Is log(x) well-defined on [-1,1]?

[docnet]

Homework Statement

is log(x) well-defined on [-1,1]?

Relevant Equations

is log(x) well-defined on [-1,1]?

True/False: The logarithm is undefined on $[-1,0]$ and well defined on $(0,1]$. Thus log(x) is well-defined on $[-1,1]$.

1. True. Undefined does not mean not well defined.

2. False. Undefined means not well defined.

 

[DaveE]

Let's try this again...
Read this post about homework questions and ask us to help you learn to figure out the answer yourself.
Tell us what you think. What have you tried? What is confusing you about this?
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686783/

 

[Delta2]

In my opinion undefined is a special case of not well defined (not well defined is a generic notion and can mean other things too) so it is 2.False.

 

[HomogenousCow]

$\log(x)$ is multi-valued for negative values of $x$ and so I think it should be false.

 

[docnet]

In my opinion undefined is a special case of not well defined (not well defined is a generic notion and can mean other things too) so it is 2.False.

I read on Wiki that "A function that is not well defined is not the same as a function that is undefined. For example, if f(x) = 1/x, then the fact that f(0) is undefined does not mean that the f is not well defined – but that 0 is simply not in the domain of f." Does it mean that log(x) is defined on [0,1]?

$\log(x)$ is multi-valued for negative values of $x$ and so I think it should be false.

Could you please explain why log(x) is multivalued for negative values of x? what about the case when x is a complex number, is the logarithm undefined over $\{x|Re(x)<0\}$ or over the origin?

 

[Delta2]

I read on Wiki that "A function that is not well defined is not the same as a function that is undefined. For example, if f(x) = 1/x, then the fact that f(0) is undefined does not mean that the f is not well defined – but that 0 is simply not in the domain of f." Does it mean that log(x) is defined on [0,1]?

In my opinion f=1/x is undefined on 0 means the f is not well defined on R but f is well defined on R-{0}. It depends which set you consider the domain set in order to say if f is well defined or not well defined on that set.

 

[Office_Shredder]

What are the definitions of well defined and undefined in your class? Does it match Wikipedia?

 

[docnet]

In my opinion f=1/x is undefined on 0 means the f is not well defined on R but f is well defined on R-{0}. It depends which set you consider the domain set in order to say if f is well defined or not well defined on that set.

The question is in context of contour integration, while trying to use the fundamental theorem of calculus to evaluate $$\int_C \frac{1}{z}dz$$
where $$\mathcal C=\{z=(\cos(t),\sin(t))|0\leq t\leq 2\pi\}$$

And I was trying to see if the antiderivative $F(z)=\log(z)$ is well-defined on $\mathcal C$ because in the class notes my professor says the antiderivative should to be well defined over the contour to use the fact that if $\mathcal C$ is a closed curve, then the contour integral zero, but only if the antiderivative $F(z)$ is well behaved.

What are the definitions of well defined and undefined in your class? Does it match Wikipedia?

My class notes say that undefined means multivalued. I might be just making a big deal out of nothing, because we can still compute the integral using FTC even if $F(x)$ is not well behaved.

 

[vela]

And I was trying to see if the antiderivative $F(z)=\log(z)$ is well-defined on $\mathcal C$ because in the class notes my professor says the antiderivative should to be well defined over the contour to use the fact that if $\mathcal C$ is a closed curve, then the contour integral zero, but only if the antiderivative $F(z)$ is well behaved.

Well behaved where? On the entire complex plane, on the contour, or in some other region?

My class notes say that undefined means multivalued. I might be just making a big deal out of nothing, because we can still compute the integral using FTC even if $F(x)$ is not well behaved.

As you know, by definition, a function is single-valued. It unambiguously maps a point $a$ in its domain to a point $b$ in its codomain. If a "function" $f$ maps a point $a$ to two different points $b_1$ and $b_2$, it is not a well-defined function because you can't unambiguously say either $f(a) = b_1$ or $f(a)=b_2$. It's in that sense that $f(a)$ is undefined.

Consider the point $z=1$ in the complex plane. You can represent $z$ in polar form as $z_1 = e^{0i}$ or $z_2 = e^{2\pi i}$. What do you get for $\log z_1$ and $\log z_2$?

 

[docnet]

Well behaved where? On the entire complex plane, on the contour, or in some other region?As you know, by definition, a function is single-valued. It unambiguously maps a point $a$ in its domain to a point $b$ in its codomain. If a "function" $f$ maps a point $a$ to two different points $b_1$ and $b_2$, it is not a well-defined function because you can't unambiguously say either $f(a) = b_1$ or $f(a)=b_2$. It's in that sense that $f(a)$ is undefined.

