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How is the expansion of √x√x√x..∞ equal to x?
Please try to understand that the first square root is enclosing the second one which in turn is enclosing the next...so on till infinity.
BA to be given.
The first square root is enclosing all the square roots and the second one too is enclosing all the square roots but the first one.
1 Answer
- ?Lv 65 years agoFavorite Answer
If the expansion has only 1 radical sign then it becomes
√x = x^(1/2)
If the expansion has 2 radical signs then it becomes
√(x√x) = (x*x^(1/2))^(1/2) = (x^(3/2))^(1/2) = x^(3/4)
If the expansion has 3 radical signs then it becomes
√(x√(x√x)) = (x*x^(3/4))^(1/2) = (x^(7/4))^(1/2) = x^(7/8)
And so on .
Therefore , if the expansion has n radical signs , it becomes x^(((2^n) - 1) / (2^n)) .
If n increases then ((2^n) - 1) / (2^n) = 1 - 1 / (2^n) approaches 1 .
So if n goes to inf , x^(((2^n)-1) / (2^n)) goes to x .
