Divided by itself equals 1?

  • Thread starter The Rev
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  • #1
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Is it a reasonable statement to say either:


1) [tex]\frac{\infty}{\infty} = 1[/tex] ?


2) [tex]\frac{0}{0} = 1[/tex] ?


[tex]\phi[/tex]

The Rev
 

Answers and Replies

  • #2
lurflurf
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No. Those are not well defined operations, and infinity is not a number. In classical analysis this problem is dealt with by considering limits.
Such as
[tex]\lim_{x\rightarrow a}\frac{f(x)}{g(x)}[/tex]
in the case when if f and g both go to 0 (or infinity) as x goes to a
further analysis is needed to decide the limit
These are called indeterminate forms since the rule about a limit of a combination of functions being equal to the individual limits combined the same way does not apply.
There are several forms of indeterminate forms that commonly occur
0/0
infinity/infinity
0^0
1^infinity
infinity-infinity
0*infinity
1/0-1/0
 
  • #3
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Is this how they cancel out infinities in renormalization in QM?
 
  • #4
matt grime
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Is what how they cancelled out infinities in renormalization? WIth infinite stuff physicists can and do simply ignore things (nothing to do with cancelling), and annoyingly it all seems to work. Hre is a link explaining it in non-specific language:

http://math.ucr.edu/home/baez/renormalization.html
 
  • #5
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Well, he lost me at Lagrangian, but I did get the feeling that renormalization isn't about cancelling out infinities with simple division. (I know nothing of this higher level math, yet.)

I was impressed, however, that singer Joan Baez is also an expert on Quantum Mechanics.

[tex]\psi[/tex]
 
  • #6
matt grime
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gosh, i bet jon's never head that one before. if i ever bump itno him again i'll be sure to pass on the joke.
 
  • #7
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matt grime said:
gosh, i bet jon's never head that one before. if i ever bump itno him again i'll be sure to pass on the joke.
I think they're cousins.
 

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