can the use of cauchy's theorem of contour integrals lead to a proof of the riemann hypothesis?
460547
This looks close enough to the techniques used to show that the roots are close to the critical line. The actual estimates are difficult, though, and nobody has been able to actually give a proof using these techniques (obviously). Go ahead and try it. Perhaps you can see through to a proof. If so, you will be famous. If not, you will have learned somthing.
116360
well i dont know how much there would be in the line of proofs steming from Fermats Last Theorem.\n\nbut i can tell you that Andrew Wiles (the man who proved fermats last theorem) developed a fair amount of new mathematics in the process of the proof, it is speculated that the new math he developed will be studied for hundreds of years.  his new math may lead to new lines of thinkings and ways to do proofs\n\nLook at it this way fermats last theorem was proved using the shimura taniyama conjecture...basically he proved that conjecture and said therefore Fermats Last theorem is true.  But he was able to say that anythign that happened in one world happened in another.  the math is way above my head...for now.\n\nSo yes, a new proof would be important.  It has been said that the answer to a question isnt nearly as imporant as the means to get there.  Implying the new developments made by proving something stem to new ideas and proofs...such as the new math used in the proof of Fermats Last Theorem
1057306
Here's a copy of the actual paper. Good luck understanding it:\n\nhttp://math.stanford.edu/~lekheng/flt/wiles.pdf
388327
I'm not sure what you mean by "Liavally" but I'll prove this using the definition of being analytic in a region.\n\nWrite f(x+iy) = u(x,y) + i v(x,y) in terms of real functions u and v. Ths function f(z) is analytic if and only if u and v satisfy the Cauchy-Riemann equations:\n\n{partial u}/{partial x} = {partial v}/{partial y}\n{partial u}/{partial y} = -{partial v}/{partial x}\n\nNow since f(z) >= 0 we must have v(x,y) := 0.  Hence {partial u}/{partial x} = {partial u}/{partial y} = 0 so that u(x,y) is a (nonnegative) constant.  This shows that f(z) must identically be a (nonnegative) constant.
998133
A conformal mapping, also called a conformal map, conformal transformation, angle-preserving transformation, or biholomorphic map, is a transformation  that preserves local angles. An analytic function is conformal at any point where it has a nonzero derivative. Conversely, any conformal mapping of a complex variable which has continuous partial derivatives is analytic. Conformal mapping is extremely important in complex analysis, as well as in many areas of physics and engineering. \n\nA mapping that preserves the magnitude of angles, but not their orientation is called an isogonal mapping (Churchill and Brown 1990, p. 241). \n\n \n \nSeveral conformal transformations of regular grids are illustrated in the first figure above, and are implemented as ComplexMap in the Mathematica add-on package Graphics`ComplexMap` (which can be loaded with the command <<Graphics`) . In the second figure above, contours of constant  are shown together with their corresponding contours after the transformation. Moon and Spencer (1988) and Krantz (1999, pp. 183-194) give tables of conformal mappings. \n\nA method due to Szegö gives an iterative approximation to the conformal mapping of a square to a disk, and an exact mapping can be done using elliptic functions (Oberhettinger and Magnus 1948; Trott 2004, pp. 71-77). \n\nLet  and  be the tangents to the curves  and  at  and  in the complex plane, \n\n (1) \n\n (2) \n\nThen as  and , \n\n (3) \n\n (4) \n\nA function  is conformal iff there are complex numbers  and  such that \n\n (5) \n\nfor  (Krantz 1999, p. 80). Furthermore, if  is an analytic function such that \n\n (6) \n\nthen  is a polynomial in  (Greene and Krantz 1997; Krantz 1999, p. 80). \n\nConformal transformations can prove extremely useful in solving physical problems. By letting , the real and imaginary parts of  must satisfy the Cauchy-Riemann equations and Laplace's equation, so they automatically provide a scalar potential and a so-called stream function . If a physical problem can be found for which the solution is valid, we obtain a solution--which may have been very difficult to obtain directly--by working backwards. \n\nFor example, let \n\n (7) \n\nthe real and imaginary parts then give \n\n   (8) \n   (9) \n\n \nFor , \n\n   (10) \n   (11) \n\nwhich is a double system of lemniscates (Lamb 1945, p. 69). \n\n \nFor , \n\n   (12) \n   (13) \n\nThis solution consists of two systems of circles, and  is the potential function for two parallel opposite charged line charges (Feynman et al. 1989, §7-5; Lamb 1945, p. 69). \n\n \nFor , \n\n   (14) \n   (15) \n\n gives the field near the edge of a thin plate (Feynman et al. 1989, §7-5). \n\n \nFor , \n\n   (16) \n   (17) \n\ngiving two straight lines (Lamb 1945, p. 68). \n\n \nFor , \n\n (18) \n\n gives the field near the outside of a rectangular corner (Feynman et al. 1989, §7-5). \n\n \nFor , \n\n   (19) \n   (20) \n   (21) \n\nThese are two perpendicular hyperbolas, and  is the potential function near the middle of two point charges or the field on the opening side of a charged right angle conductor  (Feynman 1989, §7-3).
