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Nevertheless
Posts:
28
From:
Norman, Oklahoma
Registered:
3/11/08


Re: Auto composition yields exponential
Posted:
Aug 11, 2010 6:27 PM


> Rufus Isaacs, "Iterates of fractional order", > Canadian > Journal of Mathematics 2 #4 (1950), 409416. > http://books.google.com/books?id=fQU06O9d0wC&pg=PA409
This reference appears to be what is desired; however, as I work through it to see how it applies, I will show the forum something that I have done which might lead to a solution:
Assuming that g(x) exists and that it has a Maclaurin series:
g(x) = a + bx + cx^2 + dx^3 + ...
g(x)^2 = aa + abx + acx^2 + adx^3 + ... + abx + bbx^2 + bcx^3 + ... + acx^2 + bcx^3 + ... + adx^3 + ...
= aa + 2abx + (2ac+bb)x^2 + 2(ad+bc)x^3 + ...
g(x) = a + bx + cx^2 + dx^3 + ... (rewritten here to aid in multiplication)
g(x)^3 = aaa + 2aabx + (2aac+abb)x^2 + (2aad+2abc)x^3 + ... + aabx + 2abbx^2 + (2abc+bbb)x^3 + ... + aacx^2 + 2abcx^3 + ... + aadx^3 + ...
= aaa + 3aabx + 3(aac+abb)x^2 + (3aad+6abc+bbb)x^3 + ...
g(g(x)) = a + bg(x) + cg(x)^2 + dg(x)^3 +
= a + + ab + bbx + bcx^2 + bdx^3 + ... + aac + 2abcx + (2acc+bbc)x^2 + 2(acd+bcc)x^3 + ... + aaad + 3aabdx + 3(aacd+abbd)x^2 + (3aadd+6abcd+bbbd)x^3 + ...
= a + ab + aac + aaad + (bb+2abc+3aabd)x + (bc+2acc+bbc+3aacd+3abbd)x^2 + (bd+2acd+2bcc+3aadd+6abcd+bbbd)x^3 + ...
Since 2^x = e^(x ln 2), by substuting x ln 2 into the Maclaurin series for e^x, we get a series that we can equate to the above expression and then solve for a, b, c, and d using the four equations in four unknowns.
Generating more terms and solving more equations we should be able to extend this as far as we like, although it would be tedious, for sure. It seems to me that a software solution involving these ideas would be a nightmare.



