking on responsibility for this learning environment, they need to be empowered as stakeholders in the classroom. For students to take on the ownership of their learning, it is imperative to involve them in the process of skillful discourse in order to establish their own standards and norms in the learning environment (Freire, 1970). According to Freire's theory (1970), students can only overcome their status as an oppressed citizen if actively engage in the process of decision-making and reflect on those decisions.

Therefore, this lesson aims directly at empowering students to make decisions about the learning environment of which they are a part. One way to actively engage students in the process of decision-making is to hold authentic discussions about group norms of behavior in the classroom. In a recent newsletter from The Smart & Good Schools Initiative (2009), four keys for increasing students' performance character were defined. The 4 Keys include:

 * 1) A Community That Supports & Challenges: building group norms
 * 2) Self-Study: assessing and monitoring for improvement opportunities
 * 3) Other-Study: investigating positive and negative examples
 * 4) Public Performance/Presentation: sharing products and progress

These 4 Keys are another way for teachers to create the environmental conditions necessary for critical thinking and ownership of learning. While, it may not be necessary to focus on all four keys at the same time, a teacher should be mindful of how each of the keys can work together in fostering authentic discussion and classroom norms.

For the purpose of this lesson, we will focus on Key #1: A Community That Supports & Challenges and Key #2:  Public Performance/Presentation. Students will be required to utilize the Classroom Norms Guidelines and publish their page to class Wikispaces. The lesson below describes how you might go about using Wikispaces to have students have an authentic discussion and utilize technology as a publication tool in order to build classroom norms. The underlying objective is to carry these classroom norms over to on-line rules of etiquette. Rules of on-line etiquette will be covered in a preceding lesson. However, the same student pages will host those later additions. Wikispaces are meant to be continuously updated and evolve as students take on new learning opportunities.
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Lesson Plan
Engage: Create a T-chart with the class to define RESPECT. What does it look like? What does it sound like?

Explore: Have students work in triads or small groups to build a classroom norms page to post to the classroom Wikispaces. Students should utilize the Classroom Norms Guide.

Explain: Students present their Wikispace page to the class.

Evaluate: Provide students feedback using a rubric. You may want to refer to this sample rubric.

Extend: Build on classroom norms by discussing how this looks in an on-line format. Ask students what were some of the strategies they found helpful when working with their partner on building the classroom norms page? What were some of the challenges? How did they overcome these challenges?

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text:
User:Matthewmgregoire

I am from Canada I have have little education

title: User:Matthewmgregoire


text:
Introduction to Italian/Lesson 1

COME PRESENTARSI/HOW TO INTRODUCE YOURSELF
Ciao - Hi, hello Buon giorno - Good morning Buon pomeriggio - Good afternoon (less used) Buona sera - Good evening Buona notte - Good night Arrivederci - bye bye - see you soon Addio - Goodbye

Come ti chiami? - What's your name? Mi chiamo Marco - My name is Marco Chi sei? - Who are you? (informal) Chi è Lei? - Who are you? (formal) Sono Marco - I am Marco Molto piacere - Nice to meet you (more literally "Much pleasure [to meet you]") Come stai? - How are you? (informal) Come sta? - How are you? (formal) Bene - Well, fine Benissimo - Very well Abbastanza bene - Fairly well Male - Badly Malissimo - Very badly Così così - So-so Parli italiano? - Do you speak Italian? Non parlo molto italiano - I don't speak much Italian. Non parlo italiano - I don't speak Italian. Parlo italiano abbastanza bene - I speak Italian pretty well. Chi parla inglese? - Who speaks English? Mi piace... - I like...

