Log Expansion Formula

I note down a log expansion formula that seems to hold MAIN It seems that for odd n > 0 we have -\log\left(^{n-1}x^k\right)=^\infty \left(2^{(n-1)/2} \cos\left({n}\right)\right) {k} and for even n > 0 we have -\log\left(^{n-1}x^k\right)=^\infty \left( e^{ik\pi}+ 2^{(n-1)/2} \cos\left({n}\right)\right) {k} we have ^\infty x^k = {1-x} for convergent x, and also that -\log\left({1-x}\right)= -^\infty {k} which would then imply 2^{(n-1)/2} \cos\left({n}\right) = -1, \; \forall k \in ^+ EXPONENTIATE We can turn this into an expression for all n, with an iverson bracket [ ⋅ ] -\log\left(^{n-1}x^k\right)=^\infty \left([n\bmod2 \equiv 0]e^{i \pi k}+ 2^{(n-1)/2} \cos\left({n}\right)\right) {k} and convert into ^{n-1}x^k=\exp\left(-^\infty \left( [n\bmod2 \equiv 0]e^{i \pi k}+ 2^{(n-1)/2} \cos\left({n}\right)\right) {k}\right) then ^{n-1}x^k=^\infty \exp\left(- [n\bmod2 \equiv 0]e^{i \pi k} -2^{(n-1)/2} \cos\left({n}\right) {k}\right) ^{n-1}x^k=^\infty \exp\left(- [n\bmod2 \equiv 0]e^{i \pi k}\right)\exp\left( -2^{(n-1)/2} \cos\left({n}\right) {k}\right) ^{n-1}x^k=^\infty \exp\left(- [n\bmod2 \equiv 0]e^{i \pi k}\right) ^{{2}}\exp\left(-2\cos\left({n}\right) {k}\right) This formula doesn’t seem to work for n = 1 and n = 2, however beyond that we get relationships such as 1+x+x^2 = \exp\left(x+{2}-{3}+{4}+{5}-\cdots\right) This isn’t too surprising, This is purely a statement of 1+x+x^2 = \exp(\log(1+x+x^2)) however this is a nice way to think about polynomial division, when we take the ratio of polynomials we can now equate this to their exponential series expansions {1+2x+x^2+x^3} = {2}-{3}+{4}+{5}-\cdots\right)}{\exp\left(2x-x^2+{3}-{2}+{5}-\cdots\right)} but due to the nature of exponentials we can write this as an infinite product {1+2x+x^2+x^3} = e^{-2x^3/3}e^{x^4/4}\cdots}{e^{2x}e^{-x^2}e^{5x^3/3}e^{-5x^4/2}\cdots}