Thanks Ray and andrewkirk.
The terms involving Y_0 go to 0 as T goes to infinity because |\rho| < 1. Each element ends up being the covariance of the squared errors and the whole thing equals to 0.
Thanks for you help!
Homework Statement
Calculate: PLIM (probability limit) \frac{1}{T} \sum^T_{t=2} u^2_t Y^2_{t-1}
Homework Equations
Y_t = \rho Y_{t-1} + u_t, t=1,...T, |\rho| <1 which the autoregressive process of order 1
E(u_t) = 0, Var(u_t) = \sigma^2 for t
cov(u_j, u_s) = 0 for j \neq s
The Attempt...
Alright, so if I get this right, to find the first 3 nonzero terms, I plug in my results for n=1,2,3 in equation 6 and disregard the A_{n} cos (nx) terms since I am asked for the sine series, and A_{0} vanishes so I'm left with B_{n} sin (nx) (A_{n} in my answer). Thanks again
Thank you for your reply.
So that would be sin (n\pix/l) for each corresponding n? Do I need multiply the coefficients by sine because I am asked for a sine series?
Homework Statement
Consider the function φ(x) ≡ x on (0, l). Find the sum of the first three (nonzero) terms of its Fourier sine series.
Reference: Strauss PDE exercise 5.1.3
Homework Equations
The Attempt at a Solution
I have found the coefficient without difficulty, it is...
Homework Statement
By trial and error, find a solution of the diffusion equation du/dt = d^2u / dx^2 with
the initial condition u(x, 0) = x^2.
Homework Equations
The Attempt at a Solution
Given the initial condition, I tried finding a solution at the steady state (du/dt=0)...
So that would be du/dt - d^2u/dx^2 = -1?
Then I have Lu = -1, L(cu)= d(cu)/dt - d^2(cu)/dx^2 = c du/dt + c d^2u/dx^2 = c L(u), L(u+v) = d(u+v)/dt - d^2(u+v)/dx^2 = du/dt + d^2u/dx^2 + dv/dt + d^2v/dx^2 = Lu + Lv, so it's linear and inhomogenous.
Thank you.
Homework Statement
Check du/dt + d^2u/dx^2 + 1 = 0
Homework Equations
L is a linear operator if:
cL(u)=L(cu) and L(u+v)=L(u)+L(v)
The Attempt at a Solution
L = d/dt + d^2/dx^2 + 1
L(cu) = d(cu)/dt + d^2(cu)/dx^2 + 1 = c du/dt + c d^2(u)/dx^2 + 1 ≠ cL(u) = c du/dt + c...
This reminds of a problem I had to do recently. Here are my thoughts:
Pick 2 sequences (x_{n})_{n} and (y_{n})_{n} such that lim x_{n} = infinity as n tends to infinity and lim y_{n} = infinity as n tends to infinity. Then show that lim f(x_{n}) ≠ lim f(y_{n}), therefore the limit does not exist.
Homework Statement
Let (u_{n})_{n} be a real sequence such that lim u_{n} = 0 as x→∞ and let (v_{n})_{n} be a bounded sequence. Show that lim (u_{n})_{n}(v_{n})_{n} = 0 as x→∞
Homework Equations
The Attempt at a Solution
Since (v_{n})_{n} is bounded then it has a least upper bound and...
OK I figured this one out thanks to you guys:
from the I.V. theorem, there exists c=0 belonging to [a, b] with a=1 and b=-1 in the first case, second case with a=1 and b=2. therefore f(a)=-1<c=0<f(b)=1 and f(a)=-1<c=0<f(2)=13 so we have two real values for x.
Thanks y'all for your help!
Yes I am allowed to use the intermediate value theorem.
as tends to inf: lim f(x) = inf
as tends to -inf: lim f(x) = -inf
f(0)=-1
f(1)=-1
f(-1)=1
I'm not sure on how to find the extremum though
Homework Statement
Let f(x)=x^4 - x - 1. Show that f(x)=0 has two real roots.
Homework Equations
None
The Attempt at a Solution
x(x^3 - 1 - 1/x) = 0 which gives x=0 and x^3 - 1 - 1/x=0, x^2 - 1/x - 1/x^2=0, but WolframAlpha says x~~0.724492 and x~~-1.22074. I kept dividing by x it but...