what is the value of e^-j(infinity)

iVenky

I want to know the value of

e^(-j∞)

From one angle this seems to be equal to zero. But if you expand using Euler's theorem you can see that we get

cos∞ - j sin∞

which clealry ≠ 0

So what's the correct answer?

Thanks in advance.

Hurkyl

To explain what you actually mean by exp(-j infinity). At that point, what value it should have (if any) should be clear.

If you don't know what you mean, then explain what you are trying to do that this comes up, and maybe we can figure out what you mean.

chiro

In electrical engineering, j is used instead of i as i is reserved for current so they use j to represent the square root of -1.

cocopops12

hmm interesting

what if we go back to the power series expansion definition of Euler's identity?



Now take the limit as X --> Infinity
we would get indeterminate forms in both cases where it's +j or -j

that's how i understand it....let's wait for the experts to comment on this

HallsofIvy

Yes, Hurkyl surely knows that. It was not "j" that is the problem but the fact that "[itex]\infty[/itex] is not a number that he was referring to. You cannot simply put "[itex]\infty[/itex] into a formula and expect it to mean anything. Perhaps iVensky meant this as a limit. That is how cocopops12 treated it- but still got no value.

Millennial

The limit is zero. It would be indeterminate, as you said, if we had only one trigonometric function. However, expansion as power series and then simplification (even without Euler's formula) yields the limit as zero.

Hurkyl

The limit form [itex]\cos -\infty[/itex] is indeed an indeterminate form. However, it doesn't make sense to say that a limit is indeterminate. [itex]\lim_{x \to -\infty} \cos x[/itex] simply does not exist.


A simplification cannot possibly have that effect. If the limit of <something> doesn't exist, then neither does <simplification of something>, since <something> and <simplification of something> are the same thing, just expressed differently.

One can modify a problem in order to change it into a different problem, but without knowing the context of the original question, it's hard to guess what modified problems would be appropriate for his purpose.

Millennial

Hurkyl, it can have that effect and it does. It would not if the terms of the sum had a limit. However, when you have two terms, you can't say the same. I will give an example. Take the limit [itex]\displaystyle \lim_{x\to 0}\frac{1}{x}\frac{x}{1}[/itex]. It is obvious as day this limit is one, but to achieve that we perform a step of simplification: We eliminate the x terms. From your perspective, this limit does not exist because 1/x is indeterminate at x=0.

algebrat

If j is i, this is not zero.

algebrat

This does not fall under the definition of indeterminate. (EDIT: if you are thinking of [itex]\infty-\infty[/itex], this doesn't work for [itex]\infty-i\infty[/itex];also cos and sin do not go to infinity.) The limit is simply undefined.

algebrat

Which limit is zero. Why would one trig function be indeterminate, I don't agree, see my previous comment, and check definition of indeterminate.

EDIT: And limit is not zero.

algebrat

What do you mean? (DOUBLE EDIT: There is an imaginary number in the way, and cos does not go to infinity to result in [itex]\infty-\infty[/itex])

Sure it does.

I agree.

algebrat

No it doesn't, neither limit exists.

I'm not sure where to begin.

Perhaps you are all having misunderstandings related to 3 things below.

1. what is the value of e^-j(infinity)
2. what does inderminate mean
3. why can we make cancellations in a limit

algebrat

Your limit [tex]\lim_{n\to\infty}e^{-in}[/tex] does not exist, since [itex]e^{i\theta}[/itex] is rotation around the unit circle in the complex plane.

Dickfore

The limit does not exist. Indeed, take two sequences:
[tex]
x^{1}_{n} = 2 n \pi \Rightarrow \exp( -j \, x^{1}_{n} ) = 1
[/tex]
[tex]
x^{2}_{n} = (2 n + 1) \pi \Rightarrow \exp( -j \, x^{2}_{n} ) = -1
[/tex]
So, as you take the limit [itex]n \rightarrow \infty[/itex], you have two sequences that tend to the infinite point in the complex plane, but the sequences of the values of the function, being constant, trivially tend to 1, and -1, respectively.

Hurkyl

I can't figure out where you got that idea, so I'll just state some facts.

• The partial function 1/x is not indeterminate at x=0: it is undefined at x=0.
  
• When looking for a limit of a (partial) function as x approaches 0, the value it has at 0 (if any) is wholly irrelevant.
  
• x/x is a constant partial function whose domain is the nonzero real numbers.
  
• Replacing x/x with 1 is not a simplification: it is a modification, because they are different partial functions. In particular, the former is undefined at x=0 and the latter is defined at x=0.
  
• Replacing [itex]lim_{x \to 0} x/x[/itex] with 1 is a simplification, because it's fairly evident that the limit as x approaches 0 of a constant function (defined on a punctured neighborhood of 0) is that constant.
  
• Replacing [itex]\lim_{x \to 0} x/x[/itex] with [itex]\lim_{x \to 0} 1[/itex] is a simplification (not a modification), because we know the limits of two different partial functions are equal if they agree everywhere except for one point.
  
• The limit form 0/0 is an indeterminate form, which means that [itex]\lim_{x \to 0} x/x[/itex] with [itex]\lim_{x \to 0} x / \lim_{x \to 0} x[/itex] is a modification: in this case, replacing the number 1 with an undefined expression.

micromass

Your reasoning is probably

[tex]\lim_{x\rightarrow 0} \frac{x}{x}=\frac{\lim_{x\rightarrow 0} x}{\lim_{x\rightarrow 0} x}[/tex]

But this is not allowed. It is only allowed if the two limits exists (which is the case here) and if the denominator is not 0 (which is not the case).