Cauchy-Schwarz Inequality

• Jun 26th 2010, 11:47 AM
Jukodan
Cauchy-Schwarz Inequality
I am supposed to prove that for all real numbers $\displaystyle a,b, \theta$ that
$\displaystyle (acos\theta+bsin\theta)^2\leq a^2+b^2$
I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

$\displaystyle (acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2$

Am I on the right track?
• Jun 26th 2010, 12:19 PM
dwsmith
$\displaystyle <acosx+bsinx,acosx+bsinx>=<acosx,acosx+bsinx>+<bsi nx,acosx+bsinx>$

$\displaystyle =<acosx,acosx>+2<acosx,bsinx>+<bsinx,bsinx>=.....$

Also, I am assuming inner product is defined as such:

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)dx$ because if this is the case, everything works out just fine.
• Jun 26th 2010, 12:41 PM
Jukodan
It does not specify the inner product. I have posted all the information that was given to me.
• Jun 26th 2010, 12:43 PM
dwsmith
Quote:

Originally Posted by Jukodan
It does not specify the inner product. I have posted all the information that was given to me.

Try finishing the problem with that inner product in mind and you will have your problem completed.
• Jun 26th 2010, 12:52 PM
Plato
Quote:

Originally Posted by Jukodan
I am supposed to prove that for all real numbers $\displaystyle a,b, \theta$ that
$\displaystyle (acos\theta+bsin\theta)^2\leq a^2+b^2$

$\displaystyle \left( {a\sin (t) - b\cos (t)} \right)^2 = a^2 \sin ^2 (t) - 2ab\sin (t)\cos (t) + b^2 \cos (t)$ Note the minus.
So we get $\displaystyle \left( {a\cos (t) + b\sin (t)} \right)^2 = a^2 + b^2 - \left( {a\sin (t) - b\cos (t)} \right)^2$
Notice that $\displaystyle a^2 + b^2 - \left( {a\sin (t) - b\cos (t)} \right)^2 \le a^2+b^2$
• Jun 26th 2010, 01:32 PM
dwsmith
$\displaystyle <acosx+bsinx,acosx+bsinx>=<acosx,acosx+bsinx>+<bsi nx,acosx+bsinx>$

$\displaystyle =<acosx,acosx>+2<acosx,bsinx>+<bsinx,bsinx>=.....$

Spoiler:
$\displaystyle a^2<cosx,cosx>+2ab<cosx,sinx>+b^2<sinx,sinx>$

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}(a^2cos^2x)dx=a^2$

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}(2ab*cosx*sinx)dx=0$

$\displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}(b^2sin^2x)dx=b^2$

$\displaystyle a^2<cosx,cosx>+2ab<cosx,sinx>+b^2<sinx,sinx>$

$\displaystyle a^2+b^2$

$\displaystyle \Rightarrow (acosx+bsinx)^2\leq a^2+b^2$
• Jun 26th 2010, 02:04 PM
Jukodan
Ah-ha!
• Jun 27th 2010, 12:28 PM
Opalg
Quote:

Originally Posted by Jukodan
I am supposed to prove that for all real numbers $\displaystyle a,b, \theta$ that
$\displaystyle (acos\theta+bsin\theta)^2\leq a^2+b^2$
I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

$\displaystyle (acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2$

Am I on the right track?

The C–S inequality says that $\displaystyle x_1y_1+x_2y_2\leqslant (x_1^2+x_2^2)^{1/2}(y_1^2+y_2^2)^{1/2}$. Apply that with $\displaystyle x_1=a$, $\displaystyle x_2=b$, $\displaystyle y_1=\cos\theta$, $\displaystyle y_2 = \sin\theta$. Then square both sides.