Squigonometry by W. Wood (Disclaimer: I Do Not Own This Research Paper)

4026 WordsMar 10, 201317 Pages
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 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR,…show more content…
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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