Cauchy Sequence Inequality

**Nichelle14** · June 3rd 2006, 07:03 PM

I am to prove CS inequality by mathematical induction. Don't even know where to start. Any help?

**malaygoel** · June 4th 2006, 07:03 AM

Originally Posted by Nichelle14

I am to prove CS inequality by mathematical induction. Don't even know where to start. Any help?

tell me the CS inequality.
I don't care much about names so I can't figure it out.

**rgep** · June 4th 2006, 08:26 AM

Perhaps it's Cauchy-Schwartz?

**Nichelle14** · June 4th 2006, 08:30 AM

yes. sorry you are right it is Cauchey-Schwartz.

**TD!** · June 4th 2006, 09:46 AM

By induction? Induction on what then?

**ThePerfectHacker** · June 4th 2006, 09:56 AM

Originally Posted by TD!

By induction? Induction on what then?

The dimension on the vector space.

As I understand it she IS NOT talking about $\mathbb{R}^3$ but any dimesnion.

**rgep** · June 4th 2006, 11:13 AM

The proof I usually use for Cauchy-Schwartz, in the form $\Vert u \Vert^2 \Vert v \Vert^2 \ge (u\cdot v)^2$ , is to consider the quadratic $f(x) = \Vert u + xv \Vert^2 \ge 0$ and observe that the requirement ("b^2-4ac") that f has at most one real root is precisely C-S. This doesn't depend on dimension.

If I had to do it by induction I think I'ld start by observing that it is trivial for n=1. The case n=2 follows from $(x^2+y^2)(u^2+v^2) = (xu+yv)^2 + (xv-yu)^2 \ge (xu+yv)^2$ . The induction step would be something like $((u_1^2+\cdots+u_{n-1}^2)+u_n^2)((v_1+\cdots+v_{n-1}^2)+v_n^2) \ge$ $\big\vert\sqrt{u_1^2+\cdots+u_{n-1}^2}\sqrt{v_1+\cdots+v_{n-1}^2}+u_nv_n \big\vert \ge$ $\big\vert(u_1,\ldots,n_{n-1})\cdot(v_1,\ldots,v_{n-1})+u_nv_n\big\vert$ , thus using the cases 2 and n-1.

**ThePerfectHacker** · June 4th 2006, 03:19 PM

I understand the problem as,
for any $n$ ,
$|x_1y_1+...+x_ny_n|\leq \sqrt{x_1^2+...+x_n^2}\cdot \sqrt{y_1^2+...+y_n^2}$
For any $x_i.y_i\in \mathbb{R}$

Thread: Cauchy Sequence Inequality

LinkBack

Thread Tools

Display

Cauchy Sequence Inequality

Similar Math Help Forum Discussions

General Cauchy inequality

Proving the triangle inequality using the Cauchy-Schwartzs inequality

[SOLVED] Subsequence of a Cauchy Sequence is Cauchy

cauchy sequence, contractive sequence

Cauchy-Schwartz Inequality

Search Tags