kingjewel1 Posted March 15, 2005 Share Posted March 15, 2005 how would i show that the values of x at which sinxe^(-sinx) has stationary points form an arithmetic sequence. I've so far worked out dy/dx=cosxe^-sinx(1-sinx)=0 so cosx.e^-sinx= and sinx=0 arcsin0=pi,2pi,3pi,4pi,etc so cospi.e^-sinpi=0 cos2pi.e^-sin2pi=1 cos3pi.e^-sin3pi=0 cos4pi.e^-sin4pi=1 therefore it is a repeating sequence and common difference 1,-1 but i don't think this is correct as again (im not sure how to solve (cosxe^-sinx(1-sinx)=0)) any ideas? thanks for any help Link to comment Share on other sites More sharing options...
Guest entropy Posted March 17, 2005 Share Posted March 17, 2005 how would i show that the values of x at which sinxe^(-sinx) has stationary points form an arithmetic sequence. I've so far worked out dy/dx=cosxe^-sinx(1-sinx)=0 so cosx.e^-sinx= and sinx=0 arcsin0=pi' date='2pi,3pi,4pi,etc so cospi.e^-sinpi=0 cos2pi.e^-sin2pi=1 cos3pi.e^-sin3pi=0 cos4pi.e^-sin4pi=1[/b'] therefore it is a repeating sequence and common difference 1,-1 but i don't think this is correct as again (im not sure how to solve (cosxe^-sinx(1-sinx)=0)) any ideas? thanks for any help dy = cosx(1 - sinx)e-sinx is correct. dx dy = 0 => cosx = 0 or 1 - sinx = 0 or e-sinx = 0. dx cosx = 0 => x = pi/2 + n*pi, where n is an integer. sinx = 1 => cosx = 0, so this adds no new solutions. e-sinx is never equal to zero, as the exponential function never equals zero. So the turning points occur at pi/2 + n*pi; i.e., they form an arithmetic sequence. Link to comment Share on other sites More sharing options...
kingjewel1 Posted March 17, 2005 Author Share Posted March 17, 2005 thanks very much for your help! Link to comment Share on other sites More sharing options...
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