How do you find the domain of this function?

Assuming that you are restricting your attention to *real* numbers, then the domain of (sin x)^x should consist of:

* The closed intervals [2πk, 2π(k + 1)], when k is an integer

* All other *rational* numbers which have odd denominator when reduced to lowest terms.

You should delete zero from this domain if you prefer to leave 0^0 undefined.

---

In real analysis, the expression u^v is defined:

* When u ≥ 0

* When u < 0, v is rational, and the denominator of v is odd when v is reduced to lowest terms

The expression 0^0 is left undefined by some sources in some contexts. At this point in my life, I prefer to define 0^0 = 1.

---

Thus, the domain of f should include:

* Everywhere that sin (x) ≥ 0

* Everywhere that sin (x) < 0, x is rational, and the denominator of x is odd when x is reduced to lowest terms

This is precisely what I've stated in my answer above; note that sin (x) ≥ 0 just on intervals of the form [2πk, 2π(k + 1)], where k is an integer.

---

Perhaps I'll say a few words about why (and how) u^v is defined only as above. (My comments here apply to real analysis only--exponents are handled differently in complex analysis.)

If v is an integer, then we all know how u^v is defined.

The natural way to define something like u^(1/n) for integer n is as "the nth root of u," meaning "a number which, when multiplied by itself n times, gives u."

For u^(1/n):

* When u > 0 and n is even, then there are two such numbers; the positive one is usually selected as the value for u^(1/n).

* When u > 0 and n is odd, then there is one such number.

* When u < 0 and n is even, then there are no such numbers, so we must leave u^(1/n) undefined.

* When u < 0 and n is odd, then there is one such number.

A natural way to define something like u^(p/q) is as [u^(1/q)]^p. To make this definition, we require that p/q be in lowest terms; otherwise, for (-1)^(2/6), we would end up with something like [(-1)^(1/6)]^2, which would be undefined, whereas if we had reduced 2/6 to 1/3 we would have obtained [(-1)^(1/3)], which is defined.

Thus, the expression u^(p/q) is not defined when u < 0 and p/q has even denominator when reduced to lowest terms, but it's defined in all other situations.

The way that u^v is defined when u > 0 and v is non-rational is

sup {u^w | w < v and w is rational}.

When u > 0, this supremum always exists (one can show that the set is a bounded set of real numbers, and such a set always has a least upper bound). This definition does not make sense when u is negative, because then u^w is not always defined.

So u^v is left undefined when u is negative and v is either not rational, or is rational and has even denominator when reduced to lowest terms.

---

@ Randy: Some sources identify a "principal cube root" in the complex numbers, meaning "the cube root with the smallest possible nonnegative argument" ("argument" meaning angle with the positive x-axis). I agree that, in the complex numbers, the cube root of -1 which has the smallest possible nonnegative argument is not -1; it's 1/2 + √3i/2, as you have said.

However, in the complex numbers, I do not think that there is really any reason to prefer one cube root over another. I do not think that the notion of "principal root" above is a particularly useful one, because principal roots have no important properties that other roots don't have.

Most properly speaking, in the complex numbers, the expression (-1)^(1/3) is a multi-valued expression, with the values 1/2 + √3i/2, 1/2 - √3i/2, and -1. If you define (-1)^(1/3) to be e^(1/3 ln (-1)), then we get all three values, because ln (-1) is multi-valued with the values πi + 2πik, for k an integer. It turns out that e^(1/3 * (πi + 2πik)) only assumes three values when k is an integer--namely, the three cube roots above.

However, I gather from the way this question was posed that we want the domain of [sin (x)]^x as a real function, not a complex function.

In the context of the real numbers, there are (perhaps) two reasonable ways we could try to define (-1)^(1/3):

* We could say "The complex cube root of -1 with the smallest possible nonnegative argument doesn't happen to be real, so we will leave (-1)^(1/3) undefined."

* We could say "Since there happens to be exactly one cube root of -1 which is real, then when working within the real numbers, we will define (-1)^(1/3) to be that number."

I believe that the second position makes more sense when working in the context of real numbers, and I also believe that it is more standard. For example, all real-number-only elementary algebra textbooks that I am aware of define (-1)^(1/3) to be -1. Also, the TI-89 Titanium, when in "real" mode, gives -1 as the value of (-1)^(1/3). (In "complex" mode, it gives the value 1/2 + √3i/2.)

