Recent content by TelusPig

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    Double Summation: Computing Sum with Dependent Indexes

    Oh it did turn out to be the same answer. I think I confused one of my "i"s with a "1" on paper. Thanks a lot though for the supplemental info above! :)
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    Double Summation: Computing Sum with Dependent Indexes

    I tried going back to this and elaborated more, by thinking that the sum of j is just like the sum of integers formula \sum_{i=1}^n \left( \sum_{j=i+1}^{n}i +2\sum_{j=i+1}^{n}j \right) \sum_{i=1}^n \left(i(n-(i+1)+1) +2\sum_{j=1}^{n-i}(j+i) \right) by index shifting \sum_{i=1}^n...
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    Double Summation: Computing Sum with Dependent Indexes

    May I ask how you are computing the "2j" sum? I'm guessing it's the 2 last terms n(n+1) - i(i+1) since (n-i)i sums the "i" term.
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    Double Summation: Computing Sum with Dependent Indexes

    Homework Statement How can I compute the sum An example to calculate \sum_{i=1}^n\sum_{j=i+1}^n(i+2j)?? I only have an example where n=1 and it gives a sum of 0 (why?) Maybe with n=3, what would the expanded form look like? Homework Equations I know how to do double sums, but...
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    Proving variance with moment generating functions

    OH obviously! LOL omg, I can't believe I didn't see that and thought I had to do a bunch of integrals ._. Thanks! :D
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    Proving variance with moment generating functions

    Does that mean it's \frac{M_X''(t)M_X(t)-(M_X'(t))^2}{(M_X(t))^2} Do I have to then substitute each M(t), M'(t), M''(t) with it's integral definition then? and somehow simplify that big mess o.o?
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    Proving variance with moment generating functions

    It would be for 1st derivative: \frac{1}{f(t)}*f'(t) then differentiate again for the 2nd derivative, that would be: \frac{f''(t)f(t)-(f'(t))^2}{(f(t))^2}
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    Proving variance with moment generating functions

    Moment generating functions: How can I show that Var(X)=\frac{d^2}{dt^2}ln M_X(t)\big |_{t=0} Recall: M_X(t)=E(e^{tx})=\int_{-\infty}^{\infty}e^{tx}f(x)dx E(X^n)=\frac{d^n}{dt^n}M_X(t)\big |_{t=0} Var(X)=E(X^2)-[E(X)]^2=E[(X-E(X))^2] ------------ I tried just applying the equation...
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    Proving Markov's Inequality (Probability)

    Homework Statement If g(x)\ge 0, then for any constant ##c>0, r>0##: P(g(X)\ge c)\le \frac{E((g(X))^r)}{c^r} Homework Equations I know that E(g(X))=\int_0^\infty g(x)f(x)dx if g(x)\ge 0 where ##f(x)## is the pdf of ##X##. The Attempt at a Solution I tried following a similar...
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    Find k for this probability density function to be valid.

    Never mind! I finally figured it out... I had a -infinity instead. All is good :) I got k = 1/theta if anyone was interested
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    Find k for this probability density function to be valid.

    Ok I see what you mean. But even if I do that, the upper bound of infinity is giving me problems because e^(infinity) is infinity :S
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    Find k for this probability density function to be valid.

    Homework Statement Find k such that the function f(x)=ke^{-\frac{x-\mu}{\theta}} is a probability density function (pdf), for x > \mu, \mu and \theta are constant. Homework Equations The property of a pdf says that the integral of f(x) from -\infty to \infty equals 1, that is...
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    Objects of equal mass, but different terminal velocities?

    Well I know that having a bigger area wold mean that the denominator increases, so that the overall result is smaller, as I said in my first post "I understand from this formula that the terminal velocities would be different since objects with a higher surface area would have lower terminal...
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    Objects of equal mass, but different terminal velocities?

    Sorry: m = mass, g = 9.81m/s², p = density, A = projected surface area, C = drag coefficient
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    Objects of equal mass, but different terminal velocities?

    Why do different objects having the same mass but different projected surface areas have different terminal velocities? On wikipedia, the formula for terminal velocity is Vt = sqrt (2mg/pAC) http://en.wikipedia.org/wiki/Terminal_ve ... I understand from this formula that the terminal velocities...
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