# Tag Info

1

It means no solution. Be sure you've coded the problem correctly.

-1

I used this method to find practical approximation in the problem of the comparison of 2 binomial samples.

1

Using the notations of your Mathematica command, you should notice that variables $a,b,c$ can be eliminated using the first three equations. This leads to $$a=\frac{-1}{d+e}$$ $$b=-\frac{1}{2 (d+e-1)}$$ $$c=\frac{1}{2 (d+e)}$$ Now, replacing these in the next equations, the fourth becomes $$\frac{5}{4 (d+e)^2}+\frac{1}{d+e-1}+\frac{1}{4 (d+e-1)^2}-2=0$$ and ...

0

Since $${dy\over dx}={{dy\over dt}\over {dx\over dt}}={y(5-y-x)\over x(7-x-2y)},\tag{*}$$ solution trajectories have slope zero (i.e., are horizontal) when $y=0$, $5-y-x=0\implies y=5-x$ (shown in red below), have undefined slope (i.e., are vertical) when $x=0$, $7-x-2y=0\implies y={7\over 2}-{1\over 2}x$ (shown in blue below). If you want to proceed ...

2

In the context of the complex plane there is a very useful notion of a "point at infinity" but it is in infinity "in every direction". This gives rise to a notion of the Riemann Sphere, where two copies of the complex plane give coordinate charts with transition $z\mapsto 1/z$. Complex functions with poles can then be viewed as functions from the Riemann ...

2

In Mathematica and/or WolframAlpha: Input: "(1/64^2)*Sum[Sum[Sin[(i+0.5)/64+Sqrt[(j+0.5)/64]],{j,0,63}],{i,0,63}]" Output: "0.857956" You can replace the 64 and 63 with whatever value of n and n-1 you want. Also, using the substitution $y = u^2$, we can compute the exact value of the integral: $\displaystyle\int_0^1\int_0^1\sin(x+\sqrt{y})\,dx\,dy = ... 3 In Maple: [> f := (x,y)-> sin(x+sqrt(y)): for k from 0 to 5 do evalf(1/2^(2*k)*Sum(Sum(f(1/(2*2^k)+i/2^k, 1/(2*2^k)+j/2^k), i = 1 .. 2^k-1), j = 1 .. 2^k-1), 5) end do; 0.84016 0.84016 0.84016 0.84016 0.84016 ... 0 First, note that the first two terms of their solution are equivalent to your first two terms under the identification$(C_1,C_2)=\frac{1}{2}(N_1+N_2,N_1-N_2)$. So the only potential problem is in the second halves of the two results; even there, we can recognize a common factor of$-i q_y/8\pi q^2\$, so the only issue is the exponentials themselves. To see ...

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