1
Home Assignment 3 / Re: Problem5
« on: October 10, 2012, 09:40:57 PM »
q5
Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Pages: [1]
3
Home Assignment 3 / Re: Problem 1
« on: October 09, 2012, 10:59:24 PM »
The two forms are actually the same,
\begin{equation*}
erf(z)=\sqrt{\frac{2}{\pi}}\int_0^ze^{-y^2/2}\,dy
\tag{Erf}\label{eq-Erf}
\end{equation*}
let $\frac{y}{\sqrt{2}}= t$, you can transform the above formula into the form in Wolfram
\begin{equation*}
erf(z)=\frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}\,dt
\end{equation*}
\begin{equation*}
erf(z)=\sqrt{\frac{2}{\pi}}\int_0^ze^{-y^2/2}\,dy
\tag{Erf}\label{eq-Erf}
\end{equation*}
let $\frac{y}{\sqrt{2}}= t$, you can transform the above formula into the form in Wolfram
\begin{equation*}
erf(z)=\frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}\,dt
\end{equation*}
4
Home Assignment 2 / Re: Problem 2
« on: October 01, 2012, 09:55:17 PM »
for part d), I calculated the general form of solution u(0,t)
5
Home Assignment 1 / Re: Problem 5
« on: September 23, 2012, 03:39:55 PM »
I guess that there is a typo in the assignment for 5.(c) about the Goursat problem:
the formula (9) should be as below to be a Goursat problem
\begin{equation} u|_{x=3t}=t, \quad u|_{x=-3t}=2t. \end{equation}
instead of that given in the assignment: (this is the IVP)
\begin{equation} u|_{x=3t}=t,\quad u_t|_{x=-3t}=2t. \end{equation}
the formula (9) should be as below to be a Goursat problem
\begin{equation} u|_{x=3t}=t, \quad u|_{x=-3t}=2t. \end{equation}
instead of that given in the assignment: (this is the IVP)
\begin{equation} u|_{x=3t}=t,\quad u_t|_{x=-3t}=2t. \end{equation}
6
Technical Questions / test for math
« on: September 23, 2012, 02:38:04 AM »
$x>b^2$, is it the same as latex?
$$local\ mean\ level\ \qquad S_t = \alpha(\dfrac{X_t}{D_{t-s_1}W_{t-s_2}})+
(1-\alpha)(S_{t-1} + T_{t-1})$$
$$trend\ \qquad T_t = \gamma (S_t - S_{t-1}) + (1-\gamma)T_{t-1}$$
$$seasonality\ 1 \ \qquad D_t = \delta (\dfrac{X_t}{S_t W_{t-s_2}}) + (1-\delta)D_{t-S_1}$$
$$seasonality\ 2 \ \qquad W_t = \omega (\dfrac{X_t}{S_t D_{t-S_1}}) + (1-\omega)W_{t-s_2}$$
$$\hat{X_t} = (S_t + kT_t)D_{t-s_1+k}W_{t-s_2+k}$$
$$local\ mean\ level\ \qquad S_t = \alpha(\dfrac{X_t}{D_{t-s_1}W_{t-s_2}})+
(1-\alpha)(S_{t-1} + T_{t-1})$$
$$trend\ \qquad T_t = \gamma (S_t - S_{t-1}) + (1-\gamma)T_{t-1}$$
$$seasonality\ 1 \ \qquad D_t = \delta (\dfrac{X_t}{S_t W_{t-s_2}}) + (1-\delta)D_{t-S_1}$$
$$seasonality\ 2 \ \qquad W_t = \omega (\dfrac{X_t}{S_t D_{t-S_1}}) + (1-\omega)W_{t-s_2}$$
$$\hat{X_t} = (S_t + kT_t)D_{t-s_1+k}W_{t-s_2+k}$$
Pages: [1]