3
$\begingroup$

Below, $\Delta$ means taking the derivative, $\frac{d}{dx}$. For $n\in\mathbb{Z}$, $n\geq 0$, we have $$\Delta^n\sin{x}=\sin{(x+n\tau/4)} \\ \Delta^n\cos{x}=\cos{(x+n\tau/4)}$$ I found that out while thinking about $\sin$, $\cos$ and $\Delta$. I understand what $\Delta$ does (or at least, I think I do), but I'm wondering if it works if you raise it to a negative power, and if you can do it then what does it even mean to raise it to a negative power?

I've heard that an integral is in some way the opposite to a differential, so maybe it would have something to do with integrals? I'd just like to know if you can raise $\Delta$ to a negative power and if so, what exactly it means to do so. I am not sure what tags would best suit this, so feel free to suggest some.

$\endgroup$
7
  • 2
    $\begingroup$ What is $\Delta$? What is $\tau$? $\endgroup$
    – anomaly
    Oct 29 '15 at 15:25
  • 1
    $\begingroup$ @anomaly, i think $\tau=2\pi$. it's this newfangled idea to use $2\pi$ as THE constant for circles/trigonometry, etc., with the idea that formulas are more compact and sensical, i.e., there are $\tau$ radians in a circle, Fourier integrals have a single factor of $\tau$ out front or in the argument, etc. $\endgroup$
    – rajb245
    Oct 29 '15 at 15:36
  • $\begingroup$ @anomaly $\Delta$ is the rate of change of the variable it is applied $\endgroup$ Oct 29 '15 at 15:39
  • $\begingroup$ Yeah, it seems $\Delta$ is $\frac{d}{dx}$ $\endgroup$
    – rajb245
    Oct 29 '15 at 15:42
  • $\begingroup$ @rajb245 I just prefer $\tau$ because it makes more intuitive sense to me :P $\endgroup$ Oct 29 '15 at 15:47
4
$\begingroup$

Short answer: The operation $\Delta^{-1}$ you want is indeed integration, but there are some complications to constructing it as a well-defined operator.

Longer answer: For a continuous function $f$ on an interval $[a, b]\subset \mathbb{R}$, we have \begin{align*} \frac{d}{dx}\int^x_a f(t)\, dt = f(x) \end{align*} for $x\in [a, b]$. (In particular, the integral above is differentiable.) This is the fundamental theorem of calculus, and it's usually the main result in a first-year calculus class. As you imply in your question, though, we can think of differentiation as a linear operator $\Delta:C^\infty(X) \to C^\infty(X)$, where $C^\infty(X)$ denotes the space of smooth (i.e., infinitely differentiable) functions $f:X\to \mathbb{R}$ for some fixed compact interval $X\subset \mathbb{R}$. Does this map have an inverse? That is, is there some operator $\Delta^{-1}$ such that $\Delta \Delta^{-1} f = f$ and $\Delta^{-1} \Delta f = f$ for all $f\in C^\infty(X)$?

Well, no: $\Delta (1) = 0$, so $\Delta^{-1} \Delta(1)$ must vanish by linearity. On the other hand, we know that the only functions $f\in C^\infty(X)$ with $\Delta f = 0$ are constants, so it turns out we can define $\Delta^{-1}$ up to a constant. The relation $\Delta \Delta^{-1} f = f$ still holds; the left $\Delta$ kills the arbitrary constant. On the other hand, we now have $\Delta^{-1} \Delta f = f + C$ for some arbitrary constant $C$. (Once you get to integration in your class, the teacher will inevitably nag you about adding in the "$+C$" to indefinite integrals. That's the reason why; they're only defined up to an additive constant.)

Another complication is the fact that we're working in $C^\infty(X)$ above, even though we can still apply $\Delta$ to functions that are only differentiable once. We don't need differentiability to integrate; we can define $\Delta^{-1} f$ perfectly well for any continuous $f$ (and, in fact, we can reduce that assumption even further). If $f$ is continuous, then $\Delta^{-1} f$ is differentiable, and $\Delta \Delta^{-1} f = f$; the expression $\Delta^{-1} \Delta f$ is not defined if $f$ is merely continuous rather than differentiable, though. If you're familiar with linear algebra, the source of the complication here is that we're working over an infinite-dimensional space rather than a finite-dimensional one, so injectivity and surjectivity are not equivalent.

$\endgroup$
2
$\begingroup$

Negative powers of a differential operator are possible, and can be well defined. Your intuition is correct, it works out to an integral. Use negative numbers in your formula, and you'll see that $\Delta^{-n}$ applied to sine or cosine gives you the $n^\text{th}$ integral of that sine or cosine. There are theoretical ways to show this, maybe using integral transforms like the Laplace transform, or fancy operator theory.

To summarize: yes, you can have negative powers of differential operators, and it means that you're applying the inverse operator of differentiation. You can prove that the inverse operator of differentiation is integration.

$\endgroup$
0
$\begingroup$

Generalized Differential for polynomials:

Given $$f\left(x\right)=\sum\limits_{\epsilon>0}^{1}\left(\sum\limits_{i\in\mathbb{Z}}^{}a_{i}x^{i+\epsilon}\right)$$ then $$f^{(k)}\left(x\right)=\sum\limits_{\epsilon>0}^{1}\left(\sum\limits_{i\in\mathbb{Z}}^{}\frac{\Pi\left(i+\epsilon\right)}{\Pi\left(i+\epsilon-k\right)}a_{i}x^{i+\epsilon}\right)$$

As anomaly noted, we need to adjust for the constant of integration, and those occur where $\Pi\left(n\right) = \Gamma\left(n+1\right)$ functions are undefined.

Edit: Forgot the parentheses to denote a kth derivative.

$\endgroup$
2
  • $\begingroup$ I'm having a hard time seeing how you answered the question. $\endgroup$ Jul 22 '16 at 18:10
  • $\begingroup$ The kth derivative could be any real number, including non-integers. I agree that my answer is still limited to power series and not all possible functions. $\endgroup$
    – Abraham Le
    Jul 22 '16 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.