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I find $Δ^n y(x)$ to be quite confusing and i dont know if i have understood it correctly. Does it mean simply raising the $Δy(x)$ difference to a power, or something else entirely? I know the formula of finding $Δ^n y(x)$ but im having trouble conceptualizing what it actually means.

The information about it on the internet is also very sparse, except for some advanced sources that my current math knowledge dont allow me to comprehend

Edit more context:

I saw it in a chapter which is about recursive functions. In short, it says that recursive equations and difference equations are equivalent. E.g. it says that $$\Delta^3 y(x)−2\Delta^2y(x)+5\Delta y(x)+7y(x)=3\cos x$$ can be expressed as $$y(x+3)−5y(x+2)+12y(x+1)−y(x)=3\cos x$$ But thats how the chapter starts and doesnt provide further context

Joe
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Than1
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  • Can you provide the context of where you have seen it? Do you have a particular equation it was used in? – Joe Sep 19 '21 at 11:49
  • I saw it in a chapter which is about recursive functions. In short, it says that recursive equations and difference equations are equivalent. E.g. it says that $Δ^3 y(x) -2Δ^2 y(x) + 5Δy(x) +7y(x)=3cosx$ can be expressed as $y(x+3) - 5y(x+2) + 12y(x+1) -y(x) = 3cosx$ But thats how the chapter starts and doesnt provide further context – Than1 Sep 19 '21 at 11:59

2 Answers2

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Just a guess, as your context is rather scarce, but I suppose this means: "to the nth difference". That means: $$\Delta y(x) = y(x+1) - y(x)$$ \begin{align} \Delta^2 y(x) &= \Delta y(x+1) - \Delta y(x) \\ &= y(x+2) - y(x+1) - y(x+1) + y(x) \\ &= y(x+2) - 2y(x+1) + y(x) \end{align} Which means that in recursive terms, $$\Delta^n y(x) = \Delta^{n-1} y(x+1) -\Delta^{n-1}y(x)$$

Now lets plug that in your equation: \begin{align} 3cos(x) &= \Delta^3 y(x) - 2\Delta^2 y(x) + 5\Delta y(x) + 7y(x) \\ &=(\Delta^2 y(x+1) - \Delta^2 y(x)) - 2(y(x+2) - 2y(x+1) + y(x)) \\& ~~~~+5y(x+1) - 5y(x) + 7y(x) \\ &= y(x+3) - 3y(x+2) + 3y(x+1) - y(x) - 2y(x+2) + 4y(x+1) - 2y(x) \\&~~~~+5y(x+1)+2y(x)\\ &=y(x+3)-5y(x+2)+12y(x+1)-y(x) \end{align} Which matches your example. Therefore it is fair to assume this definition is the correct one.

Petrus1904
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This is the "Nth Difference", similar to an "Nth Derivative". If, for example, y(x)= x^3 and we are focusing on integers, y(0)= 0, y(1)= 1, y(2)= 8, y(3)= 27, y(4)= 64, y(5)= 125, etc.

The "first differences" are $\Delta y(1)= 1- 0= 1$, $\Delta y(2)= 8- 1= 7$, $\Delta y(3)= 27- 8= 19$, $\Delta y(4)= 64- 27= 37$, $\Delta y(5)= 125- 64= 61$, etc.

The "second differences" are the differences of those. $\Delta^2 y(2)= \Delta y(2)- \Delta y(1)= 7- 1= 6$, $\Delta^2 y(3)= \Delta y(3)- \Delta y(2)= 19- 7= 12$, $\Delta^2 y(4)= \Delta y(4)- \Delta y(3)= 37- 19= 18$, $\Delta^2 y(5)= \Delta y(5)- \Delta y(4)= 61- 37= 24$.

The "Third differences are the differences of those- $\Delta^3 y(3)= \Delta^2 y(3)- \Delta^2 y(2)= 12- 6 = 6$,$\Delta^3 y(4)= \Delta^2 y(4)- \Delta^2 y(3)= 18- 12= 6$, $\Delta^3(5)= \Delta^2(5)- \Delta^2(4)= 24- 18= 6$.

Notice that all of those are the same, 6, so that the fourth, and then all succeeding, differences are 0. That is because this is a third degree polynomial.

user247327
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    You might want to note that you are using backward difference equations, whereas it seems that in the equations the OP posted, forward difference equations were used. – Joe Sep 19 '21 at 13:02