when the function of a contour integral is analytic the value of that integral is zero, can we use this in a way to manipulate the form of riemann's formula for the zeta function to seperate the gamma function in such a way that the infinite series can be partioned and then formulized with a supplemented 0 or 1/2 for the zeta function or real part of s respectivley?
208066
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.
198397
The integral of a function can be defined as the limit of those rectangles as they get smaller and smaller, but I think what you are asking is--why is this connected to the anti-derivative?  Why is it that you can take the anti-derivative of a function, plug in at the endpoints, and you get the area under the curve?\n\nThis is a hard question to explain; it's generally called the "Fundamental Theorem of Calculus", and most textbooks prove it in detail.  You can also check out the wikipedia article below, although it's certainly not easy to understand.\n\nIf I could sum it up in a nutshell I would say that the reason why it works is that taking area under a curve is like adding up lots of little pieces, but taking the derivative (slope) is like subtraction...little tiny pieces at a time.  Just as addition and subtraction are opposites, differentiation (taking slopes) and integration (taking areas) are opposites.\n\nIt also helps to play around with the two functions f(x)=1, and f(x)=x.  Sometimes experimenting with these simple functions can help give you a feel for why this crazy connection really works.
363974
This problem is quite amusing, as knowledge of the most well-known Pythagorean triple --- 3,4,5 --- enables the final result to be evaluated easily.\n\nYou CAN NOT simply work out the average value of a NON-LINEAR FUNCTION by just averaging its end values (Tut, tut, gopal and Aegor R)!  You have to evaluate the function's integral, and then divide by the length of the range (in this case, 3).\n\nSo, the indefinite integral of this nicely chosen function is: \n\n(x^2 + 16)^(1/2) + C; NOTE that this is the same as\n(x^2 + 4^2)^(1/2) + C (Hint, hint!)\n\nWhen working out the value, the C's will cancel out.\n\nFor x = 3, the value is obviously 5 (Pythagorean triple).\n\nFor x = 0, the value is 4.\n\nSo, the integral is just 5 - 4 = 1 (It's also 1!; a factorial joke)\n\nThus, the average value is 1/3 or 0.33333... for those fractionally challenged.\n\nCHECK: To check on the reasonableness of this result, note that the form of the function near x = 0 is the straight line y = x/4.  Were this maintained throughout the full range, the mean would be 3/8 or 0.375.  But the function slowly curves down from this straight line.  At x = 3, its value is 3/5, which would correspond to an intersection with the analogous straight line through the origin, y = x/5; the mean value for this latter function would be 3/10 or 0.300.\n\nThe original function always lies between these two comparison straight lines in the given range; so the mean value of the original function (0.333...) clearly must lie between 0.300 and 0.375 --- and it does.  BINGO!  \n\nLive long and prosper.
98834
What goes wrong is the loss of information.  All _n_ divided by zero goes into the same "infinite quality".\n\nIt is possible to admit different infinite numbers; this is Robinson's approach with "nonstandard analysis", and then there are a myriad of choices as what to admit and what to disallow.  If you take that approach, it is still easier to disallow n/0 as undefined, and permit other "zeroes" where n/d is defined, where d is one of those classes of zeroes.\n\nIt is exactly equivalent, however, to the following:\nlim(d->0) n/d.  In this case the limit doesn't exist, but there are other "zero approaching denomenators" where the limit does exist.  One very important one:\n\nlim(d->0) sin(d)/d = 1
528477
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
518767
Let the first term of the series be a.Therefore,the second term would be ar,r being the common ratio.and the fifth term is ar^4\nDividing ,the fifth term  ar^4 by the second term ar,we get.\nar^4/ar=(81/64)/3  or,r^3=27/64.Therefore r=3/4.\nPutting the valu of r in the secon term i.e. ar=3,we get a=3/r=3/(3/4)=4.\nTherefore,the series is 4,3......81/64.......\nit is apparent that the trms of the series would have the values increasing and it would not come to an end because it has attained the value of  more than an unity.\nThe sum of the series would be infinity.\nAs the sum of the series is infinity,the sum of firs 10 trms of sum of any number of trms for that matter,and the sum of the complete sries would be infinity as any number subtratd from infinity gives  result of infinity.
