This reminds me of the Gamma function, which extends the factorials to all sorts of places in the complex plane but stubbornly refuses to continue factorials to the negative integers. Here the logarithm refuses to be extrended to the negative reals!
Negative reals still works for logarithms. ln(z)=ln(IzI)+i*Arg(z) Let z=-1 ln(-1)=ln(I-1I)+i*Arg(-1) [I-1I=1, Arg(-1)=pi] ln(-1)=ln(1)+pi*i [ln(1)=0] ln(-1)=pi*i
I think a distinction should be made between two different types of slitting the plane. There's slitting the plane in the sense of "removing {-infinity, 0]", and there's slitting the plane in the sense of "cutting along the negative real axis so that the top left quadrant is no longer joined to the bottom left quadrant". When you're creating the Riemann surface at the end, it's the latter idea that's necessary, otherwise you're missing the part of each sheet that glues onto the next sheet. In either case, you do still have to remove the origin though.
Hey, Michael! In the two boxes (the yellow and the pink), you had the limit as z goes to 1 of Log(z). It should be the limit as z goes to -1 of Log(z).
And if they ask about whether or not there's multiple values of ln|z|, you can add the 2πki from it to the 2πmi of the second term to make 2πni, and leave the rest defined as the integral from 1 to |z| of dt/t, then just take ln|z| to be that integral with a possibly different value of the added multiples of 2πi.
i enjoy how the angular argument of a complex number is effectively a modular value, in the sense of modular arithmetic i don't enjoy how the magnitude of a complex number is also called a "modulus", which could easily be confusing, and suggests the people who came up with these names should have workshopped them a little longer
The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards.
I think that the log(-1) would be well behaved if we had taken the Arg(z) to be from 0 to 2pi as the problem stems from the Arg(-1) being pi and negative pi in the chosen values. So by using 0 to 2pi, the Arg(z) would be the same leading to the same value. (Correct me if I'm wrong)
If you take (0, 2pi) then you'll have problems on the positive real axis. Usually we want to take the branch cut on the negative reals because then it turns into the analytic continuation of the natural logarithm (on real numbers). But it's true you could take the branch cut along any ray if you want.
I was noticing you can trace out a circle counterclockwise from 1 on the complex plane just by saying (-1)^t with t from 0..2. That feels a lot simpler than writing e^pi*i*t and similar things.
It looks simpler but it's not really; -1 is e^ipi, and writing in that form makes it obvious what's going on - instead of t being some variable doing something arcane to -1, it's now the argument of the sum of sines and cosines
@@marcushendriksen8415 with t=0..2 we get just with four discrete points, (-1)^0 = 1, (-1)^(1/2) = i, (-1)^(1) = -1, (-1)^(3/2)=-i, (-1)^2=1 again. all very basic
I always enjoy your videos, all of them challenge me personally and I learn. The blackboard brings back memories from the 1980’s and wishing my lecturer was transparent.😁 How about a series on ‘generally useful tools’ used in solving a range of real world problems..on the blackboard..
I think the “log” and “Log” he’s using both mean natural log and he’s just using that notation to differentiate it from the natural log of a real number.
How does removing the negative real axis from the domain help? Now instead of being discontinuous, it's just got a gaping infinite hole. I don't see how this changes the fact that's the complex logarithm is still discontinuous, now it's just a different kind, right?
So the _principle_ value of Log(-1) isn't well defined (given the definition of the principle branch used here). But what about the multi-value log? Would it be fine to say log(-1)=(1+2n)πi for n ∈ Z ? Given the divergence of the limits is conveniently 2π apart from each other?
In effect, yes. That’s fine. As long as you “glue” the planes together, the multivalued log has a value. However, you can’t take the limit as n goes to infinity, because those values don’t exist by this construction
How does this extend to log_n(z)? The expansion leads to log_n(r)+log_n(e)*(log_n(θ)+log_n(i)), but that would require defining log_n(i), since it would otherwise recur ad infinitum. I'd imagine that the identity log_a(n)/log_b(n)=ln(b)/ln(a) could be used to define log_n(i) as (1/ln(n))*iπ/2, but I'm not sure if that's standard practice in mathematics or not. Using that definition, log_n(z) would be log_n(r)+log_n(e)*(log_n(θ)+(1/ln(n))*iπ/2), but again, I'm not sure if there's already a standard definition in use.
in this case we should left the [0, +infinity) as undefineable, the same way the negative real axis is not included in Log(z) in case of theta in (-п, п]
22:20 Aren’t those two limits not equal just because you chose the interval (-pi,pi] as possible values for theta? If you had chosen [0,2pi) instead, the two limits would have been equal as theta would have approached pi in both cases.
