Raising Sqrt(2)Date: 10/14/97 at 18:49:15 From: Benno Subject: Algebra Dear Dr. Math, What is the value of x in the following equation: .... x x x x x = 2 Thank you. Benno Date: 10/14/97 at 19:15:09 From: Doctor Tom Subject: Re: algebra There's a trick to solving this problem instantly. Since the exponents go on forever, the exponent of the lowest and leftmost x is the same as x to that power, so replace the equation by 2 x = 2 So x is the square root of 2, or 1.4142135... You can see that it makes sense by repeatedly raising sqrt(2) to what you've got so far. To make typing simple, let's call sqrt(2) Q: Q^Q = 1.63252691943 Q^1.63252691943 = 1.76083955587 Q^1.76083955587 = 1.84091086928 Q^1.84091086928 = 1.89271269681 You can do this as long as you want, and the result will get closer and closer to 2, indicating that you've got the right answer. A problem arises if you replace the "2" in the original equation by 4 and try the same thing: x^4 = 4, so x + fourth-root(4) = sqrt(2). That can't be right, since we just checked that the infinite tower of sqrt(2)s converges to 2, not to 4. It turns out you can use the trick only if the value on the right of the equation is less than e = 2.718281828459045... Above that it gives bogus answers. There are in fact some deep mathematical things going on here, and to understand them you'll need to know a lot about convergence of infinite series, among other things. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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