Difference between revisions of "Cauchy-schwarz inequality"

 
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<math>(a_1x+b_1)^2+(a_2x+b_2)^2+...(a_nx+b_n)^2 = 0</math>.
 
<math>(a_1x+b_1)^2+(a_2x+b_2)^2+...(a_nx+b_n)^2 = 0</math>.
 
Expanding, we find the equation to be of the form <math>Ax^2+Bx+C</math>, where <math>A=\sum_{i=1}^n a_i^2</math>, <math>B=2\sum_{j=1}^n a_jb_j</math>, and <math>C=\sum_{k=1}^n b_k^2.</math>
 
Expanding, we find the equation to be of the form <math>Ax^2+Bx+C</math>, where <math>A=\sum_{i=1}^n a_i^2</math>, <math>B=2\sum_{j=1}^n a_jb_j</math>, and <math>C=\sum_{k=1}^n b_k^2.</math>
The equation will have a solution when the discriminant is greater than or equal to 0, so <math>B^2-4AC \geq 0</math>.  Substituting the above values of A, B, and C leaves us with the '''Cauchy-Schwarz Inequality''', which states that
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Since the equation is always greater than or equal to 0, <math>B^2-4AC \leq 0</math>.  Substituting the above values of A, B, and C leaves us with the '''Cauchy-Schwarz Inequality''', which states that
<math>(a_1b_1+a_2b_2+...+a_nb_n)^2 \geq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)</math>,
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<math>(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)</math>,
 
or, in the more compact [[sigma notation]],  
 
or, in the more compact [[sigma notation]],  
<math>\left(\sum a_ib_i\right) \geq \left(\sum a_i^2\right)\left(\sum b_i^2\right)</math>
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<math>\left(\sum a_ib_i\right) \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)</math>

Revision as of 00:55, 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, $\left(\sum a_ib_i\right) \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right)$