Exploring the Relationship between Infinity and Zero in Feynman Path Integrals

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In summary, the conversation discusses whether e^(iS) equals 0 when S is infinity. The real and imaginary parts of the limit do not exist as it oscillates between -1 and 1. However, there may be a related concept that is practical to use in this situation, possibly the Feynman Path Integral at the graduate level.
  • #1
kakarukeys
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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.
 
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  • #2
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
That's what my math teacher told me,
but my physics teacher told me entirely different thing.
 
  • #4
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
Do you mean, for computation purpose, we can define the limit to be zero
it can't be proven?
 
  • #6
:eek:
this is no college level!
Feymann Path Integral!
 
  • #7
kakarukeys said:
:eek:
this is no college level!
Feymann Path Integral!


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

1. What is the meaning of "Exp(i infinity) = 0"?

The expression "Exp(i infinity) = 0" is a mathematical statement that represents the value of the exponential function raised to the power of infinity, where i is the imaginary unit. This expression is also known as Euler's identity and is a fundamental result in complex analysis.

2. How can "Exp(i infinity) = 0" be true when both infinity and i are not real numbers?

While infinity and i are not real numbers, they can still be used in mathematical calculations. The concept of infinity is often used in calculus and limits, while the imaginary unit i is a fundamental concept in complex numbers. When combined, they result in the complex number 0, which is the solution to the exponential expression "Exp(i infinity) = 0".

3. Can "Exp(i infinity) = 0" be proven mathematically?

Yes, the equation "Exp(i infinity) = 0" can be proven mathematically using complex analysis and Euler's formula. This formula states that for any real number x, the expression "Exp(ix) = cos(x) + i*sin(x)". When x is equal to infinity, the cosine function approaches 0 and the sine function oscillates between -1 and 1, resulting in the complex number 0.

4. What is the significance of "Exp(i infinity) = 0" in mathematics?

The expression "Exp(i infinity) = 0" is significant in mathematics because it connects three fundamental mathematical concepts: the exponential function, the imaginary unit, and infinity. It also has many applications in fields such as physics, engineering, and finance.

5. How is "Exp(i infinity) = 0" related to complex numbers?

The expression "Exp(i infinity) = 0" is closely related to complex numbers, as it is a fundamental result in complex analysis. It also demonstrates the behavior of complex numbers when raised to extreme powers, such as infinity. Additionally, it shows the connection between trigonometric functions and complex numbers, as seen in Euler's formula.

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