Meaning of Irrational ExponentsDate: 03/17/97 at 21:11:51 From: Jonah Knobler Subject: Meaning of Irrational Exponents I am a student in an Algebra II class, and I'm wondering a bit about exponents and roots. We know that 2^2 is the product of 2 factors of 2, and that y^x is the product of x factors of y. I also know that 2^(1/2) is the product of 1/2 a factor of 2 (i.e. it needs to be taken twice to even reach the base) and that y^(1/x) is the product of 1/x factors of y (i.e. after taking y an x number of times, we reach y^1). That's understandable. But where do irrational exponents fit in? Can you raise 2 to the sqrt(2) power? Is there any definition for this? I know there must be, because Euler's Formula (about which I've only heard vague whispers as yet) says something about it. But how can you raise a number to an irrational power? How can you have an irrational number of factors? I can't fathom it. Similarly, is there a sqrt(2) root of 2? (i.e. irrational roots)? Can you find the pi root of seven? How? What is the definition? A final corollary: what about imaginary and complex roots? How would you evaluate i to the i power? Is it real or imaginary? How can you have a number multiplied by itself an IMAGINARY or worse, a COMPLEX number of times, when you can't even count to i or (i+1)? - Jonah Knobler Date: 03/18/97 at 13:45:48 From: Doctor Rob Subject: Re: Meaning of Irrational Exponents >Can you raise 2 to the sqrt(2) power? ... How can you have an >irrational number of factors? Yes, you can raise 2 to the sqrt(2) power. You do it by approximating the irrational exponent by rational exponents that are closer and closer to the irrational one. In your example, you want the exponent sqrt(2). You know that, to one significant figure, sqrt(2) = 1, so the first approximation to 2^sqrt(2) is 2^1 = 2. To two significant figures, sqrt(2) = 1.4 = 14/10 = 7/5, so the second approximation to 2^sqrt(2) is 2^(7/5) = the 5th root of 2^7 or the 7th power of the 5th root of 2, or approximately 2.639. To three significant figures, sqrt(2) = 1.41 = 141/100, so you have to take a 100th root of 2 and raise it to the 141st power, or else take the 100th root of 2^141, to get approximately 2.65737 Continuing this way, we get the sequence of approximations 2, 2.639, 2.6574, 2.66475, 2.665119, 2.6651375, 2.66514310, ... These converge to a limit (this is a calculus idea) because the function 2^x is continuous (another calculus concept). That limit is what we call the value of 2^sqrt(2) ~=~ 2.665144143. We can't write down all the digits in the decimal expansion, but then neither can we do so for sqrt(2), so that doesn't bother us. Presumably you have not yet learned about logarithms, because what you are asking about is just the question of how do you compute the antilogarithm of sqrt(2) to the base 2. Patience, you should see this before you finish Algebra II. >Similarly, is there a sqrt(2) root of 2? (i.e. irrational roots)? >Can you find the pi root of seven? How? What is the definition? Roots are just powers with exponents which are reciprocals. Your first quantity would be 2^(1/sqrt(2)) = 2^(sqrt(2)/2) = sqrt(2^sqrt(2)) which we computed above. Answer, approximately 1.632526919. Your second quantity would be 7^(1/pi), and 1/pi ~=~ 0.318309886. Find rational numbers which get closer and closer to this decimal, and raise 7 to each of them, and they will get closer and closer to some number which is defined to be your second quantity. In this case, about 1.857817549. Of course, I haven't told you how to take the 100th root of 2^141. There are various numerical ways to do that. A simple one conceptually is to guess and check: Try 2.1: raise it to the 100th power and compare to 2^141. I got 1.66*10^32, whereas 2^141 = 2.78759*10^42, so our guess is too small. Try 2.8. 2.8^100 = 5.1976*10^44, so 2.8 is too big. Split the difference, and try 2.45, getting 8.25*10^38, again too small. Split the difference, and try 2.625, getting 8.18*10^41, still too small. Try 2.7, 1.37*10^43, too big. Try 2.66, 3.077*10^42, a hair too big. Try 2.65, 2.111*10^42, too small. So the answer is between 2.65 and 2.66, as we found above. This relies on knowing that the function 2^x is an increasing function of x (more calculus ideas). >A final corollary: what about imaginary and complex roots? How would >you evaluate i to the i power? Is it real or imaginary? How can you >have a number multiplied by itself an IMAGINARY or worse, a COMPLEX >number of times, when you can't even count to i or (i+1)? This really has to wait to calculus and the theory of infinite series, unless you are willing to take DeMoivre's Law on faith. It says e^(i*x) = cos(x) + i*sin(x) where e is 2.718281827459..., the base of natural logarithms, and i is the square root of -1. Using this, it is possible to evaluate things like r = i^i. If we put x = pi/2 in DeMoivre's Law, we are told that e^(i*pi/2) = i. Now just raise both sides to the i-th power, using the laws of exponents, and find that r = i^i = e^(i^2*pi/2) = e^(-pi/2), which is a real number. Bizarre, no? This is not the whole story, however! Try using DeMoivre's Law with x = 5*pi/2, and do the same calculation over. <GRIN> Keep up the good questions! -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/