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What is the similar equation for 2n+1?

For 2n | n is an element of the integers,

FullSimplify[(1 - Sqrt[2])^(2 n) == (1 + Sqrt[2])^(-2 n), 
     Assumptions -> n \[Element] Integers]

Yields True

What ts the similar equation for 2n+1?

Marvin Ray Burns

POSTED BY: Marvin Ray Burns
3 Replies

Your original expression is equivalent to determining how the following behaves for various conditions on n:

In[1]:= FullSimplify[(1 - Sqrt[2])^(2 n) (1 + Sqrt[2])^(2 n)]

Out[1]= (-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[2]:= FullSimplify[(1 - Sqrt[2])^(2 n + 1) (1 + Sqrt[2])^(2 n + 1)]

Out[2]= (-1)^(1 + 2 n)

In this case, if for example n is an integer, then the value is -1.

POSTED BY: David Reiss

Do you mean the following? (Replacing 1 with -1 and 2n with 2n+1 doesn't produce a true statement.)

In[13]:= m := 
 2 n + 1; FullSimplify[((-1) - Sqrt[2])^(m) == (1 + Sqrt[2])^(- m), 
 Assumptions -> n \[Element] Integers]

Out[13]= False

POSTED BY: Marvin Ray Burns
In[2]:= FullSimplify[(Sqrt[2] - 1)^(2 n + 1) == (Sqrt[2] + 1)^-(2 n + 1), Assumptions -> Element[n, Integers]]

Out[2]= True
POSTED BY: Ilian Gachevski
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