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can you give a more detailed example?

it's finite.. all of them have to be positive integers

what about uniqueness/knowing that all solutions are found

write k for floor(n^2/10) then we know that a_n = k(k+1)/2 we also know that n^2/10 - 1 < k <= n^2/10 so (n^2/10)(n^2/10 - 1)/2 < a_n <= (n^2/10)(n^2/10 + 1)/2 both bounds ~ n^4/200 and so does a_n

you're looking at something along the lines of (everything above the line) http://www.wolframalpha.com/input/?i=plot+%283%2F4^%281%2F3%29%29+*+%28x^2%29^%281%2F3%29+%2B+2 for a = 1. Don't think this is suitable for 4U.. essentially you need to figure out which parameters would make a cubic...

arg z[z-(root3 +i)] = pie/6 write a = sqrt(3) + i let p+iq = w = z - a/2 = (x - sqrt(3)/2 ) + (y-1/2)i arg( (z - a/2)^2 - a^2 /4 ) = pi/6 arg( w^2 - a^2 /4 ) = pi/6 arg( (p^2 - q^2 - 1/2) + i(2pq - sqrt(3)/2 ) = pi / 6 (2pq - sqrt(3)/2 ) / (p^2 - q^2 - 1/2) = 1/sqrt(3)...

If you are talking about complex valued functions (f:R --> C) then it's relatively straight forward because you can consider f(x) = g(x) + i*h(x) where g, h are regular real functions. What gets interesting is when you consider complex functions ie f:C --> C. for one \int_a^b f(z) dz where...

the original question: http://www.wolframalpha.com/input/?i=plot+sqrt%283%29%28x^2+-+y^2+-+1%29+%3D+2xy remove the parts in the 2nd and 4th qudarant

(2)'s wrong.. in (1) you assumed that (x1,y1) is on the circle, so you can't use it as a point on the hyperbola. you should solve (1) and x^2 - 2y^2 = 4 simultaneously and set the discriminant to 0. Actually.. the better way: Here's what I tried: Equation of tangent to circle: x[x1] + y[y1] =...

I don't have the paper, but you might want to specify the year too.

slightly more elementary: (1 - \omega_1) \times \left(\sum_{i=1}^{n} \omega _i\right) =\sum_{i=1}^{n} \omega _i - \sum_{i=1}^{n} \omega _i = 0 and since (1-\omega_1) is not 0, you know that the sum is 0

you don't really need to multiply through by the denominator.. (2x-3)/(x-2)>1 (2x-4 + 1)/(x-2)>1 2 + [1/(x-2)]>1 1/(x-2) > -1 ==> x-2 > 0 or x-2 < -1 x > 2 or x < 1

not that simple, you need some analysis to go about these things

depends on the order your techers taught you guys

The first and third lines gives the definitions of U_n, your job is to show taht the middle one is equivalent to the def.,

your method is also ok. but "Then using z^n + 1/z^n = 2cosnθ" is wrong, it only works for z on the unit circle. (z^3 + 1/z^3) + 15(z + 1/z) = 0 (z + 1/z)^3 + 12(z + 1/z) = 0 (z + 1/z)((z + 1/z)^2 + 12) = 0 (z + 1/z)((z + 1/z) - 2isqrt(3))((z + 1/z) + 2isqrt(3)) = 0 (z^2 + 1)(z^2 - 2isqrt(3)z...

you can just substitute and verfy for this q. if you want to solve it.. there are other ways: alternate method: (z+1)^6 = (-1)*(z-1)^6 taking 6ths roots we get, z+1 = w*(z-1) where w is a 6th root of -1 so z = (1+w)/(1-w) let w' be the conjugate of w and w = cos(a) + isin(a) z =...

Not really.. There's an algebraic proof.

One reminder .. the integral of 1/x is (ln |x|) not (ln(x))

almost there :) arg(a-b) -arg(c-b) is the angle at b