Square Roots and LimitsDate: 03/12/2003 at 21:39:19 From: Jonathan Coveney Subject: Square Roots Given a value of x, find the value of the expression sqrt(x sqrt(x sqrt(x... and so on, and explain why. Date: 03/13/2003 at 03:41:32 From: Doctor Ian Subject: Re: Square Roots Hi Jonathan, The easiest way to deal with this is to make use of the fact that sqrt(x) and x^(1/2) are just two ways of saying the same thing. If that doesn't seem familiar to you, take a look at Properties of Exponents http://mathforum.org/library/drmath/view/57293.html So, _______ | __ 1/2 1/2 \| x \| x = (x * x ) 3/2 1/2 = (x ) (3/2 * 1/2) = x 3/4 = x and ____________ | _______ | | __ 3/4 1/2 \| x \| x \| x = (x * x ) 7/4 1/2 = (x ) (7/4 * 1/2) = x 7/8 = x At each step, the exponent changes from (k-1)/k to ((k + (k-1))/k)/2 = (2k - 1)/(2k) More generally, for n square roots, the exponent is (2^n - 1)/(2^n) So you're right that as n approaches infinity, the exponent approaches 1, and the value of the expression approaches x. However, note that the denominator of the exponent will always be a power of 2, so x^(19/20) isn't really part of the sequence. Now for a bit of terminology. The way that we express this situation is to say Limit (2^n - 1)/(2^n) = 1 n->inf "The limit, as n goes to infinity, of [the expression] is equal to 1." This is somewhat subtle. Note that we _don't_ say that the expression itself, (2^n - 1)/(2^n) is ever actually equal to 1, because that's never the case. And we _don't_ say that n ever _becomes_ 'infinity', because infinity isn't a number, so we can't use it in arithmetical expressions. We're saying that we can make the value of the expression as close to 1 as we wish, by choosing a suitably high value for n. You'll learn more about limits when you get into calculus. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 03/13/2003 at 08:29:19 From: Jonathan Coveney Subject: Thank you (Square Roots) Thanks, I wondered what limits were, exactly. But now it makes plain sense. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/