Most Italian words end with a vowel. All the vowels must be pronunced. Most Italian words are stressed on the next-to-last syllable (as in "Arrivederci"). Double consonants are always pronounced (as in "Arrivederci"). There are a lot of regional dialects, so the pronunciation could be different by the local standard.

title: Introduction to Italian/Lesson 1


text:
User talk:Philosophia~enwikiversity

 Hello Philosophia, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Philosophia. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Philosophia~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

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Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:17, 19 April 2015 (UTC)

title: User talk:Philosophia~enwikiversity


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User talk:Sbrown1992

 Hello Sbrown1992, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 08:18, 4 November 2009 (UTC)

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User talk:Gurmit

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 08:18, 4 November 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 4 November 2009 (UTC)

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User talk:Xiaoyubaobei18

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text:
Introduction to Italian/Lesson 2

The Italian Alphabet
The Italian alphabet is very similar to the English alphabet:

Notice that it lacks the letters j, k, w, x, and y. These are "foreign letters" used only in loanwords.

It is important to notice that letters are all feminine nouns.

Vowel pronunciations
Each vowel has basically one pronunciation. They are as follows: a - pronounced roughly like the a in father or class, like the sound a dentist might have you make when you open your mouth wide. It is usually pronounced in a relaxed way, but hardens when an accent mark is added (à). e - pronounced like the e in pest or Edward. When it has an accent (è) it is pronounced like the a in plate or the ei in eight. i - pronounced like the ee in feed or the ea in peak. o - pronounced like the oa in boat or the oe in toe. u - pronounced like the o in to or who.

Note on accented vowels
In some languages accented letters are counted as their own letters (for instance, in Swedish the letter ä is considered different from the letter a). In Italian, à and è are just variations on a and e. So in a dictionary, the words ‘‘e’’ and ‘‘ è’’ will appear immediately after one another.

Consonant pronunciations
Most Italian consonants are pronounced the same as they are in English:

title: Introduction to Italian/Lesson 2


text:
Introduction to Italian/Lesson 3

Basic Vocabulary
There are a few very important words in Italian that you will definitely need to know. They are classified here by part of speech.

Pronouns
io - I tu - you lui/lei - he/she (lei can also be the formal version of tu) noi - we voi - you (collective; used for addressing multiple people directly) loro - they

In elementary schools when you start studying verb tenses, the third person singular and third person plural are translated with "egli" and "essi", which are personal pronouns too, the use is still quite common in written Italian but rarely in spoken Italian.

Simple prepositions
con - with in - in (don't forget to pronounce it differently even though it's spelled the same) su - on a - at, to di - of da - from per - for tra - within (used interchangeably with fra)

Conjunctions
e - and ma - but o/oppure - or

title: Introduction to Italian/Lesson 3


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User talk:Matthewmgregoire

 Hello Matthewmgregoire, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 08:27, 4 November 2009 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

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Introduction to Italian/Lesson 4

In Italian, as in most languages, there are two genders: masculine and feminine.

Singulars
Every singular noun ends in -a, -o, or -e; with very few exceptions, the nouns that end in -o are masculine and the nouns that end in -a are feminine (-e is generally masculine but not always). A few examples: tavolo (table, masculine), sedia (chair, feminine), figlio (son, masculine), figlia (daughter, feminine), dottore (male doctor, masculine), calcolatrice (calculator, feminine)

Plurals
With very few exceptions, when a noun ending in -o or -e is pluralized the -o or -e is changed to an -i (tavolo changes to tavoli) and when a noun ending in -a is pluralized the -a is changed to an -e (sedia changes to sedie).

The Indefinite Articles
In English, the indefinite articles are a before a consonant and an before a vowel. In Italian the indefinite article agrees with gender and quantity. This chart should explain:

Note: Technically, dei, delle, and degli all translate to of the. However, in Italian they are used to mean ‘‘some’’. ‘‘Some’’ can also be used with certain singular nouns: ‘‘del pane’’ (‘‘some bread’’), ‘‘del latte’’ (‘‘some milk’’), etc.

Get Used to Indefinite Articles
Whenever you learn a new noun and are practicing saying it, always say it with the indefinite article. It really does help. For example, take the word mela, which means apple and is, of course, feminine. It should be natural to you to say ‘‘una mela’’.

