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Taylor series of i[x]

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Taylor series of i[x]
Xorter Offline
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#10
04/06/2017, 03:42 PM
Okey, I have some news!
I got a little bit closer than before.
Let us get the length of the curve of points of the base units from i[exp(0)] to i[exp(2ipi)]:
In polar-coordinate system:
L = .5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi

In cartesian coordinate system:
L = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi

So
.5 [ sqrt(i[x]'^2 - 1) ] from 0 to 2pi = [ sqrt( i[exp(ix)]'^2 + 1 ) ] from 0 to 2pi
It looks a promising integral and differential equation.
What do you think, what should be the next step?
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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