# How to compute limit of removable singularity?

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• aheight
In summary: I need to finish this>>wDeriv=w'[t]==((-(D[theFunction,z]/D[theFunction,w])/.{z->myz[t],w->w[t]})(myz'[t]/w'[t]//FullSimplify)+(myz'[t]/w'[t] D[-(D[theFunction,z]/D[theFunction,w]),t]/.{z->myz[t],w->w[t]})/(D[theFunction,w]/D[theFunction,z]^2/.{z->myz[t],w->w[t]}));sol = NDSolve[wDeriv/.{w[0]==wstart,w'[0]==0},w,{t, t
aheight
I am given an implicit expression for an algebraic function, ##w(z)## as:##f(z,w)=(1-3 z+3 z^2-z^3)+(-4+8 z-4 z^2)w+(6-6 z)w^2+(-4)w^3+(1)w^4=0## with ##\displaystyle \frac{dw}{dz}=-\frac{\frac{df}{dz}}{\frac{df}{dw}}=\frac{6 w^2+8 w z-8 w+3 z^2-6 z+3}{4 w^3-12 w^2-12 w z+12 w-4 z^2+8 z-4}##. And note at the point ##(1,0)## we have: ##\frac{dw}{dz}=\frac{0}{0}##. However, I know ##w(z)=1+z^{1/4}+z^{1/2}+z^{3/4}## and so obviously the point ##(1,0)## is a removable singularity. In fact, ##\frac{dw}{dz}\biggr|_{(z=1)}=\{-1/2,-1/2+1/2 i, -1/2-1/2i,3/2\}##

But I only know this because I have the (fractional) polynomial solution and can compute the derivatives directly.

If I didn't know the solution but knew the point ##(1,0)## was a removable singularity, how could I compute:

##\lim_{(z\to 1, w\to 0)} \frac{dw}{dz}##?

I've thought about spiraling-in to the singularity numerically and maybe squeeze-out the (approx) limits but that process necessarily involves computing a numerical derivative which becomes increasingly indeterminate as I approach the singular point. I could try it I guess. Is numerically, the only way to find the limit if the solution is not known?

Perhaps I should also point out we can write the solution as:
##w(z)=\begin{cases}
1-z^{1/4}+z^{1/2}-z^{3/4} \\
1-i z^{1/4}-z^{1/2}+iz^{3/4} \\
1+iz^{1/4}-z^{1/2}-iz^{3/4} \\
1+z^{1/4}+z^{1/2}+z^{3/4}
\end{cases}
##
where we use the principal value of ##z^{1/4}## in the expressions.

Updated by mfb on request from aheight - July 2018

Last edited by a moderator:
Can l'Hopital's Rule help? According to this it is is still valid for complex numbers.

When I apply it to your ratio and take the limit as ##z\to 1## with ##w=0##, I get
$$\lim_{z\to 1}\frac{6z-6}{-8z+8}=\lim_{z\to 1}\frac{6(z-1)}{-8(z-1)}=lim_{z\to 1}-3/4=-3/4$$

On the other hand when I fix ##z=1## and take the limit as ##w\to 0##, l'Hopital gives me -1/2.

andrewkirk said:
Can l'Hopital's Rule help? According to this it is is still valid for complex numbers.

When I apply it to your ratio and take the limit as ##z\to 1## with ##w=0##, I get
$$\lim_{z\to 1}\frac{6z-6}{-8z+8}=\lim_{z\to 1}\frac{6(z-1)}{-8(z-1)}=lim_{z\to 1}-3/4=-3/4$$

On the other hand when I fix ##z=1## and take the limit as ##w\to 0##, l'Hopital gives me -1/2.

Thanks. I haven't reviewed, but I'm pretty sure, the limit over the complex plane cannot in general be determined by taking the limit in one direction; we'll get different values depending upon the path. For example, the value of -3/4 is not one of the values of the derivative of the functions w(z) at z=1. And all four are analytically-continuous at z=1.