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
74348
If you look at the standard epsilon/delta proof that G(x) = x^2 is continuous 0, it all works for F(x) too. \n\nThat is, given epsilon (e), you can find delta (d) such that /G(x) - G(0)/ < e when /x - 0/ < d.\n(Since for -1 < x < 1,  x^2 < |x|, just pick d = min(e, 1))\n\nNow looking at F(x), given d, pick that same e you would use with G(x). If x is rational, F(x) = G(x)\nso /G(x) - G(0)/ = /F(x) - F(0)/ < e.  If x is irrational, /G(x) - G(0)/ = /0 - 0/ < e.
585433
The Jacobian determinant for the suggested transformation is:\n\n|d(u,v,w)/d(x,y,z)| = |y x 0 , z 0 x , 0 z y| = -2xyz\n==>\ndudvdw = dxdydz(-2xyz)\n==>\ndxdydz = -dudvdw/(2xyz) = -dudvdw/(2*sqrt(uvw))\n\n(The sign just depends on the order of variables and is not significant.) All variables are positive. The volume is the integral:\n\n&#92;int_1^9 du &#92;int_1^4 dv &#92;int_1^16 dw 1/(2*sqrt(uvw)\n=\n(1/2)\n* &#92;int_1^9 du/sqrt(u) \n* &#92;int_1^4 dv/sqrt(v)\n* &#92;int_1^16 dw/sqrt(w)\n=(1/2) * 2^3 * (3-1)*(2-1)*(4-1) = 24
16666
it called the Cayley-Hamilton Theorem, type it in google and its proved in many places;\n\n\nhttp://delta.cs.cinvestav.mx/~mcintosh/comun/summer99/mcintosh/node9.html  <---this one is good :)\n\nLet B = x*I - A, where I is the identity matrix of the same size as A.  \nLet C be the classical adjoint of B, so that BC = (det B)I. Recall \nf(x) = det(B), by definition. We wish to show that f(A) = 0. It \nsuffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the \nk-th unit vector, so fix a value of k in that range.\n\nNow x = A implies that for all j,\n\n   x*e(j) = A*e(j),\n\n             n\n          = SUM A(i,j)*e(i),\n            i=1\n   \n      n                     n\n   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),\n     i=1                   i=1\n\n    n\n   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,\n   i=1\n\n    n\n   SUM B(i,j)*e(i) = 0,\n   i=1\n\n           n\n   C(j,k)*SUM B(i,j)*e(i) = 0,\n          i=1\n\n    n\n   SUM B(i,j)*C(j,k)*e(i) = 0.\n   i=1\n\nThen, summing over all j, reversing the order of summation, and using\nthe definition of the classical adjoint,\n\n    n   n\n   SUM SUM B(i,j)*C(j,k)*e(i) = 0,\n   j=1 i=1\n\n    n    n\n   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,\n   i=1  j=1\n\n    n\n   SUM [det(B)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(x)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(A)*delta(i,k)]*e(i) = 0,\n   i=1\n\n         n\n   f(A)*SUM [delta(i,k)]*e(i) = 0,\n        i=1\n\n   f(A)*e(k) = 0.\n\nThis proves that f(A) annihilates all the unit vectors, so must be\nthe zero matrix, and f(A) = 0.\n\n\nsecond one taken off the dr math website, but the link above might be easier to understand.
202670
There is no closed form of this integral. Expressions of this form are power towers, and you can use series methods to evaluate this integral. The value of this integral can be found in the source link. :)\n(To a poster above: All integrals are real, it's just that some may not be expressible in closed form.)
