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 Taylor series of i[x] Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 04/11/2017, 12:18 PM 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.) Xorter Unizo Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 07/10/2017, 04:07 PM (This post was last modified: 07/10/2017, 04:43 PM by Xorter.) First of all, I am afraid, that it is necessary to get the derivatives of the logical gates. 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? Xorter Unizo Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 02/20/2018, 09:55 PM 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? Xorter Unizo « Next Oldest | Next Newest »

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