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Difference between revisions of "Cauchy-schwarz inequality"
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or, in the more compact [[sigma notation]],  
 
or, in the more compact [[sigma notation]],  
 
<math>\left(\sum a_ib_i\right) \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)</math>
 
<math>\left(\sum a_ib_i\right) \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)</math>
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''This page should be deleted as it has become obsolete with the more complete [[Cauchy-Schwarz Inequality]]'' page with a capital S in shwarz"

Revision as of 00:12, 18 June 2006

Consider the quadratic, $(a_1x+b_1)^2+(a_2x+b_2)^2+...(a_nx+b_n)^2 = 0$. Expanding, we find the equation to be of the form $Ax^2+Bx+C$, where $A=\sum_{i=1}^n a_i^2$, $B=2\sum_{j=1}^n a_jb_j$, and $C=\sum_{k=1}^n b_k^2.$ Since the equation is always greater than or equal to 0, $B^2-4AC \leq 0$. Substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality, which states that $(a_1b_1+a_2b_2+...+a_nb_n)^2 \leq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)$, or, in the more compact sigma notation, $\left(\sum a_ib_i\right) \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)$ This page should be deleted as it has become obsolete with the more complete Cauchy-Schwarz Inequality page with a capital S in shwarz"

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