Nested Square Roots
Date: 07/17/98 at 11:15:32 From: Natasha K. Subject: Number series question Dear Dr. Math, I am in the 8th grade and I am on my school's math team. I have been practicing over the summer for next year's math league. Here is a question that I came across that I couldn't solve. I didn't even know where to begin: Solve for n: n = square root (6 + square root (6 + square root 6 + .. How can you prove what this series converges to? Thank you.
Date: 07/17/98 at 13:07:20 From: Doctor Nick Subject: Re: number series question Hi Natasha, The trick here is to notice that some of the stuff on the righthand side looks like the whole expression. That is: n = square root ( 6 + n ) One might object to this since it appears that the stuff inside the second square root has one fewer 6 than the whole expression; however, since there is an infinite number of 6's, the number of 6's in each expression is the same. So, if we square both sides we get: n^2 = 6 + n It might be clearer to get to this another way. Notice that if you square the right side and subtract 6, you get the expression you started with, n. This gives us: n^2 - n - 6 = 0 so: (n-3)(n+2) = 0 Thus n is either -2 or 3, but n can't be -2, since n has to be positive (it's at least as big as the square root of 6). Thus n is 3. You have to be careful with problems like this to make sure that the expression you're dealing with actually defines a number. When working with infinite expressions, it's especially important to be careful to know what's going on. It's not too hard to show that your expression does define a real number, but it takes a little care. Notice that we could replace 6 by any other positive number. Can you work out a general formula for: square root(a + square root(a + squareroot(a + ... where a is any positive number? (You'll probably need the quadratic formula). Have fun, - Doctor Nick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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