Wednesday, 19 November 2014

trigonometry - Maxima and minima of operatornamesinc function



The function sincπx has maxima and minima given by the function's intersections with cosπx, or alternatively by ddxsincπx=0.



Mathematica tells me that




ddxsincπx=π(cosπxπxsinπxπ2x2)



So question 1, how do I prove this?



And question 2, how do I derive an equation for all maxima and minima?


Answer



Set derivative equal to 0.



You will after some manipulation like multiplying (πx)2 and dividing by π get πxcos(πx)=sin(πx) and equivalently by dividing both sides by πx

cos(πx)=sin(πx)πx
Now the right hand side is sinc(πx) and left hand side is the function you want to show it should intersect.



So we are done showing where the extrema are.






Now to show which are max and which are min.



Sinc as a function is a multiplication between 1πx and sin(πx)




On R+ the first of these is monotonically decreasing and positive. Sin is periodic and alternating +1 -1. Both functions are continuous. We can now use an argument with theorem of intermediate value to show it will be alternatingly max and min with as many maxes as mins.


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