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Counting Extrema


Date: 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   
    
Associated Topics:
College Calculus
College Trigonometry

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