Recent content by Tony Hau

Thank you for all your responses. Let me think about it more carefully, to be honest this is my first time learning about the term "equiprobable".

Can you explain why it is so? There are four events as listed in the table and so the denominator should be 4. Plus if ## P(G_B \cap X_A) = \frac{1}{3} ##, then the resulting probability of ## P(G_B \mid X_A) = 1 ##

So this question arises from an online course on Bayes Theorem and Total Probability. The question says that there are three prisoners A, B and C. The King decides to release two of them and sentence the remaining one. Supposed that you were A and so your chance of being released is ##...

thank you for your answer and let me try to figure that out for a while...

Hello thank you for your answer. However I just read about the topic at wikipedia (this) and here is the screenshot of this It seems like frequency deviation can be held constant while increasing the message frequency as shown in the 2nd paragraph. But I have another idea after reading this...

So I am trying to learning FM from textbooks and it says that the modulation index of FM is given by ##m=\frac{\Delta}{f_{m}}##, where ## f_{m}## is the message signal bandwidth and ##\Delta## is the peak frequency deviation of the FM modulated signal. What I do not understand is that normally...

The del operator is not a vector that crosses with vectors, although it resembles the property of vectors. I am sorry I am not an expert who can explain clearly to you about it, but here is a derive of the curl in other...

The general formula for cylindrical coordinate is as follows: $$\vec{\nabla} \times \vec{V} = \frac{1}{r}\begin{vmatrix} \hat{r}\ & r\hat{\theta} & \hat{z} \\ \partial/\partial r & \partial/\partial \theta & \partial/\partial z \\ V_{r} & V_{\theta} & V_{z} \end{vmatrix} $$ Because ##V_r## and...

I am sorry. I thought the approximation only refers to ##\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi##; with regards to the approximation ##\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi## , the following ##\psi = Ae^{-\frac{\xi^2}{2}}+ Be^{\frac{\xi^2}{2}}## is exact and is not an approximation.

By the way, it would be very nice if you could explain what asymptotic form is in the second picture. I don't understand the transition from ##\psi = Ae^{\frac{-\xi^2}{2}}+Be^{\frac{\xi^2}{2}}## to ##\psi = h(\xi)e^{\frac{-\xi^2}{2}}##

The solution is same as I stated. Actually this is from a proof on my textbook on quantum mechanics, where the proof is about solving the simple harmonic oscillator. ##\psi(x)## refers to the wave function; ##\xi## refers to the constant ##\sqrt{\frac{m\omega}{\hbar}}x##

This is a very simple question: I would like to solve for ##\psi## in this equation $$\frac{d^{2}\psi}{d\xi^2} =\xi^2\psi$$ I so apply ##y=c_{1}e^{-kx}+c_{2}e^{kx}## and ##\psi## should be equal to ##\psi=c_{1}e^{-\xi^2}+c_{2}e^{\xi^2}##, because ##(D^2-\xi^2)\psi=0##. However the answer is...

At least the transform covered in the Boas book...

Yes, I have found the online version of your book. I will work through it later because I have just finished the Fourier series and transform. The problem of Hong Kong's educational system is not at the university level, but at the secondary school level. At secondary school, except for a few...