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12/23/08 3:43 PM
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Spinelock
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Member Since: 1/17/07
Posts: 393
 
Ok OG geniuses, here is a problem I can;t figure out:

solve for x:

sin(sin(sin(sin(x))))=cos(cos(cos(cos(x))))

Now i'll tell you from the get-go that their is no REAL solution, If someone really wants I'll do the proof. I suspect that there is no imaginary solution either, but can;t dissprove it.

Anywhoo, I gave up on it but would still like to see a solution.
12/25/08 9:31 PM
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Voolf
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Member Since: 10/4/05
Posts: 178
Interesting problem! I just did a few numerical investigations:

I graphed the two sides of the equation and it indeed appears they don't cross anywhere on the real line.

I plugged the equation into Mathematica, as well. It won't solve it algebraically; however, it will solve it numerically, and given an initial guess of i, it gives a solution:

FindRoot[Sin[Sin[Sin[Sin[x]]]] == Cos[Cos[Cos[Cos[x]]]], {x, \[ImaginaryI]}]

0.756887 + 0.610155*i

So there do seem to be imaginary roots. But I have no idea how to solve this analytically, truly no clue! Are you sure that it is even possible?
12/25/08 9:37 PM
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Spinelock
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^^^^ hmmmmmmmmm I don't know if its solvable analyticly.

But that gives me some hope. I'm gonna work a little on it tomorrow, I'll let ya know of any insights.
2/15/09 8:53 PM
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AlbertEinstein
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Edited: 02/15/09 9:20 PM
Member Since: 6/15/07
Posts: 1650
Did you ever figure out how to answer this problem. I'd be very interested to know how you did it.
9/15/09 1:05 AM
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victorbomba
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Member Since: 8/31/07
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A little late but...

Sin[u] = Cos[u] => u = Pi/4 where u = Sin[Sin[Sin[x]]]]

So solve for Sin[u] = Pi/4

(Exp[iu] - Exp[-iu])/2i = Pi/4
this leads to the quadratic equation

Exp[2iu] - (i pi/2) Exp[iu] - 1 = 0
which has the solution

u = ArcTan[Pi/Sqrt[16 - Pi^2]]

So Sin[Sin[x]] = ArcTan[Pi/Sqrt[16 - Pi^2]]

Carry out the identical procedure twice more substituting the result from the previous step each time and you get

x = Pi/2 - i Ln[b + Sqrt[b^2 - 1]] where

b = ArcTan[(ArcTan[Pi/Sqrt[16 - Pi^2])/Sqrt[1 - (ArcTan[Pi/Sqrt[16 - Pi^2])^2 ]

according to Mathematica the result is approximately

1.5708 - 0.499742 I

Taking the Sine of this iteratively 4 times gives back approximately Pi/4

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