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]

So what's the correct answer?

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]

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.

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]

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.

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.

 

[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]

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

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.

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.

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

 

[algebrat]

I want to know the value of

e^(-j∞)

From one angle this seems to be equal to zero.

If j is i, this is not zero.

 

[algebrat]

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

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.

 

[algebrat]

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.

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]

$\cos -\infty$ is indeed an indeterminate form.

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

However, it doesn't make sense to say that a limit is indeterminate.

Sure it does.

$\lim_{x \to -\infty} \cos x$ simply does not exist.

I agree.

 

[algebrat]

Hurkyl, it can have that effect and it does.

No it doesn't, neither limit exists.

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.

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

 

[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.

 

[Hurkyl]

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.

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.

 

[micromass]

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.

Your reasoning is probably

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

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).

 

[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 ∞ - ∞

 

[micromass]

∞ - i∞
can someone explain why this isn't an indeterminate form?

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

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

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

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

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

 

[Mentallic]

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

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.

 

[Millennial]

Your reasoning is probably

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

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).

My reasoning isn't even remotely related to that. I am saying that the given limit (in a precise sense) can equal 0 - this does not mean that it is. However, if we were obliged to assign a specific value to this limit, it would most righteously be 0.

I will explain this further because I think there might be gaps left. The divergence of the series 1+1+1+1+1... is obvious. However, if one was to assign a value to this sum, one can use zeta function regularization to land on the value -1/2, and this value is employed in physics with several applications.

 

[micromass]

My reasoning isn't even remotely related to that. I am saying that the given limit (in a precise sense) can equal 0 - this does not mean that it is. However, if we were obliged to assign a specific value to this limit, it would most righteously be 0.

I have no idea what you mean. How can a given limit equal 0 if it is not 0??

I will explain this further because I think there might be gaps left. The divergence of the series 1+1+1+1+1... is obvious. However, if one was to assign a value to this sum, one can use zeta function regularization to land on the value -1/2, and this value is employed in physics with several applications.

This is a completely different topic. I don't see what that has to do with the previous discussion??

 

[iVenky]

Let's expand e^iy

e^iy= 1 + iy + (iy)²/2! + (iy)³/3!+ ...

As you can clearly see if y→-∞ then we would clearly have the above expression equate to

1 + 0 + 0 ...

and hence the limit should become 1 if you find out the answer using the above expansion.

 

[micromass]

Let's expand e^iy

e^iy= 1 + iy + (iy)²/2! + (iy)³/3!+ ...

As you can clearly see if y→-∞ then we would clearly have the above expression equate to

1 + 0 + 0 ...

and hence the limit should become 1 if you find out the answer using the above expansion.

How would that be "clearly" true?? I don't see it.

 

[Mentallic]

e^iy= 1 + iy + (iy)²/2! + (iy)³/3!+ ...

As you can clearly see if y→-∞ then we would clearly have the above expression equate to

1 + 0 + 0 ...

Are you confusing y with e^y?

 

[pwsnafu]

Let's expand e^iy

e^iy= 1 + iy + (iy)²/2! + (iy)³/3!+ ...

As you can clearly see if y→-∞ then we would clearly have the above expression equate to

1 + 0 + 0 ...

and hence the limit should become 1 if you find out the answer using the above expansion.

But the forth term is $\frac{(iy)^4}{4!}$. Then as y→-∞...

 

[iVenky]

Sorry guys. I have actually confused y with e^y