The second position has the advantage that it is what we get if we take the full multi-valued (sin (z))^z and restrict both its domain and range to be the real numbers. (In this case, to get a single-valued function, we still need to make some kind of choice for nth roots of positive numbers when n is even; but this is the only choice we have to make, because it's the only time (sin (z))^z has two distinct real values.)

(sin(x))^x

= e^(x * ln(sin(x)))

This function will clearly be defined whenever sin(x) is postive, which is when 2nπ < x < (2n + 1)π, where n is any integer. Now, all that's left to find out is the places where the function will be defined when sin(x) is negative.

When sin(x) is negative, let sin(x) = -a, where a = |sin(x)|. So, the function becomes:

e^(x * ln(-a))

= e^(x * (iπ + ln(a)))

= e^(xiπ + xln(a))

= e^(xiπ) * e^(xln(a))

This function is defined whenever e^(xiπ) is a real number, which only happens when x is an integer.

So, the domain of the function will be {(2πn, (2π + 1)n); n ∈Z} ∪{k; k∈Z}.

Hope this helps.

Edit : Yeah, its true that this fuction doesn't have a continuous domain either for positive or negative x. As for the rational powers, I'm having my doubts. Is the cube root of -1, for instance, real or not? This has three values, of which one is real and the other two are not. Well, it all depends on how you define exponents. So, now we need to look for a definition for a^b, where a and b can be any complex numbers. We can define it as 'a multiplied b times' but this definition works only when b is a positive integer. We can extend it for negative integers and finally all rational numbers, but it stops there. We can't use this definition to evaluate something like 2^π. That's where the exponential function comes in. If we write 2^π as e^(πln(2)), we can readily evaluate its vaue using the power series expansion for e^x or something like that. So now we have definted the exponential function for all complex numbers a and b. Now, we can evaluate (-1)^(1/3) using this defintion. Write (-1)^(1/3) as e^(1/3 * ln(-1)) = e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + √3i/2, which is not real. So, according to this definition for exponents, the primary cube root of -1 is not real. So, (sin(x))^x doesn't have a domain for all rational x with odd denominators and is defined only for integral values of x.

This is a fairly complicated function. To begin, let's look at the function

( |Sin(x)| )^x

which is positive and real everywhere, with infinites for x = -πn, where n is a positive integer, everywhere else being finite. Whenever Sin(x) is positive, it is this function just described. When Sin(x) is negative, then it becomes:

(Sin(x))^x (-1)^(x) = (Sin(x))^x Cos(πx) + i (Sin(x))^x Sin(πx)

Thus, it is real only when (Sin(x))^x Sin(πx) = 0, which occurs when x = n, where n is any integer, and x = πn, where n is any positive integer. Then it has the real values:

0 for all x = πn, where n is a positive integer, and

(Sin(n))^n (-1)^n

for all integers n. A messy function! Maybe later on I can provide something of a plot for this function. It is instructive to point out that it doesn't have a simple domain either for negative x OR positive x. Here, I think many texts have this one flat out wrong.

It is obvious for the positive number and 0^0.

It is obvious for negative exponents that are integers. [(sin(-3))]^(-3) = (1 / (sin(-3)))^3

However the following is a counterexample. Since my favorite (laziness) fraction is (-1/2)

That leaves us with the task of determining: [sin(-1/2)]^(-1/2), but that involves taking the square root of a negative number. Hence this is complex. number and not in the domain that is restricted to the reals.

Brian, I leave you to generalize this last one.

Gerry

.

Edited original answer:

For all values of x, positive or negative, lying in the third and fourth quadrant sinx will be negative. In addition if the values of x are such fractions that they become even roots of (sinx), then the value of (sinx)^x will be complex.

Edit:

You sent me an email with the value of [sin(-11.9)]^(-11.9)] is real which I had doubted in my original answer making mistake in analysis and I agree that you are right. However, I have to explain that almost one-half of all real numbers will result in sinx being negative as x will be in 3rd or 4th quadrant. If x also happens to be a fraction such that it results in an even root of sinx, then its value will not be defined. This is true even for positive values of x.

For example, if x = 4.5 radian, [sin(4.5)]^(4.5) will not have a real value. This also means that even a large number of positive fractions of x which gives a negative value of sinx and even root will not have real values. So even the conclusion that for positive x, zero to infinity is the domain is not correct.