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

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

Quote by cocopops12

∞ - 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

Quote by cocopops12

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

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

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

Millennial

Quote by 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).

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

Quote by Millennial

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

Quote by 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.

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

Mentallic

Quote by iVenky

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

Quote by 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.

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

iVenky

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

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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.