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Maclaurin's series don't always work...?
By Brad Rodgers (P1930) on Friday, February 9, 2001 - 07:31 pm:

Why don't power series seem to work for imaginary numbers some of the time. e.g.

ln(-1) = -2 - 4/2 - 8/3 - 16/4 - ... = ip

I may have made a slip up with Maclaurin, but whatever the series is, it doesn't give a result that would give imaginary numbers for negative values. Why can't it deal with imaginary numbers? Are there other series like this (I realize that it isn't going to give all results for ln, as solutions can have an ik×2p added to them)

Thanks,

Brad

By Anonymous on Friday, February 9, 2001 - 07:52 pm:

Using the Maclaurin expansion for log(1+x) gives a series composed only of real terms. Yet for x = -2 log(1+x) = ip(1+2n)


Power series expanded about a point in the complex plane converge (and are thus valid) only in an (open) disc about the point. The radius of this disc is called the radius of convergence of the power series. The radius of convergence may not be defined, then the power series converges all of the complex plane. Obviously this is true for any polynomial, for example. The radius of convergence of a power series of a function about a given point is precisely the distance of that point from the nearest singularity of the function. For the Taylor series of log(1+x) the radius of convergence is 1, because we have a singularity at x = 0.

Roughly speaking a singularity is a point in the complex plane around which the function can be as large as we like. (i.e. the function goes to infinity)

By Michael Doré (Md285) on Friday, February 9, 2001 - 07:55 pm:

There could be other ways though of expanding out ln(-1) apart from substituting in x = -2 to the ln(1 + x) expansion (which of course doesn't work for the reason stated by Anonymous).

By Anonymous on Friday, February 9, 2001 - 07:59 pm:

In other words, if you want a Taylor expansion for log which is convergent at x = i. You must expand about a point closer to i than to 0. You could try this but doubtless the resulting series would be evil and you'd need to be a Professor of combinatorics to evaluate the coefficient of a general term.

By Brad Rodgers (P1930) on Friday, February 9, 2001 - 08:02 pm:

I'm still not sure I entirely understand the reason why this method doesn't work though.

By Michael Doré (Md285) on Friday, February 9, 2001 - 08:06 pm:

OK, so the problem is the one Anonymous states. Quite simply a Maclaurin series expansion won't always converge. Another example:

f(x) = 1/(1 + x)

So f(n)(x) = (-1)nn!/(1 + x)n+1

So using the Maclaurin series we get:

f(x) = 1 - x + x2 - ...

Now this is only convergent if modulus(x) > 1 yet 1/(1 + x) is defined everywhere except x = -1.

So the Maclaurin formula:

f(x) = f(0) + f'(0)x/1! + ...

only holds under certain conditions. I am not absolutely certain what these conditions are, but normally it is a safe bet that if the expansion converges, and the function is nicely behaved then the Maclaurin expansion will work.

(Obviously the definition of a "nicely behaved function" is one which the Maclaurin expansion holds for.)

By Anonymous on Friday, February 9, 2001 - 08:07 pm:

Taking liberties with notation at x = -1, log(1+x) = infinity. So the function log has a 'singularity' at 0 in the complex plane.

If you draw a circle about the point 1 (the point about which we are expanding the Taylor series) through 0 then the power series converges only within the disc drawn.

By James Lingard (Jchl2) on Friday, February 9, 2001 - 08:10 pm:

Or at least if it does converge, it won't converge to the correct value.

By Anonymous on Friday, February 9, 2001 - 08:12 pm:

I am not sure, but I don't think a series will converge at all outside its radius of convergence will it?

By Michael Doré (Md285) on Friday, February 9, 2001 - 08:18 pm:

James - I don't believe it does converge for x = -2. The series for ln(1 + x) is actually:

ln(1 + x) = x - x2/2 + x3/3 - ...

and if you substitute in x = -2, then the terms are tending to infinity.

By James Lingard (Jchl2) on Friday, February 9, 2001 - 08:21 pm:

I think you're right actually, I think they do always diverge outside the radius of convergence; at a point on the boundary they may or may not - is that right?

By Anonymous on Friday, February 9, 2001 - 08:25 pm:

Yeah, I reckon thats the way things are.

By Michael Doré (Md285) on Friday, February 9, 2001 - 08:28 pm:

That they may or may not converge on the boundary is certainly true. (For example 1/(1 - x) at x = 1, or ln(1 + x) at x = 1 do and don't converge respectively.) Proving the other statement looks like an interesting challenge...

By Michael Doré (Md285) on Friday, February 9, 2001 - 08:33 pm:

Aha - the ratio test gives a straightforward way of proving James' assertion.