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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(2): Partially done but help needed (Spoiler) | Comment 3 of 6 |
(In reply to re: Partially done but help needed (Spoiler) by JayDeeKay)

well its not a normal Bessel function if thats what the ;-) is refering to :-).  It is actually the modified Bessel function I(0,2*sqrt(1-x)) where I(a,x)=i^(-a)*J(a,i*x) with i=sqrt(-1) and J(a,x) the Bessel function.

So we end up with (3) being

i^(0)*J(0,2i*sqrt(1-x))=J(0,2i*sqrt(1-x))

and the series expansion of J(a,x) is the summation of

[(-1)^t/(t!*(t+a)!]*(x/2)^(2t+a)

so for J(0,x) we get

[(-1)^t/(t!^2]*(x/2)^(2t)

and with J(0,2i*sqrt(1-x)) we get

[(-1)^t/t!^2]*(i*(1-x)^(1/2))^2t

[(-1)^t/t!^2]*(-1)^t*(1-x)^t

[1/t!^2]*(1-x)^t

(-1)^t*(x-1)^t/[t!^2]

which agrees with the taylor series formulated from the derivative series given :-)

*edited to fix mistake made by forgetting Gamma(x)=(x-1)! and not x!  :-D  *

Edited on March 31, 2009, 10:11 pm
 Posted by Daniel on 2009-03-31 21:03:02

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