Taylor series of i[x] - Printable Version

Taylor series of i[x] - Printable Version

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RE: Taylor series of i[x] - Xorter - 04/11/2017

Hi, again!
Maybe I have found something related to the Nth derivative of i[x] (I may or may not know how to get the derivative of a non-analytic function like y = 1 if x=rational otherwise -1). Really interesting. But first of all, I would like to check it myself. But I will need your help.
Here is a produktum:
\( \prod_{k=0}^{N-1} k + {{1+(-1)^k} \over 2} \)
I would like to interpolate this expression/function, but how? Any idea?
(If I will be able to get the expansion of the Taylor series of i[x], then I promise I will share it with you.)

RE: Taylor series of i[x] - Xorter - 07/10/2017

First of all, I am afraid, that it is necessary to get the derivatives of the logical gates.

[Image: BitXor_1000.gif]

That would the 0th step to our target.
Okey, it is not hard to make contiously, just like this:
x xor y = lim (2^h x xor 2^h y)/2^h
And we know that
x xor y = (x and -y) or (-x and y)
But how to derivate?

RE: Taylor series of i[x] - Xorter - 02/20/2018

Okay, let us sign i[x] to just i and I[x] to just I where I[x] = int i[x] dx.
D := d/dx

So
D I = i
D i = i'
D i I = i' I + i^2 = i' I - 1
D I i = i^2 + I i' = I i' - 1

Therefore
i' I = 1 + (i I)'
and
I i' = 1 + (I i)'

D 1 I = D I 1 = 0 + i = i

But
D i÷I = (i' I - i^2)÷I^2 = (i' I + 1)÷I^2 = (2 + (i I)')÷I^2
D I÷i = (i^2 - I i')÷i^2 = I i' = 1 + (I i)'
D I÷i = -D i I = 1 - i' I
Thus
D I i = - i' I = I i' - 1
furthermore
i' I = 1 - I i'

I do not have to talk about i' I is not I i', right?

D 1÷I = (0 - i)÷I^2 = - i÷I^2

What is next?
The only useful information is that 1 ÷ i = -i ... or anything else?