# Square of a second derivative is the fourth derivative

I have a simple question for you guys, if I have this:

$$\left(\frac{d^2}{{dx}^2}\right)^2$$

Is it equal to this:

$$\frac{d^4}{{dx}^4}$$

Such that if I have an arbitrary function $f(x)$ I can get:

$$\left(\frac{d^2 f(x)}{{dx}^2}\right)^2 = \frac{d^4 f(x)}{{dx}^4}$$

Sorry if it's a pretty simple question, but I was trying to simplify something like this:

$$\left(a \cdot \frac{d^2}{{dx}^2} - f(x)\right)^2 g(x)$$ that's where it came up.

• It is true that $$\left(\frac{d^2}{dx^2}\right)^2=\frac{d^4}{dx^4}$$ What is not true is that $$\left(\frac{d^2}{dx^2}\right)^2$$ acting on $f(x)$ gives you $$\left(\frac{d^2 f(x)}{dx^2}\right)^2$$ Nov 23, 2015 at 14:25
• The first thing you state is true. But this: $$\left(\frac{d^2 f(x)}{{dx}^2}\right)^2 = \frac{d^4 f(x)}{{dx}^4}$$ is false. It's about like asking if $g(g(x))=(g(x))^2$ in general. Nov 23, 2015 at 14:26
• @Dylan So regarding the binomial, I should first square it before letting it act on the function outside, giving me $\frac{d^4 g(x)}{{dx}^4}$ for the first term, or is it still wrong?
– mopy
Nov 23, 2015 at 14:32
• @Aldon You have to be careful about how you square it, but otherwise yes, that should be fine. You can't use the "identity" that $(a+b)^2=a^2+2ab+b^2$, since that assumes that $a$ and $b$ commute, while it is not the case that $\frac{d^2}{dx^2}$ and $f(x)$ commute. Nov 23, 2015 at 14:35
• For anyone else who wondered what functions actually satisfy $f^{(2)}(x) = f^{(4)}(x)$, Wolfram Alpha gives the answer, in terms of the Weierstrass $\sigma$-function. I don’t know enough about differential equations to see how one might find that solution, other than by solving as a power series and recognising the result… Can anyone give a general heuristic that would cover this problem? Nov 24, 2015 at 15:50

It depends on what you mean by "square" and it's basically a problem of notation. When you write $\left(\frac{d^2}{{dx}^2}\right)^2$, implicitly the "square" means that you compose the operator $\frac{d^2}{{dx}^2}$ with itself, i.e. you consider $\frac{d^2}{{dx}^2} \circ \frac{d^2}{{dx}^2}$. This is of course equal to $\frac{d^4}{{dx}^4}$: differentiating four times is the same thing as differentiating twice then differentiating twice again. Applied to some function $f$, this then gives $\frac{d^2}{{dx}^2} \left( \frac{d^2 f(x)}{{dx}^2} \right) = \frac{d^4 f(x)}{{dx}^4}$, which is true.

On the other hand, where you write $\left(\frac{d^2 f(x)}{{dx}^2}\right)^2$, the "square" is implicitly multiplication, i.e. you're considering $\left(\frac{d^2 f(x)}{{dx}^2}\right) \cdot \left(\frac{d^2 f(x)}{{dx}^2}\right)$. This is not equal to the first thing, as simple counterexamples show (e.g. $f(x) = x^3$: $f''(x) = 6x$ thus $(d^4f)/(dx^4)(x) = 0$ while $((d^2f)/(dx^2)(x))^2 = (6x)^2 = 36x^2$). So you need to be careful with your notations.

• It's worth to note that for linear operators like differentiation, composition is, in a pretty solid sense, a multiplication operation – most obvious to see if you approximate the domain to a finite-difference mesh, because then the differential operator is just a matrix. The only reason this leads to confusion is that people keep mixing up functions with values of functions: multiplying two operators and then applying them to some argument is not the same thing as first applying each operator individually and then multiplying the results. Nov 23, 2015 at 20:48
• Of course, the way maths is usually written invites this kind of ambiguity. Especially unhelpful is ridiculous notation such as $\sin^2(x)$... Nov 23, 2015 at 20:54
• $\frac{d}{dx}$ is a operator or operation of differentiation?....@leftaroundabout @Najib Idrissi Aug 27, 2016 at 7:08
• @Najib "This is not equal to the first thing", what first thing are you referring to? Jun 28 at 22:34

I'm afraid not. Consider calculating the two quantities for

$$f(x)=x^3.$$

• So it's just equal to $\left(\frac{d^2}{{dx}^2}\right)^2$? Calculate the second derivative first before squaring it?
– mopy
Nov 23, 2015 at 14:25
• Check the reply of @dylan to your original post. Nov 23, 2015 at 14:27
• There is a difference between the second derivative of the second derivative of a function (the fourth derivative if it exists), and the square of the second derivative of a function (typically not the fourth derivative) Nov 24, 2015 at 0:15

No. If $f(x) = x^4$ then $\frac{d^4f}{dx^4} = 24$ whereas $\left(\frac{d^2f}{dx^2}\right)^2 = (12x^2)^2 = 144x^4$.

It turns out that the first fact you cited actually does apply. The difficulty appears to be confusion between the expressions $\left(a \cdot \frac{d^2}{{dx}^2} - f(x)\right)^2$ and $\left(a \cdot \frac{d^2}{{dx}^2} f(x)\right)^2$

The expression in parentheses has two terms, and expands like this:

\begin{align} \left(a \frac{d^2}{{dx}^2} - f(x)\right)^2 g(x) &= \left(a \frac{d^2}{{dx}^2} - f(x)\right) \left(a \frac{d^2}{{dx}^2} - f(x)\right) g(x) \\ &= \left(a \frac{d^2}{{dx}^2} - f(x)\right) \left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) \\ &= a \frac{d^2}{{dx}^2} \left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) - f(x)\left(a \frac{d^2}{{dx}^2} g(x)- f(x)g(x)\right) \\ &= a^2 \frac{d^4}{{dx}^4} g(x) - a \frac{d^2}{{dx}^2} \left(f(x)g(x)\right) - a f(x) \frac{d^2}{{dx}^2} g(x) + (f(x))^2 g(x) \end{align}