Solution to Problem 20 submitted by Micheal Shtilman

Problem 20

Conduct a complete investigation of the following equation, parametrized by a: 4x + 2 = a2x sin πx.


Re-write the equation as 2x + 2*2-x = a*sin(πx). The left size is always greater or equal to 2*sqrt(2) (arithmetic and geometric averages). It is equal to its minimum only if 2x = 2*2-x, i.e., if x = 0.5. The left side is also monotonically decreasing for x < 0.5 and monotonically increasing for x > 0.5.

Now, the right side is always less or equal to a, and equal to a only if x = 0.5 + 2*n for integer n. Thus,

When a < 2*sqrt(2) there is no solution to this equation.

When a = 2*sqrt(2) there is only one solution x = 0.5 to this equation.

When a > 2*sqrt(2) there are always at least 2 roots for this equation: one between 0 and 0.5 and the other between 0.5 and 1 (sin(π*0) = sin(π*1) = 0). There could be other roots, not greater than log(a), where the base of log is 2 (notice that the left hand side of the eqation is assimptotically 2x for positive x and 2*2-x for negative x). For each concrete a it is easy to find n such that 2n+1 < log(a). Now for each segment [2k, 2k+1] k = -n, -n+1, ... 0,1, ... n, the equation has exactly 2 roots (for k = 0 we have segment [0,1] mentioned above).

Last revised August 2003