Please help correct my work

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- Sep 21st 2012, 08:26 AMangelmePlease help check my solutions for vector space questions
**Please help correct my work**

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Attachment 24870 - Sep 21st 2012, 10:46 AMPlatoRe: Please help check my solutions for vector space questions
From looking at the images you posted it seems that you are proving:

$\displaystyle |V\cdot U|\le\|V\|\|U\|$

The standard proof follows from the fact $\displaystyle \frac{V\cdot U}{\|V\|\|U\|}=\cos(\theta)$ where $\displaystyle \theta$ is the angle between the two non-zero vectors $\displaystyle V~\&~U$.

Then note that $\displaystyle 0\le |\cos(\theta)|\le 1~.$ - Sep 21st 2012, 01:34 PMDevenoRe: Please help check my solutions for vector space questions
first of all, you have $\displaystyle \|U\|$ and $\displaystyle \|V\|$ wrong.

$\displaystyle \|U\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}$

$\displaystyle = \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}$

$\displaystyle = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}$

$\displaystyle = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}$

$\displaystyle = \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}$

$\displaystyle = \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}$

$\displaystyle = \frac{\sqrt{\pi}}{2}$

while:

$\displaystyle \|V\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}$

$\displaystyle = \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx} $

$\displaystyle = \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}$

$\displaystyle = \frac{\sqrt{\pi}}{2}$, as well.

thus:

$\displaystyle \|U\|\|V\| = \frac{\pi}{4}$.

on the other side of the inequality we have:

$\displaystyle |U \cdot V| = \left|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right| $

$\displaystyle = \left| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right| = \left|\frac{1}{2} \right| = \frac{1}{2} $

since $\displaystyle 1 < \frac{\pi}{2} $, we indeed have for these particular U,V:

$\displaystyle |U \cdot V| \leq \|U\|\|V\|$.

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for your second problem, you are also calculating the norms of U and V incorrectly. use the definition of the inner product you are given! - Sep 21st 2012, 06:05 PMangelmeRe: Please help check my solutions for vector space questions
Thank you so much!! Deveno! I missed the obvious point

Could you please re-check my second answer again?

Attachment 24877 - Sep 22nd 2012, 12:30 PMDevenoRe: Please help check my solutions for vector space questions
still incorrect. for starters, you have an extra 2

^{-n}term in $\displaystyle \|U\|$ and $\displaystyle \|V\|$.

secondly, you don't seem to get that the inner product of two vectors should be a SCALAR, not an "expression".

so for U, you should have:

$\displaystyle \|U\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1$

and for V, you should have:

$\displaystyle \|V\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}$

while for the dot product, you should have:

$\displaystyle |U \cdot V| = \left| \sum_{n=1}^{\infty} \frac{1}{4^n} \right| = \frac{1}{3}$