Consider the point $z=1$ in the complex plane. You can represent $z$ in polar form as $z_1 = e^{0i}$ or $z_2 = e^{2\pi i}$. What do you get for $\log z_1$ and $\log z_2$?

I think the antiderivative has to be well behaved in the region enclosed by the contour. But, if the singular point(s) of $f(z)$ are all enclosed by the contour, as in the case with $f(z)=\frac{1}{z}$ then one can still evaluate the integral using the F.T.C using an involution about a circle...?

With $z=e^{2\pi i k}$ for integer values of $k$, $\log(z)$ has infinitely many values, i.e. $2\pi i k$, unless you define the polar form $z=r\cdot e^{i\theta}$ with $\theta=\theta mod(2\pi)$. Then $log(z)$ would be single valued everywhere except the origin.

 

[vela]

I think the antiderivative has to be well behaved in the region enclosed by the contour. But, if the singular point(s) of $f(z)$ are all enclosed by the contour, as in the case with $f(z)=\frac{1}{z}$ then one can still evaluate the integral using the F.T.C using an involution about a circle...?

Right, but you just need to be careful. For your problem, the contour begins and ends at $z=1$, but it would be wrong to say
$$\int_C \frac 1z dz = \log z \big\vert_{z=1}^{z=1} = 0.$$

With $z=e^{2\pi i k}$ for integer values of $k$, $\log(z)$ has infinitely many values, i.e. $2\pi i k$, unless you define the polar form $z=r\cdot e^{i\theta}$ with $\theta=\theta mod(2\pi)$. Then $log(z)$ would be single valued everywhere except the origin.

Correct again. With no conditions on $\theta$, $\log z$ isn't a well-defined function. Sometimes it's referred oxymoronically as a multi-valued function. You can deal with this using Riemann sheets, branch cuts, etc., to effectively make the complex log single-valued.

 

[docnet]

Right, but you just need to be careful. For your problem, the contour begins and ends at $z=1$, but it would be wrong to say
$$\int_C \frac 1z dz = \log z \big\vert_{z=1}^{z=1} = 0.$$Correct again. With no conditions on $\theta$, $\log z$ isn't a well-defined function. Sometimes it's referred oxymoronically as a multi-valued function. You can deal with this using Riemann sheets, branch cuts, etc., to effectively make the complex log single-valued.

Is it acceptable to write
$$\int_C \frac 1z dz = \log z \big\vert_{z_1}^{z_2} = \log(z_2)-\log(z_1).$$ and introduce the branch cut $z_2=z_1=1$?

Is the contour integral being zero related to the mean value theorem in PDE (the mean value theorem for the LaPlace equation) that says that the integrals of harmonic functions over ball boundaries are zero?

 

[vela]

Is it acceptable to write
$$\int_C \frac 1z dz = \log z \big\vert_{z_1}^{z_2} = \log(z_2)-\log(z_1).$$ and introduce the branch cut $z_2=z_1=1$?

The cut would start at the singularity at $z=0$ and extend along the $+x$ axis. That way the log is continuous along the entire contour except at the endpoints.

Is the contour integral being zero related to the mean value theorem in PDE (the mean value theorem for the LaPlace equation) that says that the integrals of harmonic functions over ball boundaries are zero?

No, because the integral isn't 0. Try parameterizing the contour as $z=e^{it}$ and evaluating it that way.

 

[docnet]

The cut would start at the singularity at $z=0$ and extend along the $+x$ axis. That way the log is continuous along the entire contour except at the endpoints.No, because the integral isn't 0. Try parameterizing the contour as $z=e^{it}$ and evaluating it that way.

If one introduces a branch cut that starts at the singularity at $z=0$ and extends along the $+x$ axis, then does it imply that the domain of t is $[0,2\pi]$ and $2\pi \not\equiv 0$? If that's true then $\log(e^{2\pi i})=2\pi i \neq 1 = e^0$ ?

Using this new information,

Evaluating the integral using the F.T.C:

$$\log(e^{2\pi i})-\log(e^0)=2\pi i-0$$

Check:

$$\int_0^{2\pi}f(z(t))z'(t)dt = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt=\int_0^{2\pi}idt=2\pi i $$

 

[docnet]

So if one parametrizes the unit circle using values of t such that $2\pi k\leq t\leq 2\pi (k+1)$ for any real valued k and define the branch cuts $0\leq r, t=2\pi (k+1)$, the integral is still equal to $2\pi i$. Do I still have the right idea? Thank you @vela.

 

[vela]

Yup, you got it.