kingjewel1 Posted March 15, 2005 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
Guest entropy Posted March 17, 2005 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.
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