I'm unclear on what they are asking in this homework problem.

Suppose we know a function f(z) is analytic in the finite z plane apart from singularities at z = i and z=-1. Moreover, let f(z) be given by the Taylor series:

$$f(z)=\displaystyle\sum_{j=0}^{\infty}a_{j}z^{j}$$

where aj is known. Suppose we calculate f(z) and its derivatives at z = 3/4 and compute a Taylor series in the form

$$f(z)=\displaystyle\sum_{j=0}^{\infty}b_{j}\left(z-\frac{3}{4}\right)^{j}$$

1. Where would this series converge?
2. How could we use this to compute f(z)?
3. Suppose we wish to compute f(2.5); how could we do this by series method?

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Do they want the radius of convergence here? Or do they want some general statement about the function it would converge to, that is some kind of expression in z using aj?

Since we know the value of the function at all of the derivations when z=.75 I said that

f(.75) = a, f'(.75) = b, f''(.75) = c ... so now I have a sequence of constants: a, b, c, d...

But, I don't know how to proceed because I'm unclear on what they are asking. When they say "Where would this series converge?"

EnumaElish
Homework Helper
"What is the limit (sum) of the Taylor series?"

"What is the limit (sum) of the Taylor series?"

f(z)?

But, is there enough information here to say more about f(z) than that?

I'm just trying random things at this point. ratio tests says that it will converge when
$$\frac{\left| b_{n+1}\right|}{\left| b_{n}\right|}\left| z- \frac{3}{4}\right|<1$$

that's not saying much...

Is

$$b_{j}=\frac{f^{(j)}\left( \frac{3}{4}\right)}{j!}$$ ?

EnumaElish
Homework Helper
Based on the given info, I am guessing it is asking for a simple affirmation, e.g., "the series converges to function f(z)."

Avodyne
The answer to part 1 is, "inside a circle centered on z=3/4 with radius R=???" Can you figure out what R is?

part 2 is ambiguous. I think they want a general discussion of how to get the value of f(z) at any z other than i and -1.

Okay, if aj and bj are constants dependent on j but not z then the radius should be |z| < 1 for the first one so it stays the same and is 1, but centerted at 3/4ths.

How can we use the derivatives to compute the function? Well, if I'm right that

$$b_{j}=\frac{f^{(j)}\left( \frac{3}{4}\right)}{j!}$$

then if we divide each f(.75) = a, f'(.75) = b, f''(.75) = c, f'''(.75) = d ... by their respective j!-vaule.

$$\frac{a}{0!}+\frac{b}{1!}\left(z-\frac{3}{4} \right)^{1}+\frac{c}{2!}\left(z-\frac{3}{4} \right)^{2}+\frac{c}{3!}\left(z-\frac{3}{4} \right)^{3}+\cdots+\frac{f^{(j)}}{j!}\left(z-\frac{3}{4} \right)^{j}+\cdots$$ so that's how we could compute f(z) ?

HallsofIvy
Homework Helper
1. Where would this series converge?

Every power series converges within its radius of convergence so, yes, when they say "where does it converge" they are asking about the radius of convergence.

In general, a Taylor series for a function will converge "as long as it can"- as long as radius of convergence includes only points where the function is analytic. Here you are told that the function is analytic everywhere except at z= i and z= -1. Since your (new) Taylor's series is centered on z= 3/4, the radius will be the distance from 3/4 to the nearer of those. In particular, |i- 3/4|= 5/4 and |-1-3/4|= 5/4 so those two points are equi-distant from 3/4. The radius of convergence is 5/4 and the Taylor's series will converge for all z such that |z- 3/4|< 5/4.

2. How could we use this to compute f(z)?
Calculate z- 3/4 and use the series of course- as long as |z-3/4|< 5/4.

3. Suppose we wish to compute f(2.5); how could we do this by series method?
We can't using this series. The distance from 2.5 to 3/4 is 1.75> 3/4 and so 1.75 is not within the radius of convergence.

Thank you for the help, I'm still pretty confused, let me ask a few questions... Also, I see that I made a mistake in typing up the problem, it should be i and -i. not i and -1.

In general, a Taylor series for a function will converge "as long as it can"- as long as radius of convergence includes only points where the function is analytic.
Okay so the original sum:

$$f(z)=\displaystyle\sum_{j=0}^{\infty}a_{j}z^{j}$$

converges everywhere except at i and -i?

Here you are told that the function is analytic everywhere except at z= i and z= -1. Since your (new) Taylor's series is centered on z= 3/4, the radius will be the distance from 3/4 to the nearer of those. In particular, |i- 3/4|= 5/4 and |-1-3/4|= 5/4 so those two points are equi-distant from 3/4. The radius of convergence is 5/4 and the Taylor's series will converge for all z such that |z- 3/4|< 5/4.

There's no way to go around these points via analytic continuation? If the new version is restricted to a circle of radius of 5/4 then why was the original sum good everywhere except at i and -i? Or was I wrong about that and, because it's centered at zero, for the original one the radius would be 1?

