Recent content by braindead101

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    Joint probability density function problem

    Suppose that h is the probability density function of a continuous random variable. Let the joint probability density function of X, Y, and Z be f(x,y,z) = h(x)h(y)h(z) , x,y,zER Prove that P(X<Y<Z)=1/6 I don't know how to do this at all. This is suppose to be review since this is a...
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    Deduce taylor series

    can anyone help me out, i am still stuck in same place
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    Deduce taylor series

    I have tried to do this, but I am stuck and cannot get them to equal each other. For f(x)=(1+(-4x))^(-1/2) The taylor series about 0 is: sum (-1/2 choose n) (-4x)^n expanding binomial coefficients: sum -1/2(-1/2-1)(-1/2-2)...(-1/2-n+1) / n! x (-4)^n (x)^n sum -1/2(-3/2)(-5/2)...(1/2-n)/n! x...
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    Deduce taylor series

    Deduce that the Taylor series about 0 of 1/sqrt(1-4x) is the series summation (2n choose n) x^n. From this conclude that summation (2n choose n) x^n converges to 1/sqrt(1-4x) for x in (-1/4,1/4). Then show that summation (2n choose n) (-1/4)^n = 1/sqrt(1-4(-1/4)) = 1/sqrt(2) What I know...
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    Variational calculus - dual problem

    the primal problem was: min (x^T)Px i found g(r) and the partial derivative of g(r) w.r.t. x to be: x=-1/2(P^-1)(A^T)r i have found the dual problem to be: max -1/4(r^T)A(P^(-1))(A^T)r - (b^T)r subject to r>= 0 I am told to find x* and r* (which i think is just x and r): i have not...
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    Radius of convergence

    I am looking for radius of convergence of this power series: \sum^{\infty}_{n=1}a_{n}x^{n}, where a_{n} is given below. a_{n} = (n!)^2/(2n)! I am looking for the lim sup of |a_n| and i am having trouble simplifying it. I know the radius of convergence is suppose to be 4, so the lim sup...
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    Showing it is orthogonally diagonalizable

    Suppose that the real matrices A and B are orthogonally diagonalizable and AB=BA. Show that AB is orthogonally diagonalizable. I know that orthogonally diagonalizable means that you can find an orthogonal matrix Q and a Diagonal matrix D so Q^TAQ=D, A=QDQ^T. I am aware of the Real Spectral...
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    Finding the function, given the gradient.

    ok, so i can get 1/px^p for the x>0 case. but for the x<0 case: i am struggling i have, integ( (-x)^(p-2) x dx) can i write this as: = integ( (-1)^p (x)^(p-2) x dx ) so, = (-1)^p integ (x^(p-2) x dx) which is just = (-1)^p 1/p x^p now how can i put the two together... to make x into |x|?
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    Finding the function, given the gradient.

    can you tell me how to integrate this? or at least start, so i can get 1/p|x|^p , i need this small part for a bigger problem and this is making me stuck. i have thought about what you said about the piecewise, but that confuses me even more as i have to deal with not one but 2 functions now
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    Finding the function, given the gradient.

    so which way should i be working? gradient f(x) -> f(x)?
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    Finding the function, given the gradient.

    ok, should i be working from the gradient f(x) -> f(x) or vice versa. as well , i am getting confused. is this correct: to work from gradient f(x) -> f(x) we integrate. and f(x)-> gradient f(x) we differentiate. working from f(x) -> gradient.. i dont see how i can get gradient f(x). and going...
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    Finding the function, given the gradient.

    no the function is this: gradient f(x) = x |x|^(p-2) maybe this is more clear way to write it. and somehow get f(x) = 1/p |x|^p from it.
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    Finding the function, given the gradient.

    yes, i have the gradient f(x)= |x|^p-2 x, and i need to find f(x), in class, the definition of gradient is just the derivative w.r.t x of f(x) so i am asking why 1/p |x|^p is the answer because i don't see how you can use this, to find the gradient function |x|^p-2 x. so I thought the function...
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    Finding the function, given the gradient.

    the gradient function is |x|^p-2 x and i need to find the function, which apparantly is 1/p |x|^p but i can't figure out how to show this. This is for a bigger problem where the function must be convex. and also p>1 I tried, finding the derivative of 1/p |x|^p , but i don't get the gradient...
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    Gateaux derivative

    lim ep->0 <A(u+epv),u+epv> - <Au,u> / ep lim ep->0 <A(u+epv),u>+<A(u+epv),epv>-<Au,u> / ep lim ep->0 <Au,u> + <Aepv,u> + <Au,epv> + <Aepv,epv> - <Au,u> / ep lim ep->0 <Aepv,u> + <Au,epv> + <Aepv,epv> / ep associativity: lim ep->0 ep*<Av,u> + ep<Au,v> + ep*<Av,epv> / ep so ep*(ep) = |ep|^2...
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