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Impossible equation to solve?

  1. Jun 26, 2009 #1
    Hi, I know its possible to get an answer by graphing the curves and finding the point of intersection but I was wondering if there was a way to do them algebraically.

    sin(2x) = [tex]\frac{x}{2}[/tex]

    or [tex]e^{\frac{x}{8}}[/tex] = x

    Also, do these types of equations have a name?
  2. jcsd
  3. Jun 26, 2009 #2


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    Your first equation is not solvable (in terms of elementary functions or any I know of). The equation simply is not written in a way in which it is possible to express it as a solution x=something.

    The second equation at first glance appears the same, but is actually solvable (although not in terms of elementary functions). The solution is:
    x=-8 W(-1/8)
    Where W is the Lambert W function. ( http://en.wikipedia.org/wiki/Lambert's_W_function )

    I don't know of a name for these [unsolvable] equations, I'd call em non-algebraic (not sure if that's official though).
    Last edited: Jun 26, 2009
  4. Jun 26, 2009 #3


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    They are called transcendental equations.
  5. Jun 26, 2009 #4
    Thanks for the replies.
  6. Jun 26, 2009 #5
    haha try this one. Let me know if you're interested in the problem, I actually got it from physicsforums.com.

    [tex]\left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi[/tex]

  7. Jun 26, 2009 #6
    And what are the 0 and 6 ??
  8. Jun 26, 2009 #7


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    I assume that he means that formula evaluated between 6 and 0: its value at 6 minus its value at 0 is equal to [itex]e+ \pi[/itex].
  9. Jun 26, 2009 #8
    of course. Its an integral remember?

    [tex] \int_0^{6}\sqrt{1 - (nx)^2}dx = e + \pi[/tex]
  10. Jun 26, 2009 #9
    I'm still in high school so is that possible to solve using my level of mathematics?

    After I subbed the numbers in, I did a quick sketch on graphmatica and it didn't show any intersections between

    [tex] y = \sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -1 [/tex]

    and y = 2e + 2pi

    so have I done something wrong already?
  11. Jun 27, 2009 #10
    I agree that there is no solution to

    [tex]\left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi[/tex]

    However, there is a solution to

    [tex] \int_0^{6}\sqrt{1 - (nx)^2}\,dx = e + \pi[/tex]

    camilus is mistaken in asserting that these are equal.
  12. Jun 27, 2009 #11
    The equations in the first post can't be solved explicitly for x, but you could use a method of iteration to get as accurate a solution for x as you require.
  13. Jun 27, 2009 #12
    wtf? explain..?

    [tex] \int_0^{6}\sqrt{1 - (nx)^2}\dx = \left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6} = e + \pi[/tex]

    now you gotta find what value of n will make will make the integral equal to e + pi. I have found n to several digits, although Im under the impression that n is nonalgebraic.
  14. Jun 27, 2009 #13


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    [tex]\int_0^{6}\sqrt{1 - (nx)^2}\dx \neq \left [{1 \over 2}\sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n} \right ]_{0}^{6}[/tex]

    Now, I can't actually explain because I don't know how to evaluate the integral, but I did go about it backwards and derived the result. They're not equal from what I can see.
  15. Jun 28, 2009 #14
    the integral should be

    \int \sqrt{1-(nx)^2}\,dx =\frac{1}{2}\;x\;\sqrt{1-(nx)^2} +

    notice the extra [tex]x[/tex]?
  16. Jun 28, 2009 #15


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    It is relatively easy to show that:

    [tex]\int \sqrt{1 - (nx)^2}\, dx = {1 \over 2} x \sqrt{1 - n^2x^2} + {\arcsin (nx) \over 2n}[/tex]

    Two methods come to mind.

    Method 1.

    Substitute [itex]nx = \sin(u)[/tex] and it transforms into [itex]1/n \int \cos^2(u) du[/itex]. Then just use the trig identity cos^2(u) = 1/2 + 1/2 * cos(2u) and it's straight forward to get the above stated result.

    Method 2. (A little more complicated but avoids need for trig identities and "trig of inverse trig" simplifications).

    Note that [tex]\sqrt{1-(nx)^2} = 1/ \sqrt{1-(nx)^2} \, - \, (nx)^2 / \sqrt{1-(nx)^2} [/tex]. The first term is pretty much a standard inverse sin integral and the second term is amenable to integration by parts. This is actually one of those integration by part problems where you end up with the original integral on both the left and right hand side of the equals sign and it simplifies down pretty easily (again to give the above stated result).

    Edit. Yes thanks g_edgar. I didn't notice that missing x before.
    Last edited: Jun 28, 2009
  17. Jun 28, 2009 #16


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    No uart, you are wrong here. g_edgar posted the correct integral and I was able to confirm it for myself by again differentiating the result.

    Funny how it can be relatively easy to show the wrong answer :tongue:
  18. Jun 28, 2009 #17


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    Yes it's a relatively easy integral. I posted two ways to do it (and have done it many times in the past) but no I didn't actually do the calculations this time as the expression looked correct (as in it looked that same as what I've got in the past when I've actually done the integral rather than just describing the method to use). Sorry I didn't notice that the "x" was missing in the expression you posted.
  19. Jun 28, 2009 #18
    Actually, the task is to solve for n

    [tex]\sqrt{1-36n^{2}} +\frac{arcsin (6n)}{n} -\frac{1}{2}-\frac{arcsin(0)}{2n}= e+\pi



    [tex]\sqrt{1-sin^2(x)} + \frac{6x}{sin(x)} - \frac{1}{2} = e+ \pi[/tex]

    [tex]cos(x)+\frac{6x}{sin(x)}=e+\pi + \frac{1}{2}[/tex]
  20. Jun 28, 2009 #19


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    Yes we knew that. :)

    And with the corrected equation that would be,

    [tex]3 \sqrt{1-36n^2} + {\frac {\sin^{-1}(6n)} {2n} = \pi + e [/tex]

    Which is still a transendental equation. (that is, no closed form solution in terms of standard functions). Numerically n=0.0617 to 3 significant figures.
  21. Jun 28, 2009 #20
    Look at my way of solving, it's much easier. Yes, I know that it is still transcendental equation.
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