What is the ln (0)??

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.

Reply 5

happyheart

OP

hang so say you were doing an intergration with a limit 0 and one of the terms was ln x what would you do??

Reply 6

It seems unlikely. I guess you hope it's either a mistake or it cancels out magically.

Reply 7

M15T3R CH4RL35

what donkey gave you a question like that to do?!

Reply 8

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

Reply 9

Ah, but who says we can't extend the domain (and range) of ln? After all, we know

e^{i \pi} = -1

, so, if we define

\ln(-1) = i \pi

, we can extend ln to the negative reals:

\ln(-x) = i \pi + \ln x

. 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.

Reply 10

generalebriety

16

Zhen Lin

Ah, but who says we can't extend the domain (and range) of ln? After all, we know

e^{i \pi} = -1

, so, if we define

\ln(-1) = i \pi

, we can extend ln to the negative reals:

\ln(-x) = i \pi + \ln x

. 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.

Reply 11

Zhen Lin

Ah, but who says we can't extend the domain (and range) of ln? After all, we know

e^{i \pi} = -1

, so, if we define

\ln(-1) = i \pi

, we can extend ln to the negative reals:

\ln(-x) = i \pi + \ln x

. 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

i \pi

. In fact, since

e^{2ki \pi} = 1

for all k and

e^{(2k+1)i \pi} = -1

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.

Reply 12

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

e^w = z

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.

Reply 13

Zhen Lin

e^w = z

Are you limiting z to the x axis? Such that you get a three dimensional trace

Reply 14

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?

Reply 15

edward_wells90

Zhen Lin

we know

e^{i \pi} = -1

How would I be able to do that on my calculator?

Reply 16

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

Reply 17

edward_wells90

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:

\mathrm{log}_a a = 1

Since

\mathrm{log}_e x = \ln{x}

then

\ln{e} = 1

BTW, is there anyway to do the e exponential in latex or are we just expected to type in the letter e?

Reply 18

Dystopia

10

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

x^{m} (\ln x)^{n}

, which tends to zero as x tends to zero provided that m is positive.

edward_wells90

How would I be able to do that on my calculator?

I suspect that you wouldn't. You would use

e^{ix} = \cos x + i \sin x

Reply 19