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Thread: New Math Function Redux

  1. #271
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    Quote Originally Posted by tusenfem View Post
    you should get a result in the range of cot-1, which is usually [-pi/2 pi/2], it looks like you used (cot(a))^-1 instead of acot in your plot.
    I think you are right: when I changed it from cot-1() to cot()-1 it didn't make any difference. I will try it again in a moment. See if it recognises acot().

    That fixed it. Thanks. The graph now looks like:
    Untitled.jpg
    (with y and x instead of alpha and lambda, for values of v = pi/2.2, pi/3, pi/4, pi/6)

    Although what it all means ...

    There is an odd discontinuity in the curve (I haven't cut it off at y=0 and y=2*pi)
    Last edited by Strange; 2017-Oct-30 at 04:03 PM.

  2. #272
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    Quote Originally Posted by Strange View Post
    ...There is an odd discontinuity in the curve (I haven't cut it off at y=0 and y=2*pi)
    The two-dimensional representation can be normalized in the same manner (both conceptually and mathematically) as the unit vector is normalized using conventional techniques.

    For your x axis, let

    This gets rid of the length associated with the representation of that angle. Once that is accomplished then the whole thing might make a little more sense. In the same way that removing or "normalizing" length from a vector leaves only direction information, this method also only leaves direction information. And since each two-dimensional graph represents only the directional component of the angle , the direction information has been quantified (defined without the use of any length).

  3. #273
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    Quote Originally Posted by Strange View Post
    That is what I expected, with a shallow slope near 0 and a steep slope at 2*upsilon -- contrary to the graphs endorsed earlier.

    Quote Originally Posted by 01101001 View Post
    Last edited by 01101001; 2017-Oct-30 at 09:48 PM.
    0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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  4. #274
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    No, there is still something wrong, because sin(pi/4)= 0.7, so for sin(x/2) > 0.7 the ratio within the arcsin becomes greater than 1, which means that you will obtain a complex number, so in general one expects a complex number for the equation shown by strange.
    If I take v = pi/4 and put the equation into Matlab, and let -2pi <= x <= 2pi I get the following with blue the real part and red the imaginary part:

    x = [-2 : 0.01 : 2] * pi;
    v = pi/4;
    y = acot( cos(v) * tan( asin( sin(x/2) / sin(v) ) ) );
    plot1.jpg
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  5. #275
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    I can explain that.

  6. #276
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    Quote Originally Posted by tusenfem View Post
    If I take v = pi/4 and put the equation into Matlab, and let -2pi <= x <= 2pi I get the following with blue the real part and red the imaginary part:

    x = [-2 : 0.01 : 2] * pi;
    v = pi/4;
    y = acot( cos(v) * tan( asin( sin(x/2) / sin(v) ) ) );
    In an earlier (article) steveupson specified the domain of his function to be 0 to 2*upsilon (or twice your v), and that keeps it real.

    Everyone needs to be cautious of how one's tool defines arccot. See Which is the correct graph of arccot x? Different tools have different values for arguments < 0 and some have discontinuities at 0.

    But for those who want to dabble, here's the WolframAlpha plot: arccot(cos(pi/4.0)*tan(arcsin(sin(lambda/2)/sin(pi/4.0)))), (pi/2)*(1+sin((pi/4)*lambda/(pi/4.0)+pi)) from 0 to 2*(pi/4.0)

    f(lambda).png

    That is the function with a translated, scaled sin(lambda) for reference. The curve is a problem the proponent can address.

    Here's the Google Calculator version (lacking arccot), same curves: arctan(1/(cos(pi/4.0)*tan(arcsin(sin(x/2)/sin(pi/4.0))))), (pi/2)*(1+sin((pi/4)*x/(pi/4.0)+pi)) from 0 to 2*(pi/4.0)
    0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...
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  7. #277
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    Quote Originally Posted by 01101001 View Post
    In an earlier (article) steveupson specified the domain of his function to be 0 to 2*upsilon (or twice your v), and that keeps it real.

    Everyone needs to be cautious of how one's tool defines arccot. See Which is the correct graph of arccot x? Different tools have different values for arguments < 0 and some have discontinuities at 0.

    But for those who want to dabble, here's the WolframAlpha plot: arccot(cos(pi/4.0)*tan(arcsin(sin(lambda/2)/sin(pi/4.0)))), (pi/2)*(1+sin((pi/4)*lambda/(pi/4.0)+pi)) from 0 to 2*(pi/4.0)

    f(lambda).png

    That is the function with a translated, scaled sin(lambda) for reference. The curve is a problem the proponent can address.

