1617-257/Riddle Repository: Difference between revisions

Revision as of 11:43, 22 September 2016

Date	Riddle	Solutions, Discussion, etc...
Sept 12	We want to compute [math]\displaystyle{ (x^x)' }[/math]. Prof. A claims [math]\displaystyle{ (x^n)'=nx^{n-1} }[/math], so [math]\displaystyle{ (x^x)' = xx^{x-1} = x^x }[/math] Prof. B claims [math]\displaystyle{ (a^x)' = a^x\log(a) }[/math], so [math]\displaystyle{ (x^x)' = x^x\log(x) }[/math] Smart student says [math]\displaystyle{ (x^x)' = x^x + x^x\log(x) }[/math]. Why is the derivative the sum of the Prof's solutions?
Sept 14	Can all of [math]\displaystyle{ \mathbb{R}^2 }[/math] be covered by a set of disjoint, non-degenerate, circles? What about [math]\displaystyle{ \mathbb{R}^3 }[/math]? [math]\displaystyle{ \mathbb{R}^4 }[/math]?
Sept 16	Can you find uncountably many disjoint subsets of [math]\displaystyle{ \mathbb{R} }[/math]?
Sept 19	Can uncountably many Y shapes be fit into [math]\displaystyle{ \mathbb{R}^2 }[/math]
Sept 21	On any pair of potatoes, can you draw a pair of 3D congruent curves?
Sept 23

@@ Line 10: / Line 10: @@
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 |Sept 12
+|We want to compute <math>(x^x)'</math>.
-|
+Prof. A claims <math>(x^n)'=nx^{n-1}</math>, so <math>(x^x)' = xx^{x-1} = x^x</math>
+Prof. B claims <math> (a^x)' = a^x\log(a)</math>, so <math>(x^x)' = x^x\log(x)</math>
+Smart student says <math>(x^x)' =  x^x + x^x\log(x)</math>. ''Why is the derivative the sum of the Prof's solutions?''
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 |Sept 14
-| Can all of <math>\mathbb{R}^3</math> be covered by a set of disjoint, non-degenerate, circles?
+| Can all of <math>\mathbb{R}^2</math> be covered by a set of disjoint, non-degenerate, circles? What about <math>\mathbb{R}^3</math>? <math>\mathbb{R}^4</math>?
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