@@jakubmazur1317 So that means the logarithmic function isn't actually discontinuous for a specific set of values. It just depends on our choice of theta.
Hello... Can someone give me some guidance? In high school, I hit a "wall" once mathematical concepts got more complex. As a result, I diverged from math altogether and focused on history/philosophy... I promised myself that I would come back to math, which I did in my 50s... However, I always encounter the same obstacles when introduced to new concepts. I understand the math but I cannot delve deep in a subject-matter w/o good contextual and conceptual understanding of the subject-matter, i.e., new to sequences/series, not hard, but every lecture or text just launch into divergence-converge w/o context (Why is divergence-convergence important? What does it do? How does it fit into the larger calculus landscape?)... Are there resources out there for me that provide a full explanation, not just "here's a recipe, follow it"? Or do I have to modify my approach?
"Convergence" is a concept that bundles up a lot of ideas that boil down to "a thing behaves nicely" (for some definition of "nice"). Here the issue with convergence comes from the discontinuity forced by the logarithm function. The log is not defined on negative numbers because it's not continuous. There's a lot of handwaving in what I'm saying but I hope you were able to follow the explanation of *why* it's discontinuous around the 18:00 mark... essentially continuous means that no matter how you come at a thing, you always get the same value. If things behave "nice" when functions are continuous, they "converge" on the same value. As example of "convergence" where you don't have continuity, imagine the sequence of reciprocals of the integers (1/1, 1/2, 1/3, 1/4, 1/5)... As you go farther and farther down the list (1/100, 1/10000, 1/1000000, etc), the values of the sequence get as close to zero as you can imagine. So, even though the value is never *actually* zero (because there is no integer such that when you take its reciprocal, you get zero), you can get so close that it doesn't matter, and we say that the sequence *converges* to zero. As far as online resources you can tap into for self-study, there are probably good videos on youtube you can find. Dig through 3Blue1Brown's channel and blackpenredpen's... 3B1B has a very wide array of topics he speaks to, while BPRP is more calculus focused, but they both explain their concepts at a fairly "low" (compared to the grad-school level stuff that Penn's channel does at lot of the time). Just beware that BPRP will often cite theorems and algorithms that are common knowledge for someone who has just walked out of 21st century HS calculus class so you might have to look up some things. (I say that as someone who taught HS calc in the 20th century and there are definitely a few shibboleths laying around.)
I don't know whether any suitable resource exists, but it sounds as if it would be helpful for you to have something that takes a historical approach to the development of maths - i.e. explaining what mathematical questions or issues each generation of mathematicians worked on, and how this influenced the way that they thought and the maths that they developed.
Seems like all that he proved was that log(-1) was equal to different values at the same time though right? Isn’t that the case with all the other points on the graph? Just started complex numbers and this didn’t make sense to me, would appreciate any help.
First he tries to define log so that it is a well-defined function (it only gives exactly one result for every input z (different from zero)). He manages to do that by simply choosing that the result of the arg function has to be in (-pi;pi]. (That's a rather arbitrary, but rather usual, choice.) And then he shows that this leads to a problem with the continuity of this (single-valued) function log, i. e. if one wants to have a continuous log ''function'' which is defined on the whole complex plane (without zero), on has to accept that it has to be multi-valued.
