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I Derivation of pi using calculus

  1. Sep 20, 2017 #1
    I tried to derivate pi using calculus but i just found a quite different result. Can you spot my wrong please?

    First i started with equlation of a circle which is:
    I am assuming circle's center stands on the center of origin.
    To reach pi we shoud consider the situation that diameter = 1 because when diameter equals 1 circumference equals ##\pi##.Then:
    Simplify the equlation to get:
    $$y=\sqrt{\frac{1}{4} - x^2}$$
    We can use the formula which says that our arc lenght is
    $$ L =\int_a^b \sqrt{1-(f^\prime(x))^2}$$
    Take the derivative of our function which is:
    $$-\frac{x}{\sqrt{\frac{1}{4} - x^2}}$$
    Stick this equlation to the formula to found arclenght.(and multiply by two to get full arclenght we just find the half of the circle)
    $$L=\int_a^b \sqrt{1-\frac{4x^2}{1 -4x^2}}$$
    The circumference is the twice of the value of this integral between ##-\frac{1}{2}## and ##\frac{1}{2}##.
    So the equlation comes to:

    $$L=\int_{-\frac{1}{2}}^\frac{1}{2} \sqrt{1-\frac{4x^2}{1 -4x^2}}$$

    That's where i got stuck. So i used our friend Wolfram Alpha to reach result(taking indefinite integral):
    $$\frac{1}{2} E(sin^{-1}(2x)|2)$$
    I've absolutely no idea what that means and i simply take the definite integral and found the result as(by using the WolframAlpha of course):
    Well i have no idea what does this mean and i ask for some help from you. Can someone explain what this result is and how did i get to this result? By the way E(2) equals:
    0.59907011736779610371996124614016193911360633160782577913... +
    0.59907011736779610371996124614016193911360633160782577913... i
  2. jcsd
  3. Sep 20, 2017 #2


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    I have not tried to follow your math because I think that any "proof" of the value of pi is going to, somewhere along the way, use, implicitly or explicitly, the fact that pi is what it is so you will end up having proved nothing. That is, pi is an empirical value.
  4. Sep 20, 2017 #3


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    You got the formula for arc length wrong. It should be:

    ##L = \int_a^b \sqrt{1+(f'(x))²}dx##

    If you want to calculate ##\pi## using integrals (using numerical methods, of course), I suggest you the following alternative:

    ##\pi = 4\int_0^1 \frac{1}{1+x²}dx##
  5. Sep 20, 2017 #4

    I like Serena

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    After correcting the plus sign as @Math_QED explained, W|A finds:
    $$L = \sqrt{1+\frac{4x^2}{1-4x^2}}dx = \frac 12 \sin^{-1}(2x)\Big|_{-1/2}^{1/2} = \frac\pi 2$$
    as expected.

    I'm not familiar with the E function, but W|A does show that it is apparently related to the Gamma (##\Gamma##) and Pi (##\Pi##) functions, which are generalized versions of the factorial function.

    Note that we're not getting around the ##\sin^{-1}## function, meaning we're effectively using trigonometry regardless of the attempt to use calculus.
  6. Sep 20, 2017 #5
    Thanks for all your help.
  7. Sep 20, 2017 #6
    Euler's textbook on Calculus is interesting because he was the one who popularized the symbol π for the circle constant and he shows the proper way to go about this kind of analysis. Here's the relevant chapter:


    He begins with π as a numerical approximation for the semi-circumference of the unit circle (he cites it to 127 decimal places!) and then develops definitions of the circular functions sin, cos, tan, and so on with this constant. He eventually defines the functions as power series. Then he uses those power series definitions to give justification to infinite series that converge to π. And now the value of π can be calculated to an arbitrary degree of precision.
  8. Sep 21, 2017 #7


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