The Definite Articles
In English, the definite article is the. One of the few ways in which English is easier to learn than other languages is that we have only one definite article. However, in Italian there are as many definite articles as indefinite, and once again they are used according to gender and quantity. Let's recycle the other chart:

Numbers
Let's just do up to one hundred for now: 0 - zero 1 - uno 2 - due 3 - tre 4 - quattro 5 - cinque 6 - sei 7 - sette 8 - otto 9 - nove 10 - dieci 11 - undici 12 - dodici 13 - tredici 14 - quattordici 15 - quindici 16 - sedici 17 - diciasette 18 - diciotto 19 - dicianove 20 - venti (yes, like the coffee size) 21 - ventuno 22 - ventidue 23 - ventitre 24 - ventiquattro 25 - venticinque 26 - ventisei 27 - ventisette 28 - ventotto 29 - ventinove 30 - trenta

Every other number up to one hundred follows the same pattern as the 20’s: ‘‘venti’’+’’uno’’ = ‘‘ventuno’’, ‘‘venti’’+’’due’’ = ‘‘ventidue’’, ‘‘venti’’+’’tre’’ = ‘‘ventitré’’, etc., so ‘‘trenta’’+’’uno’’ = ‘‘trentuno’’, ‘‘trenta’’+’’due’’ = ‘‘trentadue’’, ‘‘trenta’’+’’tre’’ = ‘‘trentatré’’, and so on. Notice that the numbers that end in ‘‘uno’’ and ‘‘otto’’ leave off the last letter of the tens digit. That applies to every number. Here are all the tens: 10 – dieci (irregular; see above) 20 – venti 30 – trenta 40 – quaranta 50 – cinquanta 60 – sessanta 70 – settanta 80 – ottanta 90 – novanta 100 - cento

Some Basic Nouns
Here are a few nouns to help build a fundamental vocabulary. You don’t need to memorize them all immediately, but be sure to reference them often.

Animals
un cane - a dog un gatto - a cat un uccello – a bird

Homes and Rooms
una casa - a house un appartamento – an apartment un bagno - a bathroom (don't forget the gn diphthong) una camera (da letto) – a room (bedroom) una cucina – a kitchen un soggiorno – a living room

People
un uomo - a man (irregular pluralization: uomini; don't be too concerned with it) una donna - a woman un ragazzo - a boy una ragazza - a girl un bambino - a baby boy una bambina - a baby girl

Family
una famiglia – a family un marito – a husband una moglie – a wife un figlio – a son una figlia – a daughter un padre – a father una madre – a mother un fratello – a brother una sorella – a sister un nonno – a grandfather una nonna – a grandmother

Foods
una mela – an apple una banana – a banana un’arancia – an orange della carne – some meat del prosciutto – some ham del pollo – some chicken del latte – some milk un gelato – an ice cream del caffè – some coffee del pane – some bread un panino – a sandwich (usually on a roll) un tramezzino – a triangular sandwich del formaggio – some cheese un uovo – an egg (plural is "uova", NOT "uovi", and is feminine, while the singular is masculine)

Electronics
una macchina – a machine una calcolatrice – a calculator una radio – a radio ("una" is used because "radio" is an abbreviation of a feminine word) un telefono – a phone un telefonino/un cellulare – a cell phone un computer/un elaboratore – a computer un televisore/una televisione/una TV – a television/a TV una ricetrasmittente/un walkie-talkie – a walkie-talkie

Transportation
una macchina – a car un camion/un autocarro – a truck un furgone/un furgoncino – a lorry un autobus (un bus) – a bus una bicicletta (una bici) – a bicycle (a bike)

One thing to which attention should be paid is that there are multiple words for "car" in Italian (including "auto", "automobile" and "autovettura", all feminine), although "macchina" is the most common, but unformal.

Odds and Ends
un libro – a book una matita – a pencil una penna – a pen un foglio di carta – a paper sheet della carta – some paper un giornale – a newspaper una rivista – a magazine una chiave – a key

Clothes
un capotto - a coat un cappello - a hat una maglia - a shirt una maglietta - a T-shirt una camicia - a longsleeve shirt dei pantaloni - some pants un vestito - a dress una gonna - a skirt un mantello - a mantle

title: Introduction to Italian/Lesson 4


text:
Introduction to Italian/Lesson 5

As in most languages, in Italian verbs are conjugated to agree with their tense.

