Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with binomial-coefficients
Search options not deleted
user 101288
For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.
-1
votes
2
answers
1k
views
Binomial expansion for noncommutative operator
Is it possible to find a closed formula for $(A^\dagger -kA)^n$ with $[A,A^\dagger]=1$ ?
I am looking for the normal ordinate form: $\sum (A)^{n-j}(A^\dagger)^j$— possibly something to do with the co …
- 35.5k