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transformers revenge of the fallen actress - Anadea

📅 May 16, 2026 👤 scraface

May 16, 2026

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📌 Third differences: constant at $2 - 2 = 0$? Wait — only two second differences. Compute one more step.

📌 Assume third difference is constant at $d$. Since second differences increase by 2 each time, the first difference at $x=5$ should be $7 + 2 = 9$, so $f(5) = f(4) + 9 = 17 + 9 = 26$. Then the second difference becomes $9 - 7 = 2$, still consistent. So third difference is $2 - 0$? Wait — first second difference is 2, so next must be $2 + 2 = 4$, so $f(5) - f(4) = 4$, so $f(5) = 17 + 4 = 21$? Contradiction.

📌 But this is quadratic — yet it satisfies the values:

📌 $f(1)=2$, $f(2)=5$, $f(3)=10$, $f(4)=17$ — yes! But cubic? Yes, with leading coefficient zero. But the problem says cubic, so perhaps we assume non-degenerate. But the interpolation is unique for degree at most 3, and this fits.

📌 But wait — are there other cubic polynomials passing through these points? No — a cubic is determined uniquely by four points. So even if its technically quadratic, the cubic coefficient happens to be zero. So $f(5) = 26$

📌 But lets double-check with finite differences:

📌 Alternatively: third difference is 0, so second differences increase by 0, so next second difference is 7 + 0 = 7? No:

📌 Then $f(5) = f(4) + 9 = 17 + 9 = 26$

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