sketchucation logo sketchucation
    • Login
    ⚠️ Attention | Having issues with Sketchucation Tools 5? Report Here

    An argument of infinite proportions (pi)

    Scheduled Pinned Locked Moved Corner Bar
    24 Posts 10 Posters 539 Views 10 Watching
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • J Offline
      JuanV.Soler
      last edited by

      My thinking is that pi, must in fact, end .
      (for it must provide a shape).
      Whilst 1/3 does not have to provide a shape.
      it is not his work.
      can dream.

      JuanV.Soler

      ,))),

      1 Reply Last reply Reply Quote 0
      • M Offline
        Masta Squidge
        last edited by

        JuanV.Soler wrote:
        My thinking is that pi, must in fact, end .
        (for it must provide a shape).
        Whilst 1/3 does not have to provide a shape.
        it is not his work.
        can dream.

        Smile

        ooh circular reasoning (pun intended)

        That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)

        I would be interested in running that by someone in the profession of calculating stupid numbers to places beyond any usable level of accuracy. Something about that statement, while it makes sense right now, doesn't exactly sit right in the back of my mind and I cant place it.

        Anyways, I'm off to bed so I can survive work tonight, please by all means continue this because I love reading these sorts of things.

        Masta Squidge

        1 Reply Last reply Reply Quote 0
        • J Offline
          JuanV.Soler
          last edited by

          Quote:
          That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)

          Sorry, I dont get you.

          JuanV.Soler

          ,))),

          1 Reply Last reply Reply Quote 0
          • soloS Offline
            solo
            last edited by

            So who was the genius that coined the phrase "easy as pi"?

            solo

            http://www.solos-art.com

            If you see a toilet in your dreams do not use it.

            1 Reply Last reply Reply Quote 0
            • soloS Offline
              solo
              last edited by

              A number isn't just something made up it's the answer to an equation. Under that definition if you take that answer to equations can sometimes be quantified as between a real & imaginary number and infinity then numbers can go on forever. Numbers can be limitless.
              You should tell him 2 isn't really 2 and see what he says.
              Pi is limitless, Pie is not... especially around guys from the south.

              solo

              http://www.solos-art.com

              If you see a toilet in your dreams do not use it.

              1 Reply Last reply Reply Quote 0
              • J Offline
                Jackson
                last edited by

                Okay, maths is very much my weakness, but I'm confused by your stance. In the other forum you wrote:

                "I say they never end as pi is an irrational (which by definition never ends) and that 1/3 also never ends as it is a non terminating repeating decimal."

                But in your post here you state:

                "Now, anyone who paid attention in math class in middle school knows that he is wrong, that pi is called an irrational number for a reason, and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."

                For the record I agree that neither number ends- especially 1/3 as a decimal. Your "opponent" ain't worth the argument if he seriously reasons that 1/3 expressed as a decimal logically ends in a 3!

                Jackson

                Jackson

                1 Reply Last reply Reply Quote 0
                • 3 Offline
                  3eighty
                  last edited by

                  What are you guys smoking? E Grin Pi r not Sq...Pi art round! Thumbs up

                  3eighty

                  1 Reply Last reply Reply Quote 0
                  • R Offline
                    Ross Macintosh
                    last edited by

                    Masta Squidge wrote:
                    someone in the profession of calculating stupid numbers...

                    Recently it was reported that some university professor here in Canada completed a massive seven or eight year long calculation using a supercomputer. He was able to absolutely prove that a game of checkers will always end in a draw provided neither player ever makes a mistake or cheats. It is only the mistakes and cheats that swing the outcome away from a draw.

                    Its great to see my tax dollars at work.

                    I can't understand why anyone would think 1/3 would ever end. The repeating character of the ratio seems easy enough to comprehend as unending. I hope that university professor doesn't decide to tackle the 1/3 number next.

