Wikipedia:Reference desk/Archives/Mathematics/2010 September 30

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September 30

Poisson Process

I have a non-homogeneous Poisson process where the arrival rate is a function of time; $\lambda (t)$. I want to find the expected time of the first arrival.

I know that in a regular Poisson process arrival times have an exponential distribution with parameter $\lambda$. However with the non-homogeneous process I cannot take the expected value (${\frac {1}{\lambda }}$) because this parameter is a function of time and is not constant. What can I do instead? —Preceding unsigned comment added by 130.102.158.15 (talk) 00:00, 30 September 2010 (UTC)

The solution T to the equation $\int _{0}^{T}\lambda (t)dt=1$ is some kind of mean time to the first arrival. (Perhaps not the kind you asked for, but probably the kind you want.) Bo Jacoby (talk) 07:32, 30 September 2010 (UTC).

The expected time of first arrival is $\int _{0}^{\infty }e^{-\int _{0}^{x}\lambda (t)dt}dx$. I think what Bo had in mind was that the median T is the solution to $\int _{0}^{T}\lambda (t)\ dt=\ln 2$. -- Meni Rosenfeld (talk) 08:29, 1 October 2010 (UTC)

Use de Moivre's theorem and the Pythagorean identity?

Use de Moivre's theorem and the Pythagorean identity to show that $\cos {3x}=4\cos ^{3}{x}-3\cos {x}$.
The problem is trivially easy if you allow the identity $\cos {x}={\frac {e^{ix}+e^{-ix}}{2}}$ but unfortunately they don't want it solved that way.MrMahn (talk) 00:14, 30 September 2010 (UTC)

First work out what happens when you treat x as the angular part of a complex number written in polar coordinates, and do the multiplications with de Moivre's theorem. 67.122.209.115 (talk) 00:18, 30 September 2010 (UTC)

Just think of $e^{i3x}$ in two ways. In one way, it's $\cos(3x)+i\sin(3x)$. On the other hand, it is $(e^{ix})^{3}=(\cos x+i\sin x)^{3}$. You multiply out this right side and then compute real and imaginary parts. You should get a formula for $\sin(3x)$ at no extra charge. StatisticsMan (talk) 02:05, 30 September 2010 (UTC)

StatisticsMan appears to have accidentally left out the "i" in e^i3xand (e^ix)³ above -- not that Euler's formula is necessary here. As SM states, just take de Moivre's formula with n=3, multiply out the cube, equate the respective real and imaginary parts, and simplify with the Pythagorean identity. -- 111.84.196.147 (talk) 13:04, 3 October 2010 (UTC)

Oops, you are correct, thank you. I fixed it now I believe. StatisticsMan (talk) 03:04, 4 October 2010 (UTC)

Projectile motion problem

"A particle is projected, at an angle α and speed V, from the edge of a cliff of height h. When it hits the ground, its path forms an angle A = arctan(2 tan α) with the horizontal. Find its horizontal range R and the value of V."

I've had trouble interpreting what the angle of impact tells us about either R or V. Where y and x represent the vertical and horizontal components of motion, and g is gravity, it is usual to derive the following equations:

$x=Vt\cos \alpha ,\qquad (1)\,\!$

$y=-(g/2)t^{2}+Vt\sin \alpha +h.\qquad (2)\,\!$

Then I reasoned that, viewing the terminal trajectory as a tangent to the particle's parabolic motion,

${\frac {dy/dt}{dx/dt}}={\frac {-gt+V\sin \alpha }{V\cos \alpha }}={\frac {dy}{dx}}=\tan A=2\tan \alpha .$

However, this quickly leads to an absurdity. Solving for t yields a time of flight T of

$T=-{\frac {V}{g}}\sin \alpha .$

Where V is positive, g has been chosen as positive, and a is in the interval [0, π/2), this implies T < 0 (which is ridiculous). So I think my initial intuition is wrong. Thanks for any help. —Anonymous Dissident^Talk 03:23, 30 September 2010 (UTC)

