incremental computation of standard deviation

I think the easiest way to do this is with an orthogonality trick. I'll show how to incrementally compute the variance instead of the standard deviation. Let $X_1, X_2, ...$ be an iid sequence of random variables with $\bar X = n^{-1} \sum_{j = 1} ^ n X_j$ and $s^2_n$ defined similarly as the $n$'th sample variance (I use a denominator of $n-1$ instead of $n$ in your picture to keep things unbiased for the variance, but you can use the same argument just adding $1$ to all the terms with $n$ in them). First write $$ s^2_n = \frac{\sum_{j = 1} ^ n (X_j - \bar X_n)^2}{n - 1} = \frac{\sum_{j = 1} ^ n (X_j - \bar X_{n - 1} + \bar X_{n - 1} - \bar X_n)^2}{n - 1}. $$ Expand this to get $$ s^2_n = \frac{(n - 2)s^2_{n - 1} + (n - 1) (\bar X_{n - 1} - \bar X_n)^2 + 2 \sum_{j = 1} ^ {n - 1} (X_j - \bar X_{n - 1})(\bar X_{n - 1} - \bar X_n) + (X_n - \bar X_{n})^2}{n - 1} $$ and it is easy to show that the summation term above is equal to $0$ which gives $$ s^2_n = \frac{(n - 2)s^2_{n - 1} + (n - 1)(\bar X_{n - 1} - \bar X_n)^2 + (X_n - \bar X_{n})^2}{n - 1}. $$

EDIT: I assumed you already have an incremental expression for the sample mean. It is much easier to get that: $\bar X_n = n^{-1}[X_n + (n-1)\bar X_{n-1}]$.

The standard deviation is a function of the totals $T_{\alpha}=\sum_{i=1}^{N}x_{i}^{\alpha}$ for $\alpha=0,1,2$, each of which can be calculated incrementally in an obvious way. In particular, $E[X]=T_{1}/T_{0}$ and $E[X^2]=T_{2}/T_{0}$, and the standard deviation is $$ \sigma = \sqrt{\text{Var}[X]} = \sqrt{E[X^2]-E[X]^2} = \frac{1}{T_0}\sqrt{T_{0}T_{2}-T_{1}^2}. $$ By maintaining totals of higher powers ($T_{\alpha}$ for $\alpha \ge 3$), you can derive similar "incremental" expressions for the skewness, kurtosis, and so on.

What you refer to as an incremental computation is very close to the computer scientist's notion of an online algorithm. There is in fact a well-known online algorithm for computing the variance and thus square root of a sequence of data, documented here.