Exp(i infinity) = 0 ?

  • Thread starter kakarukeys
  • Start date
  • #1
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:grumpy: :yuck:
Hello there,

please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]

S is the action of a particle [tex] = \int L dt[/tex]
e.g. when the particle goes to infinity and comes back, kinetic energy blows up,
then S blows up.
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
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kakarukeys said:
:grumpy: :yuck:
Hello there,

please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]
It does'nt:
[tex]e^{ix}=\cos{x}+i\sin{x}[/tex]
[tex]\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}[/tex]
But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.
 
  • #3
190
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That's what my math teacher told me,
but my physics teacher told me entirely different thing.
 
  • #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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I would suspect both your math teacher and physics teacher are right!

The map x → e^(ix) is, indeed, discontinuous at +∞, and thus one cannot continuously extend this map to have a value at +∞. (continuous extension is the process that is generally used to justify statements like arctan +∞ = π / 2 or x/x = 1).


However...


As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.
 
  • #5
190
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Do you mean, for computation purpose, we can define the limit to be zero
it can't be proven?
 
  • #6
190
0
:eek:
this is no college level!
Feymann Path Integral!
 
  • #7
574
0
kakarukeys said:
:eek:
this is no college level!
Feymann Path Integral!

College=university. Don't get freaked out.
 
  • #8
190
0
Feynmann Path Integral is at graduate level :rofl:
 

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