How come we can compute the Taylor series for the function below about x = 0, despite the fact that it is not defined at x = 0?

$\displaystyle \frac {\cos (2x) -1 }{x^2}$

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- Jan 13th 2013, 08:17 AMMathCrusaderTaylor series: (cos(2x)-1)/x²
How come we can compute the Taylor series for the function below about x = 0, despite the fact that it is not defined at x = 0?

$\displaystyle \frac {\cos (2x) -1 }{x^2}$ - Jan 13th 2013, 08:52 AMPlatoRe: Taylor series: (cos(2x)-1)/x²
- Jan 13th 2013, 10:03 AMMathCrusaderRe: Taylor series: (cos(2x)-1)/x²
- Jan 13th 2013, 10:24 AMPlatoRe: Taylor series: (cos(2x)-1)/x²
- Jan 13th 2013, 10:52 AMSworDRe: Taylor series: (cos(2x)-1)/x²
The principle behind your question is that the discontinuity at x=0 is a removable one. This means that it is defined everywhere else

*and*that we can "fill in" the value at x=0 with a value that it is "supposed to" have, and then it will be continuous. For example, consider the simple example:

$\displaystyle f(x) = \frac{x^2}{x}$

The function is not defined at x = 0 yet clearly everywhere else it is equal to just x, so the taylor series is x.

A less trivial example is:

$\displaystyle f(x) = \frac{\sin(x)}{x}$

The function is not defined at x=0. However, the taylor series can be obtained by using the series for sin(x), and dividing everything by x.

$\displaystyle \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...$

$\displaystyle \frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} ...$

Your example is fundamentally the same. Can you get the series for cos(x) and then perform the necessary steps? - Jan 13th 2013, 10:56 AMSworDRe: Taylor series: (cos(2x)-1)/x²
- Jan 13th 2013, 11:04 AMPlatoRe: Taylor series: (cos(2x)-1)/x²
- Jan 13th 2013, 11:14 AMMathCrusaderRe: Taylor series: (cos(2x)-1)/x²
Thank for that expository answer. Please note that I am not having trouble

*finding*the Taylor series, but rather understanding the rationale behind doing it in spite of the function not being defined at x = 0.

Your answer was illuminating indeed, but I wonder*why*the discontinuity would be a removable one. We are certainly not working with limits (or are we?). - Jan 13th 2013, 11:29 AMSworDRe: Taylor series: (cos(2x)-1)/x²
The original example is the same in principle as finding the taylor series of $\displaystyle f(x) = \frac{x^2}{x}$. The only difference is that it is masked/obfuscated by trigonometric functions. If you understand why this simpler function x^2/x has a taylor series even though its undefined, you should understand your example as well.

It does have something to do with limits: if the discontinuity is removable, that means the limit at that point exists, and the value of the limit is precisely what the value of the function is "supposed" to be.

If you want a more complicated but informative response, the discontinuity was removable because cos(2x)-1 has a zero of order 2 at that point, therefore you can divide out an x^2. - Jan 13th 2013, 11:36 AMfarzeroRe: Taylor series: (cos(2x)-1)/x²
Well I think that because a taylor series attempts to create a function by a limit as the number of terms approaches infinity, we are indeed working with a limit. However, I am still slightly skeptical as to why, even though the discontinuity is indeed removable, thus leaving a "hole" in the graph, if you will, it is ok to "fill in" the value. I would investigate the radius of convergence of this series and see if x=0, perhaps being in it, has something to do with the fact that we can "fill it in." The disk around x=0 might have some connectivity property. I'm not sure, sorry if my answer isn't too precise. I'm wondering myself.

- Jan 13th 2013, 11:54 AMSworDRe: Taylor series: (cos(2x)-1)/x²
No you are thinking the wrong way. Removable discontinuities are actually quite harmless. They aren't truly "supposed" to be there, but are merely the result of the way we formally define functions.

It is ok to fill in the graph because we can imagine a different function: one that is the same for all values*except*x=0. That is:

$\displaystyle f(x) \;=\; \begin{cases} \frac{cos(2x)-1}{x^2} & \text{if }x \ne 0 \\ -2 & \text{if }x=0\end{cases}$

THIS function satisfies all the criteria of having a Taylor series, being defined and differentiable at that point. And since it is coincident to the original function at all points except x=0, it is in effect a series for that function. Its radius of convergence is also infinite, by the way, because there are no actual singularities.

Notice also, the -2 is not arbitrary. That is the value of

$\displaystyle \lim_{x \to 0}\frac{\cos(2x)-1}{x^2}$

If I put any number other than -2, f(x) would be discontinuous and fail to satisfy the requirements for having a taylor series.