The discontinuity makes it difficult, so you can't get a series that works for all z, but you can for example get one that works for all complex numbers inside a unit disc centered at z=1: Log(z) = (z-1)¹/1 - (z-1)²/2 + (z-1)³/3 - (z-1)⁴/4 + ... Check out the channel "zetamath" to see how the discontinuity ruins the series
@@nicolascalandruccio Actually yeah you can start with 1-(z-1)+(z-1)²-(z+1)³+... which sums to 1/(1-(-(z-1))) which is 1/z, and then take the integral of this, the integral of 1/z is log(z) and integrating this geometric series term-by-term gives the series for log I mentioned
That depends on how you define a function. Some parts of math also say that a function must be defined on the entire domain, so a function having points where it is undefined is also an oxymoron. This is sometimes called a "partial function". The generalized notion which allows undefined points and multiple values is sometimes called a "relation". So we best just keep calling everything "function" since we understand what we mean
@@AssemblyWizard Maths shouldn't be about "let's just use this word and hope everyone understands what is meant here, although we use it with a meaning which contradicts its usual definition".
@@angelmendez-rivera351 of course that's the default, but look at the original comment you're commenting on. They asked if adding the prefix "multi-valued" to "function" is legit. Which I claim it is, like saying "vegan meat" is fine even though "meat" alone is specifically not vegan. The prefix can change the meaning of the term that follows
lol you skewed the limits z === -1 + i.0 then the limits of Log is the same as -pi^+ is NOT IN THE domain so.... FIX YOUR LIMIT!!! { [ -pi^+ 2pi] E ( -pi, pi] }
¿ por qué siguen los matemáticos hablando de imaginarias ? , están tratando con una simetría del espacio tiempo de 90 grados y darle el nombre de imaginaria crea un desvarío que hasta ahora sigue inventando cosas que no se puede explicar a la gente sencilla sin recurrir a cosas extrañas para el sentido común
This reminds me of the Gamma function, which extends the factorials to all sorts of places in the complex plane but stubbornly refuses to continue factorials to the negative integers. Here the logarithm refuses to be extrended to the negative reals!
The problem with gamma isn't that it's multivalued, it's that it is undefined/divergent at negative integers.
Never thought of that.
The Hadamard gamma function also interpolates the factorial while being defined at negative integers.
Negative reals still works for logarithms.
ln(z)=ln(IzI)+i*Arg(z)
Let z=-1
ln(-1)=ln(I-1I)+i*Arg(-1) [I-1I=1, Arg(-1)=pi]
ln(-1)=ln(1)+pi*i [ln(1)=0]
ln(-1)=pi*i
The log(0) is the other true singularity. All other discontinuities are mesomorphic and analytic continuation holds.
I think a distinction should be made between two different types of slitting the plane. There's slitting the plane in the sense of "removing {-infinity, 0]", and there's slitting the plane in the sense of "cutting along the negative real axis so that the top left quadrant is no longer joined to the bottom left quadrant". When you're creating the Riemann surface at the end, it's the latter idea that's necessary, otherwise you're missing the part of each sheet that glues onto the next sheet. In either case, you do still have to remove the origin though.
@@angelmendez-rivera351 It's not equivalent if you're adding something back in/not taking it out in the first place
31:54
your job is just to watch every single michael penn video, click to the very end, then comment the timestamp? wow
@@aaaaaa-rr8xm
Yes, there must be a good reason for that to stop, too. ;)
@@aaaaaa-rr8xm it's a bit sad, isn't it?
@@Grizzly01 I think it's a little fun to see this comment everywhere. Let's me know how long I'm really gonna be watching for.
So the non-positive x-axis is the "international date line" of the complex Log function.
Nice!
Thank you, professor. That was a very clear explanation I’ll use this year.
Wow that visualization at the end of the multivalued log function was really cool.
at 20:32 doesnt he mean as z approaches -1?
I think so
Most likely
Hey, Michael! In the two boxes (the yellow and the pink), you had the limit as z goes to 1 of Log(z). It should be the limit as z goes to -1 of Log(z).
More like theta goes to pi or minus pi, when z = exp(i*theta).
And if they ask about whether or not there's multiple values of ln|z|, you can add the 2πki from it to the 2πmi of the second term to make 2πni, and leave the rest defined as the integral from 1 to |z| of dt/t, then just take ln|z| to be that integral with a possibly different value of the added multiples of 2πi.
small thing at 22:26 , i think it's lim t->-pi+(Log e(it)) which is lim t->-pi+ (i*theta) (there's an extra exp)
Yes.
i enjoy how the angular argument of a complex number is effectively a modular value, in the sense of modular arithmetic
i don't enjoy how the magnitude of a complex number is also called a "modulus", which could easily be confusing, and suggests the people who came up with these names should have workshopped them a little longer
The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass.