Tenses
Tenses are something that people often have trouble with. In reality, they aren't that hard. First person: referring to yourself (I am a man, I don't like spinach). Second person: addressing someone directly (you are a woman, you don't like cabbage). Third person: referring to someone else (he is a man, she is a woman). First person plural: referring to a group of people including you (we are all men, we don't like asparagus). Second person plural: Addressing a group of people directly (you all are women, none of you like artichokes). Third person plural: Referring to a group of people (they are all men, they don't like broccoli). Tenses also include past, present, and future. As in, "I ate a burger," "I'm eating a burger," and, "I will eat a burger." Those are complicated and won't be discussed until later.

Regular Verbs
In Italian, verbs have an infinitive form and a conjugated form for every tense. An infinitive is basically like saying "to be", which conjugates to am, are, and is. Infinitives in Italian are only one word (instead of to be, to eat, etc.) and most end in -are, as in andare (to go), mangiare (to eat), passare (to pass), and many more. Infinitives that end in -are are considered regular verbs. Infinitives can also end in -ere and -ire, and most of those that do are irregular (meaning they don't follow a pattern when you conjugate them).

Conjugating Regular Verbs
There's a simple pattern to conjugating regular verbs. Basically, you just replace the -are suffix with another one according to the tense. This chart will show which suffixes are added for which tense: Now let's apply that to a word. Here are all the conjugations of mangiare (to eat). Important: Notice that for the second person conjugation the g is soft. This is only because it is soft in the root word, mangi-. If the root were mang (making the infinitive mangare) the second person conjugation would be manghi so as to preserve the hard g.

Some Basic Verbs
Here are some basic verbs in their infinitive form. Practice conjugating them. abitare – to live (to reside) arrivare – to arrive ascoltare – to listen (to) aspettare – to wait (for) ballare – to dance cantare – to sing cercare – to look (for) chiamare – to call comprare – to buy desiderare – to want, to wish dimenticare – to forget entrare – to enter frequentare – to attend, to frequent giocare – to play (a game) guardare – to watch, to look (at) guidare – to drive imparare – to learn incontrare – to meet insegnare – to teach lavorare – to work mandare – to send mangiare – to eat pagare – to pay (for) parlare – to speak passare – to pass (by); to spend time pensare (di) – to think (of) portare – to wear, to bring ricordare – to remember studiare – to study telefonare – to telephone (to call on a phone) tornare – to return trovare – to find usare – to use viaggiare – to travel visitare – to visit

Irregular Verbs to Memorize
Most irregular verbs will come later because they are more difficult to conjugate than regular ones. However, there are two in particular that you need to just memorize.

Essere
Essere is an infinitive meaning to be. You pretty much can't get by without knowing it. So here are it's conjugations. Memorize them. There are a few important notes on essere: 1. Sono is, in fact, the correct conjugation for both the first person and the third person plural. That's not a typo. 2. È is a very important word. It translates to is, which, of course, is used much more than am or are.

Avere
Avere is an infinitive meaning to have. It is also irregular and also needs to be memorized. Remember that the h is silent. h is made to distinguish the written form from the different meaning that may otherwise have o - conjunction ai - preposition a - preposition anno - year

Idiomatic expressions using avere
In Italian there are many idiomatic expressions that use avere in its various conjugations:

These are just things that they say. Have need of is sometimes used in English, as you may know, but not much recently. Don't forget to conjugate avere, too - it's not "io avere freddo" (if you hear it on TV or read it in books, it usually is how stereotypized foreigners speak), it's "io ho freddo".

title: Introduction to Italian/Lesson 5


text:
Introduction to Italian/Lesson 6

Preposition Contractions
Most prepositions are frequently combined with definite articles. Here is a chart showing combinations of prepositions and definite articles: These contractions often take on a life of their own. For example, the a contractions replace con (with) when talking about food qualities and tastes. So instead of un panino (a sandwich) con prosciutto (ham), it's un panino al prosciutto. Also, the di contractions, as we've already established, can mean some (the "plural" of un, una or un')

title: Introduction to Italian/Lesson 6


text:
Introduction to Italian/Lesson 7

In English, adjectives are almost always the same and come before a noun. In Italian, adjectives are modified to fit a noun's gender and quantity and they can come after or before the noun, based on the type of adjective.