Also, I'm having problems attempting to find the limits of the derivatives numerically. Need ##\displaystyle \frac{d^2 w}{dt^2} ## for a spiral path around 1 like for example ##z(t)=1+\frac{\cos (t)+i \sin (t)}{2 t}## and that ends up being very messy. This is what I get:

Given:

##f(z,w)=(1-3 z+3 z^2-z^3)+(-4+8 z-4 z^2)w+(6-6 z)w^2+(-4)w^3+(1)w^4=0## and ##z(t)=1+\frac{\cos (t)+i \sin (t)}{2 t}##, then we have:

##\frac{dw}{dt}=-\frac{\frac{df}{dz}}{\frac{df}{dw}} \frac{dz}{dt}## and so the function ##u(t)=\frac{dw}{dt}## is what I wish to investigate by numerically integrating it over an analytically-continuous path so I need ##\frac{du}{dt}## and so using the chain rule and product rule, I get:

## \frac{du}{dt}=\frac{d^2w}{dt^2}=\displaystyle-\biggr[ \frac{\frac{df}{dz}}{\frac{df}{dw}} \frac{d^2 z}{dt^2}+\frac{dz}{dt}\bigg\{\frac{\frac{df}{dw}\frac{d}{dt}\left(\frac{df}{dz}\right)-\frac{df}{dz}\frac{d}{dt}\left(\frac{df}{dw}\right)}{\left(\frac{df}{dw}\right)^2}\bigg\}\biggr]##

I think that's right so then I just start the integration on one of the four branch sheets and numerically integrate for say t=500 and I should be very close to ##\lim_{t\to\infty} \frac{dw}{dt}## and then I assume that I can then write ##\lim_{z\to 1}\frac{dw}{dz}\approx\lim_{t\to 500} \frac{\frac{dw}{dt}}{\frac{dz}{dt}}## to find the limit but not sure. However, my initial attempts to code it produces results not in agreement with the derivatives of the four functions above.

Anyone want to look at the expression and check if you agree with it?

Thanks.

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I just realized I can traverse the branch sheets of the function along an analytically-continuous path spiraling towards the point z=1, and then plug-in the (z,w) values along the way into the expression for ##\frac{dw}{dz}## above. This will give an analytically-continuous path along the derivative towards the four limiting cases at z=1. This is not at all difficult in Mathematica: Compute the derivative, numerically integrate it for say from t=(pi, 100) to obtain the analytic-continuation of the function starting at one of the four branch values, then use those values in the expression for ##\frac{dw}{dz}##. Lot easier than messing with the second derivative above. Here is the code to do this for the branch sheet having the limit of -1/2+1/2 i and a plot showing how the path over the derivative for the sheet, spirals into the limit. Similar results are obtained starting on the other branch sheets leading to the other three limiting cases.

Code:
theFunction=1-4 w+6 w^2-4 w^3+w^4-3 z+8 w z-6 w^2 z+3 z^2-4 w z^2-z^3
myz[t_]:=1+(Cos[t]+I Sin[t])/t;
startVals=w/.Solve[theFunction==0/.z->myz[\[Pi]],w]//Flatten

wstart=startVals[[4]];
tstart=\[Pi];
tend=100;

wDeriv=w'[t]==((-(D[theFunction,z]/D[theFunction,w]) D[myz[t],t])/.{w->w[t],z->myz[t]})

myazsol=First[NDSolve[{wDeriv,w[tstart]==wstart},w,{t,tstart,tend},MaxSteps->1500000,MaxStepSize->0.0001]];
myDerivTrace[t_]=Evaluate[Flatten[w[t]/.myazsol]];
theDeriv=(-D[theFunction,z]/D[theFunction,w])/.{z->myz[t],w->myDerivTrace[t]};

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##\displaystyle \frac{dw}{dz}=-\frac{\frac{df}{dz}}{\frac{df}{dw}}=\frac{6 w^2+8 w z-8 w+3 z^2-6 z+3}{4 w^3-12 w^2-12 w z+12 w-4 z^2+8 z-4}##

is not just a function of two independent variables z and w but must consider w as a function of z. That changes the limiting process: as z approaches 1, w approaches 0 by a functional dependence on z. That changes everything! We can see this by simply substituting the four expressions for w(z) into the expression for the derivative. For example, when I substitute ## w=1 - z^{1/4} + z^{1/2} - z^{3/4}## and simplify, I get: ##\frac{2 \sqrt[4]{z}-3 \sqrt{z}-1}{4 z^{3/4}}## and we see that:

##\lim_{z\to 1}\frac{2 \sqrt[4]{z}-3 \sqrt{z}-1}{4 z^{3/4}}=-1/2##

The remaining limits are obtained by substituting the other three expressions for w.

