X = TanX
Date: 03/21/2001 at 01:37:00 From: Andrew Subject: Infinite solutions Show that x = tanx has an infinite number of solutions. I am not sure where to start.
Date: 03/21/2001 at 09:18:28 From: Doctor Rob Subject: Re: Infinite solutions Thanks for writing to Ask Dr. Math, Andrew. You can show that x = tan(x) has a solution in every interval of the form ([2*k-1]*Pi/2, [2*k+1]*Pi/2), where k is any integer. That is because x is bounded in these intervals, but tan(x) is unbounded in both positive and negative directions. Thus their graphs must cross. If you look at the graph of y = tan(x), you will see that it has some vertical asymptotes. They occur at x = Pi/2, 3*Pi/2, 5*Pi/2, 7*Pi/2, ..., and also at x = -Pi/2, -3*Pi/2, ... These are the odd multiples of Pi/2. That is why you look at the interval between two consecutive odd multiples of Pi/2. Two consecutive odd numbers have the form 2*k-1 and 2*k+1 for some integer k. Thus the interval between two asymptotes is given by the form [2*k-1]*Pi/2 < x < [2*k+1]*Pi/2. Call that interval I(k). At the values of x where there is an asymptote, the value of tan(x) is undefined. Just to the right of any asymptote, the function is positive and very large in absolute value. Just to the left of any asymptote, the function is negative and very large in absolute value. The closer you get to the asymptote, the larger the absolute value of tan(x). In fact, you can make the absolute value of tan(x) as large as you like by taking x close enough to the asymptote. Now look at the graph of y = x. This is a straight line, of course. When x is in I(k), since x = y, it is also true that y is in I(k). That means that |y| < (1+2*|k|)*Pi/2. Putting this together, you can conclude the following. For values of x close enough to the left end of I(k), tan(x) < -(1+2*|k|)*Pi/2 < x. Thus tan(x) - x < 0. For values of x close enough to the right end of I(k), tan(x) > (1+2*|k|)*Pi/2 > x, so tan(x) - x > 0. Since tan(x) is continuous on I(k), there must be at least one value of x between where tan(x) - x = 0 (by the Intermediate Value Theorem), so tan(x) = x at that value. Since there is at least one such solution for each integer k, there are infinitely many solutions to tan(x) = x. I hope the above is clear. If not, write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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