# Umbral Calculus Derivation of the Bernoulli numbers

\(“(B-1)^n = B^n”\)

\begin{align*} (B-1)^2 &= B^2\ B^2 - 2B^1 + 1 &= B^2 \ -2B^1+1 & = 0\ B^1 &= \frac{1}{2} \ B_1 &= \frac12 \end{align*}

\begin{align*} (B-1)^3 &= B^3\ B^3 - 3B^2 + 3B^1 - 1 &= B^3 \ B^3 - 3B^2 + 3B_1 - 1 &= B^3 \ -3B^2 + 3(\frac{1}{2})-1& = 0\ -3B_2 + \frac{1}{2} &= 0 \ B_2 &= \frac{1}{6} \end{align*}

Thanks to Laurens Gunnarsen for showing me this strange trick.

I’ve finally understood the principle which allows us to lower the index. The step where we move (B^i) to be (B_i) is quite simple. As Rota and Roman say, one method of expressing an infinite sequence of numbers is by a transform method. That is, to define a linear transform (B) such that \(B x^n = B_n\)

So, the above “lowering of the index” is actually using the relation ((X-1)^n = X^n), and applying (B) to both sides of it. To get (B(X-1)^n = B(X^n)). Let’s look for example at the “lowering step” of the first calculation: \begin{align*} X^1 &= \frac{1}{2} \ B (X_1) &= B(\frac12) \ B_1 &= \frac12 B(1) = \frac12 \end{align*}