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A fourth degree polynomial p(x) satisfies $$Δ4P(0) = 24$$ $$Δ3P(0) = 6$$ $$Δ2P(0) =0$$ where $$ΔP(x) = P(x+1) - P(x)$$

Compute $$Δ^2P(10)$$

The answer is apparently 1140, but I do not know how to arrive at that answer. The furthest I got to was using the Newton Forward-Difference Formula and arriving to here:

$$ P(0) + ΔP(0)x + x(x-1)(x-2) + x(x-1)(x-2)(x-3) $$

but I don't know how to proceed from here.

Belphegor
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2 Answers2

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$Q=Δ^2P$ is a polynomial of second degree with $ΔQ=Δ^3P$ and $Δ^2Q=Δ^4P$. You now only need to interpolate $Q$ from the given data. Knowledge of $P$ is not required and not possible given the inputs.

Lutz Lehmann
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Here is your solution.
The formula for divided differences is: $ ΔP(x)=P(x+1)-P(x)$
As it is a degree-IV polynomial, we know that the fourth order differences will be same. Hence
$Δ^4P(0) =Δ^4P(1) =Δ^4P(2) = .....=Δ^4P(8)= 24.$
Now,
We know that $Δ^4P(0) = Δ^3P(1)-Δ^3P(0).$
$⟹ Δ^3P(1) = 24+6 = 30. $
Also, $Δ^2P(2) = Δ^3P(1) + Δ^2P(1) = 30 + 6 = 36. $
Similarly,$ Δ^3P(0)=Δ^2P(1)-Δ^2P(0) $
$⟹ Δ^2P(1) = 6+0 = 6. $
Similarly,$ Δ^4P(1)=Δ^3P(2)-Δ^3P(1) $
$⟹ Δ^3P(2) = 30+24 = 54. $
Continuing in the above pattern, we obtain all the values of $Δ^2P(i) ∀ i∈[0,10].$
Continuing in the above pattern, we obtain all the values of $Δ^3P(i) ∀ i∈[0,9].$
Now,$ Δ^3P(9) = Δ^2P(10)-Δ^2P(9) $
$⟹ Δ^2P(10) = Δ^3P(9) + Δ^2P(9) $
$⟹ Δ^2P(10) = 918+222 $
$⟹ Δ^2P(10) = 1140.$

Adarsh Krishna
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