Wikipedia:Reference desk/Archives/Mathematics/2019 April 8

= April 8 =

Fractional exponents
I am a retiree trying to fill in some holes in my education. I am trying to understand exponents. Right now I am trying to wrap my head around fractional exponents. I get 2 to the 2nd power. Multiply 2 by itself two times 2 x 2 = 4. I get 2 to the 3rd power. Multiply 2 by itself three times 2 x 2 x 2 = 8. But what of 2 to the 2.5 power? How do I multiply something by itself two and a half times? And why does my calculator give the answer 5.6569?

I have been studying the web page at https://medium.com/i-math/what-do-fractional-exponents-mean-1bb9bd2fa9a8. Everything on that page makes sense after a little studying, exponents, negative exponents, exponents of numbers smaller than one, but I have a question about fractional exponents. That page only talks about things like 2 to the 2/5 power and how to change it to a root and solve it that way. But what if I am starting with a decimal like 2 to the 2.273 power? The technique the page gives me, splitting up the numerator and the the denominator and moving them around, only works with proper fractions, not decimal fractions. do I just turn the 2.273 into 2273/1000 or is there a simpler way? Retired2786510541 (talk) 10:31, 8 April 2019 (UTC)
 * The fundamental law of exponentiation has $$a^{b + c} = a^b \cdot a^c$$. From this we can define any rational exponent in terms of nth roots; for example, $$2^{10/7} = 2^{1 + 3/7} = 2^{1 + 1/7 + 1/7 + 1/7} = 2 \cdot (2^{1/7})^3$$. In general, when a is positive (and real), we have $$a^{b/c} = (a^{1/c})^b$$ so your example can be written as $$(2^{1/1000})^{2273}$$.
 * For general real exponents, it is easiest to use the exponential function of base-e.-Jasper Deng (talk) 10:38, 8 April 2019 (UTC)
 * That is a very clear explanation. Thank you. If I turn $$2^{2.273}$$ (I mean 2^2.273) into $$(2^{1/1000})^{2273}$$, what do I turn $$2^{e}$$ (I mean 2^e) into? (By "e" I mean E_(mathematical_constant) ) The Wikipedia page for "e" says that it cannot be turned into a fraction. Do I just pretend that "e" is exactly 2.71828 and convert it to $$(2^{1/10000})^{271828}$$? That's pretty close but not exact. Or is it possible to work the problem without having to figure out what proper fraction to turn the decimal fraction into? Thanks for patiently explaining this to me. It has been 50 years since they tried to teach me math in school. It has always bothered me that I don't really understand math so I am spending a lot of time watching math videos on youtube. Math is strangely "beautiful" once you understand it. Retired2786510541 (talk) 16:51, 8 April 2019 (UTC)
 * There are two different methods that give the same answer; both require tools from calculus. (In other words, exponentiation with real number exponents really is a more complicated concept than exponentiation with rational number exponents; quite possibly, you never were introduced to a proper definition.)
 * One way is using the concept of a limit: define $$2^e$$ to be the limit of $$ 2^{a/b}$$ as the fraction $$a/b$$ approaches closer and closer to the number e. [There's something implicit here, namely, that as the numbers $$a/b$$ get closer and closer to e, the exponential $$2^{a/b}$$ gets closer and closer to some fixed value; this is true, but I wouldn't call it obvious.]  This is in the spirit of what you were suggesting: by taking the exponents to be successively the rational numbers 2.7, 2.71, 2.718, 2.7182, ... (or any other sequence that got closer and closer to e).
 * The other approach, which Jasper Deng alluded to, is the following: it turns out that the value $$e^x$$ can be defined for any real number x, in any of several ways (as a value of a certain limit, or using a certain differential equation, or ...), as can the natural logarithm function $$\ln(x) = \log_e(x)$$ (usually as the value of a certain integral). Then using exponentiation and logarithm rules, one can define an arbitrary exponentiation $$a^b$$ by $$a^b = e^{b \cdot \ln(a)}$$ -- that is, to compute a^b, first compute the natural logarithm of a, then multiply that by b, then compute the exponential function e^(that product you just computed).
 * The fact that these (very different-looking!) procedures give the same answer is, again, true but not obvious!
 * Best, JBL (talk) 20:07, 8 April 2019 (UTC)


 * Hi, JBL! Among those several ways to define exp(x) is there any way not relying on real analysis (i.e., not using any integrals or limits)? --CiaPan (talk) 21:24, 8 April 2019 (UTC)
 * Not JBL, but I don't think so. I mean, you can avoid integrals and limits, if you're willing to accept derivatives :-).  But I don't think there's anything that's going to look more elementary than calculus. --Trovatore (talk) 21:50, 8 April 2019 (UTC)
 * The infinite series definition is algebraic:
 * $$e^x = \sum_{n=0}^\infty {x^n \over n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
 * It is all simple addition and multiplication.  21:53, 8 April 2019 (UTC)
 * It requires the notion of a limit. Count Iblis (talk) 22:15, 8 April 2019 (UTC)
 * One could probably get pretty far with treating this as a formal power series definition for $$e$$ - such series can be added and multiplied, etc. But yeah, if you insisted on coercing it to a real number, that is part of analysis. -- 22:41, 8 April 2019 (UTC)
 * Of course if it's algebraic, it's infinitary algebraic (that is, it must accept infinitely long terms). I guess that's "algebraic" in some sense, but it's not the first thing I think of as a referent for that word. --Trovatore (talk) 23:04, 8 April 2019 (UTC)
 * I agree with Trovatore that I don't think there's any way to understand what real number $$2^e$$ is without invoking a limit, or some other mathematical construction (a derivative, an integral) that is itself defined in terms of limits. (This is less weird than it seems, maybe -- there's no way to talk about what value most real numbers have without invoking a limit.  It just happens that the numbers we spend most of our time working with (integers, rationals, quadratic irrationals, $\pi$) don't have this property.) --JBL (talk) 06:05, 9 April 2019 (UTC)
 * Well, I don't know if I said that, exactly. You don't necessarily have to go through limits first.  (Suprema, for example, are not exactly limits, though they can be expressed that way if you like.)  But I think you have to do something that's just as calculus-y, or just as infinitary, as limits themselves. --Trovatore (talk) 06:12, 9 April 2019 (UTC)
 * is there any way [to define the exponential function] not relying on real analysis (i.e., not using any integrals or limits)? - The two most frequent definitions are the power series (see above) and as solution of a differential equation (i.e. the exponential is the function $$f$$ such that $$f' = f$$ and $$f(0) = 1$$), but I assume both count as real analysis.
 * A third way is the functional equation $$f(x + y) = f(x)f(y)$$. You can show with few mathematical tools (only recursion) that the (non-zero) solutions verify $$f(p/q) = f(1)^{p/q}$$ for all rational $$p/q$$, and extend this to real numbers if you assume the function is continuous. However, the useful properties of the function you define thusly are only demonstrated with real analysis; for instance, the exponential function proper (i.e. f(1)=e) is only a noteworthy solution of that equation by the fact that its derivative is itself. Tigraan Click here to contact me 09:22, 9 April 2019 (UTC)
 * Nonetheless, the ability to even define exponentiation for real numbers must involve analysis since completeness is required for existence of a general real power.--Jasper Deng (talk) 11:58, 9 April 2019 (UTC)
 * And there's the concept of a number to the power zero ... Multiplied by itelf no times? →86.184.128.162 (talk) 22:20, 9 April 2019 (UTC)
 * See empty product.--Jasper Deng (talk) 22:39, 9 April 2019 (UTC)