- #26

gabbagabbahey

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- #26

gabbagabbahey

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- #27

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but-2y would mean there's some kind of dependency between the two variables?

- #28

gabbagabbahey

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[itex]\alpha=0[/itex] is a root, and is independent of [itex]\beta[/itex] (and hence independent of [itex]t[/itex]), and so values of [itex]x[/itex] where [itex]\alpha=0[/itex] will be stationary points (since the expression will be zero for all [itex]t[/itex] anytime [itex]\alpha=0[/itex]).

There will also be six complex or real roots of the factor:

[itex](2087-16500 \alpha ^2+32928 \alpha ^4-18816 \alpha ^6-16500 \beta ^2+131760 \alpha ^2 \beta ^2[/itex]

[itex]-263424 \alpha ^4 \beta ^2+150528 \alpha ^6 \beta ^2+32928 \beta ^4-263424 \alpha ^2 \beta ^4[/itex]

[itex]+526848 \alpha ^4 \beta ^4-301056 \alpha ^6 \beta ^4-18816 \beta ^6+150528 \alpha ^2 \beta ^6-301056 \alpha ^4 \beta ^6+172032 \alpha ^6 \beta ^6)[/itex]

But they will all depend on [itex]\beta[/itex]; and hence they also depend on [itex]t[/itex]; and so they are

So....the only stationary points are the points where [itex]\alpha=0[/itex]

- #29

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so to solve for the roots then, would just plain old factoring be the best way?

- #30

gabbagabbahey

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You mean solving for the 6 roots of that ugly expression in my last post?

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- #32

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- #34

gabbagabbahey

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Why do you say "they seem to be stationary"?

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Also I had the impression that the problem would've been simpler in terms of all the trig stuff...

- #36

gabbagabbahey

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Oh.....try graphing it for t=0.2 and t=0.4...do they still look stationary?

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