406687
Here's the basic idea:  We usually estimate area under the curve using Riemann Sums, ie rectangles.  The height of these rectangles is most often based off the right endpoint, left endpoint, or the midpoint (though we could use any point - and we do when trying to prove the Fundamental Theorem of Calculus).\n\nI often ask students, "Given a Riemann Sum, how could we make it more accurate?"  Well, the answer I'm generally looking for is: use more rectangles... But, another common (and worthwhile) answer is: Average the Right Riemann sum and the Left Riemann Sum.  \n\nIn fact, this simple suggestion results in the trapezoid rule.  Try drawing some figures and convincing yourself that the average of the right and left triangles can be drawn as a trapezoid...\n\nSo, suppose we have n rectangles and an interval from a to b.  I will use the following notation:\n(delta)x = (b-a)/n\nx_0 = a\nx_1 = x_0 + (delta)x\n...\nx_n = b\n\nThen, we can write the Right Riemann Sum as follows:\nR = (delta)x*f(x_1) + (delta)x*f(x_2) + ... + (delta)x*f(x_n)\n=[f(x_1) + f(x_2) + ... + f(x_n)]*(delta)x\nWe can also write the Left Riemann Sum as follows: \nL = (delta)x*f(x_0) + (delta)x*f(x_1) + ... + (delta)x*f(x_n-1)\n=[f(x_0) + f(x_1) + ... + f(x_n-1)]*(delta)x\n\nTo get the trapezoid rule, we average them: T = (R+L)/2\n\nNotice that f(x_0) and f(x_n) are each used in only one formula each, but all the others are used in both sums.  So, basically, we get:\nT =(1/2)*[f(x_0) + 2*f(x_1) + 2*f(x_2)... + 2*f(x_n-1)+ f(x_n)]*(delta)x\n\nor, just to write it different we could say:\nT =[(1/2)*f(x_0) + f(x_1) + f(x_2)... + f(x_n-1)+ (1/2)*f(x_n)]*(delta)x
266884
If f(z) is zero at any point z then |f(z)| = 0 which requires |f| to be zero everywhere ==> f(z)= 0 everywhere ==> f is constant. Therefore we need only consider the case for which f(z) is nowhere 0.\n\nLet f(z) = u + iv and denote the partial derivatives of u with repect to x and u, u_x, u_y etc. Since |f| is constant, the derivatives of |f|^2 = u^2+v^2 with respect to x and y are both zero:\n\n0 = d/dx (u^2+v^2) = 2u u_x + 2v v_x \n0 = d/dy (u^2+v^2) = 2u u_y + 2v v_y\n\nRewriting (using the Cauchy-Riemann equations u_x=v_y, u_y=-v_x to replace y derivatives by x derivatives in the second equation):\n\n0 = u u_x + v v_x \n0 = -u v_x + v u_x\n\nMultiply the first equation by u, the second by v, and add equations to get:\n\n(u^2 + v^2) u_x = 0\n\nMultiply the first equation by v, the second by u and subtract equations to give\n\n(u^2 + v^2) v_x = 0\n\nSince |f|^2 is assumed nonzero everywhere, u_x and v_x are zero, and therefore (Cauchy-Riemann relations) u_y and v_y are zero also ==> df/dz is zero everywhere ==> f is constant.
998304
From Wikipedia.org: 'Georg Friedrich Bernhard Riemann (September 17, 1826 - July 20, 1866) (pronounced REE mahn or in IPA ['ri:man]) was a German mathematician who made important contributions to analysis and differential geometry, some of them paving the way for the later development of general relativity.\n\nIn mathematics, the Riemann hypothesis (also called the Riemann zeta-hypothesis), first formulated by Bernhard Riemann in 1859, is one of the most famous unsolved problems. It has been an open question for well over a century, despite attracting concentrated efforts from many outstanding mathematicians. Unlike some other celebrated problems, it is more attractive to professionals in the field than to amateurs.\n\nThe Riemann hypothesis is a conjecture about the distribution of the zeros of the Riemann zeta-function ζ(s). The Riemann zeta-function is defined for all complex numbers s ≠ 1. It has zeros at the negative even integers (i.e. at s = −2, s = −4, s = −6, ...). These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that:\n\nThe real part of any non-trivial zero of the Riemann zeta function is ½. \nThus the non-trivial zeros should lie on the so-called critical line ½ + it with t a real number and i the imaginary unit. The Riemann zeta-function along the critical line is sometimes studied in terms of the Z-function, whose real zeros correspond to the zeros of the zeta-function on the critical line.\n\n \nThe real part (red) and imaginary part (blue) of the critical line Re(s) = 1/2 of the Riemann zeta-function. You can see the first non-trivial zeros at Im(s) = ±14.135, ±21.022 and ±25.011. \nA polar graph of zeta, that is, Re(zeta) vs. Im(zeta), along the critical line s=it+1/2, with t running from 0 to 34The Riemann hypothesis is one of the most important open problems of contemporary mathematics; a $1,000,000 prize has been offered by the Clay Mathematics Institute for a proof. Most mathematicians believe the Riemann hypothesis to be true. (J. E. Littlewood and Atle Selberg have been reported as skeptical. Selberg's skepticism, if any, waned, from his young days. In a 1989 paper, he suggested that an analogue should hold for a much wider class of functions, the Selberg class.)'
