Recent content by frenchkiki

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!

Thanks andrewkirk. I've seen somewhere in my notes that the errors (u_t's) are i.i.d. I'll use independence then.

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...

Yes it is correct.

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.

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...

A brief visit at my professor's office clarified the problem. You pick two sequences xn and yn tending to 0. In the case p=0 case f(xn)=0 f(yn)=1 with xn=1/nπ and yn=1/(2n + 1/2)π. then lim (n→∞) xn = 0 ≠ 1 = lim (n→∞) yn. Therefore lim (x→0+) xp sin (1/x) does not exist for p=0. Thank you...