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# What is the ln (0)??

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1. I have always assumed htis to be 1 is thi correct on my calculator its maths error now im really confused??? can anyone help?
2. It's undefined.

Consider y = ln(x)

So, x = e^y

What value of y can you plug in to make x = 0? There is none.
3. If you get ln(x), you are basically asking "what is the value that when you do exp(that number) you get x"

E.g. ln(1), so what power of e, gives 1. Answer 0

So for ln(0), what power of e, gives 0. Answer, there isn't one. Which is why you get an error.
4. I don't think you can have ln(0)

Anything to the power of 0 is one, and you can't have the power of something being 0.

ln(1) = 0 as e to the power 0 is 1.
ln(0) would mean that e to the power of something is 0 which never occurs.
5. However, the limit of ln x as x approaches zero from the right is negative infinity. Not sure what the limit as you approach from the left is though.
6. hang so say you were doing an intergration with a limit 0 and one of the terms was ln x what would you do??
7. It seems unlikely. I guess you hope it's either a mistake or it cancels out magically.
8. what donkey gave you a question like that to do?!
9. (Original post by Zhen Lin)
However, the limit of ln x as x approaches zero from the right is negative infinity. Not sure what the limit as you approach from the left is though.
There are no values as you approach from the left, because e^x>0 for all real x
10. Ah, but who says we can't extend the domain (and range) of ln? After all, we know , so, if we define , we can extend ln to the negative reals: . This, of course, means that the limit from the left of ln 0 does not agree with the limit from the right...

But, this may be the wrong way to extend ln to the negatives. Hmm.
11. (Original post by Zhen Lin)
Ah, but who says we can't extend the domain (and range) of ln? After all, we know , so, if we define , we can extend ln to the negative reals: . This, of course, means that the limit from the left of ln 0 does not agree with the limit from the right...

But, this may be the wrong way to extend ln to the negatives. Hmm.
ln isn't well defined then. (For example, you can have ln(-1) = -i.pi too. We then also have ln(-1) + ln(-1) = 2i.pi = ln 1, so ln has become multivalued for all real numbers.) This is the usual problem, though. There are ways of taking 'principal' values.
12. (Original post by Zhen Lin)
Ah, but who says we can't extend the domain (and range) of ln? After all, we know , so, if we define , we can extend ln to the negative reals: . This, of course, means that the limit from the left of ln 0 does not agree with the limit from the right...

But, this may be the wrong way to extend ln to the negatives. Hmm.
Then it would agree, although displaced in the imaginary direction by . In fact, since for all k and for all k, then there is an infinite number of parallel solutions (in parallel imaginary planes) to the equation y=lnx. The limit from the right approaches 2kiπ (imaginary) - ∞(real), and from the left it approaches (2k+1)iπ (imaginary) - ∞(real). I suppose you would have to limit your range a little...

That would also mean that, at x=0, there is an infinite number of parallel solutions, each as non-existent as one another, so it is still undefined.
13. After reading some articles about the complex logarithm, it seems that there is indeed no way to have a meaningful value for ln 0. If we consider all the solutions to and plot (z, w), it forms a pretty spiral sheet around z = 0, where it is discontinuous... Interestingly, if we think of the structure of it that way, it's hardly a surprise that we can arrive at different values for ln -1 by following different "paths" starting from z = 1.
14. (Original post by Zhen Lin)
Are you limiting z to the x axis? Such that you get a three dimensional trace
15. Hmm. Well, the graph of (z, w) has four (real) dimensions, or two complex dimensions. So I guess we would have to limit one of the parameters in order to plot it in three dimensions... perhaps plotting (z, Im w) or (z, |w|) would be good?
16. (Original post by Zhen Lin)
we know
How would I be able to do that on my calculator?
17. (Original post by Zhen Lin)
Hmm. Well, the graph of (z, w) has four (real) dimensions, or two complex dimensions. So I guess we would have to limit one of the parameters in order to plot it in three dimensions... perhaps plotting (z, Im w) or (z, |w|) would be good?
Well we are interested in the graph of e^y=x, then y=lnx, and we are then looking for values of y when x=0. You are sliding up and down the x axis, so x is the one dimenstional variable. y would be the 2-dimensional variable, since we talk about e^(ipi) etc

so in the above case, w is two dimensional and z is one dimensional
18. (Original post by happyheart)
I have always assumed htis to be 1 is thi correct on my calculator its maths error now im really confused??? can anyone help?
I think you might be getting it confused with this:

Since

then

BTW, is there anyway to do the e exponential in latex or are we just expected to type in the letter e?
19. (Original post by happyheart)
hang so say you were doing an intergration with a limit 0 and one of the terms was ln x what would you do??
It is possible that you may end up with , which tends to zero as x tends to zero provided that m is positive.

(Original post by edward_wells90)
How would I be able to do that on my calculator?
I suspect that you wouldn't. You would use
20. (Original post by mpd1989)
Well we are interested in the graph of e^y=x, then y=lnx, and we are then looking for values of y when x=0. You are sliding up and down the x axis, so x is the one dimenstional variable. y would be the 2-dimensional variable, since we talk about e^(ipi) etc

so in the above case, w is two dimensional and z is one dimensional
Well, if you plot it that way the spiral structure of the complex logarithm is not obvious - you would simply see a discontinuity at ln 0 where the imaginary part suddenly flips from 0 to ±π.

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