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Its a numerical analysis question where $\Delta$ is the forward difference operator. But although I have learned about the different operators I really can't understand the meaning of $0^m$. The answer options are

(a) 500 (b)515 (c)530 (d)540

I am not able to proceed because of the $0$ which I am not able to understand properly.

Xander Henderson
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Azka
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    The meaning of $0^6$ in the context of $\Delta^n0^m=\left.\Delta^nx^m\right|_{x=0}$ seems clear. It has little to do with the normal $0^6=0$. $\Delta^3f(x)=f(x+3)-3f(x+2)+3f(x+1)-f(x)$. – robjohn Sep 24 '19 at 15:43
  • While I agree with @robjohn that the notation is clear (more or less), I have voted to close this question for lacking context. Can you please share with us the source of this problem? – Xander Henderson Sep 25 '19 at 02:11
  • It is an ISS-Indian Statistical Serives Exam (Paper 1) question asked in the year 2016. As I am preparing for the exam next year, I am solving the previous year questions. – Azka Sep 26 '19 at 07:25

1 Answers1

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I don't understand the $0^m$ either, but we can still run through the thing symbolically. We know that $$\Delta^30^6 = \Delta^3 x^6\rvert_{x=0},$$ and this in turn is \begin{align*} \Delta^3 x^6 &=\Delta^2 [(x+1)^6 - x^6]\\ &=\Delta [((x+2)^6 - (x+1)^6) - ((x+1)^6 - x^6)]\\ &=[((x+3)^6 - (x+2)^6) - ((x+2)^6 - (x+1)^6)]\\&\qquad-[((x+2)^6 - (x+1)^6) - ((x+1)^6 - x^6)]\end{align*}evaluated at $x=0$ to get 540.

Still no clue what $0^6$ means in this context. Would be interested if anyone knows.

mostsquares
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  • My guess is that $\Delta^30^6=\Delta^nx^m|_{x=0, n=3, m=6}$ – B.Swan Sep 24 '19 at 15:40
  • Is that different from what I've put down? – mostsquares Sep 24 '19 at 15:42
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    Thanks a lot for your clarification. I think the way you have put it down is correct. 3 and 6 are just the values for n and m. – Azka Sep 24 '19 at 15:43
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    No, I just think that the meaning of $0^m$ in the term is that we evaluate the expression $x^m$ at $x=0$... Instad of writing down that we evaluate it, we just put it instead of $x$ in the expression. – B.Swan Sep 24 '19 at 15:45
  • Oh, now I see what you're getting at -- it's just convenient notation – mostsquares Sep 24 '19 at 15:47
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    The interval of difference you have taken as 1. Is that default? Obviously we won't reach the answer if we take it as h itself. But just needed to be sure. – Azka Sep 24 '19 at 15:47
  • Yeah, typically if $h$ is not specified $h=1$ is assumed – mostsquares Sep 24 '19 at 15:48