I agree with you up to $2\tan \alpha$. With a parabolic trajectory any expression for time will be a quadratic rather than linear function so I don't share your conclusion that:

$T=-{\tfrac {V}{g}}\sin \alpha$

When the projectile hits the ground its y co-ordinate is -h. You can find a quadratic equation that will look something like:

$t^{2}-({\tfrac {2}{g}}V\sin \alpha )t-{\tfrac {4}{g}}h=0$

You can use the quadratic formula to solve for t. You must expect to find two values of t. One will be positive and the other negative. Good luck. Dolphin (t) 03:58, 30 September 2010 (UTC)

On second thoughts, your conclusion that $T=-{\tfrac {V}{g}}\sin \alpha$ is probably correct. When the particle hits the ground the gradient of the trajectory is negative so:

$\tan A=-2\tan \alpha$

Therefore $\sin \alpha$ (and $\tan \alpha$) must be negative. Your expression also begins with a negative sign so time T will be positive. Dolphin (t) 05:45, 30 September 2010 (UTC)

If you are correct, and the equation for T is not a contradiction, then the range can be found by putting T into the equation for x:

$R=VT\cos \alpha =V(-V/g\sin \alpha )\cos \alpha =(-V^{2}/g)\sin \alpha \cos \alpha .\,\!$

However, this value for R is at odds with the textbook's answer of 2h cot α. Either these expressions for R are equivalent, and we're left to prove this, or I've been wrong all along. —Anonymous Dissident^Talk 07:01, 30 September 2010 (UTC)

The time is a red herring in the beginning of this problem. The trajectory is the parabola y=f(x) where f(x)=ax²+bx+c satisfies f(0)=h, f(R)=0, f '(0)=tan(α)=S, f '(R)=−2S. Solve to get (a,b,c,R) as functions of the parameters (h,S). Then introduce the time t and the acceleration g. Bo Jacoby (talk) 07:54, 30 September 2010 (UTC).

If you say ${\frac {dy}{dx}}=-2\tan \alpha$ you end up with $T=3{\tfrac {V}{g}}\sin \alpha$

That changes things a bit. I'm still working on it. Dolphin (t) 07:58, 30 September 2010 (UTC)

Bo Jacoby's approach yields R = 2h cot α; you don't even need to introduce t or g for this part. I suppose the key was to express the given information in a more useful way. —Anonymous Dissident^Talk 08:29, 30 September 2010 (UTC)

(ec. I have corrected this, it was in error before. Sorry.) The trajectory is y=f(x)=−(3/4)(S²/h)x²+Sx+h, and the range is R=2h/S. Bo Jacoby (talk) 08:24, 30 September 2010 (UTC).

Yes, and putting f(R) = 0 brings out the solution for R quite easily. I'm still working on V. —Anonymous Dissident^Talk 08:31, 30 September 2010 (UTC)

Then if you note x = Vt cos α, transform f from a function of x to a function of t, and equate f'(t) with the derivative of y in (2), you get V = √(2gh/3) csc α and we're done. Thanks to the both of you. My only point of concern is that this method differs significantly from the treatment of projectile motion given by my textbook, which stresses the need to derive equations (1) and (2) straight away. I wish I knew what the authors had in mind when they set the problem. —Anonymous Dissident^Talk 08:52, 30 September 2010 (UTC)

If you solve equation (1) with respect to t and substitute this solution into equation (2), then t is eliminated and you get the parabolic trajectory equation y=f(x)=−gx²/(V²cos²(α))+x tan(α)+h, which must satisfy the requirements that f(R)=0 and f '(R)=−2 tan(α). Solving these two equations gives you R and V. Bo Jacoby (talk) 09:27, 30 September 2010 (UTC).