323961
it can also be defined more geometrically this way:\n\nconsider the graph of\n\ny = 1/x\n\ne is the only point on the x-axis > 1 such that\nthe integral of y=1/x from x=1 to x=e is 1.\n\nin other words, for any number b >= 1, let A_b\nbe the area inside the region between x=1, x=b, y=0\n(the x-axis) and y=1/x. So for b=1, A_b = 0, and as b gets\nbigger, A_b gets bigger. In fact one can prove that\nA_b becomes as large as you want by making b big enough. So there's some point b=e where A_e = 1. That's e.\n\nThis is related to the logarithm, as the function\nf(b) = A_b\nis actually the natural log function, i.e.\nlog(b) = A_b.\n\nand, it's not a ratio of integers (not a rational number), i.e. there's no integers m, n such that e = m/n. this is also the case with pi. i guess when you say pi is a ratio, you mean the ratios to do with the circle. e is a ratio in that sense, for instance e = e/1, or e = (pi*e)/pi. but i'm not aware of any useful geometric ratio.\n\ne and pi also have the property (along with most real numbers) that there's no polynomial with integer coefficients p(x) such that p(e)=0 or p(pi)=0. contrast this with say sqrt(11), which isn't a rational number, but does give 0 when plugged in the polynomial\n\nP(x) = x^2 - 11.\n\nthis is the definition of transcendental versus algebraic. e and pi are trans, whereas sqrt(11) is algebraic. there are "more" transcendental numbers than algebraic numbers, in a quite reasonable sense of "more".
341308
How rare it is to see questions from complex calculus, a subject I barely got to in graduate school.\n\nI've placed a helpful link to Worlfram's MathWorld below, to start you on your way.\n\nGood luck!
153099
The function f(x)=Sigma of((f^n)(c)/n!)*((x-c)^n) is what you start with.  You can replace the Sigma(sum) with an integral.  Derivatives are the basically the opposite of Integrals.  If you derive the integral, you end up back at the start: ((f^n)(c)/n!)*((x-c)^n).  It might be easier to use a graph to see what happens. \n\nBut basically, intervals are merely used to look at a part of whatever is on the graph.  It doesn't matter what you do to that part, you are still looking at that part.  That's why the intervals don't change.\n\nDoes that help you or confuse you?
675781
Well the answer is 0.1106977 just in case you were wondering.  I did it in a CAD program.  As for integrating, hmmmmm....\n\nYeah, stick to a CAD program.  I did it symbolically just for the internal integral and you get something UNBELIEVABLY nasty.  However, if you want to give it a go, I'd say trig and u*dv  substitution since there are many terms.  \n\nAnother couple of tricks, since I think it's ridiculuous to try this by hand:\n\n1)  Use Taylor series to approximate the integral to a few terms.\n\n2)  Use binomial theorem to get another estimate.
537963
OK, I had to pull out my calc book for this.\n\nI think the "disk method" would be the "volume by cross-sections method" in my book.  Let me know if not.\n\nSo, do you have a mental picture of what's going to happen?\n\nx^4 meets y = 1 at x = 1 and x = -1, so the thing that's going to be rotated looks a bit like a squashed, colored-in letter U.\n\nThe method of cross sections (or disks) involves trying to find the area of an infinitely thin cross section, rotated around some other line.\n\nSo, if you can imagine rotating around y=2, going out from that line, you first run into y=1 and then into y=x^4.  That is, y=1 is the "inside" function and y=x^4 is the "outside" function.\n\nSo imagine for a moment that y=1 isn't there and you're just rotating y=x^4 about y=2.  Then the radius of the disk would be "2-x^4", and the area would be pi times (2-x^4)^2, right?\n\nNow subtract out the "inner ring" formed by rotating y=1 around y=2.  The radius would be 1, so the area would be just pi.\n\nSo... the function for the area of the cross section, A(x) would be:\n\nA(x) = pi[(2-x^4)^2 - 1] = pi[x^8 - 4x^4 + 4 - 1]\nA(x) = pi[x^8 - 4x^4 + 3]\n\nAnd then you would just integrate this from x = -1 to x = 1.\n\nIf you need more help, I could take it further.\n\nDisclaimer: I haven't had calc since 1991.
