# Taylor Expansion of Natural Logarithm

1. Jan 10, 2013

### golanor

Hello!
I was trying to look for a possible expansion of the ln function. The problem is, that there is no expansion that can be used in all points (like there is for e, sine, cosine, etc..)
Why do you think that is?
To clarify:
Let's say i do the MacLaurin expansion of ln(x+1):
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}...+(-1)^{n}\frac{x^n}{n}$$
And around 10:
$$\text{Log}[11]+\frac{x-10}{11}-\frac{1}{242} (x-10)^2+\frac{(x-10)^3}{3993}-\frac{(x-10)^4}{58564}+\frac{(x-10)^5}{805255}...$$

Are there any other (smooth) elementary functions which cannot be expanded to a taylor series?

2. Jan 10, 2013

### lurflurf

That is because of a pole at zero. Other elementary functions with poles like 1/sin(x) and 1/(x^2+1) do not have Taylor series valid everywhere.

3. Jan 10, 2013

### Staff: Mentor

In general, functions with any pole in the complex numbers cannot have a Taylor series everywhere (not even for all real numbers). The point where you calculate the expansion has a fixed distance to the closest pole, and the convergence radius cannot be larger than this distance. Even worse, you need an infinite set of taylor series to "cover" all points in at least one series ;).

4. Jan 10, 2013

### golanor

But the power series converges for any expansion i do for both $$\frac{1}{\sin x}$$
and for: $$\frac{1}{1+x^{2}}$$
The power series for the taylor expansion of ln doesn't converge.

5. Jan 10, 2013

### HallsofIvy

Staff Emeritus
Yes, ln(x) has a singularity at x= 0. No, it is NOT a "pole", it is an "essential singularity". The difference is that the Laurent series for a function about a pole has a finite number of negative powers- for some n, multiplying by $(z- z_0)^n$ gives an analytic function. The Laurent series for a function about an essential signgularity has an infinite number of negative powers and is much harder to work with.

6. Jan 10, 2013

### Staff: Mentor

@golanor: In the full real or complex numbers, with a Taylor series? I doubt that.
A Laurent series is more powerful, but it cannot have more than a single singularity (or no singularity -> Taylor series) to be convergent everywhere. Your functions have more than one singularity.

7. Jan 10, 2013

### golanor

So, the ln function doesn't have a taylor series expansion that fits the function on all points because of something in complex analysis?
In other words, no infinite series of polynomials can be plotted to fit the ln function, because it has an essential singularity?
Maybe it's because i have yet to study complex analysis, but i'm not seeing the connection.
@mfb - i'm talking about real numbers. I'm not fluent in complex analysis.

8. Jan 10, 2013

### Staff: Mentor

You can show this in real analysis as well, as the singularity at x=0 is in the real numbers and the concept of a convergence radius applies to real numbers as well.

It cannot fit the ln function in the whole (or whole positive) real numbers. It can give the ln function in some open interval (0,a) for any a, if you develop the taylor series around a/2.

There is no Taylor series for $\frac{1}{1+x^2}$ which covers all real numbers.
You can show this easily with complex analysis, but the statement is true for both.

There is no Taylor series for $\frac{1}{sin(x)}$ which covers all real numbers except integer multiples of pi. You can show this in the real numbers.

9. Jan 10, 2013

### golanor

So the mathematical proof of the fact that there is no possible taylor series expansion for ln(x+1) in the open interval (0,∞) is because there is a singularity in x=-1?
But if i expand the series around a different point, say, 9, the power series converges on the interval [4,19], so x=0 doesn't play a role here. Why isn't there some point $$x_{0}$$ that if we expand the series there the radius of convergence for the power series will be R=∞?
i'm not saying that for every point ln (x+1) will equal that power series, just that the power series we get from using taylor expansion will always converge.

10. Jan 10, 2013

### Staff: Mentor

Every power series, for any point x0, has a convergence radius R (this can be 0 or "infinity" or any positive real number). It converges within this radius, in the open interval ( x0-R, x0+R ). Its convergence is unclear at the borders (x0-R and x0+R) and has to be checked manually there. It diverges everywhere else.

Therefore, if you calculate the power series of ln(x) around x0 and require that it converges to ln(x)*, the power series has to diverge at x=0 (together with the log) and the maximal convergence radius is x0. This gives a maximal** interval with convergence: ( 0 , 2x0 ). It could be convergent for 2x0, too, but it has to be divergent everywhere else.

*In complex analysis, we can drop this requirement: If the function is analytic ("differentiable"), the power series converges to the function itself in its convergence radius.

**If you use complex analysis again, the maximal radius happens to be the actual convergence radius.

ln(x+1) is the same as ln(x), just shifted by 1:
If you expand ln(x+1) at x=0, you get convergence in (-1,1).
If you expand ln(x) at x=1, you get convergence in (0,2).
If you expand ln(x) at x=100, you get convergence in (0,200).

11. Jan 10, 2013

### golanor

If you expand at x=100, you get convergence in (0,200), but it doesn't converge to ln(x) around (0,10), for example.
Obviously, ln(x) converges only in positive intervals.
I was just wondering what is the reason it doesn't converge in real-number analysis. not ln, but a power series, which in some interval also converges to ln (but no necessarily always converges to ln)

12. Jan 10, 2013

### Staff: Mentor

(0,10) is a subset of (0,200), it converges to ln(x) there as well.

Edit:
WolframAlpha evaluation at x=1, the deviation from 0 is a rounding error.
WolframAlpha evaluation at x=2, giving ln(2).

Last edited: Jan 10, 2013
13. Jan 10, 2013

### golanor

Series expnasion for ln(x) at x=100:
$$\text{ln}(x) = ln(100)-\sum _{n=1}^{\infty } \frac{(100-x)^n}{n*100^n}$$
(The (-1)^n is ignored because i switched the top part.)

You are right. it does converge only for (0,200) and always to ln(x) in that interval.
Thanks alot, this really helped!

14. Jan 13, 2013

### golanor

I just realized something - can we even talk about a taylor series in this case?
The Lagrangian remainder doesn't approach 0 as x approaches 0 (or even 1).
The interval in which the remainder approaches 0 is (50,200), and not (0,200).

15. Jan 13, 2013

### Staff: Mentor

Why? The lagrange remainder just states "there is a $\xi$ in the interval [...], it does not tell you where this $\xi$ is. For $n \to \infty$, it goes to zero.

16. Jan 13, 2013

### golanor

That's the thing.
$$\lim_{n\to \infty } \frac{(x-100)^n}{c^{n+1}*(n+1)}$$
goes to 0 only for x≥50. x≤c≤100
for x<50 it cannot be determined.

17. Jan 13, 2013

### Staff: Mentor

That c is not constant, it can go to 100.
I don't see why x=50 should be special here.

18. Jan 13, 2013

### golanor

That's why it is undetermined and not infinity.
x=50 is the smallest x for which the limit will always be 0.
anything lower than 50, and the nominator could be larger than the denominator => the limit will diverge.

19. Jan 13, 2013

### Staff: Mentor

"It could be larger" => you cannot use the general version of the lagrangian error to show convergence
It converges, but you cannot use this way to show it.

With c arbitrary small, it would diverge for other x as well, by the way.

20. Jan 13, 2013

### golanor

yes, but since c is between x and 100, if x is bigger than 50 or smaller than 200 it won't diverge for any x and for any c.
when i pick x=1, for example, for 1<c<99 the limit diverges.
you cannot talk about a series in that case, because the remainder does not necessarily converge to 0.

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