In mechanics there is a place for equations like your (1) and (2). My guess is that your textbook has covered that approach to problem solving, and then given this question as an example to be solved using this approach. Bo Jacoby's method is a simpler method but it doesn't exercise the student's skills at using equations like your (1) and (2). However, having sweated over your equations you have improved your skills in this area. Good luck with the remainder of the course! Dolphin (t) 12:33, 30 September 2010 (UTC)

Finding function definition from description

I was reading this part of the article about Srinivasa Ramanujan. One of the first problems he posed in a mathematical journal was finding the value of

${\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.$

here's my solution:

$f(n)={\sqrt {1+(n+1){\sqrt {1+(n+2){\sqrt {1+\cdots }}}}}}$

$f(1)={\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}$

$f(n)={\sqrt {1+(n+1)f(n+1)}}$

$f(n)^{2}=1+(n+1)f(n+1)$

after some trial and error I found $f(n)=n+2$.

$(n+2)^{2}=1+(n+1)((n+1)+2)$

which means $f(1)=1+2=3$. The problem is I guessed f(n). What is a better way to find a function from its description ? Example: $f(n+1)^{2}=n(f(n)^{2}-1)$. George (talk) 04:12, 30 September 2010 (UTC)

There's not a universal way to solve recurrence relations (the article discusses some forms that are solvable). Some techniques to try would be guessing the form you think might work and then plugging in a generic solution and solving for the coefficients. For example if you wanted to try a polynomial solution to f(n)² = 1 + (n+1)f(n+1), it's clear that any terms beyond the linear term would have to be zero, so plugging in f(n) = an + b you get a²n² + 2abn + b² = 1 + an² + (2a+b)n + a+b. So a² = a, 2ab = 2a + b, b² = 1 + a + b, which does have a solution. I don't know how you would arrive at the idea of trying a polynomial except by using some intuition or trying a bunch of things.

To solve f(n+1)² = n(f(n)²-1), note that letting f² = g lets us write g(n+1) = n(g(n)-1). Supposing we start with g(1) which is a positive integer, g(n) is then

$g(n)=(n-1)!\left(g(1)-\sum _{i=0}^{n-2}{\frac {1}{i!}}\right)$

which I doubt has a closed form. Then f(n) is the square root of that. Rckrone (talk) 05:06, 30 September 2010 (UTC)

Edit: Had to fix that formula a bunch of times... Rckrone (talk) 05:30, 30 September 2010 (UTC)

If g(0) = 1, then $g(n)=n!-\lfloor e*n!\rfloor +1$, based on [1]. Dragons flight (talk) 06:21, 30 September 2010 (UTC)

Of course your sum has a rather closed form! Based on formula number (2) from here,

$\sum _{i=0}^{n-2}{\frac {1}{i!}}={\frac {\Gamma (n-1,1)e}{(n-2)!}}$

where $\Gamma$ is the incomplete gamma function. — Pt _(T) 22:45, 30 September 2010 (UTC)

Thanks for the answers everyone.

I originally intended to write $f(n+1)=n(f(n)^{2}-1)$ but wrote $f(n+1)^{2}=n(f(n)^{2}-1)$ instead. anyway, now I have a few ideas about what to do. --George (talk) 05:05, 2 October 2010 (UTC)

---

the basic details regarding fourier transform..... —Preceding unsigned comment added by Priyabujji27 (talk • contribs) 14:12, 30 September 2010 (UTC)

{0,1} support RV

Is any random variable with support {0,1} describable as Bernoulli? That is, can any random variable with support {0,1} be written as a Bernoulli RV for some p? Thanks, --TeaDrinker (talk) 23:04, 30 September 2010 (UTC)

Yes. There's a certain probability that it is equal to 1. Call that p. Then the probability that it is equal to 0 is 1 − p. Michael Hardy (talk) 23:49, 30 September 2010 (UTC)

Ah, perfectly clear. Many thanks! --TeaDrinker (talk) 04:48, 1 October 2010 (UTC)