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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

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Keep trying | Comment 5 of 6 |
(In reply to Keep trying by Jer)

Oh dear, too much hyperbole with item (4), so we'll keep it simple and go for sin x instead. Thanks for the tip.

I can't see anything wrong with (5), unless you think it would be less ambiguous written as (2^x) ln2.

In (3), my allusion to Bessel resulted from the squared factorial in the denominator of each of the Taylor terms (even though I seemed to miss it out in my general term!). I was having trouble with the alternating signs, but Daniel's reply showed how a modified Bessel function solved this difficulty, and I agreed with his conclusion: Bessel I(0,2sqrt(1-x)). Perhaps we're missing something here.

 Posted by Harry on 2009-04-03 00:06:20

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