Why does this trick to derive the formula for $[A^n,B]$ in terms of repeated commutators work so well?

It is a known result that, given generically noncommuting operators $ A,B$ , we have $ $ A^n B=\sum_{k=0}^n \binom{n}{k} \operatorname{ad}^k(A)(B) A^{n-k}.\tag A $ $ This can be proved for example via induction with not too much work.

However, while trying to get a better understanding of this formula, I realised that there is a much easier way to derive it, at least on a formal, intuitive level.

The trick

Let $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ (standing for “shift” and “commute”, respectively) denote operators that act on expressions of the form $ A^k D^j A^\ell$ (denoting for simplicity $ D^j\equiv\operatorname{ad}^j(A)(B)$ ) as follows:

\begin{align} \hat{\mathcal S} (A^k D^j A^\ell) &= A^{k-1} D^j A^{\ell+1}, \ \hat{\mathcal C} (A^{k-1} D^{j+1} A^\ell) &= A^{k-1} D^j A^{\ell+1}. \end{align} In other words, $ \hat{\mathcal S}$ “moves” the central $ D$ block on the left, while $ \hat{\mathcal C}$ makes it “eat” the neighboring $ A$ factor.

It is not hard to see that $ \hat{\mathcal S}+\hat{\mathcal C}=\mathbb 1$ , which is but another way to state the identity $ $ A[A,B]=[A,B]A+[A,[A,B]].$ $ Moreover, crucially, $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ commute. Because of this, I can write

$ $ A^n B=(\hat{\mathcal S}+\hat{\mathcal C})^n (A^n B)=\sum_{k=0}^n\binom{n}{k} \hat{\mathcal S}^{n-k} \hat{\mathcal C}^{k}(A^n B),$ $ which immediately gives me (A) without any need for recursion or other tricks.

The question

Now, this is all fine and dandy, but it leaves me wondering as to why does this kind of thing work? It looks like I am somehow bypassing the nuisance of having to deal with non-commuting operations by switching to a space of “superoperators”, in which the same operation can be expressed in terms of commuting “superoperators”.

I am not even sure how one could go in formalising this “superoperators” $ \hat{\mathcal S},\hat{\mathcal C}$ , as they seem to be objects acting on “strings of operators” more than on the elements of the operator algebra themselves.

Is there a way to formalise this way of handling the expressions? Is this a well-known method in this context (I had never seen it but I am not well-versed in this kinds of manipulations)?

Is this string approach some way better

I usually store strings as const before use them inside a for loop, I don’t know why but someone told me that type the string inside the loop is creating the same string every iteration, is that true? I mean:

private const string ADDRESS = "/User/Data/data";  private string[] Example() {     var userData = new string[data.Length];      for (int i = 0; i < data.Length; i++)     {         userData[i] = xmlDocument.DocumentElement.SelectSingleNode(ADDRESS + i);     }      return userData; } 

VS

private string[] Example() {     var userData = new string[data.Length];      for (int i = 0; i < data.Length; i++)     {         userData[i] = xmlDocument.DocumentElement.SelectSingleNode("/User/Data/data" + i);     }      return userData; } 

Meaning of strange Pointer in assembly code – is this a vector table address?

Apologies if this is a silly question, but I’m confused. I’m working on a reverse engineering assignment. While looking at a disassembled dll of possibly malicious code, I found these lines:

push nsize                  ;  makes sense push offset Security_Attr   ;  makes sense push 80h                    ;  this address does not make sense push offset read_buf        ;  makes sense call CreatePipe             ;  makes sense 

This is calling a Windows function called CreatePipe. “80h” should point to a buffer that the pipe writes to. The value just seems way too small! Is this address pointing to the user_interrupt section of the vector table? If so, is this pipe overwriting user_interrupt handlers in the vector table?

Any pointers are appreciated.

What information about a monster can a PC expect to know (be told really) if this is the 1st ever encounter with this type of monster? [duplicate]

This question already has an answer here:

  • How much do I tell new players about new monsters? 7 answers

As an example, I’m considering vulnerabilities, resistances and immunities. If a pc is unfamiliar with a monster and asks for this information, is it reasonable to provide this.
Another way to ask this maybe is: Is there information that is typically not provided on a 1st ever encounter?