[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 556
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
Alsu, correct me if I'm being obtuse.
sX = x0 + x1rn
dX = x0 - x1rn
sX+dX = x0+x1rn+x0-x1rn = 2*x0
after normalisation we get x0 again
)))
sX = x0 + x1rn
dX = x0 - x1rn
sX+dX = x0+x1rn+x0-x1rn = 2*x0
after normalization we get x0 again
)))
Right, missed the sum and difference normalisation in the interval.
sX = x0 + x1rn
dX = x0 - x1rn
sX->sXn; dX-> dXn;
sXn+dXn = x0+x1rn+x0-x1rn = 2*x0 = X1
after normalizing (dividing by the square root of 2) we get x1 :)
......... It's a bit more complicated, it doesn't work that easy. After obtaining xi vector at each step, one should first "add-subtract-normalize" it with the next input vector and so on until the input vectors are exhausted. Something like this.
MetaDriver, alsu, sorry for interrupting the 'orthogonal vector set' discussion.
On your knees !!!
;)
Well yes, missed the sum and difference normalisation in the interval.
sX = x0 + x1rn
dX = x0 - x1rn
sX->sXn; dX-> dXn;
sXn+dXn = x0+x1rn+x0-x1rn = 2*x0 = X1
After normalizing (dividing by the root of 2) we get x1 = what we need. :)
still does not work
Example
x0 = (1/sqrt(2), 1/sqrt(2)), x1rn = (-1/sqrt(2), 1/(sqrt(2))
sX = (0, sqrt(2)), sXn = (0,1)
dX = x1rn-x0 = (sqrt(2), 0), dXn = (1,0)
sXn+dXn = (1,1) - this vector is orthogonal to neither x0 nor x1
although both were initially orthogonal))) but we can give an example without this
I'm asleep already)))) It's working out, of course)))
still doesn't work
Example
x0 = (1/sqrt(2), 1/sqrt(2)), x1rn = (-1/sqrt(2), 1/(sqrt(2))
sX = (0, sqrt(2)), sXn = (0,1)
dX = x1rn-x0 = (sqrt(2), 0), dXn = (1,0)
sXn+dXn = (1,1) - this vector is not orthogonal to either x0 or x1
although both were orthogonal initially)))) but you can give an example without that
not true. it is orthogonal. :) the result after normalization is equal to the first vector, and as you correctly noted - it is orthogonal to the second. :)
OK, go to sleep now. )))
Not true, it is orthogonal. :) the result after normalisation is equal to the first vector and, as you correctly pointed out, it is orthogonal to the second. :)
OK, go to sleep now. )))
It still doesn't work shit, those just worked out because I originally took a couple of orthogonal ones:
Example
x1rn = (0.6, 0.8), x0 = (1, 0)
it's an approximation, but you can see everything.
It still doesn't work, those just worked out because I initially took a pair of orthogonal ones:
Example
x1rn = (0.6, 0.8), x0 = (1, 0)
figure is approximate, but everything is visible
Tex. Looks like you are right. The solution is close, but the formula has to be corrected.
After calculating sX and dX, we don't need to normalise them, but exchange their modules. i.e. we calculate |sX| and |dX|,
and then transform sXtr = sX*|dX|/|sX| ; dXtr = dX*|sX|/|dX|
Then they can then be added together and expanded with the correct output result.
No? Again ???
After calculating sX and dX, we do not need to normalise them, but exchange their modules. i.e. we calculate |sX| and |dX|,
and then transform sXtr = sX*|dX|/|sX| ; dXtr = dX*|sX|/|dX|
Then they can then be added together and expanded with the correct output result.
It goes something like this:
Here a=x0, b=x1rn