It seem then I must conclude: without knowing the functional dependence of w on z, the limit cannot be evaluated. Even using the numerical approach above, we are implicitly determining that dependence and then substituting the dependence into the expression for the derivative.

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Maybe a stupid idea, but if the functions w(z) and f(z,w) are sufficiently smooth (would have to be seen what exactly is required), then I would expect that the directional derivative of f is zero exactly if you hit the "right direction" - the direction of w(z). Those directions are linear relations between w and z, so you can plug them in and see when the derivative of f gets zero?

mfb said:
Maybe a stupid idea, but if the functions w(z) and f(z,w) are sufficiently smooth (would have to be seen what exactly is required), then I would expect that the directional derivative of f is zero exactly if you hit the "right direction" - the direction of w(z). Those directions are linear relations between w and z, so you can plug them in and see when the derivative of f gets zero?

Hi Mfb. Afraid I don't understand this. Can you elaborate this? I just don't see how it's possible to compute this limit (exactly) without knowing what w(z) is.

Thinking more about it: it won't work in general, but it could work for a large class of functions.

If your functions are sufficiently smooth, you can approximate ##w=1+\frac{dw}{dz} \cdot z##.

Let's assume for now that this approximation is exact, i.e. w and z have a linear relationship, and define ##c=\frac{dw}{dz}##. Then f is zero iff this equation is satisfied. You can calculate the directional derivatives of f.
If you calculate the 1-dimensional derivative of f along the line ##w=1## and evaluate it at z=0, you'll get some value. In general, this value won't be zero, because f(1,z)=0 only at z=0.
If you calculate the 1-dimensional derivative of f along the line ##z=0## and evaluate it at w=1, you'll get some other value. In general, this value won't be zero, because f(w,1)=0 only at w=1.
If you calculate the 1-dimensional derivative of f along the line ##w=z+1## and evaluate it at z=0,w=1, you'll get yet another value. In general, this value won't be zero, because f(z+1,z)=0 only at z=0.

If you calculate the 1-dimensional derivative of f along the line ##w=1+cz## and evaluate it at z=0,w=1, you'll get zero, because f(1+cz,z)=0. You have to find the direction where the derivative is zero.
By calculating the general 1-dimensional derivative and setting it to zero, you can find c if there is a unique solution. Otherwise you get multiple candidates for c.

Your w(z) is not linear, but if the functions are sufficiently smooth, I would expect that the approach still works.

The approach will fail in cases like f(w,z)=(w-z)2 where all 1D derivatives will be zero.

## 1. How do I identify a removable singularity in a limit?

A removable singularity in a limit can be identified by a discontinuity in the function that is removable by redefining the function at that point. This means that the function can be made continuous by filling in the hole at the singularity.

## 2. What is the process for computing the limit of a removable singularity?

The process for computing the limit of a removable singularity involves simplifying the function to eliminate the singularity, evaluating the limit at the point where the singularity was removed, and then checking if the limit exists at that point.

## 3. Can a removable singularity have a limit?

Yes, a removable singularity can have a limit as long as the limit exists at the point where the singularity was removed. This means that the function is continuous at that point and the limit can be evaluated.

## 4. How can I tell if the limit of a removable singularity exists?

The limit of a removable singularity exists if the function is continuous at the point where the singularity was removed. This can be determined by checking if the limit from both sides of the point is equal.

## 5. Are there any special cases when computing the limit of a removable singularity?

Yes, there are some special cases to consider when computing the limit of a removable singularity. For example, if the function is undefined at the point where the singularity was removed, the limit does not exist. Additionally, if the function is not continuous at that point, the limit cannot be evaluated.

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