Counting ExtremaDate: 10/23/96 at 2:40:56 From: Pia Kurimo Subject: tan(Na)=Ntan(a) Fraunhofer Diffraction I was trying to find the number of minima (and maxima) of the function [sin(Na)/sin(a)]^2 on the interval [0,pi]. I took the derivative of the function and got that tan(Na) = Ntan(a) or that sin(Na) = 0. I don't, however, know how to get the correct answer that on this interval, the number of minima is N-1 and the number of maxima is N-2. Please help me! Yours, Pia Date: 01/04/97 at 19:00:41 From: Doctor Yiu Subject: Re: tan(Na)=Ntan(a) Dear Pia, Your question is a very interesting one. An approach using calculus, as you have already attempted, would lead to the following equation (as well as other related equations): n Tan x - Tan nx = 0 Doctor Ken has drawn the graphs of the functions (n Tan x) and (Tan nx) on the same axes to help you count the number of roots: To distinguish the maxima from the minima, we inevitably run into an analysis of the SECOND derivative of (Sin Na/Sin a)^2, or the SIGN DISTRIBUTION of the first derivative. These can be handled somewhat more efficiently by using implicit differentiation to avoid excessive calculations. ------------------------------------ I would like to offer an answer without too much computation. I prefer to make a slight change of notation. Change your a into an x, and your N into an n, so that your statement now reads: (sin nx/sin x)^2 This function has (n-1) MINIMA and (n-2) MAXIMA for x between 0 and pi. ------------------------------------ Let's begin with a somewhat more general situation. OBSERVATION (I): If y is a polynomial in t, of DEGREE (n-1), and y has (n-1) distinct REAL ROOTS, then y has exactly (n-2) MAXIMA and MINIMA, and these occur ALTERNATELY, each STRICTLY between two neighbouring roots of y. ------------------------------------ To see this, let's begin by marking a number of DISTINCT points on a line, the t-axis. (I shall say that there are (n-1) points; to fix ideas, you might take 5 points.) Starting from a point on the far LEFT side (but not on the line), try to trace toward the RIGHT side, a continuous, smooth curve (graph of a POLYNOMIAL) cutting the line ONLY at these (n-1) points. Then, STRICTLY between every two neighbouring points on the line, there is either a MAXIMUM or a MINIMUM. These account for (n-2) DISTINCT points which are the ROOTS of the DERIVATIVE of the polynomial. Since the derivative has degree (n-2), it has NO OTHER ROOTS and these (n-2) points are precisely ALL of the maxima and minima. (Note that none of these maxima or minima can occur at the roots of the polynomial. If so, one or more of the roots would have MULTIPLICITY greater than 1, and there would be more than (n-1) roots of the polynomial of degree (n-1). This is clearly impossible.) ------------------------------------ OBSERVATION (II): Let y be a polynomial in t, of degree n-1, as in (I) ABOVE. Then y^2 is a polynomial of degree (2n-2); it has exactly (n-1) MINIMA and (n-2) MAXIMA. ------------------------------------ First of all, the degree of y^2 is TWICE that of y. This means that y^2 has degree (2n-2). Begin with the graph of y, sketched as above, and try to sketch the graph of y^2. SQUARE the value of each y to form the graph of y^2 and you will see that: (i) y^2 is a MINIMUM at each root of y, and there are (n-1) of these; (these minima are all ZERO.) (ii) y^2 is a MAXIMUM at each of the EXTREMA of y, (whether maxima or minima; the minima, after squaring, are REFLECTED about the x-axis and become maxima), and there are (n-2) of these, as explained in (I) above. These are all different from the roots of those points in (i) above. These MINIMA and MAXIMA account for (n-1) + (n-2) = (2n-3) DISTINCT ROOTS of (y^2)'. Since (y^2)' is a polynomial of degree 2(n-1)-1 = 2n-3, these are ALL of the MAXIMA and MINIMA of (y^2)'. ------------------------------------ Now, the essential link of these OBSERVATIONS (I) and (II) to your question is the following: OBSERVAYION (III) The function y_n = sin nx/sin x is a POLYNOMIAL in cos x, of degree (n-1). Note that I have used y_n to indicate the dependence on n. ------------------------------------ For example, y_1 is identically equal to 1, a polynomial of degree ZERO. y_2 = sin 2x/sin x = (2sin x cos x)/sin x = 2cos x. If you write t = cos x, then y_2 = 2t, a polynomial in t of degree ONE. You can of course proceed to check a few more cases, like: y_3 = sin 3x/ sin x = (3sin x - 4sin^3 x)/sin x = 3 - 4sin^2 x = 3 - 4(1-cos^2 x) = 4cos^2 x - 1 = 4t^2 - 1, a polynomial in t of degree TWO. But it is enough to examine the general situation. ------------------------------------ Now, in TRIGONOMETRY, there are formulas that help convert a sum into a product. I mean, specifically, there is an identity: sin A + sin B = 2sin ((A+B)/2) cos ((A-B)/2) With A = (n+2)x and B = nx, we have: sin (n+2)x + sin nx = 2sin (n+1)x cos x ------------------------------------ If we move the terms around: sin (n+2)x = 2cos x sin (n+1)x - sin nx, and divide both sides by sin x, then, in terms of the y's and t = cos x, we have: y_(n+2) = 2t y_(n+1) - y_n SUPPOSE y_n and y_(n+1) are polynomials in t, of degrees (n-1) and n respectively, then this last equation shows that y_(n+2) is a polynomial in t, of degree n+1. Since y_1 and y_2 are indeed polynomials in t, of degree 0 and 1 respectively, we conclude, by mathematical induction, that for every integer n = 1, 2, ..., y_n = sin nx / sin x is a polynomial in t = cos x, of degree n-1. ------------------------------------ Now, in order to apply OBSERVATIONS (I) and (II), we need to try to understand the roots of y_n. Note that y_n = sin nx/sin x is equal to ZERO exactly when sin nx is zero and sin x is NONZERO. Thus, y_n is zero when: nx = Pi, 2Pi , ...., (n-1)pi or in the range of x between 0 and Pi: x = Pi/n, 2Pi/n, ....., (n-1)Pi/n ------------------------------------ Note that the "t-"values, that is, the COSINES of these x's, are all distinct since the COSINE function is strictly DECEASING from 0 to Pi. This justifies that y_n does indeed have exactly (n-1) roots. ------------------------------------ Therefore, by OBSERVATION (II), (y_n)^2 has exactly (n-1) MINIMA and (n-2) MAXIMA. This completes the proof of your statement. ------------------------------------ I hope you agree that this analysis sheds more light into the situation than would a routine calculation of derivatives. -Doctor Yiu, The Math Forum Check out our web site! http://mathforum.org/dr.office/help |
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