Hi all ... yes i know that this is not an irrationality related

group but you guys have proven yourselves a great number of trust so

I can really trust your math skills.

I was doing Taylor series yesterday and i figured something

interesting out ...

Define: F(x) = x^e ==> F(1) = 1

F'(x) = e*x^(e-1) ==> F'(1) = e

F''(x) = e(e-1)*x^(e-2) ==> F''(1) = e(e-1)

F'''(x) = e(e-1)(e-2)*x^(e-3) ==> F'''(1) = e(e-1)(e-2)

If f(x) is the Taylor series of F(x)=x^e at a=1 ==>

f(x) = 1 + e(x-1)/1 + e(e-1)(x-1)^2/2! + e(e-1)(e-2)(x-1)

^3/3! + ...

f(2) = 1 + e + e(e-1)/2 + e(e-1)(e-2)/6 + e(e-1)(e-2)(e-3)/24 + ...

Let f_1(2) = 1 + e

f_2(2) = 1 + e + e(e-1)/2 = 1+e/2+e^2/2

f_3(2) = 1 + e + e(e-1)/2 + e(e-1)(e-2)/6 = 1 + 5e/6 +e^3/6

==> f_n(2) = 1 + p_1*e/q_1 + p_2*e^2/q_2 + ... + p_n*e^n/q_n,

where p_k and q_k are integers.

F(2)=2^e= lim n->inf: fn(2)

Let us assume that 2^e is rational, that is 2^e = p/q ==> F(2) = p/q

= lim n->inf: f_n(2)

Let y_n(z) = 1 + p_1*z/q_1 + p_2*z^2/q_2 + ... + p_n*z^n/q_n = p/q

==> 1 + p_1*z/q_1 + p_2*z^2/q_2 + ... + p_n*z^n/q_n - p/q =0 ==> z=e

is a solution to this equation, but e is a transcendental number

which means that it can never be such a solution ==> our assumption

about the rationality of f_n(2) was incorrect ==> 2^e is irrational.

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What do you think ? I have probably made a mistake somewhere because

such a simple thing is too easy to overlook over the years by many

great mathematicians. However I am interested to know what is wrong

with that proof. Also if there is something wrong - can you please

say whether you think it is "fixable" ?

Thank you.

Rusi Kolev