let $$f(x) = \begin{cases} \frac{\cos x -1}{x^2} & \text{for } x \neq 0 \\ \\ \\ -\frac{1}{2} & \text{for } x = 0 \end{cases} $$

The Taylor series for this is

$$\dfrac{1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \ldots + \left(-1\right)^n \dfrac{x^{2n}}{\left(2n\right)!} - 1} {x^2} $$

$$\Rightarrow -\dfrac{1}{2!}+\dfrac{x^2}{4!}-\dfrac{x^4}{6!}+ \ldots + \left(-1\right)^{n-1}\frac{x^{2n}}{\left(2n+2\right)!}$$

Now determine whether $f$ has a relative maximum, relative minium, or neither at $x = 0$.

$f^{\prime}(0) $ is the first derivative of $-\frac{1}{2}$ which is zero.

Here is where my confusion lies:

1) The second derivate of $f\left(x\right)$ is $\dfrac{-(x^2-6) \cos(x)+4 x \sin(x)-6}{x^4}$ and when you put in zero for $x$, you get $0/0$ which is undefined, thus because of the cases it should be the second derivative of when $f(x)$ is defined at zero which is a constant, thus zero.

2) From the Taylor series: $\dfrac{f^{\prime\prime}(0)}{2!} = \dfrac{1}{4!}$ which is greater than zero.

Which way is the proper method for understanding the problem.


Our function, as you observed, is given by the series $$ -\dfrac{1}{2!}+\dfrac{x^2}{4!}-\dfrac{x^4}{6!}+ \cdots.$$ For $x$ not far from $0$, in fact for $0\lt |x|\le 1$, this is an alternating series.

So if we truncate at the term $-\dfrac{1}{2!}$, the error is positive, that is, our function is $\gt -\dfrac{1}{2!}$ for $x$ near $0$ but not equal to $0$. It follows we have a local minimum at $x=0$.

Remark: The "second derivative test" is often not the best way to determine whether we have a local max or min at a certain point. But if you want to use the second derivative, take advantage of the fact you have a well-behaved series for your function, and differentiate term by term. The first derivative of our function at $0$ is $0$. Thus there is a critical point at $x=0$.

The second derivative at $x=0$ is $\frac{2}{4!}$. This is positive, and you can derive the conclusion that there is a local minimum at $x=0$ from that.

  • $\begingroup$ 1) I don't exactly understand what you mean by `truncate at the term $-\frac{1}{2}$. 2) About the error being positive and the function being positive at that point, always means that its a local minimum? Sorry, if my question are convaluted, my knowledge level is low. $\endgroup$ – yiyi Apr 16 '13 at 4:44
  • 1
    $\begingroup$ We often use partial sums of a series as to approximate the full sum. By truncate at the term $-\frac{1}{2!}$ we mean that we do not add any other terms. In alternating series, the error when you do that has absolute value less than the absolute value of the first "neglected" term. The sign of the first neglected term tells you whether your truncated estimate is too small (that happens when the first neglected term is positive) or too big (first neglected negative). This is very useful information about alternating series. But the answer above also has a derivative-oriented approach. $\endgroup$ – André Nicolas Apr 16 '13 at 5:12

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