Infinite square rootDate: 6/4/96 at 13:29:14 From: Anonymous Subject: Infinite square root I know this is an easy question, but I do not know how to figure it out. For instance, if y = sqrt(2+ sqrt(2+ sqrt(2+ sqrt(2+ ..., y = 2, and this happens always, because if k = e+ sqrt(e+ sqrt(e+ ..., and it goes indefinitely, k = e. However, how can I prove that this is true, using normal properties of roots? Date: 6/4/96 at 16:14:46 From: Doctor Darrin Subject: Re: Infinite square root If we think about the properties that y has to have, we can find its value. What do we get if we square y? We get 2+sqrt(2+sqrt(2+..., so we see that y^2-2=sqrt(2+sqrt(2+sqrt(2+... . Thus, y^2-2=y, so y is a root of the quadratic equation y^2-y-2==0. You can work out what the roots of this equation are, and this gives two possible values of y (one is positive and one is negative). Since y is defined as a square root, it makes sense to take the positive value. I am not sure what you mean when you say that "this happens always". In the definition you give for y, if we replace 2 by another number (say n), we can use essentially the same procedure to find what y is, but in general, y does not equal n. When n=3, y=2.302..., (the positive root of y^2-y-3=0) and when n=4, y=2.56... The definition you give for k is different, and in fact k does not equal e in general. For e=3, k=5.302... In fact, there is no value of e (except e=0) for which k=e. Thanks for asking the question, it is very interesting -Doctor Darrin, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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