Re: Inverse function solution
- To: mathgroup at smc.vnet.net
- Subject: [mg132658] Re: Inverse function solution
- From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
- Date: Wed, 30 Apr 2014 01:39:55 -0400 (EDT)
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Solve[ {x == Cos[u], y == Cos[u + v] }, {u, v} ]
Its closed/analytic solution is not possible, even numerically.
The known solutions are ellipses from sine waves with a phase difference, having x^2, x y and y^2 terms, as also sketched in Lissajous curves:
ParametricPlot[{Cos[u], Cos[u + v]}, {u, -Pi, Pi}, {v, -Pi, Pi}]
Can there be a work around?
Narasimham
Hi, Narasimham,
It can be solved numerically, though, since it is double-periodic it should be done carefully. This is a fast shot of how it can be done:
Here I put y=1 and vary only x:
lst = Table[{x,
FindRoot[{x == Cos[u],
1 == Cos[u + v]}, {{u, 0.1}, {v, 0.1}}] /. {x_ -> a_,
y_ -> b_} -> {a, b}}, {x, 0., 0.4, 0.05}]
This is the outcome:
{{0., {7.85398, -7.85398}}, {0.05, {7.80396, -7.80396}}, {0.1, \
{7.75381, -7.75381}}, {0.15, {7.70341, -7.70341}}, {0.2, {7.65262, \
-7.65262}}, {0.25, {7.6013, -7.6013}}, {0.3, {7.54929, -7.54929}}, \
{0.35, {7.49641, -7.49641}}, {0.4, {5.12391, -5.12391}}}
Let us check, if it is right:
lst /. {x_, {y_, z_}} -> {x == Cos[y], 1 == Cos[y + z]}
{{False, True}, {True, True}, {True, True}, {True, True}, {True,
True}, {True, True}, {True, True}, {True, True}, {True, True}}
So, the result is right except the one in the first parentheses. Let us plot it. Try this:
ListPlot[{lst /. {x_, {y_, z_}} -> {x, y},
lst /. {x_, {y_, z_}} -> {x, z}}]
Have fun, Alexei
Alexei BOULBITCH, Dr., habil.
IEE S.A.
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