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Given $f(x) = \cos(x^5)$

And that $f'(0) = 0$

What does $f^{(90)}(0) = -\dfrac{90!}{18!}$ really tell us? (Note, my previous question solved for this value).

How can the rate of change of $f$ at $0$ be $0$, but the rate of change of the range of change of the rate of change of the rate of change...(90 times) end up being some ridiculously large negative integer?

How does that make sense and why is that information useful?

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user17753
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    Can your speed be zero, but still be accelerating? That's the case of $f'(0)=0$ and $f''(x)$ not being zero. Can your speed and acceleration be zero and your acceleration be changing? When you first hit the gas in your car, your acceleration was zero, and it changes, so $f'(0)=f''(0)$ but $f'''(0)\neq 0$. – Thomas Andrews Dec 18 '15 at 15:16
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    In general, the "real" measure of the size of your $n$th derivative is $\frac{f^{(n)}(0)}{n!}$. So in your case, that measure would be $\frac{1}{18!}$, which isn't that large... – Thomas Andrews Dec 18 '15 at 15:19
  • It´s just what you get when you calculate $f^{90};$ at 0. I dont get your question tbh – XPenguen Dec 18 '15 at 15:20
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    Note, notationally, we usually write $f^{(90)}$ for the 90th derivative, to distiguish it from other possible meanings of $f^{90}$. – Thomas Andrews Dec 18 '15 at 15:20

2 Answers2

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If you know the first $90$ derivatives at $0$, then you know something about the difference:

$$g(x)=f(x)-\sum_{k=0}^{90} \frac{f^{(k)}(0)}{k!}x^k$$

when $x$ is "close to zero." Specifically, you know that $\frac{g(x)}{x^{90}}\to 0$ as $x\to 0$.

Another way of saying this is that $h(x)=\sum_{0}^{90}\frac{f^{(k)}(0)}{k!}x^k$ is universally the best polynomial approximator for $f$ of degree $90$ or less near $0$ - that is, if you give me another polynomial $p$ of degree $90$ or less, there is some neighborhood of $0$ where $|f(x)-h(x)|\leq |f(x)-p(x)|$ for all $x$ in the neighborhood.

Also, consider the a question of units. If $f$ is a function of time returning a position in meters, then $f'(x)$ has units $m/s$, $f''(x)$ has units $m/s^2$, etc. $f^{(90)}$ has units $m/s^{90}$. So there is no point in comparing derivatives, because their values implicitly have different units. If you change your units of time from seconds to milliseconds, then $1 m/s^{90}$ is $\frac{1}{10^{270}}m/ms^{90}$, while $1 m/s$ only scales to $\frac{1}{1000}m/ms$. So trying to compare these values is actually at heart a mistake.

Let's look at an easier function:

$$f(x)=\frac{1}{1-x}$$

This has $f^{(n)}(0)=n!$. That's big. And since the power series for this function converges when $|x|<1$, we see that these terms do start adding up when $x$ gets nearer and near to $1$. If we write:

$$f(a+x)=\frac{1}{1-a-x} = \frac{1}{1-a}\frac{1}{1-\frac{x}{1-a}}$$

we see that $f^{(n)}(a)=\frac{n!}{(1-a)^{n+1}}$. So, as $a$ approaches $1$, these values are just getting terribly huge, until the function completely explodes.

Another way to think of it: Velocity is relative. In Newtonian physics, if we observe a particle moving, our frame of reference affects the measurement of that particle's velocity. But, at least if we aren't accelerating ourselves, that particle's acceleration measurement is exactly the same. Indeed, the exact measurements of the $n$th derivatives for $n>1$ really tells us nothing about the size of the first derivative, because we could change our frame of reference, measure exactly the same higher derivatives, but different velocities.

Thomas Andrews
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  • Can you explain why I divide $g(x)$ by $x^{90}$? And why the the "real" measure is only $\frac{1}{18!}$. I think that's the part I'm not fully getting. – user17753 Dec 18 '15 at 15:54
  • Ignore the "real value" being $\frac{1}{18!}$. There was a reason that phrase was placed in a comment, but the not in my answer, when I took time to reconsider and could edit. – Thomas Andrews Dec 18 '15 at 16:01
  • In other branches of mathematics, they would write $g(x)=o(x^{90)}$. It's just a way of estimating how well our power series for $f$ approximates $f$, and it just means that for any $\epsilon>0$ there is a $\delta>0$ such that if $|x|<\delta$ then $|g(x)|<\epsilon |x|^{90}$. Why is it $o(x^{90})$? That's a longer story, too long for a comment, but the real key is, we don't know how close to $0$ to get a small $\epsilon$. – Thomas Andrews Dec 18 '15 at 16:05
  • The bound $o(x^{90})$ could be very hard to get - you might need to take $\delta$ very small to satisfy, say, $\epsilon=1$. Then when $x$ is that small, $x^{90}$ is unbelievably small (compare, say, $x=\frac{1}{10}$ to $\frac{1}{10^{90}}$.) So even an actually large coefficient is not going to change $f$ much. – Thomas Andrews Dec 18 '15 at 16:05
  • For example, if we take $f(x)=\cos x$, then, near $0$, $f(x)=1-\frac{x^2}{2}$ is the best polynomial of degree $2$ or less to approximate $f$. Now, $p(x)=1$ is "better" when, say, $x=2\pi$, that $x$ isn't close to $0$. and we aren't making a universal claim, only that it is better that $p(x)$ at ever point in some neighbor hood of zero. – Thomas Andrews Dec 18 '15 at 16:24
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How can the rate of change of ff at 0 be 0, but the rate of change of the range of change of the rate of change of the rate of change... be non-zero.

A simple example helps to get an intuitive feel for this.

Consider a projectile fired directly upwards. At some point it reaches the top of it's climb and falls down.

Let $h(t)$ be the height at time h.

At the top of the climb the rate of change of $h(t)$ must be zero, so that $h(t)$ will stop decreasing.

So at the top of it's climb velocity is zero ( the rate of change of $h$ ).

But the rate of change of velocity has not become zero. It is still being subjected to gravity so it's rate of change is negative.

But the rate of change of velocity is also the rate of change of the rate of change of height, and that's negative at the same point the rate of change of height is zero.

So the rate of change and the rate of change of the rate of change do not have to be zero at the same time.

You can probably see how this can be extended to higher derivatives.

Of course with some functions they can both be zero, and any mix of $f^{(n)}(0) = f^{(k)}(0)$ are possible, but not necessarily the case.

For another viewpoint consider the Maclaurin Series of any function. Left as an exercise.

StephenG - Help Ukraine
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