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 Taylor series of i[x] 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 »

 Messages In This Thread Taylor series of i[x] - by Xorter - 01/12/2017, 04:50 PM RE: Taylor series of i[x] - by sheldonison - 01/12/2017, 08:50 PM RE: Taylor series of i[x] - by Xorter - 01/13/2017, 05:26 PM RE: Taylor series of i[x] - by Xorter - 01/13/2017, 07:13 PM RE: Taylor series of i[x] - by Xorter - 01/14/2017, 10:14 AM RE: Taylor series of i[x] - by mike3 - 01/23/2017, 07:38 AM RE: Taylor series of i[x] - by Xorter - 02/26/2017, 11:10 AM RE: Taylor series of i[x] - by Xorter - 03/01/2017, 03:06 PM RE: Taylor series of i[x] - by Xorter - 03/04/2017, 09:40 AM RE: Taylor series of i[x] - by Xorter - 04/06/2017, 03:42 PM RE: Taylor series of i[x] - by Xorter - 04/11/2017, 12:18 PM RE: Taylor series of i[x] - by Xorter - 07/10/2017, 04:07 PM RE: Taylor series of i[x] - by Xorter - 02/20/2018, 09:55 PM

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