    Here's the Google Calculator version (lacking arccot), same curves: arctan(1/(cos(pi/4.0)*tan(arcsin(sin(x/2)/sin(pi/4.0))))), (pi/2)*(1+sin((pi/4)*x/(pi/4.0)+pi)) from 0 to 2*(pi/4.0)
    okay i did not look at the domain, but my plot and yours are the same for the correct domain.
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  8. #278
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    The original graph was made by Strange and I furnished the labels for the axes. I think they are correct but there might be some additional explanation required here.
    Fig 5.PNG

    I rotated the graphs that I made in Excel to the same orientation as the original one done by Strange. Before rotation they look like this:

    EXCELx4.JPG
    When I look at the new graph from Strange

    Untitled.jpg

    it shows the same information except that it hasn’t been “normalized” in the same way as my graphs have been. The normalizing is important because it gets rid of the length reference that is contained in the model because of the use of a unit sphere. This was mentioned but not properly explained over several posts upthread

    https://forum.cosmoquest.org/showthr...54#post2362654

    I don’t know if this explanation helps, but it is the history of how we got here.
    Last edited by steveupson; 2017-Nov-01 at 03:50 PM.

  9. #279
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    I don't really know why I am doing this () but here is the same graph, normalised to 1 on each axis for a few more values of alpha and with the addition of cos (dotted line).
    Untitled.jpg

  10. #280
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    Quote Originally Posted by Strange View Post
    I don't really know why I am doing this () but here is the same graph, normalised to 1 on each axis for a few more values of alpha and with the addition of cos (dotted line).
    Untitled.jpg
    Would it be more clear if you were to rotate the cos 90 degrees clockwise such that the y axis is zero and the two ones from each graph coincide with each other? Or rotate the new curves 90 degrees counterclockwise and use the fourth quadrant of the cos from to ?

  11. #281
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    These graphs represent a relationship that exists in Euclidean 3-space.

    What happens to this relationship under relativistic transformation?

  12. #282
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    The graphs are 2D and so exist in a 2D Euclidean space. There is no time so there cannot be any relativistic transformation other than the trivial y = y', z = z'.

  13. #283
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    Quote Originally Posted by Reality Check View Post
    The graphs are 2D and so exist in a 2D Euclidean space. There is no time so there cannot be any relativistic transformation other than the trivial y = y', z = z'.
    In my understanding the graphs have no more relevance to the question of relativistic transformations than a ruler has when discussing length contraction.

    Can you explain how the geometric relationship expressed by the function is somehow invalidated by the fact that the function can be graphed? I don't follow your reasoning, at all.

  14. #284
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    That was my point - the graphs are irrelevant . A Euclidean 3D space is irrelevant.
    Relativity is a theory of space and time. The Lorentz transformation includes time.

  15. #285
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    Quote Originally Posted by Reality Check View Post
    That was my point - the graphs are irrelevant . A Euclidean 3D space is irrelevant.
    Relativity is a theory of space and time. The Lorentz transformation includes time.
    Yes. Lorentz transformations customarily use Cartesian coordinates to represent an n-dimensional Euclidean space, more specifically a 3-dimensional Euclidean space, which when combined with time becomes what is customarily called spacetime. The relationship that we're concerned with exists in Euclidean 3-space and cannot currently be duplicated in a surface or manifold (my money says that it will be proven to be impossible within ten years.)

    The graphs are a representation of the relationship or function that is illustrated in the mathematical model that is presented at the beginning of this thread.

    I think the relationship in question will not be affected by a Lorentz transformation since it can be expressed in the form of an equation. If you understand the math, then you will see that this creates some new issues with the model, if true.

  16. #286
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    Quote Originally Posted by steveupson View Post
    I think the relationship in question will not be affected by a Lorentz transformation since it can be expressed in the form of an equation. If you understand the math, then you will see that this creates some new issues with the model, if true.
    Of course the bold part is complete nonsense, not all equations are Lorentz invariant.
    If you want to do it "relativistically" then you will have to take your sphere and let it move, then it will contract in the direction of motion and instead of sphere you will have a "pancake", and then you will have to start all over again with your calculations.
    Not that it will bring you anything, this only shows that there a relationship can be found between two angles that are defined by tangents and planes on or through a sphere, the only thing one can say about that is "so what"?
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  17. #287
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    With the banning of steveupson, this thread is closed.

    Frankly, it is hard to imagine anyone here wanting to reopen this thread, but if you do, you know the drill; you'll need a very convincing argument to do so.
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