In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards.
I think that the log(-1) would be well behaved if we had taken the Arg(z) to be from 0 to 2pi as the problem stems from the Arg(-1) being pi and negative pi in the chosen values. So by using 0 to 2pi, the Arg(z) would be the same leading to the same value. (Correct me if I'm wrong)
Yes, but then log(1) wouldn't be well-behaved.
If you take (0, 2pi) then you'll have problems on the positive real axis. Usually we want to take the branch cut on the negative reals because then it turns into the analytic continuation of the natural logarithm (on real numbers). But it's true you could take the branch cut along any ray if you want.
@@NeilGirdhar right I didn't think about that
@@TheEternalVortex42 seems interesting
I was noticing you can trace out a circle counterclockwise from 1 on the complex plane just by saying (-1)^t with t from 0..2. That feels a lot simpler than writing e^pi*i*t and similar things.
It looks simpler but it's not really; -1 is e^ipi, and writing in that form makes it obvious what's going on - instead of t being some variable doing something arcane to -1, it's now the argument of the sum of sines and cosines
@@marcushendriksen8415 with t=0..2 we get just with four discrete points, (-1)^0 = 1, (-1)^(1/2) = i, (-1)^(1) = -1, (-1)^(3/2)=-i, (-1)^2=1 again. all very basic
I always enjoy your videos, all of them challenge me personally and I learn. The blackboard brings back memories from the 1980’s and wishing my lecturer was transparent.😁 How about a series on ‘generally useful tools’ used in solving a range of real world problems..on the blackboard..
NO IT'S NOT A GOOD PLACE TO STOP!!!
why we can't define Log(-1)= | iπ | (only positive value ) so we can get Log(-n) = log|n| + iπ
Such an awesome explanation.
Best explanation, really useful with quantum theory
How did you change to the natural logarithm at 9:05?
I think the “log” and “Log” he’s using both mean natural log and he’s just using that notation to differentiate it from the natural log of a real number.
also, for upper level mathematics, “log” usually means base e, not base 10. In comp sci it usually means base 2, which is interesting
How does removing the negative real axis from the domain help? Now instead of being discontinuous, it's just got a gaping infinite hole. I don't see how this changes the fact that's the complex logarithm is still discontinuous, now it's just a different kind, right?
So the _principle_ value of Log(-1) isn't well defined (given the definition of the principle branch used here).
But what about the multi-value log? Would it be fine to say log(-1)=(1+2n)πi for n ∈ Z ? Given the divergence of the limits is conveniently 2π apart from each other?
In effect, yes. That’s fine. As long as you “glue” the planes together, the multivalued log has a value. However, you can’t take the limit as n goes to infinity, because those values don’t exist by this construction
You mean "principal". Professor Penn misspelt it.
if we define Arg (z) to be in range (-pi,pi] than Log(-1) is i*pi, and im(Log(z)) is in (-pi,pi].
How does this extend to log_n(z)? The expansion leads to log_n(r)+log_n(e)*(log_n(θ)+log_n(i)), but that would require defining log_n(i), since it would otherwise recur ad infinitum. I'd imagine that the identity log_a(n)/log_b(n)=ln(b)/ln(a) could be used to define log_n(i) as (1/ln(n))*iπ/2, but I'm not sure if that's standard practice in mathematics or not. Using that definition, log_n(z) would be log_n(r)+log_n(e)*(log_n(θ)+(1/ln(n))*iπ/2), but again, I'm not sure if there's already a standard definition in use.
Excellent. Thank you.
5:03 wouldn't it be easier for theta to be in [0,2pi) range ? and just go one full circle ?
It's an equivalent convention, but usually it's more important for the region around the positive real axis to be continuous than the negative
@@romajimamulo yeah, makes sense
in this case we should left the [0, +infinity) as undefineable, the same way the negative real axis is not included in Log(z) in case of theta in (-п, п]
22:20 Aren’t those two limits not equal just because you chose the interval (-pi,pi] as possible values for theta? If you had chosen [0,2pi) instead, the two limits would have been equal as theta would have approached pi in both cases.
