Recent content by otto

Pascals identity to be exact. I've read some "proofs" if you can call them that, but they were rather not helpful. math stack exchange. When I replace my summation from k=0 to n, with the combinatorics: \sum_{k=0}^n{n+1 \choose k} = \sum_{k=0}^n {n \choose k} +{n \choose k-1} I...

So look at what I've done: {n+1 \choose k} = \frac {(n+1)!} {(n+1-k)! \cdot k!} = \frac {(n+1)\cdot n!}{(n-(k-1))!\cdot k \cdot (k-1)!} = \frac {(n+1)}{k} \cdot \frac { n!}{(n-(k-1))!\cdot (k-1)!} = \frac {(n+1)}{k} \cdot {n \choose k-1} [SIZE="1"]oops I accidentally posted this...

thank you for your posts, all the answers help, reading the answers written in different ways really gives more perspective on Leibniz's notation than a single answer would have. I still have a few questions anyhow, (it takes quite a while to formulate them so Ill just post them one at a time)...

First off: I think I understand the chain rule and how it derives from \lim_{h \to 0} \frac{ f(x+h)-f(x)}{h} and how to apply the chain rule when taking the derivative of an implicit function. The textbook I am reading Applied Calculus (by B. Rockett) uses the following example on...