# What is the ln (0)??Watch

Announcements
#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?
0
11 years ago
#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.
0
11 years ago
#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.
0
11 years ago
#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.
0
11 years ago
#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.
0
#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??
0
11 years ago
#7
It seems unlikely. I guess you hope it's either a mistake or it cancels out magically.
0
11 years ago
#8
what donkey gave you a question like that to do?!
0
11 years ago
#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
0
11 years ago
#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.
0
11 years ago
#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.
0
11 years ago
#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.
0
11 years ago
#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.
0
11 years ago
#14
(Original post by Zhen Lin)
Are you limiting z to the x axis? Such that you get a three dimensional trace
0
11 years ago
#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?
0
11 years ago
#16
(Original post by Zhen Lin)
we know
How would I be able to do that on my calculator?
0
11 years ago
#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
0
11 years ago
#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?
0
11 years ago
#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
0
11 years ago
#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.
0
X

new posts

### 1,079

people online now

### 225,530

students helped last year
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Aberdeen
Tue, 27 Aug '19
• Norwich University of the Arts
Sat, 31 Aug '19
• University of Lincoln
Guardian Offices, Kings Cross, London Postgraduate
Mon, 2 Sep '19

### Poll

Join the discussion

#### How are you feeling about GCSE Results Day?

Hopeful (220)
12.46%
Excited (166)
9.4%
Worried (308)
17.44%
Terrified (389)
22.03%
Meh (183)
10.36%
Confused (39)
2.21%
Putting on a brave face (244)
13.82%
Impatient (217)
12.29%