Recent content by Ray Vickson

I am sorry to report the death of Ray Vickson who was a member of this forum. He enjoyed assisting members with their problems during his retirement. Lynne Vickson

Your solution makes no sense: you have a small triangle with area 9 next to a much larger parallelogram with area 1, next to a smaller triangle with area 2, etc. Anyway, you need to show your work. Just saying "Through symmetry of parallelogram,I have come to this:" is not an explanation.

You will get nowhere using the above, because you do not have a valid dictionary. Your last equation has ##x_5## on the left of the "=" sign, but the same ##x_5## appears on the right-hand-side of the first and third equations. Remember, in a "dictionary" one group of variables are on the left...

So, in the current dictionary you have ##u = 10- 3x_1 - 3x_2 + x_4 + x_5##, and you want to minimize ##u##. Currently, when you have all the non-basic variables ##x_1 = x_2 = x_4 = x_5 = 0## you have ##u=10.## Is that really the minimum possible value of ##u##? If you increase some non-basic...

An Erlang distribution is just a Gamma distribution with an integer value of ##\alpha##. It is true that an Erlang random variable is the sum of ##\alpha## iid exponential (##\lambda##) random variable; usually these are referred to as stages or phases, not steps. (I have also not heard of them...

There were no ##u, v_1, v_2## in your original system, so where did they come from? Also, what happened to ##a_1, a_2##? It is true that after you have found a basic solution with variables ##a_1, a_2=0## non-basic, you can then drop ##a_1, a_2## from all subsequent equations, etc. However...

Every Bernoulli experiment is finite, because the probability that you need more than ##n## trials (to get your first "success") is ##(1-p)^n##, which goes to zero in the limit ##n \to \infty.## It is true that you cannot put a guaranteed bound on the number of trials to reach success, but it...

No, this is definitely not the case. Consider the two linear systems: $$\begin{array}{lc} x_1 + x_2 - x_3 &=4\\ 2x_1 - x_2 - x_4 &=6\\ -x_1 + x_2 + x_5 &=1 \\ x_1, x_2, x_3, x_4, x_5 &\geq 0 \end{array} \hspace{3em}(1) $$ and $$ \begin{array}{lc} x_1+x_2-x_3+a_1 &=4\\ 2x_1 - x_2 - x_4 +...

First: in this Forum, LaTeX needs # # (no space) at the start and at the end of each in-line equation, and needs $ $ (no space) at the start and the end of each "displayed" equation. So, your problem is $$\begin{array}{clc} \min&x_1+x_2 &\\ &x_1 + x_2 - x_3 &=4\\ &2x_1 - x_2 - x_4 &=6\\ & -x_1...

I just answered the questions as I read them: "I want to figure out the distribution of number of trials for a given fixed number of successes and given probability for success for Bernoulli trials." ---- no mention of a finite bound on the total number of trials. "Let's say the probability...

$$\sum_{s=0}^\infty s p^s = p \frac{d}{dp} \sum_{s=0}^\infty p^s = p \frac{d}{dp} \frac{1}{1-p}$$

Yes, it is all standard Probability 101 material. The number of Bernoulli trials ##N## until ##n=1## success is the so-called geometric distribution ##P(N=k) = p (1-p)^{k-1}, \;k = 1, 2, \ldots.## Note that there is no upper limit on ##k##; possibly you might need 10 billion coin tosses until...

I get something different. First, write $$1 \cdot 3 \cdots (2n+1) =\frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n) \cdot (2n+1)}{2 \cdot 4 \cdots (2n)}\\ \hspace{3em} = \frac{(2n+1)!}{2^n n!}$$ so your ratio is $$\text{ratio} = \frac{2^n (n!)^2}{(2n+1)!}$$ Now one can apply Stirling's formula ##k...

You can render the equation separable by changing to ##P = Q- r/k##, so that ##dP/dt +kP = 0## (because, of course, ##dP/dt = dQ/dt##).

None of those remarks are rude; they are meant to be helpful, and you would find them to be so if you were less defensive and more willing to work with helpers.