This is from something I was doing more than 4 years ago that I found going through an old laptop, and I have spent alot the night trying rewrite the results that I obviously thought to be important at the time, I apologize in advance for lack clarity and detail.
I was looking for all possible solutions of $(x,y) \in \mathbb R^2$ to the equation:
$$E:\,y^{ixy}+y^{-ixy}=y^{ix(y-2)}+y^{-ix(y-2)}$$
So the first set I found are as expected upon inspection of the equation:
$$(x,y) \in {\{(0,t):t \in \mathbb R}\} \cup {\{(r,1):r \in \mathbb R}\}$$
But then I have given a second solution set as follows, and this is the part that I am unsure of my working out:
$$(x,y) \in {\{(n \pi,e^t):n \in \mathbb N \land t \in \mathbb N}\}$$
So I then substituted this into E to give:
$$ \left( {{\rm e}^{t}} \right) ^{in\pi \,{{\rm e}^{t}}}+ \left( { {\rm e}^{t}} \right) ^{-in\pi \,{{\rm e}^{t}}}= \left( {{\rm e}^{t}} \right) ^{in\pi \, \left( {{\rm e}^{t}}-2 \right) }+ \left( {{\rm e}^ {t}} \right) ^{-in\pi \, \left( {{\rm e}^{t}}-2 \right) }$$
Which maple (The CAS I currently use) confirms is true with a Boolean function that checks for consistency either side of an equality symbol.
But then I simplified this using Euler's formula to:
$$\cos \left( n\pi \,{{\rm e}^{t}}t \right) =\cos \left( n\pi \, \left( {{\rm e}^{t}}-2 \right) t \right) $$
Which the same Boolean function maple has returns a value of false, and the results to ten decimal place float approximations are clearly indeed false, for example for $t=20$, and $n \leq 20$, I get:
But when I force the float approximation to over 100 digits of precision, then take a 10 digit float approximation of those values, the values imply that the equality is true:
Evaluating the individual trigonometric terms to their formal power series is another means to verify it is true:
$$\sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k} \left( n\pi \,{ {\rm e}^{t}}t \right) ^{2\,k}}{ \left( 2\,k \right) !}}-\sum _{k=0}^{ \infty }{\frac { \left( -1 \right) ^{k} \left( n\pi \, \left( {{\rm e} ^{t}}-2 \right) t \right) ^{2\,k}}{ \left( 2\,k \right) !}}=\cosh \left( \pi \,\sqrt {-{n}^{2}{{\rm e}^{2\,t}}{t}^{2}} \right) -\cosh \left( \pi \,\sqrt {-{n}^{2}{t}^{2} \left( {{\rm e}^{t}}-2 \right) ^{ 2}} \right) $$
And substitution of either $n=t$ or $t=n$ will produce the arithmetic zero function: $$\cos \left( {n}^{2}\pi \,{{\rm e}^{n}} \right) -\cos \left( {n}^{2} \pi \,{{\rm e}^{n}} \right) \left( \cos \left( {n}^{2}\pi \right) \right) ^{2}-\sin \left( {n}^{2}\pi \,{{\rm e}^{n}} \right) \sin \left( {n}^{2}\pi \right) \cos \left( {n}^{2}\pi \right) =0 \,\,\forall n \in \mathbb N$$
So I would just appreciate help from people here as to what level of concern I should apply in terms of my mathematical understanding of an important thing the software is telling me to acknowledge, or I can just dismiss this as another bug?
I really don't mind, I am just curious as to why the float approximations are so horribly not correct at 10 digits of precision, the default setting for the software, intuitively one would expect the inaccuracy to be proportional to the amount of digits to which we are truncating, but, I am not an expert in the float approximation software, because it is inbuilt code and user are not allowed to debug inbuilt code.
Thanks in advance.
