"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.
I want to know why the equation gives an approximate square.
I was just playing around with functions and I wanted to see if (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
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 , the value of sinϕ can to be approximately . 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 and b being and 1 respectively.
The equation came out to be :
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.
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.
Answer & Explanation
First, let's determine the possible values for x.
Since is non-negative, we have:
For the decimal approximation used above, see this WolframAlpha computation.
Note that for we have and hence y=0.
When we find that and . The table below shows the result of several similar calculations.
Thus, using the fact that is a decreasing function of for 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 and , reaches a maximum height above the x-axis at the point (0,1), and visually it will look like a horizontal segment for along with a pair of vertical segments, one at and the other at . 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).