applied math problem
robert bristow-johnson
pbjrbj at viconet.com
Thu Jun 11 01:29:59 EDT 1998
Let 0 <= p <= 1 and -1 <= t <= +1
Can anyone express (hopefully in closed form) a class of real
and non-negative functions, a(t), such that
0 <= a(t) <= +1
and
[a(t)]^2 + 2*p*a(t)*a(-t) + [a(-t)]^2 = 1
or
[a(t) + a(-t)]^2 = 1 - 2*(p-1)*a(t)*a(-t)
for any constant p: 0 <= p <= 1 ?
I have manipulated the problem statement to another form:
Let z(t) = ln( a(t) ) + ln(2)/2 = ln( sqrt(2) * a(t) )
z(t) <= ln(2)/2
p = exp( -[z(t) + z(-t)] ) - cosh( z(t) - z(-t) )
or
p = -[exp( -z(t) )*sinh( z(-t) ) + exp( -z(-t) )*sinh( z(t) )]
where exp(x) = e^x .
I have tried:
Let exp( -z(t) )*sinh( z(-t) ) = exp( -z(-t) )*sinh( z(t) ) = -p/2
which can't work for p > 0 because z(t) can = 0.
Also tried:
Let sinh( z(t) ) = -p/2 * exp( -z(t) )
and same for z(-t) which leads to
p * cosh( z(t) ) = 0 which can't be true for p > 0.
There are solutions for the two extremes: p = 0 and p = 1.
p = 0 ---> a(t) = sqrt( (1+t)/2 ) ---> z(t) = ln(1+t)/2
p = 1 ---> a(t) = (1+t)/2 ---> z(t) = ln(1+t) - ln(2)/2
Of course the "t" in any solution can be replaced with the any odd
function, f(t), such that
f(-t) = -f(t), -1 <= f(t) <= 1 when -1 <= t <= 1, and f(1) = 1 .
I have not been able to find a solution for any p: 0 < p < 1 .
--
r b-j
pbjrbj at viconet.com a.k.a. robert at audioheads.com
a.k.a. robert at wavemechanics.com
"Don't give in to the Dark Side. Boycott intel and microsoft."
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