August 4, 2010 6:50 PM Subscribe

I need help understanding the "principle domain" (a term only my professor seems to use) of a polar function. That is, how do I find the smallest value delta such that [0,delta] plots all of the unique points on the curve; any values greater than delta re-trace points. I suspect that if MeFi could identify the more common name for this, uh, procedure, I'd be set.

posted by phrontist to Education (10 answers total)

posted by phrontist to Education (10 answers total)

Ah, yes, probably is. Still no luck googling though.

This is for a Calculus II class, if it's relevant.

posted by phrontist at 6:58 PM on August 4, 2010

This is for a Calculus II class, if it's relevant.

posted by phrontist at 6:58 PM on August 4, 2010

Well, if you imagine a picture of an arbitrary differentiable function, a portion of the function which is one-to-one (has only one of any particular y-value) is one in which the derivative of the function does not change sign. As soon as the derivative passes through zero, a maximum or minimum occurs, which means the function will start repeating y-values.

So, to find a domain where the function is one-to-one, you could take the derivative, and select a region between two zero-crossings of the derivative which spans the entire range of the function. For instance, for sin(x)

d(sin(x))/dx = cos(x)

cos(x) has zeros at x = -pi/2 and pi/2

Between those values, sin(x) spans the whole range of sin(x) (from -1 to 1)

Therefore the domain [-pi/2, pi/2] is the domain you're looking for - I guess you could call it a "principal domain".

posted by Salvor Hardin at 7:04 PM on August 4, 2010

So, to find a domain where the function is one-to-one, you could take the derivative, and select a region between two zero-crossings of the derivative which spans the entire range of the function. For instance, for sin(x)

d(sin(x))/dx = cos(x)

cos(x) has zeros at x = -pi/2 and pi/2

Between those values, sin(x) spans the whole range of sin(x) (from -1 to 1)

Therefore the domain [-pi/2, pi/2] is the domain you're looking for - I guess you could call it a "principal domain".

posted by Salvor Hardin at 7:04 PM on August 4, 2010

[0, 2*pi] wouldn't fulfill the uniqueness criteria, since sin(x) repeats y-values for that domain.

posted by Salvor Hardin at 7:05 PM on August 4, 2010

posted by Salvor Hardin at 7:05 PM on August 4, 2010

Salvor has a more general approach, though I'll point out that technically one end of the interval should be open, e.g., [-π/2, π/2).

posted by jedicus at 7:13 PM on August 4, 2010

posted by jedicus at 7:13 PM on August 4, 2010

The original poster seems to be talking about *polar* functions, e.g. r= sin(theta), for which the 'principal domain' would indeed be [0, 2*pi); or r = cos(2*theta), in which case we cover all the unique points on the polar curve in the domain [0, pi). If I'm wrong, disregard the next paragraph (which is unhelpful anyway).

In general I don't know of a non-tedious procedure for doing this. You could potentially 'take the easy way out' and graph it with a graphing device or software and determine a domain for theta in that way. But mainly all I can think of is to graph it yourself with pencil and paper: for different values of theta, you get different values of r to plot -- at some point you start retracing the curve. And certainly determining the periodicity of your function is helpful. And sometimes, if you can convert your polar function to rectangular (e.g., 1/(sin(theta) + cos(theta)) is much easier to work with in cartesian coordinates) that can help with the graphing process.

posted by evinrude at 7:43 PM on August 4, 2010

In general I don't know of a non-tedious procedure for doing this. You could potentially 'take the easy way out' and graph it with a graphing device or software and determine a domain for theta in that way. But mainly all I can think of is to graph it yourself with pencil and paper: for different values of theta, you get different values of r to plot -- at some point you start retracing the curve. And certainly determining the periodicity of your function is helpful. And sometimes, if you can convert your polar function to rectangular (e.g., 1/(sin(theta) + cos(theta)) is much easier to work with in cartesian coordinates) that can help with the graphing process.

posted by evinrude at 7:43 PM on August 4, 2010

Uh, wait -- the principal domain of r = sin(theta) would just be [0, pi) after all -- as theta goes from pi to 2*pi, we traverse the circle a second time. And the principal domain of r=cos(2*theta) is [0, pi/2). (And I only knew that by actually sketching the graph in my head, not by following some general procedure -- sorry)

posted by evinrude at 7:46 PM on August 4, 2010

posted by evinrude at 7:46 PM on August 4, 2010

Look in a complex analysis book. What your prof is trying to get at is also known as the principle value of a complex valued function (which, as you may know, can be visualized as a polar function by using the euler formula, r * exp(i*phi) =r*( cos(phi) + i * sin(phi)))

see: Principal Value from Wolfram Mathworld

posted by scalespace at 7:47 PM on August 4, 2010 [1 favorite]

see: Principal Value from Wolfram Mathworld

posted by scalespace at 7:47 PM on August 4, 2010 [1 favorite]

See p. 71 of this text (PDF), also this doc, p. 1 here, here.

Or in fact look at all these.

In short, I think "principal domain" is in fact a very common name for this concept--part of the trouble is it also seems to be a very common phrase so when you google it you find tons of non-math related results unless you also add some other math-related search terms (trig, polar coordinates, etc.).

posted by flug at 11:06 PM on August 4, 2010

Or in fact look at all these.

In short, I think "principal domain" is in fact a very common name for this concept--part of the trouble is it also seems to be a very common phrase so when you google it you find tons of non-math related results unless you also add some other math-related search terms (trig, polar coordinates, etc.).

posted by flug at 11:06 PM on August 4, 2010

This thread is closed to new comments.

posted by box at 6:56 PM on August 4, 2010