601173
What everyone has said is correct, but look at it this way: Consider a square. We know from the theorem of pythagoras that the ratio of the hypotenuese to the length of one side = sqrt(2) or, if you like, the ratio of one side to the hypotenuese = 1/sqrt(2). These constant ratios are independent of the size of the square. We have the ratio of the perimeter of the square to the hypotenuese = 4/sqrt(2) = 2*sqrt(2), which is independent of the size of the square.  Similarly the ratio of the circumference of a given circle to the longest chord(the diameter) has an exact value, and similarly this value is independent of the size of the circle. As has been mentioned, for any regular n-gon(square, pentagon, hexagon etc) the ratio of the perimeter to the length of the longest line segment that can be drawn within the n-gon will have a specific value depending on n, and this value can be determined. As we let n approach infinity, the n-gon approaches a circle and the ratio of the perimeter of the n-gon to the longest "chord" approaches a specific limit and this limit is pi, the ratio of the circumference to the diameter.  And, as mentioned above, analytic techniques can (and have) determined the value of the constant pi to a trillion or so digits.\n\nFootnote: SK, determining the limit of a series or even that the limit exists does not necessarily require calculus.\nFor example, the sum of a geometric series such as Sum[1/(2^n)] = 1/2 + 1/4 + 1/8+....+1/2^n is proved without calculus to be 1- 1/2^n and the limit of it (without calculus) as n---> inf. = 1. Check out the below link regarding pi.\n\n http://mathworld.wolfram.com/ArchimedesAlgorithm.html
552849
The basic idea of a fourier series is to express a function - in this case the sawtooth wave - as a sum of sine and cosine functions of different frequencies (i.e. n omega0).  The result gotten for a sawtooth wave is the thing you described (the sideways 8 means "infinity").  That is, it's an infinite sum of sin curves, each one weighted by 1/n.  See a picture here: http://mathworld.wolfram.com/FourierSeriesSawtoothWave.html
370213
the first thing is to parametrize the segment:\nx= 1 + t(0)\ny = 2+ t(3)\nz=1+ t(4),\nnow when t=0, then x=1,y=2 and z=1, and when t=1: x=1,y=5 and z=5,\nso\nnow\nthe line integral is:\nintegral from 0 to 1 (3x dx+y dy+(x^2z–y^3)dz) \n=integral form 0 to 1 3(1) (0) + (2+3t)3dt - (1(1+4t)- (2+3t)^3 )(4)dt\n= integral form 0 to 1 (6 +9t -4 - 16t +4(2+3t)^3 ) dt\n= integral from 0 to 1 (2 - 7t + 4(2+3t)^3 ) dt\n= 2t - 7t^2 + 4 (2+3t)^4 (1/4) (1/3) evalutated at 1 and 0\n= 2 -7 + (1/3) (5)^4 - ( 0 -0 + (1/3) 2^4 )\n'
80191
I haven't worked with this in a long time and I'm not sure this will work, but try starting w/ this:\n1) Define an equation that describes the " ∞ " with polar coordinates (something like r = C1 - C2*theta);\n2) If L is lengh, delta.L = delta.theta / (2 pi) * r\nand dL = r/(2 pi) dtheta  ;\n3) substitute for r and integrate from 0 to 2pi (or the necessary for one complete " ∞ ".\n \nProbably you'll find more than one mistakes here, but I think the general idea is correct.
214541
A complex function is differentiable at a point z0 if the limit z -> z0  (f(z) - f(z0))/(z - z0) exists. However, the differivative no longer has any geometrical interpretation.\n\nThe differentiation can be done in a fashion that resembles real differentiation using the definition:\n\nlim dz -> z   ( f(z0 + dz) - f(z0) ) / dz\n\nHowever since z is a complex plane, the limit as dz -> z can approach from any direction. If it is differentiable, the value of the derivative is the same when approached from any direction, so a path is chosen that makes the calculations easy, typically keeping the real or imaginary part constant and just approaching in a direction parallel to either axis. The effect is that the limit calculation is done on a single variable. But this only finds the derivative at a single point.\n\nEquations for the derivative of a function can be found by using the Cauchy-Riemann Equations, where valid, if \nw = f(z) = u(x,y) + i v(x,y)\n\nin which case if partial du/dx = dv/dy AND partial dv/dx = -du/dy AND all theses derivatives are continuous, then \n\ndw/dz = partial du/dx + i dv/dx = partial dv/dy - i du/dy
358424
let theta be represented by 't' and integral sign be replaced by I. Then\n\nI[1 - cos(2t)dt\n= I[1*dt - cos(2t)*dt]\n= I(dt) - I[cos(2t)]dt\n= t - [sin(2t)]/2 + C ___________(1)\nwhere C = arbitrary constant of integration\n\nFrom (1)\nWhen t = 0, I(0) = 0 - 0 + C = C \nWhen t = T, I(T) = T - [sin(2T)]/2 + C\n\nFrom the Fundamental Theorem of Integral Calculus we can write that \nthe reqd. integral = I(T) - I(0)\n= T - [sin(2T)]/2 + C - C\n= T - [sin(2T)]/2\n= T - {sin(T)}*{cos(T)} [since sin(2T) = 2*{sin(T)}*{cos(T)}]