Linear Algebra Help - Pythagorem Theorem

eunhye732
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I have been trying to figure this out but can't. I even went to the tutorial but the lady kept mumbling so it was obvious that she didn't really know how to do it either. So I really hope someone could help me. The only thing he told me was to look at the pythagorem theorem. Thanks
 

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what is the question? i cannot read the scan.
 
The question is to show that norm(v over norm(v))= 1 for v a vector.
That is, that
\left|\left|\frac{v}{\left|\left|v\right|\right|}\right|\right|= 1
eunhye732, one of the things you certainly should have learned about the "norm" is that ||av||= |a| ||v|| for a any number and v a vector. In particular, if a is a positive number then ||av||= a ||v||.
Think about that with a= 1/||v||.
 
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HallsofIvy said:
The question is to show that norm(v over norm(v))= 1 for v a vector.
That is, that
\left|\left|\frac{v}{\left|\left|v\left|\left|}\left|\left|= 1
eunhye732, one of the things you certainly should have learned about the "norm" is that ||av||= |a| ||v|| for a any number and v a vector. In particular, if a is a positive number then ||av||= a ||v||.
Think about that with a= 1/||v||.

Oh i see how to do it now.
That helped a lot.
Thanks so much!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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