I want to know why the equation y 2 = 1 − 4 x 10 12 π 2 gives an approximate square.BackgroundI was just playing around with functions and I wanted to see if y = | sin ⁡ ( π x 2 ) | (radians) would give a semicircle for the interval [0,2] as the distance of (1,0) is the same from (0,0), (2,0) and (1,1), all of which will lie on the curve. The equation of a unit semicircle with its centre at (1,0) is y = 2 x − x 2 I know that the curves of both the equations don't resemble each other much but I still thought of approximating the sine function using this because I thought that it could still be combined with another approximation to make a better approximation. Anyway, I did it and for ϕ = x r a d i a n s , the value of sinϕ can to be approximately 2 π π x − x 2 . It looked like a semi-ellipse and so I verified it to find that it was a semi-ellipse. I thought of using this to derive the equation for an ellipse with it's centre at the origin and the value of a and b being π 2 and 1 respectively.The equation came out to be : y 2 = 1 − 4 x 10 12 π 2 Finally, I thought of playing with this equation and changed the exponent of x. I observed that as I increased the power, keeping it even, the figure got closer and closer to a square. y 2 = 1 − 4 x 10 12 π 2 gave a good approximation of a square. For the exponent of x being some power of 10 greater than 1012, a part of the curve began to disappear.

Question

I want to know why the equation       y    2    =  1  −            4              x                              10                          12                                                  π      2       gives an approximate square.BackgroundI was just playing around with functions and I wanted to see if   y  =      |    sin    ⁡                  (                                      π          x                2                            )              |   (radians) would give a semicircle for the interval [0,2] as the distance of (1,0) is the same from (0,0), (2,0) and (1,1), all of which will lie on the curve. The equation of a unit semicircle with its centre at (1,0) is   y  =      2    x    −          x      2      I know that the curves of both the equations don&#039;t resemble each other much but I still thought of approximating the sine function using this because I thought that it could still be combined with another approximation to make a better approximation. Anyway, I did it and for   ϕ  =  x         r    a    d    i    a    n    s  , the value of sinϕ can to be approximately             2      π            π    x    −          x      2      . It looked like a semi-ellipse and so I verified it to find that it was a semi-ellipse. I thought of using this to derive the equation for an ellipse with it&#039;s centre at the origin and the value of   a and b being             π      2       and 1 respectively.The equation came out to be :       y    2    =  1  −            4              x                              10                          12                                                  π      2      Finally, I thought of playing with this equation and changed the exponent of x. I observed that as I increased the power, keeping it even, the figure got closer and closer to a square.      y    2    =  1  −            4              x                              10                          12                                                  π      2       gave a good approximation of a square. For the exponent of x being some power of 10 greater than 1012, a part of the curve began to disappear.

cenjene9gw · Accepted Answer

First, let&#039;s determine the possible values for x.Since       y    2   is non-negative, we have:  1    −              4              x                                            10                                      12                                                          π            2          ≥    0      x                            10                          12                        ≤                      π            2        4    −            (                                    π                    2                4            )                                10                          −          12                        ≤    x    ≤              (                                    π                    2                4            )                                10                          −          12                      −  1.0000000000009031654105793  …    ≤    x    ≤    1.0000000000009031654105793  …For the decimal approximation used above, see this WolframAlpha computation.Note that for   x  =  ±            (                                    π                    2                4            )                                10                          −          12                                        =                    def                ±  β  , we have       y    2    =  0  , and hence y=0.When   x  =  ±    0.999999  , we find that         y    2    ≈  1    –              10              −      434      ,      000         and     y  ≈  ±      (    1        –                      10                    −        217        ,        000              )  . The table below shows the result of several similar calculations.                    x                              y          2                            y                                                      0                    1                    ±                1                            ±                0.9                    1        −                              10                                −            45            ,            700            ,            000            ,            000                                      ±                  (          1          −                                    10                                      −              22              ,              900              ,              000              ,              000                                )                                    ±                  (          1          −                                    10                                      −              6                                )                        =                ±                0.999999                    1        −                              10                                −            434            ,            000                                      ±                  (          1          −                                    10                                      −              217              ,              000                                )                                    ±                  (          1          −                                    10                                      −              10                                )                        =                ±                0.9999999999                    1                −                2.5        ×                              10                                −            44                                      ±                  (          1                    −                    1.2          ×                                    10                                      −              22                                )                                    ±                  (          1          −                                    10                                      −              12                                )                            0.8509        …                    ±                0.9224        …                            ±                  (          1          −                                    10                                      −              15                                )                            0.5951        …                    ±                0.7714        …                            ±                1                    0.5947        …                    ±                0.7711        …                            ±                1.000000000000903                    0.000165        …                    ±                0.012860        …                            ±                β                    0                    0            Thus, using the fact that       y    2   is a decreasing function of       |    x      |   for   −  β  &amp;lt;  x  &amp;lt;  β  , it follows that the points (x,y) on the graph form two nearly horizonal arcs and two nearly vertical arcs. The upper arc is concave down, has endpoints   (  −  β  ,  0  ) and   (  β  ,  0  )  ,, reaches a maximum height above the x-axis at the point (0,1), and visually it will look like a horizontal segment for   −  β  ≈  −  1  &amp;lt;  x  &amp;lt;  1  ≈  β along with a pair of vertical segments, one at   x  =  1  ≈  β and the other at   x  =  −  1  ≈  −  β  .. The lower arc is the reflection of the upper arc about the x-axis.Visually, the upper arc will look like the upper horizontal and two vertical sides of a rectangle whose vertices are (−1,0) and (−1,1) and (1,1) and (1,0). Visually, the lower arc will look like the lower horizontal and two vertical sides of a rectangle whose vertices are (−1,−1) and (−1,0) and (1,0) and (1,−1). Together, these two arcs will visually look like the four sides of a square whose vertices are (−1,−1) and (−1,1) and (1,1) and (1,−1).

"Why does the graph of y2=1−(4x^10^12)/(pi2) look so much like a square? I want to know why the equation y2=1−(4x^10^12)/(pi2) gives an approximate square.

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