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Where is x^Sin(y)=y^Cos(x)?

  1. May 4, 2005 #1

    saltydog

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    Don't wish to double-post, but last week or so, someone asked a question I've worked on and I was unable to find the thread:

    Solve:

    [tex]x^{Sin(y)}=y^{Cos(x)}[/tex]

    I've attached a plot of the solution for the first quadrant. Surprising isn't it? The only way I could do this is to brute-force check each point in the plane and choose it if close enough (<0.01) to make the expression zero. I can see no other way to do this.

    Also, it becomes interesting when negative values are considered since the trig expressions return almost all irrational numbers and a negative number raised to an irrational number is not Real. Right?

    There is a nice explanation for the form of the plot below. Note how the behavior changes as one moves up the y-axis: from curve to eye-shaped, to x-shaped, to inequality-shaped. Anyway, if you guys wish, I'll delete this post and link my comments to the original one if you tell me where it is.
     

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    Last edited: May 4, 2005
  2. jcsd
  3. May 4, 2005 #2
    umm x = y = pi/4

    am i missing something?

    -- AI
     
  4. May 4, 2005 #3

    saltydog

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    Hello Tenali,

    I've attached a plot showing:

    [tex]y(x)=6^{Sin(x)}[/tex]

    and:

    [tex]y(x)=x^{Cos(6)}[/tex]

    Where they meet is one solution, that is, (6,2.60325) approx.
    The plot above for the first quadrant is a check of 5 million points. Each black pixel is a solution.
     

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  5. May 5, 2005 #4
    Well i did miss something, i missed what you were trying to say :biggrin:
    As i look at that first quadrant diagram, it reminds me of the electric field line diagrams i used to see in my physics textbook. You think there is something spooky going on here? Would like to know what u are thinking in that case.

    -- AI
     
  6. May 5, 2005 #5

    Zurtex

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    I'm assuming you want a real solution?

    I mean this works:

    [tex]x = -\frac{47}{10} + \frac{181}{10}i \quad \text{and} \quad y = \frac{91}{10} + \frac{122}{5}i[/tex]
     
  7. May 5, 2005 #6

    saltydog

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    Thanks Zurtex.

    I think I might like to know what are all the loci of points in the complex plane satisfying the equation. You know, that's an interesting problem in itself: what are all the complex solutions?

    As far as the appearance of the Real solution set in the first quadrant, that's easily explained by considering how the curves:

    [tex]y(x)=a^{Sin(x)}[/tex]

    [tex]y(x)=x^{Cos(a)}[/tex]

    interact with each other as "a" is varied. Consider when a=6 and look at the 1st quadrant solution set at x=6 and draw a vertical line there. You'll note it crosses a black point only once, that one is shown by the two curves above. As "a" is varied, the second equation above changes and intersects the other curve more often giving rise to multiple solutions. It is precisely this behavior that gives rise to the appearance of the first quadrant plot.
     
  8. May 5, 2005 #7

    saltydog

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    Ok Zurtex, it's not happening for me. When I attempt to prove that your solution works, I get really large numbers which overflow Mathematica.

    Could you please show how this solution satisfies the equation or at least provide a clue to doing such?

    Also, just how do you solve the equation for complex numbers? I'll spend time on it but it may take a while.
     
  9. May 5, 2005 #8

    Zurtex

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    Haha, I have to admit I can't actually prove it. But type this in to mathematica:

    FindInstance[x^Sin[y] == y^Cos[x], {x, y}]

    :tongue:

    Oh, you can expand that function to say get 20 solutions:

    FindInstance[x^Sin[y] == y^Cos[x], {x, y}, 20]
     
    Last edited: May 5, 2005
  10. May 5, 2005 #9

    saltydog

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    Zurtex, I must question your solution. Here's why:

    Using your values of x and y, I can reduce it to:

    [tex]x^{Sin(y)}=e^{\alpha}[Cos(\beta)+iSin(\beta)] \quad\text{with:}\quad[/tex]

    [tex]\alpha=1.43079*10^{11} \quad\text{and}\quad \beta=-1.17792*10^{11}[/tex]

    and:

    [tex]y^{Cos(x)}=e^{j}[Cos(p)+iSin(p)] \quad\text{with:}\quad[/tex]

    [tex]j=4.30262*10^{7} \quad\text{and}\quad p=-1.20099*10^{8}[/tex]

    I see no way for these expressions to be equivalent.

    You know, I've tried an exhaustive search and it's too CPU-intensive since for each point in the complex plane for x, I would need to check each point in a portion of another plane for y. I'm disappointed that this may remain unresolved for complex numbers without access to a super computer to run the algorithm. Anyone here have access to a fast computer to check it?

    Or if there is a better way, that too. :smile:
     
  11. May 5, 2005 #10

    Zurtex

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    Why not just look at this with the values of y and x given:

    [tex]\sin y \ln x = \cos x \ln y[/tex]

    And you're quite right, they are not equal. Guess we better write in to Wolfram :eek:
     
  12. May 6, 2005 #11

    saltydog

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    Ohhhh. Suppose that would be easier . :blushing:

    As far as Wolfram, I don't think they like me since I called them "math junkies" and besides I don' t have a support contract and they don't like me bothering them. Also "FindInstance" must be a ver. 5.0 feature, and I have 4.x. Thanks though. Think I'll work some more with it anyway.
     
  13. May 6, 2005 #12

    saltydog

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    I've figured out a graphical means of searching for the complex solutions for:


    [tex]x^{sin(y)}=y^{Cos(x)}[/tex]

    First convert the expression to:

    [tex]\frac{Cos(x)}{ln(x)}\quad\text{(1)}\quad[/tex]

    [tex]\frac{Sin(y)}{ln(y)}\quad\text{(2)}\quad[/tex]

    Now, consider how equation (1) maps coordinates in the complex plane of the form (5+bi) as b ranges from -2.2 to 2.2. This mapping is shown in the first plot.

    Now consider how equation (2) maps coordinates in the complex plane of the form (1+bi) as b goes from -2.2 to 2.2. This the the second plot.

    Superimposing the two plots show that the curves intersect. Two solutions are at the intersections. In this case, they are (approx):


    [tex]x=5+1.992i\quad\text{ and }\quad y=1-0.494i[/tex]

    [tex]x=5-1.992i\quad\text{ and }\quad y=1+0.494i[/tex]

    Substituting these expressions into the original equation, yields in the first case:

    [tex]5.43055-0.594184i\approx 5.43153-0.596194i[/tex]

    And the second:

    [tex]5.43055+0.594184i\approx5.43153+0.596194i[/tex]

    One could conceivably then search the entire plane in this way to obtain the solution set. However, since I'm not very good in Complex Analysis, I would bet a dollar there is some relation somewhere that would make this an easier task.
     

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    Last edited: May 6, 2005
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