Modification
Some types of adjectives (such as colors) take on the suffix of the noun they modify (-o for a singular masculine noun, -e for a plural feminine noun, etc.). Others, such as sizes, do not. It varies on a situational basis and you will just have to memorize them. Here are some example sentences: I cani grandi sono bianchi (the big dogs are white). I cani bianchi sono grandi (the white dogs are big). Le mele sono gialle (the apples are yellow). Le mele sono rosse (the apples are red). La carne è gialla (the meat is yellow). La carne è rossa (the meat is red). Although there isn't really any yellow meat, you should get the idea.

Possessive Adjectives
Possessive adjectives (in English) are my, your, his/her, etc. In English they are not preceded by a definite article, but in Italian they are (except when used to talk about your family members in the singular, including padre, madre, sorella, fratello, zia, zio, etc). For example: Il mio cane è nero directly translates to the my dog is black, but in Italian simply means my dog is black. Normally you would say the big dog is black, so it really makes sense that since my is an adjective you would say the my dog is black. This chart shows all of the possessive adjectives according to gender and quantity (by now you should have noticed that almost everything relates to those two things): Loro is both the pronoun and the possessive adjective for the third person plural. It doesn't change based on gender or quantity.

Using Possessive Adjectives
The gender and quantity of a possessive adjective is determined by the noun it modifies, not the possessor. Examples: Giulia gioca con il suo cane (Giulia is playing with her dog). Even though Giulia is a girl, the possessive adjective modifies cane, so it is masculine. Giovanni incontra la sua ragazza al ristorante (Giovanni is meeting his girlfriend at the restaurant). Mio padre ha voglia del mio gelato (my father wants my ice cream).

Possessive Adjectives and Family Members
With family members the definite article is not used: Mia madre è un architetto (my mother is an architect). Mio padre è un dottore (my father is a doctor). Mio fratello è un insegnante (my brother is a teacher). Mia sorella è una studentessa (my sister is a student). But the definite article is used when plurals or informal names are used: I miei fratelli sono insegnanti (my brothers are teachers). Le mie sorelle sono studentesse (my sisters are students). Il mio papà è un dottore (my dad is a doctor). La mia mamma è un architetto (my mom is an architect).

The Act of Belonging
This is not technically an adjective but should be included with the possessives. In English, to describe an apple as belonging to Giacomo we would say Giacomo's apple. In Italian, however, there is no suffix like the apostrophe-s, so they would say la mela di Giacomo, which, of course, translates to the apple of Giacomo.

Some Basic Adjectives
grande – big (doesn’t change for singular nouns, does for plurals) piccolo – little (always changes) verde – green (always changes) giallo – yellow (always changes) rosso – red (always changes) azzurro – blue (always changes) nero – black (always changes) bianco – white (always changes) seduto – seated (always changes) molto – can mean very, many, or much depending on the situation (always changes) questo – this (always changes) Note: Questo can also be used as a pronoun, as in questo è un gatto (this is a cat) or questa è una mela (this is an apple).

title: Introduction to Italian/Lesson 7


text:
Introduction to Italian/Lesson 8

C'è, ci sono, and ecco all mean there is/there are but in different ways.

C'è and ci sono
Basically, c'è and ci sono indicate presence. So c'è un libro sul tavolo basically means there exists a book on the table, and ci sono molti libri alla biblioteca means there exist many books at the library. Of course, it doesn’t sound as awkward in Italian. Think of them as there’s a book on the table and there are lots of books at the library respectively. Note: The word c'è is basically just a contraction of the words ci and è.

