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# What is the similar equation for 2n+1?

Posted 9 years ago
 For 2n | n is an element of the integers, FullSimplify[(1 - Sqrt)^(2 n) == (1 + Sqrt)^(-2 n), Assumptions -> n \[Element] Integers]  Yields True What ts the similar equation for 2n+1? Marvin Ray Burns
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Posted 9 years ago
 In:= FullSimplify[(Sqrt - 1)^(2 n + 1) == (Sqrt + 1)^-(2 n + 1), Assumptions -> Element[n, Integers]] Out= True 
Posted 9 years ago
 Do you mean the following? (Replacing 1 with -1 and 2n with 2n+1 doesn't produce a true statement.) In:= m := 2 n + 1; FullSimplify[((-1) - Sqrt)^(m) == (1 + Sqrt)^(- m), Assumptions -> n \[Element] Integers] Out= False
Posted 9 years ago
 Your original expression is equivalent to determining how the following behaves for various conditions on n: In:= FullSimplify[(1 - Sqrt)^(2 n) (1 + Sqrt)^(2 n)] Out= (-1)^(2 n) which, for integer n, is one. But you can also evaluate it for various other values of n. The 2n+1 case can be treated similarly: In:= FullSimplify[(1 - Sqrt)^(2 n + 1) (1 + Sqrt)^(2 n + 1)] Out= (-1)^(1 + 2 n) In this case, if for example n is an integer, then the value is -1.