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

Re-write the equation as 2^{x} +
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 2^{x} = 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 2^{x} 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).

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Last revised August 2003