2. How could we use this to compute f(z)?
Calculate z- 3/4 and use the series of course- as long as |z-3/4|< 5/4.
so take this:

$$\frac{a}{0!}+\frac{b}{1!}\left(z-\frac{3}{4} \right)^{1}+\frac{c}{2!}\left(z-\frac{3}{4} \right)^{2}+\frac{c}{3!}\left(z-\frac{3}{4} \right)^{3}+\cdots+\frac{f^{(j)}}{j!}\left(z-\frac{3}{4} \right)^{j}+\cdots$$

but only when |z-3/4|< 5/4?

I need to do another problem like this one becuse I still don't get it, is it the same with real numbers? Can I look at the examples on Taylor series in my advanced calc book to get this idea a little better?

Office_Shredder
Staff Emeritus
Gold Member
Thank you for the help, I'm still pretty confused, let me ask a few questions... Also, I see that I made a mistake in typing up the problem, it should be i and -i. not i and -1.

Okay so the original sum:

$$f(z)=\displaystyle\sum_{j=0}^{\infty}a_{j}z^{j}$$

converges everywhere except at i and -i?

No, it converges on a ball centered at 0 with the largest possible radius R such that neither i nor -i are in the ball. So it would have a radius of convergence of 1

There's no way to go around these points via analytic continuation? If the new version is restricted to a circle of radius of 5/4 then why was the original sum good everywhere except at i and -i? Or was I wrong about that and, because it's centered at zero, for the original one the radius would be 1?

You can extend the function to a larger region than a ball of radius 1, but at those points your original series wouldn't converge (note that Taylor series doesn't guarantee a series that converges to the function everywhere, just a series that converges to the function near a point)

I need to do another problem like this one becuse I still don't get it, is it the same with real numbers? Can I look at the examples on Taylor series in my advanced calc book to get this idea a little better?

It's similiar to real numbers, except you don't always "see" the poles that would bound your radius of convergence. For real numbers you'd use the ratio test, or be clever and look at the function as one on the complex plane (for example, 1/(1+x2) has no singularities in the set of real numbers, but has a radius of convergence of 1. Looking at it as a function of complex numbers, you see immediately that it has poles at +/-i). In general, a chapter about taylor series for real functions won't be very helpful when looking at taylor series for complex functions I think

HallsofIvy
Homework Helper
Thank you for the help, I'm still pretty confused, let me ask a few questions... Also, I see that I made a mistake in typing up the problem, it should be i and -i. not i and -1.

Okay so the original sum:

$$f(z)=\displaystyle\sum_{j=0}^{\infty}a_{j}z^{j}$$

converges everywhere except at i and -i?
No, I didn't say that. A power series converges everwbere INSIDE IT RADIUS OF CONVERGENCE. What is said was that the radius of convergence is the shortest distance from the point about which you are calculating the series and i and -i.

There's no way to go around these points via analytic continuation? If the new version is restricted to a circle of radius of 5/4 then why was the original sum good everywhere except at i and -i? Or was I wrong about that and, because it's centered at zero, for the original one the radius would be 1?
No, "analytic continuation" cannot take you past points where the fuction is not analytic!

so take this:

$$\frac{a}{0!}+\frac{b}{1!}\left(z-\frac{3}{4} \right)^{1}+\frac{c}{2!}\left(z-\frac{3}{4} \right)^{2}+\frac{c}{3!}\left(z-\frac{3}{4} \right)^{3}+\cdots+\frac{f^{(j)}}{j!}\left(z-\frac{3}{4} \right)^{j}+\cdots$$

but only when |z-3/4|< 5/4?

I need to do another problem like this one becuse I still don't get it, is it the same with real numbers? Can I look at the examples on Taylor series in my advanced calc book to get this idea a little better?

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Avodyne
No, "analytic continuation" cannot take you past points where the fuction is not analytic!

Um, I think that's misleading. The original series converges inside a circle of radius 1 centered at the origin. I can use it to get f(z) and all its derivatives at z = 3/4. From these, I can construct a new series that converges inside a circle of radius 5/4 centered at z = 3/4. I can now take a point inside this circle (but outside the original one), such as z = 1/2 + i, and compute f(z) and all its derivatives there. Then I can construct a new series that will converge inside a circle of radius 1/2 centered at z =1/2 + i. Continuing in this way, I can cover the whole plane, except for the points +i and -i.

Um, I think that's misleading. The original series converges inside a circle of radius 1 centered at the origin. I can use it to get f(z) and all its derivatives at z = 3/4. From these, I can construct a new series that converges inside a circle of radius 5/4 centered at z = 3/4. I can now take a point inside this circle (but outside the original one), such as z = 1/2 + i, and compute f(z) and all its derivatives there. Then I can construct a new series that will converge inside a circle of radius 1/2 centered at z =1/2 + i. Continuing in this way, I can cover the whole plane, except for the points +i and -i.

That makes sense, but in each case we're talking about a new series, not the same one, the radius is fixed for the original series right?

By the way: THANK YOU to everyone who has responded, I've found this really helpful!

Avodyne