Recent content by QuantumLuck

oops I am dumb. clearly to find x i take \xi - \eta to find that \xi - \eta \ = 2x and as such \ x = (1/2)\xi - \eta so i then plug in x and find that \ y =\sqrt{\xi\eta} at which point i then showed that kinetic energy \ T=(m/8)(\xi+\eta)(\dot{\xi}^2/\xi +\dot{\eta}^2/\eta). the final thing...

so yeah i think this is right. so here is what i did; \xi = \sqrt{x^2+y^2}+x and then i squared both sides to obtain \xi^{2} = 2x^{2} + y^{2} + 2x \sqrt{x^2+y^2} and \eta^{2} = 2x^{2} + y^{2} -2x \sqrt{x^2+y^2} so now that i have these equations i have been playing around with them trying to...

Homework Statement I am told that parabolic coordinates in a plane are defined by \xi = r + x and \eta = r - x. after this i am then asked to show that this leads to a given expresion for the kinetic energy (if i knew x and y i could find this without a problem). From this I am then told to...

when you say strip you mean just that piece of the real line right? because we are only varying x. so it look like a standard sine function, oscillating between -1 and 1...which is the part of our real line that still exists in my omega. but i fail to see what the connection is between this fact...

while interesting, the problem that i see with this is that the zeroes of sin(z) are points on the real line so it does not relate to my original omega. after a little discussion and searching we found a simplified form of the bilinear mapping when one of the points is -inf or +inf. however...

while interesting, the problem that i see with this is that the zeroes of sin(z) are points on the real line so it does not relate to my original omega. after a little discussion and searching we found a simplified form of the bilinear mapping when one of the points is -inf or +inf. however...

Homework Statement Let B1 = {z element C : abs(z) < 1}, f be a holomorphic function on B1 with Re f(z) > greater than or equal to 0 and f(0) =1. then show that: abs(f(z)) less than or equal to [(1+abs(z))/(1-abs(z))] Homework Equations Schwarz's Lemma: Suppose that f...

Homework Statement Let Omega = C\((-inf,-1]U[1,inf)), find a holomorphic bijection phi:omega-->delta, where delta is the open unit disk Homework Equations Reimann Mapping Theorem Special Mapping formulas: can map wedges onto wedges, with deletion of real line from zero to infinity in...

Homework Statement find the Laurent series for f(z) = 1/(z(z-1)(z-2)) on the annulus between 1 and 2. with the origin as center. Homework Equations The Attempt at a Solution so i found the partial fraction decomposition of this function and it turns out to be f(z) = 1/2z +...

argh. i am a fool. i forgot the parentheses; cosine is in the denominator. the integrand is dtheta/(2 + cos(theta))

so as long as the function you are tying to converge to is defined over the entire annulus any sequence which approaches that function (the Laurent series of that function) will converge uniformly? and this is relevant to my question because i am seeking a polynomial sequence which is just the...

it is exactly as i said. the integral from 0 to pi of 1/2+cos(theta) dtheta. now what i know how to evaluate is the integral from 0 to 2 pi of the previous integrand. i am just not sure how to modify the equation (if that is the correct path to take anyways).

Homework Statement Can we find a sequence, say p_j(z) such that p_j ---> 1/z uniformly for z is an element of an annulus between 1 and 2, that is 1 < abs(z) < 2? Then i am asked to do the same thing but for p_j ---> sin(1/z^2).Homework Equations Not too sure about this, maybe Taylor...

Homework Statement So I know how to evaluate the integral from 0 to 2pi of 1/2+cos theta. However, the question I am being asked to do has me calculate this integral from 0 to pi. I am not sure what adjustment is necessary to get the integral i am given (from 0 to pi) to the form I know how...

Thanks a lot Dick I was able to put together what you said with some of my professor's help to solve the problem. he didn't like the books method either. so i had to show that the power of the polynomial couldn't be even or an odd number that is three or greater. hence, it must be one.