                    Ross Macintosh

                    1 Reply Last reply Reply Quote 0
                    • M Offline
                      Masta Squidge
                      last edited by

                      Jackson Barkess wrote:
                      ...and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."

                      I believe that is what we call a typo Embarassed I believe it should read "does in fact NOT end, thats why its called a repeating, NON-terminating decimal.

                      Yeah, I slipped, it was past my bedtime.

                      Masta Squidge

                      1 Reply Last reply Reply Quote 0
                      • S Offline
                        SchreiberBike
                        last edited by

                        Masta Squidge wrote:
                        Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.

                        Yep, that's the one.

                        SchreiberBike

                        1 Reply Last reply Reply Quote 0
                        • S Offline
                          SchreiberBike
                          last edited by

                          Not that it will help, but here is a mathematical proof that pi is irrational:

                          Quote:
                          Niven's proof

                          The proof uses the characterization of π as the the smallest positive zero of the sine function. Like all proofs of irrationality, the argument proceeds by reductio ad absurdum.

                          Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function

                          f(x) = {x^n(a - bx)^n \over n!},\quad x\in\mathbb{R},

                          and denote by

                          F(x) = f(x) + \cdots + (-1)^j f^{(2j)}(x) + \cdots + (-1)^n f^{(2n)}(x). ,

                          the alternating sum of f and its first n even derivatives.

                          Claim 1: F(0) = F(π)

                          Proof: Since

                          f(x)=b^n{x^n(\pi - x)^n \over n!}=f(\pi-x),\quad x\in\mathbb{R},

                          the chain rule and mathematical induction imply

                          f^{(j)}(x) = (-1)^j f^{(j)}(\pi - x),\quad x\in\mathbb{R},

                          for all the derivatives, in particular

                          f^{(2j)}(0)=f^{(2j)}(\pi),

                          for j = 1, 2, ...,n and Claim 1 follows from the definition of F.

                          Claim 2: F(0) is an integer.

                          Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we get the representation

                          f(x)={1\over n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j}.,

                          Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n, we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,

                          f^{(j)}(0)={j!\over n!}{n \choose j-n}a^{2n-j}(-b)^{j-n}\quad\mbox{for } n\le j\le 2n.

                          Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and every derivative of f at 0 is an integer and so is F(0).

                          Claim 3:

                          \frac12 \int_0^\pi f(x)\sin(x),dx=F(0)

                          Proof: Since f (2n + 2) is the zero polynomial, we have

                          F'' + F = f.,

                          The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x, hence the product rule implies

                          (F'\cdot\sin - F\cdot\cos)' = f\cdot\sin.

                          By the fundamental theorem of calculus

                          \frac12 \int_0^\pi f(x)\sin(x),dx= \frac12 \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{x=0}^{x=\pi}.

                          Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.

                          Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 1 and 3 show that F(0) is a positive integer. Since

                          x(\pi -x) = \Bigl(\frac\pi2\Bigr)^2-\Bigl(x-\frac\pi2\Bigr)^2\le\Bigl(\frac\pi2\Bigr)^2,\quad x\in\mathbb{R},

                          and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have

                          \frac12 \int_0^\pi f(x)\sin(x),dx\le \frac{b^n}{n!}\Bigl(\frac\pi2\Bigr)^{2n+1},

                          which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0).

                          Not that I understand it, but there it is.

                          SchreiberBike

                          1 Reply Last reply Reply Quote 0
                          • GaieusG Offline
                            Gaieus
                            last edited by

                            Aha...
                            Well, thanks - at least for that very last line (now I can hope I'm not alone in the universe who does not understand it

                            Gaieus

                            Gai...

                            1 Reply Last reply Reply Quote 0
                            • 1
                            • 2
                            • 2 / 2
                            • First post
                              Last post
                            Buy SketchPlus
                            Buy SUbD
                            Buy WrapR
                            Buy eBook
                            Buy Modelur
                            Buy Vertex Tools
                            Buy SketchCuisine
                            Buy FormFonts

                            Advertisement