Talk:Cauchy-Schwarz Inequality
this inequalite is the same that this Cauchy's Inequality ?? because i learn that Cauchy's Inequality is the same that Cauchy–Schwarz and Cauchy–Schwarz-(other name) -- Gamma 15:20, 1 January 2009 (UTC)
Yes it is but I don't understand the notation on this page. I wrote a new page using conventional notation and concepts. --Matt Westwood 16:02, 1 January 2009 (UTC)
I'm not sure about the original exposition of this. I believe there's a misunderstanding as to the meaning of $\left\|{x}\right\|$ because the usual statement of it is $\left|{\left \langle {x, y} \right \rangle}\right| \le \left\|{x}\right\| \times \left\|{y}\right\|$.
More explanation is needed as to the precise meaning of all objects involved, because at the moment it's not at all solid. --Matt Westwood 19:46, 31 March 2009 (UTC)
This proof is better. Wok 03:24, 20 February 2011 (CST)
Vectors
Under which version of the C-S inequality does this form fall under? Is it implied in one of the guys up, or does it need to be added?
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \vert \mathbf{x} \cdot \mathbf{y} \vert\) | \(\le\) | \(\displaystyle \vert \vert \mathbf{x} \vert \vert \ \vert \vert \mathbf{y} \vert \vert\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
for $\mathbf{x}$, $\mathbf{y} \in \R^n$.
--GFauxPas 21:08, 26 February 2012 (EST)
- The dot product is an inner product on $\R^n$; hence you can consider it under the (semi-)inner product space version. --Lord_Farin 02:07, 27 February 2012 (EST)
- Alternatively, you could take Cauchy's inequality with $1\le i \le n$. --Lord_Farin 02:09, 27 February 2012 (EST)
