Good day, Sir/Madame! I'm currently working on the standard deviation of a particular function $\frac{2}{\pi} \ln n$, where n is the degree of certain random polynomial.
By the use of computer software I found out that the standard deviation of the logarithmic function above is somehow close to $0.778n^{0.15}$. I want to solve this mathematically, but I dunno if I'm doing it right.
I started to compute for the mean, u, $$u = \frac{\sum_{n=1}^{n} nf(n)}{\sum_{n=1}^{n} f(n)}$$ $$= \frac{\frac{2}{π} \sum_{n=1}^{n} n*ln (n)}{\frac{2}{π} \sum_{n=1}^{n} ln (n)}$$ $$= \frac{\sum_{n=1}^{n} n*ln (n)}{\sum_{n=1}^{n} ln (n)}$$ $$= \frac{\sum_{n=1}^{n} ln (n)^{n}}{\sum_{n=1}^{n} ln (n)}$$
Since $$\sum_{n=1}^{n} ln (n)^{n}$$ is $$ln (1^1) + ln (2^2) + ln (3^3) + . . . ln (n^n) = ln((1^1)x(2^2)x(3^3)x(4^4)x . . . x(n^n))$$ is approximately $$ln|1.2824n^{\frac{6n^2 + 6n + 1}{12}} e^{\frac{-n^2}{4}}|$$
and the $$\sum_{n=1}^{n} ln (n) = ln (n!)$$
hence $$u = \frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4})}{ln (n!)}$$
Substituting this result to the formula of standard deviation $$sd^{2} = \frac{\sum_{n=1}^{n} (u - f(n))^{2}}{\sum_{n=1}^{n} ln (n)}$$
it will finally become $$sd^2 = \frac{\sum_{n=1}^{n} {(u^2 - 2uf(n) + [f(n)]^2)}}{\frac{2}{π} ln (n!)}$$ $$sd^2 = \frac{\sum_{n=1}^{n} {((\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4)}}{ln (n!)})^2 - 2(\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4})}{ln (n!)})(\frac{2}{π} ln (n)) + (\frac{4}{π^{2}} ln^2(n))}}{\frac{2}{π} ln (n!)}$$ $$sd^2 = \frac{\sum_{n=1}^{n} {((\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4)}}{ln (n!)})^2 - \frac{4}{π}(\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4})}{ln (n!)})(ln (n)) + (\frac{4}{π^{2}} ln^2(n))}}{\frac{2}{π} ln (n!)}$$ $$sd = \sqrt{\frac{\sum_{n=1}^{n} {((\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4)}}{ln (n!)})^2 - \frac{4}{π}(\frac{\ln (1.2824n^{(6n^2 + 6n + 1)/12} e^{(-n^2)/4})}{ln (n!)})(ln (n)) + (\frac{4}{π^{2}} ln^2(n))}}{\frac{2}{π} ln (n!)}}$$
Verifying the above claim when n = 5 the result will be 1.04 where based on the formula 0.778*n^0.15 it is 0.90.. somehow close but when I test it when n = 10, the above claim become 1.7 but the formula 0.778*n^0.15 it is 1.099 a large discrepancy is evident? What do I miss?