Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20090206 10:11:19
Cauchy–Schwarz–Bunyakovsky inequalityToday I learned from Introduction to Metric and Topological Spaces by W.A. Sutherland a version of the Cauchy–Schwarz inequality involving integrals. It goes by the funny name of Cauchy–Schwarz–Bunyakovsky inequality. for al real numbers . Proof: The inequality is obviously true if . Hence we may assume that at least one is not 0. Then Treating the LHS as a quadratic in , we see that its discriminant cannot be positive. This is the proof given in Introduction to Metric and Topological Spaces. The integral version is as follows. The proof is similar to the Cauchy–Schwarz case, only this time you start with Last edited by JaneFairfax (20090207 02:32:01) #3 20090206 23:56:02
Re: Cauchy–Schwarz–Bunyakovsky inequality
I don't follow this part. Why can't the discriminant be positive? Wrap it in bacon #4 20090207 00:04:20
Re: Cauchy–Schwarz–Bunyakovsky inequalityBecause then the quadratic equation LHS = 0 would have two distinct real roots and the LHS would be negative between these roots. Last edited by JaneFairfax (20090207 00:36:09) #5 20090207 00:09:12
Re: Cauchy–Schwarz–Bunyakovsky inequality
The fact that the quadratic is always greater than or equal to 0 means that it must have at most one real root. Last edited by Daniel123 (20090207 00:10:35) #6 20090207 03:14:03
Re: Cauchy–Schwarz–Bunyakovsky inequalityI'm an idiot. Thanks for the explanation. Wrap it in bacon #7 20090207 06:01:18
Re: Cauchy–Schwarz–Bunyakovsky inequality
That is rather odd, normally I've heard it referred to as the CauchySchwarz special case of Holder's inequality, where Holder's inequality is: Where 1/p + 1/q = 1. Of course, CauchySchwarz is the special case with p = q = 2. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #8 20090208 08:26:59
Re: Cauchy–Schwarz–Bunyakovsky inequalityThat's nice. I remember we have a similar proof in Calculus for multivariable function's Taylor expansion. It adds in lamda as well. Such method is called "adding parameter", which shares the same delicacy as adding a line to solve geometry problems. X'(yXβ)=0 #10 20100927 23:06:44
Re: Cauchy–Schwarz–Bunyakovsky inequality...hi jane .........but you assumed that "lambda" was real but by setting discriminant to less than zero ur making lambda value imaginary:) #11 20100927 23:08:05
Re: Cauchy–Schwarz–Bunyakovsky inequality..iam really in need of a solution to this problem please reply soon #12 20101005 21:24:11
Re: Cauchy–Schwarz–Bunyakovsky inequalityJane has already stopped discussing serious topics. X'(yXβ)=0 #13 20101005 21:40:37
Re: Cauchy–Schwarz–Bunyakovsky inequalityHi George,Y; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #14 20101117 11:56:45
Re: Cauchy–Schwarz–Bunyakovsky inequalityBasically, the roots cannot be real. Consider the quadratic equation as a parabola. If the equation has real roots then it crosses the x axis twice and has negative values between them, but we know that our quadratic function cannot be negative, so the roots have to be imaginary and so the discriminant has to be less than zero. #15 20101126 23:46:50
Re: Cauchy–Schwarz–Bunyakovsky inequality
Last edited by JaneFairfax (20101222 06:37:00) 