15. The following puzzle appeared without attribution in the Spring 1989 Newsletter of the Northeastern Section of the Mathematical Association of America. What is the flaw in this argument? Euler's identity? ei = (ei)2π/2 = (e²ni) 0/2π = (1) 0/2π = 1.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter6: The Trigonometric Functions
Section6.2: Trigonometric Functions Of Angles
Problem 79E
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15. The following puzzle appeared without attribution in the Spring 1989 Newsletter of the
Northeastern Section of the Mathematical Association of America. What is the flaw in
this argument?
Euler's identity?
ei = (ei)2π/2 = (e²ni) 0/2π = (1) 0/2π = 1.
Transcribed Image Text:15. The following puzzle appeared without attribution in the Spring 1989 Newsletter of the Northeastern Section of the Mathematical Association of America. What is the flaw in this argument? Euler's identity? ei = (ei)2π/2 = (e²ni) 0/2π = (1) 0/2π = 1.
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