Quantum geometry

  • #1
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in the next link there is an article about quantum geometry:http://www.maths.qmw.ac.uk/~majid/qgeom.html [Broken]
i would like to know how was the second equation derived?:
f'(x)=f(x)-f(qx)/(1-q)x.
 
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  • #2
marcus
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Originally posted by loop quantum gravity
in the next link there is an article about quantum geometry:http://www.maths.qmw.ac.uk/~majid/qgeom.html [Broken]
i would like to know how was the second equation derived?:
f'(x)=f(x)-f(qx)/(1-q)x.
Hello Loop,

What Majid means by "quantum geometry" is very different from what Abhay Ashtekar means by it----that is, what it means in the context of LQG.

Majid's website is largely an advertisement for Majid's personality and Majid's book. He does not explain enough so that one can derive anything.

I think you left off a couple of parentheses when you copied
f'(x) = (f(x)-f(qx))/(1-q)x.

This is a finite difference approximation to the ordinary derivative f'(x)
which is the limit of that expression as q --> 1
If you make q very close to 1 then that formula will tend to approach the ordinary derivative. Like, try 0.999 for q.
So what's the big deal?

It strikes me as teasing instead of teaching----he tells just enough to get you interested and then tries to sell you his book.
 
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