Disagreed—if you know the general shape and you know the derivative at 0 is 1, then while you can’t calculate pi very accurately, you can find out that it’s closer to 3 than to 5.
Yeah, I thought about that, but this information doesn’t exactly define the curve, and so it becomes unclear which portion of the work is done by visual imagination, and which just fits the known result, taking a few obvious bounds into account. Unrolling half a circle, on the other hand...
It took me a little while to think of a definition of the sine function that does mention pi, though it turned out to be the first one taught in (my) school: “the y coordinate after going t/2pi times counterclockwise around the unit circle starting at (1,0)”. If I were to draw the curve, I’d use Euler’s method or roll a circle, both of which use the derivative going between −1 and 1 instead of pi for the frame of reference.
Disagreed—if you know the general shape and you know the derivative at 0 is 1, then while you can’t calculate pi very accurately, you can find out that it’s closer to 3 than to 5.
If you know the derivative at 0 is 1, then you know the value of pi… just sayin’.
That’s not strictly true, seeing as...
...but I agree that general-shape + derivative-at-zero is not really enough to form estimate of pi.
Yeah, I thought about that, but this information doesn’t exactly define the curve, and so it becomes unclear which portion of the work is done by visual imagination, and which just fits the known result, taking a few obvious bounds into account. Unrolling half a circle, on the other hand...
It took me a little while to think of a definition of the sine function that does mention pi, though it turned out to be the first one taught in (my) school: “the y coordinate after going t/2pi times counterclockwise around the unit circle starting at (1,0)”. If I were to draw the curve, I’d use Euler’s method or roll a circle, both of which use the derivative going between −1 and 1 instead of pi for the frame of reference.
Since the derivative is also a sine curve, it helps only very approximately.