Is it possible to calculate the sum $$ \sum _{n=0}^{\infty} \frac{\left(-\pi ^2\right)^n \cos \left(2^nb\right)}{(2 n)!} $$ in closed form?
Formal naive argument gives $$ \sum _{n=0}^{\infty} \frac{\left(-\pi ^2\right)^n}{(2 n)!}\sum _{m=0}^{\infty} \frac{\left(-b ^2\right)^m}{(2 m)!}2^{2mn}=\sum _{m=0}^{\infty} \frac{\left(-b ^2\right)^m\cos\left(2^m\pi\right)}{(2 m)!}=\cos b-2. $$ However, numerical calculation shows that this is wrong. For example, for $b=1/3$ http://www.wolframalpha.com/input/?i=N%5B1%2F(Cos%5B1%2F++++3%5D+-2)Sum%5B(-%5C%5BPi%5D%5E2)%5En%2FFactorial%5B2+n%5D+Cos%5B1%2F3+2%5En%5D,+%7Bn,+0,+150%7D%5D,+35%5D