For any polynomial function f(x) if you know f(a), f'(a), f"(a), f"'(a)... for some value of
a you can reconstruct the function. This is true even if the polynomial has an infinite number of terms.
(f' is the first derivative of f, f" is the second derivative etc.)
Define s(a) to be the sequence of f(a), f'(a), f"(a), f"'(a), ...
For each of the following, find the function:
(1) s(1) = 19, 23, 32, 18, 0, 0, 0, 0, ...
(2) s(0) = 1, 2, 4, 8, 16, 32, ...
(3) s(1) = 1, -1, 1/2, -1/6, 1/24, -1/120, 1/720, ...
(4) s(0) = 0, 1, 0, -1, 0, 1, 0, -1, ...
(5) s(0) = ln(2), ln(2ln(2)), ln(2ln(2ln(2) )), ...
Function 1 is all zeros after the fourth term, so it is a cubic polynomial.
F'''(x) = 18
F''(x) = 18(x+1) + 14
F'(x) = 9(x+1)^2 + 14(x+1)
F(x) = 3(x+1)^3 + 7(x+1)^2 + 9 = 3x^3 + 16x^2 + 23x + 9
Function 2 doubles each term. A funciton of the form e^(poly in x) seems obvious. F(x) = e^(2x) is the answer.
Funciton 3 has factorials of inverting sign, which looks like repeated differentiation of a negative exponent.
F(x) = 1/x differentiates as F'(x) = -1/x^2, F''(x) = 2/x^3, F'''(x) = -6/x^4, etc. Which makes the sequence 1, -1, 2, -6, 24, etc.
This is close, but everything is reciprocals of what is sought.
Function 4 is periodic with a repeat of 4 terms. Likely some polynomial in sin(x) and cos(x). The trivial F(x) = sin(x) is the answer.
Rewrite the terms in Function 5 as ln(2), (ln(2))^2, (ln(2))^3, etc. The ln(2) implies 2^(poly in x). F(x) = 2^x is the answer
4 of 5 solved
