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The derivation operator is $D=d/dx$, where we have: $D(f(x)g(x)) = fDg+gDf$.

Now while reading a book on difference equations, it occured to me that for the discrete operator, i.e $\Delta(y_n) = y_{n+1}-y_n$, we don't have the same identity, i.e. we have: $\Delta(z_n y_n) = z_{n+1}y_{n+1}-z_n y_n$, and $z_n\Delta y_n + y_n \Delta z_n = z_n(y_{n+1}-y_n)+y_n(z_{n+1}-z_n) = z_n y_{n+1}+y_nz_{n+1}-2z_ny_n$.

What other differences are there between these operators? I prefer those that aren't obvious, although if you were to ask me before calculating I would have said the same identity should be also true for the discrete case.

MathematicalPhysicist
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  • There is in fact a product rule $\ \Delta(f_n g_n ) = f_{n+1} \Delta g_n +(\Delta f_n), g_n,, $ and it often comes in handy, e.g. I employed it here to illustrate a 2-dimensional form of telescopy when computing the sum $,1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n., $ There is an entire theory of difference calculus that is a discrete analog of differential calculus. – Bill Dubuque Aug 05 '16 at 15:39
  • @BillDubuque yes I also noticed this identity, but it's not really a product rule, since it's different than the differential one. – MathematicalPhysicist Aug 06 '16 at 05:27
  • The difference doesn't prevent it from widely being calfed the product rule. – Bill Dubuque Aug 06 '16 at 11:55

2 Answers2

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They are pretty similar but at the same time slightly different due to the fact that:

$$\Delta y=y(x+1)-y(x)=\frac{y(x+1)-y(x)}{(x+1)-x}$$

The solution to the equation:

$$y=Dy$$

Is as we know $ce^x$.

The solution to the equation:

$$y=\Delta y$$

Is on the other hand,

$$y=c2^x$$

Given $f(x)=Dy$ then to find $y$ we integrate (a constant also appears when we integrate):

$$y= \int f(x) dx$$

Given $f(x)=\Delta y$ then to find $y$ we sum and add a corresponding constant for example,

$$\sum_{i=0}^{x-1} f(x)=\sum_{i=0}^{x-1} (y(i+1)-y(i))=y(x)+y(0) \implies y(0)+\sum_{i=0}^{x-1} f(x)=y(x)$$

We also have a pretty similar but different taylor series:

$$y(x)=\sum_{i=0}^{\infty} \Delta^i(0){x \choose i}$$

If we take $\Delta^0(0)=1$ and $\Delta^i(0)$ to represent $\Delta^i(f(x))$ when $x=0$.

The product rule similar as mentioned in the comments illustrates similarity but slight difference , but "the chain rule" is very different and more complex: Chain rule for discrete/finite calculus.

The product rule can also be written as,

$$\Delta(f(x)g(x))=g(x)\Delta(f(x))+f(x)\Delta(g(x))+\Delta(f(x))\Delta(g(x))$$

Here is the quotient rule which is again slightly different:

$$\Delta(\frac{f(x)}{g(x)})=\frac{g(x)\Delta f(x)-f(x)\Delta g(x)}{g(x)(g(x)-\Delta g(x))}$$

Community
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Ahmed S. Attaalla
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The Chain Rule is trivial for differences: $$\Delta (g\circ f)(c)= \Delta g\, (f(c)).$$

Peter Saveliev
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