1 + 1 = 2 {\displaystyle 1+1=2}
f 1 n ( t ) = e x p { 1 n ∑ m = 1 ∞ q 2 m − 1 ( q p ) 2 m − 1 ( e i t ( 2 m − 1 ) − 1 ) } {\displaystyle \textstyle f^{\frac {1}{n}}(t)=\mathrm {exp} \left\{{\frac {1}{n}}\sum _{m=1}^{\infty }{\frac {q}{2m-1}}\left({\frac {q}{p}}\right)^{2m-1}\left(e^{it(2m-1)}-1\right)\right\}}
For all n ∈ N {\displaystyle \textstyle n\in \mathbb {N} }
