# 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."

```