# What is the value of e^-j(infinity)

1. Jun 24, 2012

### 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

2. Jun 24, 2012

### Hurkyl

Staff Emeritus
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.

3. Jun 24, 2012

### 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.

4. Jun 24, 2012

### 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

5. Jun 24, 2012

### HallsofIvy

Yes, Hurkyl surely knows that. It was not "j" that is the problem but the fact that "$\infty$ is not a number that he was referring to. You cannot simply put "$\infty$ 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.

6. Jun 24, 2012

### 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.

7. Jun 24, 2012

### Hurkyl

Staff Emeritus
The limit form $\cos -\infty$ is indeed an indeterminate form. However, it doesn't make sense to say that a limit is indeterminate. $\lim_{x \to -\infty} \cos x$ 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.

8. Jun 24, 2012

### 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 $\displaystyle \lim_{x\to 0}\frac{1}{x}\frac{x}{1}$. 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.

9. Jun 24, 2012

### algebrat

If j is i, this is not zero.

10. Jun 24, 2012

### algebrat

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

Last edited: Jun 24, 2012
11. Jun 24, 2012

### 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.

Last edited: Jun 24, 2012
12. Jun 24, 2012

### 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 $\infty-\infty$)

Sure it does.

I agree.

Last edited: Jun 24, 2012
13. Jun 24, 2012

### 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

14. Jun 24, 2012

### algebrat

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

15. Jun 24, 2012

### Dickfore

The limit does not exist. Indeed, take two sequences:
$$x^{1}_{n} = 2 n \pi \Rightarrow \exp( -j \, x^{1}_{n} ) = 1$$
$$x^{2}_{n} = (2 n + 1) \pi \Rightarrow \exp( -j \, x^{2}_{n} ) = -1$$
So, as you take the limit $n \rightarrow \infty$, 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.

16. Jun 24, 2012

### Hurkyl

Staff Emeritus
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 $lim_{x \to 0} x/x$ 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 $\lim_{x \to 0} x/x$ with $\lim_{x \to 0} 1$ 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 $\lim_{x \to 0} x/x$ with $\lim_{x \to 0} x / \lim_{x \to 0} x$ is a modification: in this case, replacing the number 1 with an undefined expression.

17. Jun 24, 2012

### micromass

$$\lim_{x\rightarrow 0} \frac{x}{x}=\frac{\lim_{x\rightarrow 0} x}{\lim_{x\rightarrow 0} x}$$

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).

18. Jun 25, 2012

### cocopops12

When taking the limit as x -> infinity, i was talking about applying it to the part with polynomial expansion, not the the cosx -jsinx

∞ - i∞
can someone explain why this isn't an indeterminate form?
i know ∞ isn't a fixed number, it represents a very large quantity, and if you scale that very large quantity by a finite quantity you still get an infinite quantity

so i think you can treat ∞ - i∞ as ∞ - ∞

19. Jun 25, 2012

### micromass

It's not indeterminate. It is undefined. You could give several (non-equivalent) meaningful definitions to it though.

NO!!!!! Infinity is not a number, it is not a very large quantity. Infinity is a symbol that behaves in a certain way.

First you got to define what these two expressions even mean.

20. Jun 25, 2012

### Mentallic

Why are you even bringing this up? The question in the OP was about $$\lim_{x\to\infty}e^{-ix}$$ which is equivalent to $$\lim_{x\to\infty}\left(\cos(x)-i\sin(x)\right)$$$$=\lim_{x\to\infty}\cos(x)-i\lim_{x\to\infty}\sin(x)$$

cos and sin fluctuate between -1 and 1, so the point you're trying to make about what $\infty-i\infty$ is both irrelevant and futile.