Month: December 2016

\\(“(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*}