Then discontinuity would happen to be across theta=0, which is much worse choice, as we would have to remove positive reals
@@jakubmazur1317 So that means the logarithmic function isn't actually discontinuous for a specific set of values. It just depends on our choice of theta.
@@natashakwa yes, the matter is it have to be discontinous on some path connecting infinity and the origin, but it is never cont. everywhere
aww, I was a fan of the old title about unwinding the logarithm
At 21:00 Michael seems to have made a mistake. Shouldn't it be z --> -1 ?
High schooler: has no solution on -π, qed
Math teacher: let’s create an infinite amount of planes and we have infinite solutions
please correct 'principle' into principal value branch
Hello... Can someone give me some guidance? In high school, I hit a "wall" once mathematical concepts got more complex. As a result, I diverged from math altogether and focused on history/philosophy... I promised myself that I would come back to math, which I did in my 50s... However, I always encounter the same obstacles when introduced to new concepts. I understand the math but I cannot delve deep in a subject-matter w/o good contextual and conceptual understanding of the subject-matter, i.e., new to sequences/series, not hard, but every lecture or text just launch into divergence-converge w/o context (Why is divergence-convergence important? What does it do? How does it fit into the larger calculus landscape?)... Are there resources out there for me that provide a full explanation, not just "here's a recipe, follow it"? Or do I have to modify my approach?
Convergence vs divergence is important, as it lets you know if what you're writing makes sense. For example, the power series of 1/(1-z) for |z|
"Convergence" is a concept that bundles up a lot of ideas that boil down to "a thing behaves nicely" (for some definition of "nice"). Here the issue with convergence comes from the discontinuity forced by the logarithm function. The log is not defined on negative numbers because it's not continuous.
There's a lot of handwaving in what I'm saying but I hope you were able to follow the explanation of *why* it's discontinuous around the 18:00 mark... essentially continuous means that no matter how you come at a thing, you always get the same value. If things behave "nice" when functions are continuous, they "converge" on the same value.
As example of "convergence" where you don't have continuity, imagine the sequence of reciprocals of the integers (1/1, 1/2, 1/3, 1/4, 1/5)... As you go farther and farther down the list (1/100, 1/10000, 1/1000000, etc), the values of the sequence get as close to zero as you can imagine. So, even though the value is never *actually* zero (because there is no integer such that when you take its reciprocal, you get zero), you can get so close that it doesn't matter, and we say that the sequence *converges* to zero.
As far as online resources you can tap into for self-study, there are probably good videos on youtube you can find. Dig through 3Blue1Brown's channel and blackpenredpen's... 3B1B has a very wide array of topics he speaks to, while BPRP is more calculus focused, but they both explain their concepts at a fairly "low" (compared to the grad-school level stuff that Penn's channel does at lot of the time). Just beware that BPRP will often cite theorems and algorithms that are common knowledge for someone who has just walked out of 21st century HS calculus class so you might have to look up some things. (I say that as someone who taught HS calc in the 20th century and there are definitely a few shibboleths laying around.)
I don't know whether any suitable resource exists, but it sounds as if it would be helpful for you to have something that takes a historical approach to the development of maths - i.e. explaining what mathematical questions or issues each generation of mathematicians worked on, and how this influenced the way that they thought and the maths that they developed.
If log(z) = Log(z) + 2i π n, then doesn't that make the log(z) function continuous at z = -1?
Cool. I always thought that complex logarithms were invalid
Can someone help me understand at 9:22 how he went from log(r) to ln|z| in the first half of the expansion of the log(z)
By construction, |z| = r so log(r) = log(|z|).
And also he uses log for the complex (natural) logarithm and ln for the real-valued natural logarithm.
That's why log(r) turns into just ln|z|.
Indeed a bit weird to use log for the natural logarithm in complex case
@@RandomBurfness I think they meant how he went from log to ln
13:00 wouldn't it be better to avoid using the tangent and instead use Z / e^(i×arg(z)) being real and greater than or equal to 0?
