# A difficult series expansion (finding a limit)

• TheSodesa
In summary: Well, gee-whiz.In summary, the program calculates the series of the whole quotient and indeterminates when x = 0.
TheSodesa

## Homework Statement

Find $\lim_{x \to 0}\frac{ln(1+x^2)}{1-cos(x)}$ by using series representations. Check using L'Hospitals rule.

## Homework Equations

Taylor polynomial at x=0: $\sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(x)^{k} = f(0) + f'(0)(x) + f''(0)x^{2} +...$

## The Attempt at a Solution

Using L'Hospital's rule gives me a limit of 2. Wolfram Alpha gives the following series expansion for the quotient: http://www.wolframalpha.com/input/?i=series+(ln(1+x^2))/(1-cosx), which obviously evaluates to 2 when $x \to 0$

What I don't get is how the program calculates the series. Differentiating the quotient given in the problem statement (to find out values of the derivatives to plug into the Taylor-formula) gives very complicated results that either equal zero or are indeterminate when x = 0.

For example, $\frac{d}{dx} \frac{ln(1+x^2)}{1-cos(x)} = \frac{2x(1+x^{2})^{-1}}{1-cos(x)} - \frac{ln(1+x^{2})sin(x)}{(1-cos(x))^{2}}$ where both terms are indeterminate when x = 0. The same thing happens with the second derivative: http://www.wolframalpha.com/input/?i=second+derivative+(ln(1+x^2))/(1-cosx)

So there are indeterminates where things would otherwise seem to evaluate to something that resembles 2, and zeroes everywhere else. It also doesn't help that f(0), the first term in the series, is again indeterminate.

I really have no idea how to approach this problem beyond what I've already tried (which in addition to this approach included plugging in the series for ln(1+x2) and 1-cos(x) in their respective places in the fraction). Could someone point me in the right direction?

Last edited:
TheSodesa said:
What I don't get is how the program calculates the series. Differentiating the quotient given in the problem statement (to find out values of the derivatives to plug into the Taylor-formula) gives very complicated results that either equal zero or are indeterminate when x = 0.

For example, $\frac{d}{dx} \frac{ln(1+x^2)}{1-cos(x)} = \frac{2x(1+x^{2})^{-1}}{1-cos(x)} - \frac{ln(1+x^{2})sin(x)}{(1-cos(x))^{2}}$ where both terms are indeterminate when x = 0. The same thing happens with the second derivative: http://www.wolframalpha.com/input/?i=second+derivative+(ln(1+x^2))/(1-cosx)
You don't need to evaluate the series of the whole quotient. First write down each of the Taylor expansion of the numerator and denominator separately.

blue_leaf77 said:
You don't need to evaluate the series of the whole quotient. First write down each of the Taylor expansion of the numerator and denominator separately.
I tried doing that as well, but a similar issue arises:

If

around x = 1: $ln(1+x^{2}) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}(k-1)!\cdot x^{2k}}{k!} = -\sum_{k=1}^{\infty} \frac{(-1)^{k}(k-1)!\cdot x^{2k}}{k!} =-\sum_{k=1}^{\infty} \frac{(-1)^{k}\cdot x^{2k}}{k}$

and

around x = 0: $cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}$

then

$\lim_{x \to 0} \frac{ln(1+x^{2})}{1-cos(x)} = \lim_{x \to 0} \frac{ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}x^{2k}}{k}}{1 - \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}} = \lim_{x \to 0} \frac{ -\sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{k}}{1 - \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}} = \lim_{x \to 0} \frac{ -\sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{k}}{1 - ( 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!})}$
where the numerator and denominator still both evaluate to zero (since I can't just cancel the xs because of the series notation). Therefore I'm still stuck with an indeterminate form.

It may be more transparent if I write the expansions explicitly for first few terms
$$\frac{\ln(1+x^{2})}{1-\cos(x)} = \frac{ x^2-\frac{x^4}{2}+\frac{x^6}{3}-\cdots }{ \frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}\cdots}$$
Now what happens if you factor out ##x^2## from the numerator and denominator?

blue_leaf77 said:
It may be more transparent if I write the expansions explicitly for first few terms
$$\frac{\ln(1+x^{2})}{1-\cos(x)} = \frac{ x^2-\frac{x^4}{2}+\frac{x^6}{3}-\cdots }{ \frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}\cdots}$$
Now what happens if you factor out ##x^2## from the numerator and denominator?

Well, gee-whiz.

If you factor out x2 and take the limit as $x \to 0$, we are left with $\frac{1}{\frac{1}{2!}} = \frac{2!}{1} = 2$.

Thanks a bunch. I'm not used to working with series notation and have been simply thrown in the middle of it all. I should probably start writing the first few terms of a series whenever I'm faced with a problem like this.

## 1. What is a difficult series expansion?

A difficult series expansion is a mathematical technique used to find the limit of a function by representing it as an infinite sum of simpler functions. This method is often used when traditional algebraic methods fail to find the limit.

## 2. How do you perform a difficult series expansion?

To perform a difficult series expansion, you first need to identify the function and the value you want to find the limit of. Then, you need to express the function as an infinite sum of simpler functions, using known series expansions such as Taylor or Maclaurin series. Finally, you can manipulate the series to find the limit by taking the limit of each term in the series.

## 3. When is a difficult series expansion necessary?

A difficult series expansion is necessary when traditional algebraic methods, such as direct substitution or factoring, fail to find the limit of a function. This usually happens when the function is complex or involves variables that approach infinity.

## 4. What are the limitations of a difficult series expansion?

A difficult series expansion can only be used to find the limit of a function if the series converges. This means that the series must approach a finite value as the number of terms approaches infinity. Additionally, the series must also have a finite number of terms, which may not always be the case for complex functions.

## 5. How do you know if a difficult series expansion is accurate?

The accuracy of a difficult series expansion can be determined by comparing the obtained limit with the actual limit, if it is known. If the two values match, then the series expansion is accurate. However, if the two values differ significantly, it may be necessary to use more terms in the series to improve the accuracy of the result.

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