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Taylor series of i[x]

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Taylor series of i[x]
Xorter Offline
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#13
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
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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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