Ecco
The word ecco is generally used to draw attention to a specific object. Say your dog ran away. If you found him down the street you might alert your family by saying Ecco il cane! (There’s the dog!). Ecco doesn’t change based on quantity; ecco il cane and ecco i cani are both correct.

Notes on c'è, ci sono, and ecco
Basically, c'è and ci sono refer to an indefinite (un libro) whereas ecco refers to a definite (il libro). Also, when using ecco the object you are referring to has to be immediately available (a book on a table or a library that you can see down the street), whereas c'è and ci sono usually refer to something that’s not immediately available (a book on a table back at home or a library in another city).

title: Introduction to Italian/Lesson 8


text:
Introduction to Italian/Lesson 9

The Second Conjugation (-ere)
Infinitive verbs in the second conjugation all end in -ere. Most of them conjugate regularly (according to the chart below) but there are more irregularities than the first conjugation. Here is an example of how verbs ending in -ere conjugate.

Some Basic Verbs in the Second Conjugation
Here are a few basic -ere verbs to help build your vocabulary. chiedere – to ask (for) chiudere – to close conoscere – to know credere – to believe decidere di (+ infinitive verb) – to decide to do something (io decido di prendere un gelato – I’m deciding to get an ice cream) discutere (di) – to discuss (noi discutiamo di politica – we are discussing politics) leggere – to read mettere – to put, to place perdere – to lose prendere – to take, to have (as in have some food or have a drink), to get ricevere – to receive rispondere – to answer scrivere – to write spendere – to spend (money) vedere – to see vendere – to sell

title: Introduction to Italian/Lesson 9


text:
Introduction to Italian/Lesson 10

Here are a few new nouns to help expand you vocabulary.

Places
una città – a city un ristorante - a restaurant un parco - a park una biblioteca – a library una scuola – a school un’università – a college un negozio – a shop un mercato – a market un supermercato – a supermarket fuori città – outside the city un bar/un caffè – used interchangeably to mean café un cinema – a movie theater (the masculine article is used because cinema is an abbreviation) un teatro – a theater (for plays) una discoteca – a dance club una farmacia – a pharmacy una chiesa – a church

Furniture
un tavolo – a table una sedia – a chair una porta – a door una finestra – a window un gabinetto – a toilet un letto – a bed un lavandino – a sink (faucet)

(l')Europa – Europe
(l')Italia - Italy (la) Gran Bretagna/(il) Regno Unito – Great Britain/the United Kingdom (l')Inghilterra – England (la) Scozia – Scotland (il) Galles – Wales (l')Irlanda del Nord – Northern Ireland (l')Irlanda – Ireland (la) Francia – France (la) Svizzera – Switzerland (la) Spagna – Spain (il) Portogallo – Portugal (la) Germania/(l')Alemagna – Germany (la) Svezia – Sweden (la) Norvegia - Norway

(l')Asia – Asia
(la) Cina – China (il) Giappone – Japan

(l')Africa – Africa
(il) Sudafrica – South Africa

(l')America del Nord/(il) Nordamerica – North America
(gli) Stati Uniti – the United States (il) Canada/(il) Canadà – Canada (il) Messico – Mexico

(l')America del Sud/(il) Sudamerica – South America
(il) Brasile – Brazil (l')Argentina – Argentina

(l')Oceania – Oceania
(l')Australia – Australia (la) Nuova Zelanda – New Zealand

Studies
biologia – biology chimica – chemistry fisica – physics geografia – geography scienze naturali – natural sciences filosofia – philosophy religione – religious studies letteratura – literature lingue straniere – foreign languages scienze politiche – politcal science storia – history matematica - mathematics informatica - computer science

title: Introduction to Italian/Lesson 10


text:
Introduction to Italian/Exam 1

This test is meant to help you determine where you are in Italian and if you need to go back and review. Hover over a question with your mouse to see the answer. Below are links to every lesson if you decide you need to review some.

