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) )), ...
(1) f(x) =3x^3 + 16x^2 + 23x + 9
(2) f(x) = a^(2x)
(3)if f(x)=1/x, then, we have f(x) differentiates as: f'(x) = -1/x^2, f''(x) = 2/x^3, f'''(x) = -6/x^4, f''''(x) = 24//x^5, which makes the sequence 1, -1, 2, -6, 24, -120, 720,
We take the reciprocals of these to get to the required sequence.
(4) the trivial sequence.f(x) = sin x is the required sequence.
(5) f(x) = 2^x
Edited on March 8, 2024, 10:11 am
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