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Squigonometry Author(s): William E. Wood Reviewed work(s): Source: Mathematics Magazine, Vol. 84, No. 4 (October 2011), pp. 257-265 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/10.4169/math.mag.84.4.257 . Accessed: 09/09/2012 06:26
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Parameterizing the squircle To develop our theory of squigonometry, we must deﬁne functions that do for squircles what cosine and sine do for circles. If we are to use a coupled initial value problem to deﬁne our functions, the solutions u(t) and v(t) must make the function g(t) = u(t)4 + v(t)4 (2) constant. We design the following coupled initial value problem with that property in mind: VOL. 84, NO. 4, OCTOBER 2011 259 3 x (t) = −y(t) 3 y (t) = x(t) x(0) = 1 y(0) = 0 (3) Thus, if u(t) and v(t) are the solution functions to (3), then g (t) = 4u(t)3 u (t) + 4v(t)3 v (t) = −4u(t)3 v(t)3 + 4v(t)3 u(t)3 = 0, so g(t) must be constant. Since g(0) = 1, it follows that g(t) is identically one, as desired. We denote the unique pair of functions satisfying (3) by cq(t) = x(t) and sq(t) = y(t), the cosquine and squine functions, respectively. Again, note that Theorem 2 is what allows us to use the CIVP as a device to deﬁne these functions. E XERCISE 1. Prove that the cosquine and squine functions are even and odd, respectively. E XERCISE 2. Use a computer algebra system to ﬁnd the Maclaurin series for cosquine and squine. We also deﬁne the tanquent function tq(t) = sq(t) . cq(t) Then tq(t)4 + 1, d sq(t)4 + cq(t)4 1 tq(t) = = = 2 dt cq(t) cq(t)2 1 the last equality following from the easily veriﬁable identity tq(t)4 + 1 = cq(t)4 . We see from their graphs in F IGURE 2 that the squigonemetric functions are

Parameterizing the squircle To develop our theory of squigonometry, we must deﬁne functions that do for squircles what cosine and sine do for circles. If we are to use a coupled initial value problem to deﬁne our functions, the solutions u(t) and v(t) must make the function g(t) = u(t)4 + v(t)4 (2) constant. We design the following coupled initial value problem with that property in mind: VOL. 84, NO. 4, OCTOBER 2011 259 3 x (t) = −y(t) 3 y (t) = x(t) x(0) = 1 y(0) = 0 (3) Thus, if u(t) and v(t) are the solution functions to (3), then g (t) = 4u(t)3 u (t) + 4v(t)3 v (t) = −4u(t)3 v(t)3 + 4v(t)3 u(t)3 = 0, so g(t) must be constant. Since g(0) = 1, it follows that g(t) is identically one, as desired. We denote the unique pair of functions satisfying (3) by cq(t) = x(t) and sq(t) = y(t), the cosquine and squine functions, respectively. Again, note that Theorem 2 is what allows us to use the CIVP as a device to deﬁne these functions. E XERCISE 1. Prove that the cosquine and squine functions are even and odd, respectively. E XERCISE 2. Use a computer algebra system to ﬁnd the Maclaurin series for cosquine and squine. We also deﬁne the tanquent function tq(t) = sq(t) . cq(t) Then tq(t)4 + 1, d sq(t)4 + cq(t)4 1 tq(t) = = = 2 dt cq(t) cq(t)2 1 the last equality following from the easily veriﬁable identity tq(t)4 + 1 = cq(t)4 . We see from their graphs in F IGURE 2 that the squigonemetric functions are

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