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Pi's connection to e, and its ubiquity in math

  1. Sep 8, 2010 #1
    As a freshman in college, I was wondering, is pi's connection with the very simple differential equation [tex]\frac{d^{2}y}{dx^{2}}[/tex] = -y with initial condition that (0,0) be included (or even if that is not the case a connection can be made) the reason it is so ubiquitous in mathematics. This of course implies that a circle is just a specific case that utilizes this differential equation. Pi can be defined in my head as the difference between x's of sequential zeros in the solution above.

    So really what I am asking of people is to explain to me anything related to this in higher level mathematics perhaps that I would not have come by or thought of. And by the way this is just my own random thinking, nobody has taught me anything about pi in this manner.

    The solution of the differential equation [tex]\frac{dy}{dx}[/tex] = y with initial condition that (0,1) be included, looks very similar to the equation above, and of course the solution to this equation is the exponential function, e[tex]^{x}[/tex].

    Thank you in advanced, Matt.
     
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2

    HallsofIvy

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    I'm not sure I would say this "explains" the relationship between e and [itex]\pi[/itex] but in terms of complex numbers, [itex]e^x[/itex] and the trig functions are essentially the same function!
    That is [itex]e^{x+ iy}= e^x(cos(y)+ i sin(y))[/itex]. In particular, [itex]e^z[/itex] where z= x+ iy is exactly like [itex]e^x[/itex] on the "real axis" (where y= 0) and exactly like [itex]cos(y+ i sin(y)[/itex] on the imaginary axis (where x= 0).

    Yes, the solution to the differential equation dy/dx= y is [itex]Ce^x[/itex] and the general solution to the differentential equation [itex]d^2y/dx^2= y[/itex] is [itex]Ce^x+ De^{-x}[/itex]. If you were to "try" a solution of the form [itex]y= e^{rx}[/itex] then you would have [itex]y'= re^{rx}[/itex] and [itex]y"= r^2e^{rx}[/itex]. Putting that into dy/dx= y, you would have [itex]re^{rx}= e^{rx}[/itex] and since [itex]e^{rx}[/itex] is never 0, you can divide through by it leaving the "characteristic" equation [itex]r= 1[/itex]. Doing the same with [itex]d^2y/dx^2= y[/itex] you get [itex]r^2e^{rx}= e^{rx} for "characteristic equation" r^2= 1 which has roots 1 and -1: [itex]e^{1x}= e^x[/itex] and [itex]e^{-1x}= e^{-x}[/itex]. If you do the same with [itex]d^2y/dx^2= -y[/itex] the "characteristic equation" is [itex]r^2= -1[/itex] which has roots i and -i. That is the solutions are [math]e^{ix}[/math] and [math]e^{-ix}[/math]. But [math]e^{ix}= cos(x)+ i sin(x)[/math] and [math]e^{-ix}= cos(x)- i sin(x)[/math] so those solutions can be written in terms of sine and cosine.
     
  4. Sep 8, 2010 #3
    you can try the operator calculus where the exponentials come as a natural answer rather than clever guesses.

    I remember someone posted quite some time ago. There must be a link to it on this forum

    EDIT: Here it is :
    https://www.physicsforums.com/showthread.php?t=54055

    Read the pdf in the second post.
     
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