Lessons

 * Lezione Uno : Common Phrases
 * Lezione Due : The Italian Alphabet
 * Lezione Tre : Some Simple Words
 * Lezione Quattro : Nouns and Articles
 * Lezione Cinque : Verbs
 * Lezione Sei : Preposition Contractions
 * Lezione Sette : Adjectives
 * Lezione Otto : C’ è, Ci Sono, and Ecco
 * Lezione Nove : The Second Conjugation (-ere)
 * Lezione Dieci : Some New Nouns

Avere

 * 1) What is the second person conjugation of avere?
 * 2) What is the first person conjugation of avere?
 * 3) What is the third person conjugation of avere?
 * 4) How would you say I want... using an avere idiomatic expression?
 * 5) How would you say I need... using an avere idiomatic expression?

Essere

 * 1) What is the third person conjugation of essere?
 * 2) What is the third person plural conjugation of essere?
 * 3) What is the first person conjugation of essere?

Preposition Contractions
Give the contraction for given preposition and definite article pair.
 * 1) di and il
 * 2) in and la
 * 3) su and i

Possessive Adjectives
Give the definite article and possessive adjective for the following objects (translated noun is given in the answer but not necessary).
 * 1) Your dog.
 * 2) Your car.
 * 3) Your mother.
 * 4) Their cat.
 * 5) Their famous sandwiches.

Conjugating Verbs
Conjugate the following infinitives in the third person:
 * 1) mangiare
 * 2) cercare
 * 3) prendere
 * 4) leggere

Conjugate the following infinitives in the first person:
 * 1) chiamare
 * 2) ascoltare
 * 3) credere
 * 4) scrivere

Translation
Translate the following sentences into Italian.
 * 1) This is a black dog.
 * 2) There are many books at the library.
 * 3) I like to eat apples.
 * 4) My father is a big man.
 * 5) My mother is a small woman.
 * 6) There is a banana in the car.
 * 7) There's my pencil!
 * Hi, my name is Giorgio.

Translate the following sentences into English.
 * 1) Questo è il cellulare di Lorenzo.
 * 2) Io porto una gonna.
 * 3) Lui mangia un tramezzino al prosciutto.
 * 4) Noi abbiamo un uccello piccolo. Lei canta.
 * 5) Loro sono studenti all'università di Berkeley.
 * 6) Io ho un caffè per te.

title: Introduction to Italian/Exam 1


text: User:Proteasfanz

title: User:Proteasfanz


text:
Talk:Calculus/Limits/Exercises

if "f" is a function defined as f(x)=[x2+3x+2]/[2-mod(x)], show that left hand limit = Right hand limit, at 'x' tends to infinity?--58.65.203.47 14:08, 4 November 2009 (UTC)

title: Talk:Calculus/Limits/Exercises


text:
User:B9 hummingbird hovering/Beacon of Certainty by Mipham/aashraya

आश्रय Etymology: आ- √श्रि.

 * that to which anything is annexed or with which anything is closely connected or on which anything depends or rests
 * a recipient, the person or thing in which any quality or article is inherent or retained or received
 * seat, resting-place
 * dwelling, asylum, place of refuge, shelter, depending on, having recourse to
 * help, assistance, protection
 * authority, sanction, warrant
 * a plea, excuse (Legal)
 * the being inclined or addicted to, following, practising
 * attaching to, choosing , taking
 * joining, union , attachment
 * dependance, contiguity , vicinity
 * relation
 * connection
 * appropriate act or one consistent with the character of the agent
 * (in Gr.) the subject, that to which the predicate is annexed
 * (with Buddhists) the five organs of sense with मनस् or mind (the six together being the recipients of the आश्रित or objects which enter them by way of their आलम्बन or qualities)
 * source, origin
 * आ-श्रय depending on, resting on , endowed or furnished with (e.g. अष्ट-गुणा*श्रय » under अष्ट).

Source: enter 'aashraya' change input to ITRANS; http://www.sanskrit-lexicon.uni-koeln.de/monier/indexcaller.php