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 Derivative chains. (Posted on 2009-03-29)
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) )), ...

 No Solution Yet Submitted by Jer No Rating

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 Partially done but help needed (Spoiler) | Comment 1 of 6

With help from Taylor etc., but can anyone find a closed form for (3)?

(1) f(x) = 19 + 23(x - 1) + 32(x - 1)²/2! + 18(x - 1)³/3!
= 9 + 7x² + 3x³

(2) f(x) = 1 + 2x + 4x²/2! + 8x³/3! + ..
= 1 + (2x) + (2x)²/2! + (2x)³/3! + ..
= exp(2x)

(3) f(x) = 1 - (x - 1) + 1/2(x - 1)²/2! - 1/6(x - 1)³/3! + ..
...  + (1 - x)^r /r!  +  ...     Help
Perhaps my friend Bess'll help me with this.

(4) f(x) = x - x³/3! + x^5/5! - ..
= sinh x

(5) f(x) = ln2 +ln(2^ln2)x + ln(2^ln(2^ln2))x²/2! + ..
= ln2 + (ln2)²x + (ln2)³x²/2! + ..
= ln2[1 + xln2 + (xln2)²/2! + ..]
= ln2 exp(xln2)
= ln2(2^x)

 Posted by Harry on 2009-03-29 20:26:42

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