SIN(X)^COS(X)
11-26-2017, 03:41 PM
Post: #1
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
SIN(X)^COS(X)
In function app, the plot is what I expect...defined over [0,pi),[2pi,3pi), etc. In advanced graphing app, I get an unexpected plot. Over (pi,2pi], I get a plot symmetric to [0,pi) with respect to the y axis, and and another plot of this interval mirrored below the x axis.

Why am I seeing this?

On the July version, (on my iPad) I can trace and display x,y values on this part (what I am expecting to be imaginary numbers) of the graph. On the beta on my physical unit, the plot is as described, but it will not allow me to trace a point on the unexpected part of the plot.
11-26-2017, 08:01 PM
Post: #2
 John Colvin Member Posts: 173 Joined: Dec 2013
RE: SIN(X)^COS(X)
Sin(x) is < 0 in the 3rd and 4th quadrants and it is taken to a fractional power. So I am guessing that you are seeing the complex conjugates plotted in those quadrants in the advanced graphing application. The function app only plots functions of one variable which explains why the 3rd and 4th quadrant plots don't show anything.
11-27-2017, 03:09 AM
Post: #3
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
I'm not sure if this correct as the advanced graphing app is not always plotting the complex number plane. For example, 3*X^2 - 13*X + 17 = Y has 2 complex roots, but only the real number plane is plotted.
11-27-2017, 06:19 AM
Post: #4
 AlexFekken Member Posts: 151 Joined: May 2016
RE: SIN(X)^COS(X)
When sin(x) < 0 we have:

sin(x)^cos(x) =
exp(cos(x)*ln(sin(x)) =
exp(cos(x)*(i*pi + ln|sin(x)|)) =
exp(i*pi*cos(x)) * |sin(x)|^cos(x)

And the real part of exp(i*pi*cos(x)) is cos(pi*cos(x))

Does that explain what you see?
11-27-2017, 03:55 PM
Post: #5
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
But should I be seeing this on the real x,y plot?
11-28-2017, 02:52 PM
Post: #6
 Fortin Member Posts: 59 Joined: Jul 2015
RE: SIN(X)^COS(X)
I believe that the plot of Y=(-1)^X is caused by the same issue: a power with a negative base.
11-29-2017, 03:19 AM
Post: #7
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
I understand...there are an infinity of real solutions that are a subset of an infinity of non-real solutions (for (-1)^x). Not as intuitive (to me) for (sin x)^(cos x), especially with regard to the symmetric solutions with respect to the x axis for the non real "y" values for y=(sin x)^(cos x) l pi<x<2pi.

Again, thanks!
11-29-2017, 02:37 PM
Post: #8
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
I think what is confusing me is that the mirrored plot looks like a continuous function vs an infinity of results that are a subset of a continuous function.
12-01-2017, 01:37 AM (This post was last modified: 12-01-2017 01:39 AM by Fortin.)
Post: #9
 Fortin Member Posts: 59 Joined: Jul 2015
RE: SIN(X)^COS(X)
I suspect (but this is only a wild guess) that this plotting artifact is a result of a conversion (or bug ;-) ) done internally to deal with these negative values within the interval arithmetic.

Y=e^(COS(X)*LN(-SIN(X)))
Y=-(e^(COS(X)*LN(-SIN(X))))

Y=e^(X*LN(-(-1)))
Y=-(e^(X*LN(-(-1))))
12-01-2017, 07:40 PM
Post: #10
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
As an example, on the iPad Prime, 4.0171 on the graph produces a "y" value of +/- ~1.1843663159. On my WP34S, I get the ~1.1843663159 result when I take the absolute value of the complex number that is produced by sin(4.0171)^cos(4.0171) in radian mode...

I think the advanced graphing app should not be plotting this part of the graph on the real number plane.
12-01-2017, 11:42 PM (This post was last modified: 12-01-2017 11:42 PM by Fortin.)
Post: #11
 Fortin Member Posts: 59 Joined: Jul 2015
RE: SIN(X)^COS(X)
I think you're right that it's the abs of a C in this case, but I didn't have luck with that cause in all cases I tested. I do agree that it shouldn't be plotting those on an x/y plane.
12-04-2017, 10:17 PM
Post: #12
 chazzs Junior Member Posts: 22 Joined: Feb 2015
RE: SIN(X)^COS(X)
(12-01-2017 07:40 PM)lrdheat Wrote:  As an example, on the iPad Prime, 4.0171 on the graph produces a "y" value of +/- ~1.1843663159. On my WP34S, I get the ~1.1843663159 result when I take the absolute value of the complex number that is produced by sin(4.0171)^cos(4.0171) in radian mode...

I think the advanced graphing app should not be plotting this part of the graph on the real number plane.

Oh, I think it should. There are an infinite real values between pi and 2pi, it just so happens that there are many more complex results. The real ones are nearly impossible to find, but they do exist. Some are positive and some are negative (along with the complex.) In the advanced graphing app you are "seeing" all of these POINTS, and definitely not a continuous graph.

This is similar to plotting y=(-2)^x. Most calculators show various points for the function, i.e, (0,1), (1,-2), (2,4), etc. Also, some values are real for some rational exponents, i.e., when x=1/3 (there is a real root and two complex, most calculators will plot the real), and some are complex, i.e., x=1/2. Hence the discrete nature of the graph.

I believe that is what is happing to sin(x)^cos(x).

C
12-06-2017, 02:47 AM
Post: #13
 lrdheat Senior Member Posts: 815 Joined: Feb 2014
RE: SIN(X)^COS(X)
I am finding that I am agreeing now...that there are an infinite number of real solutions amongst a larger infinite universe.

My next question is why the regular graphing app fails to show this behavior! My CASIO Classpad 400 shows a graphing result similar to the Prime, but the non CAS CASIO fx-CG 10 does not show the discrete real solutions part of the graph.
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