If you glue the discontinuity could you get an infinitely long cylinder with radius 2pi?
did you mean limit as z -> -1? at ~21:30
could you make a 3D surface plot for real part of logz versus z and another plot of im logz versus z.
Does the gluing of the slit-planes take care of the discontinuities in the images?
What happens if you use newton's method on this?
Why Argz is different of ( -π) in
(-π,π] ? think you.
Howdy.
Mesmerizing ...
Regards.
Seems like all that he proved was that log(-1) was equal to different values at the same time though right? Isn’t that the case with all the other points on the graph? Just started complex numbers and this didn’t make sense to me, would appreciate any help.
First he tries to define log so that it is a well-defined function (it only gives exactly one result for every input z (different from zero)). He manages to do that by simply choosing that the result of the arg function has to be in (-pi;pi]. (That's a rather arbitrary, but rather usual, choice.) And then he shows that this leads to a problem with the continuity of this (single-valued) function log, i. e. if one wants to have a continuous log ''function'' which is defined on the whole complex plane (without zero), on has to accept that it has to be multi-valued.
but i thought z = - 1 for ln z : ln ( - 1 ) = ln 1 + i * 0.5*pi = 0.5*i*pi
11:11 how does log(|z|) = ln|z| ?
I think it should be more explicit that r is a nonnegative real number.
Is there a series which is equal to Log(z) just like exp(z)?
The discontinuity makes it difficult, so you can't get a series that works for all z, but you can for example get one that works for all complex numbers inside a unit disc centered at z=1:
Log(z) = (z-1)¹/1 - (z-1)²/2 + (z-1)³/3 - (z-1)⁴/4 + ...
Check out the channel "zetamath" to see how the discontinuity ruins the series
@@AssemblyWizard Thanks, I'll take a look. I guess I got it. 1/(1-t)=Sum of t^n for n>=0 is valid -1
@@nicolascalandruccio Actually yeah you can start with 1-(z-1)+(z-1)²-(z+1)³+... which sums to 1/(1-(-(z-1))) which is 1/z, and then take the integral of this, the integral of 1/z is log(z) and integrating this geometric series term-by-term gives the series for log I mentioned
Isn't ''multi-valued function'' an oxymoron? ;-)
That depends on how you define a function.
Some parts of math also say that a function must be defined on the entire domain, so a function having points where it is undefined is also an oxymoron. This is sometimes called a "partial function".
The generalized notion which allows undefined points and multiple values is sometimes called a "relation".
So we best just keep calling everything "function" since we understand what we mean
@@AssemblyWizard Maths shouldn't be about "let's just use this word and hope everyone understands what is meant here, although we use it with a meaning which contradicts its usual definition".
@@angelmendez-rivera351 of course that's the default, but look at the original comment you're commenting on. They asked if adding the prefix "multi-valued" to "function" is legit. Which I claim it is, like saying "vegan meat" is fine even though "meat" alone is specifically not vegan. The prefix can change the meaning of the term that follows
This was amazing - it felt like a lesson, taught to students in a class.
At 12:16 the condition tan(Arg(z))=Im(z)/Re(z) only determines Arg(z) up to the addition of π. For the full condition, use Arg(z)=atan2(Im(z),Re(z)).
But the (square root of 3) plus 1 is not the square root of 4.
he didn't say that - you have a right triangle with shorter sides of 1 and root 3 so the hypotenuse is 2
@@John-pn4rt that makes more sense thanks
lol you skewed the limits z === -1 + i.0 then the limits of Log is the same as -pi^+ is NOT IN THE domain so.... FIX YOUR LIMIT!!! { [ -pi^+ 2pi] E ( -pi, pi] }
Does anyone here actually like mathematics?
I do. I visit this channel for fun.
Read my username!!
i do math for fun everyday.
Yes, me for sure and probably others
kんkんklん
¿ por qué siguen los matemáticos hablando de imaginarias ? , están tratando con una simetría del espacio tiempo de 90 grados y darle el nombre de imaginaria crea un desvarío que hasta ahora sigue inventando cosas que no se puede explicar a la gente sencilla sin recurrir a cosas extrañas para el sentido común