45488
It's not Riemann integrable because it's nowhere continuous.  It's Lebesgue integrable because the rationals are countable, so the Lebesgue integral is zero.
684813
I had to refresh myself on the method of Lagrange multipliers. The following theorem is paraphrased from a problem in Spivak's "Calculus on Manifolds", an excellent calculus text for mathematicians.\nLet A be a subset of R^n. If g:A -> R^m is differentiable and has rank m on the kernel of g, and if f:R^n -> R is also differentiable and the extremum of f on the kernel of g occurs at a, then there exist m real multipliers L1, ..., Lm such that df(a) = (L1)d(g1)(a) + ... + (Lm)d(gm)(a). (d is the differential operator)\nThe proof is not that difficult, but a concise one would require the introduction of too much machinery, so we'll leave it as something to look up.\nIf g has only one component function (if g:A -> R), then the above is very simple with only one multiplier. If n=2, the kernel of g is just a curve in R^2 and the graph of f is a surface in R^3. In this case, you can visualize the constraint on the surface as a curve in the surface of f whose projection on R^2 considered as a subset of R^3 is the kernel of g.\nIn your case, we want to minimize the distance function f(x, y, z) = Sqrt[x^2 + y^2 + z^2] over the two surfaces above, or equivalently the kernel of the differentiable function g(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) (g is thus a map from R^3 -> R^2, so m = 2. The kernel of g is trivially the intersection of the two surfaces. You should show that the rank of g is 2 whenever g(x, y, z) = 0 to validate the use of Lagrange multipliers). \nWe simplify this by noting that the points corresponding to the minimum of f over the kernel of g coincides with the points corresponding to the minimum of [f(x, y, z)]^2 = F(x, y, z) over the kernel of g which is easier to deal with. The theorem tells us we will have to deal with two Lagrange multipliers, one for each component of g. We thus have the system of 5 equations:\n2x = (L1)*0 + (L2)*8x\n2y = (L1)*2 + (L2)*8y\n2z = (L1)*4 + (L2)*(-2z)\ng(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) = (0, 0)\nThese are 5 equations in 5 unknowns, so it is easily solved. Note that the actual values of the multipliers are seldom important.\n\nIn case you are not familiar with the differential operator, if f(x, y, z) = x^2 + y^2 + z^2 + xy, then df(x, y, z) is a covector (actually a cotangent bundle in more formal terms) such that df(x, y, z) = 2x dx + 2y dy + 2z dz. It is similar to notation for implicit differentiation but has a much more rigorous foundation and is used in differential geometry. The dx, dy and dz terms are treated as basis vectors, thus the above is a 3-dimensional vector, normally called the gradient of f in multivariable calculus, where it is treated as if it were a vector in R^3: (2x, 2y, 2z). You should now see where the first 3 equations come from.
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
786224
Actually, kaksi_guy seems to be onto something. Unfortunately, not only does his proof have the flaw that you noticed, but even if that were rectified, he would still have only proved that f is zero at infinitely many points on the interval, not that f is zero on every point in the interval. Still, his post is the inspiration for what follows:\n\nFirst, we need to assume that g(x) is defined everywhere on the interval between the two points. We don't need to assume that g(x) is continuous, but if we allow a function that is undefined at even one point, such as g(x)=(x-3/x)/2 (which is undefined at zero), we may easily generate exceptions to the theorem (in this case, f(x)=x²-1).\n\nWith that out of the way, we can begin our proof. Suppose f(x) = 0 at two points a and b. By the extreme value theorem, f(x) has both a maximum and a minimum on the closed interval [a, b]. Let us denote the minimum value f(x) attains on [a, b] as min(f(x)) (and analogously for max(f(x))). Now, since f(a)=f(b)=0, min(f(x))≤0 and max(f(x))≥0. Suppose that min(f(x))<0. Then the minimum does not occur on the endpoints, so it must occur on a point c in the open interval (a, b). It is readily shown that this implies f'(c)=0. Therefore, from the original differential equation, f''(c) - f(c) = 0 and thus f''(c) = f(c). However, f(c)<0, so this implies the function is concave down at c, contradicting the fact that c is a minimum. Thus min(f(x))=0. Similarly, if max(f(x))>0, then there would be a point c on (a, b) which is a maximum, but is concave up, which is also a contradiction. Thus, min(f(x))=max(f(x))=0, and a function whose minimum and maximum are both the same is a constant function, in this case f(x)=0 on the interval [a, b]. Q.E.D.