What is the expansion of sin hax in fourier series?
94920
you mean sinh(ax)?\n\nwhat is this hax
799668
For a function of the form y = a tan (bx + c)\n\nThe amplitude is a.\nThe period is pi / b.\nThe horizontal shift is c.\n\nSo, the amplitude is 2, the period is 1, and the horizontal shift is 3.
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx
830695
UPDATE: Sorry for the edit, but on further review, it IS possible using an infinite summation. I can't really write it out here, so check this wikipedia link:\n\nhttp://en.wikipedia.org/wiki/Fresnel_integral\n\nYou can't do it with straight alegebra because there is no trigonometric function to expand f(x) = cos(x²), and there is no coefficient of x (so the chain rule doesn't apply). Basically, it's impossible with basic algebra.\n\nHowever, I put this function into Maple 7.0 and this is what it gave:\n\n1/2 * sqrt(2 * pi) * FresnelC(x * sqrt(2 / pi))\n\nSo apparently there exists some Fresnel function that can be used to derive the answer. I'm not sure though, that may be much more work than you want to do.
722288
(x/2)*(100-x^2)^(1/2)+\n50*arcsin(x/10)\n\nSorry it's on two lines, yahoo cuts it off if I leave it on one. arcsin is the inverse sine function.
120506
i'm giving the final answer not hte full steps ok\n(a sin(wt)-w cos(wt))/w*(w^2+a^2)  + e^(-at)/(a^2+w^2)\nby using convolution theorem
212677
Little reply: No.\n\nP.S. 12.08 AM GMT\nThis gives the Fresnel Integral S(x), more exactly\n\nS(x)=Integral(.5Pi*x^2)\n\n(the integral is from 0 to x), and it is not possible to reduce it in a finite way.\n\nHowever S(x) has an asymptotic expansion for x>>1\n\n.5-(cos(.5Pix^2))/(x*Pi)\n\nwhich gives limit S(infinity)=0.5
146506
well, you don't have to worry about sinx and cosx as they are always between -1 and +1. so the first part of the function (1/2)* sinx is bounded.\ncos(x) is bounded.\nSo, you can rewrite the function as,\ny(x) = f(x) - x*g(x).\nwhere both f(x) and g(x) are bounded.\nso, as x goes to infinity, y(x) tends to minus infinity because of -x in the right hand side. \nSimilarly when x goes to minus infinity, y(x) goes to plus infinity\n----------------------\nResponse to the counter-example - \nin case y(x) = 2+x*(-3),\nas x -> infinity y -> -infinity.\nso y(x) is unbounded\nwhat's wrong with that??????
511652
look at the argument.  For example the argument of sin(2x) is 2x.  We know the period of sin function is 2pi.  So the period of sin(2x) is such that 0<2x<2pi, 0<x<pi, or x is pi periodic.\n\nJust stick the argument of the function into the inequality expression for the period and solve.
1045359
I hate Fourier transforms.
49580
NO WAY Why?\n\nF(t) = A Sin(wt)\n\nA called the magnitude : how high the sine function reach\nif A =1 the sine is between 1 and -1\nif A =5 the sine is between 5 and -5\n\nW = 2*pi*f where f is the frequency : how many period during 1 sec\nif w=1 f=1/(2pi) this means at t=2pi we will have one period\nsin(2pi)=sin(0), one period for each 2pi\nif w=2 >>> one period ecah pi\n\n\nSO when the function is expressed by the sum of many sines, many differents frequancies, impossible to regroup them in one frequency\n\nTHe basic frequency called fimdemental and the other frequencies called harmonics (mulpitles of fundemental)\n\nFourrier Series gives the magnitude of all harmonice of unregular from, BUT IT IS NOT POSSIBLE TO JOINT MORE THAN ONE FREQUENCY IN ONE TERM\n\nlike 3x^2+4x^5 since the powers are differents we can't sum or add them in one term
598177
First by frobenius method, you can easily obtain the series expansion of the first order of bessel function. The coefficients of series expansion of this function help you to find the degree. And you need to know that the asymptotic forms of Bessel functions are relevant to the gamma functions.  (z/2)^(-n) J_n(z)~1/gamma(n+1) is a degree of 1 \nIf you need to show that you need to use the asymptotic expansion of gamma function which can be also obtained by using the frobenius method. But you can also use the stirling formula which approximates to the gamma function. \nThen take the limit of the rho=n^n e^{-n}/log(a_n) as n goes to infinity. a_n: is the coefficient of the series expansion of bessel function , rho is the order. You will see that rho is going to be equal to 1. Hadamard's theorem will help you to understand how you define the order(rho) as above.\n \nAll routines in matlab, fortran are written using these mathematics. By the way, it is not a simple question. Good luck:)
148647
x+sinx\nLim ------        \n    x+2sinx\nx->0\n \n    1+cosx\nlim-------   [according to Hospital's theory]\n    1+2cosx\nx->0\n\n   1+cos0\n=----------\n   1+2cos0\n\n  1+1\n=-----\n  1+2\n\n  2\n=----\n  3\n ans 2/3.\nThanks.\n  3
322159
Are you kidding me? Sin(x) is awesome! I totally love sinx-ing  those sexy numbers :)\n\nI sinxed pi the other day, and it totally melted away.
454694
1\nDetails:\nLet y = lim x->0 of (sinx)^x\nTake the natural log of each side.\nln(y) = lim xln(sinx)\nWrite xln(sinx) as (ln(sinx))/(1/x)\nas x goes to 0, this goes to infinity over infinity.\nUse L'Hopital's rule \n(ln(sinx))/(1/x) becomes\n(cosx/sinx)/(-x^(-2))\n= (-x^2)(cosx)/(sinx)\nAs x goes to 0, this goes to 0/0.\nUse L'Hopital's rule  again\n(-x^2)(cosx)/(sinx) becomes\n[(x^2)(sinx) - 2x(cosx)]/(cosx)\nPlug in 0.\n[0-0]/1 = 0\nThis equals ln(y)\nln(y) = 0 means y = 1
901088
4sinxcosx =1 (assumed for 0 ≤ x ≤ 2π)\nSo 2 sin2x = 1 (since sin2x = 2sinxcosx)\nSo sin2x = ½ (0 ≤ 2x ≤ 4π)\n2x = π/6, 5π/6, 13π/6, 17π/6\nThus x = π/12, 5π/12, 13π/12, 17π/12\n\nsin4x= -2sin2x (assumed for 0 ≤ x ≤ 2π)\nSo 2sin2xcos2x + 2sin2x = 0\nie 2sin2x(cos2x + 1) = 0\nso sin 2x = 0 or cos 2x = -1/2 (0 ≤ 2x ≤ 4π)\n2x = 0, π, 2π, 3π, 4π or 2x = 2π/3, 4π/3, 8π/3, 10π/3\nwhence x = 0, π/3, π/2, 2π/3, π, 4π/3, 3π/2, 5π/3, 2π
356491
http://mathworld.wolfram.com/HarmonicSeries.html