Wikipedia:Reference desk/Archives/Mathematics/2012 April 23

= April 23 =

Pi
What would you give a proper definition of Pi?How do every circle's C is always about just more than three times of its D?


 * Pi is defined as the ratio of a circle's circumference to its diameter. Are you asking for a proof that the ratio is constant for all circles? Or are you looking for an algebraic definition? KyuubiSeal (talk) 01:45, 23 April 2012 (UTC)


 * (ec) You just gave the definition, the ratio of a circle's circumference to it's diameter. Why it's an irrational number is a question for philosophers, not mathematicians. StuRat (talk) 01:46, 23 April 2012 (UTC)

Modulus
Is the set of all ordered k-tuples mod n Zk/n or (Z/n)k? --108.206.4.199 (talk) 03:04, 23 April 2012 (UTC)
 * I think the second one is more correct or at least more common, but either one would probably get the point across. If you're considering (Z/n)k as a group, it's a quotient group of Zk, but not by n since n isn't in Zk.  But (Z/n)k is also a Z-module and is the quotient module of Zk by nZk so Zk/n works in that sense.  Rckrone (talk) 16:15, 23 April 2012 (UTC)

integrals
Using the concept of differentials, is there a meaning to things like $$\int x^2 d(cosx)$$? --85.250.73.51 (talk) 10:03, 23 April 2012 (UTC)
 * $$\int x^2 d(\cos x)=-\int x^2 \sin x dx$$

What is the problem? Bo Jacoby (talk) 10:11, 23 April 2012 (UTC).
 * Things like this, with nontrivial stuff inside the d, are basically what you are doing when you do Integration by substitution. Staecker (talk) 10:29, 23 April 2012 (UTC)
 * Indeed ...
 * $$u = \cos x$$


 * $$\Rightarrow \frac{\text{d}u}{\text{d}x} = - \sin x$$


 * $$\Rightarrow \text{d}u = - \sin x \, \text{d}x$$
 * (pause while pure mathematicians in the room are revived with smelling salts) Gandalf61 (talk) 11:09, 23 April 2012 (UTC)
 * We're not fainting, we're napping through the boring parts! Rschwieb (talk) 15:07, 23 April 2012 (UTC)

Minor details
Let $$f$$ be a uniformly continuous real valued function defined on a bounded set $$E \subset \mathbb{R}$$. Prove that $$f$$ is bounded on $$E$$.

I think I understand intuitively why this is true, and this is the proof that I have written down: Fix some $$a \in E$$ and some $$\epsilon >0 $$. Since $$f$$ is continuous at $$a$$, there exists some $$\delta$$ such that $$f(x) \in B_{\epsilon}(f(a))$$ for all $$x$$ in the set of points $$\{x \in E: d(x,a) < \delta\}$$. Note therefore that $$f$$ is bounded on this set of points. Now consider an increasing sequence $$(x_n)_{n=1}^m$$ of points in $$\mathbb{R}$$ such that $$\forall x \in E : x_0\le x \le x_m$$ and that $$d(x_{j},x_{j+1}) < \delta$$ for $$j = 1, ..., m-1$$. Since $$f$$ is uniformly continuous, such a sequence exists and it follows that each on each interval $$[x_j,x_{j+1}]$$, $$f([x_j,x_{j+1}]) \subset B_\epsilon(f(x_j))$$. Therefore the function is bounded on its whole domain.

I think I may be making a few unjustified assertions which intuitively I know are true but that I have not formally proven, such as the fact that a such a sequence exists if $$f$$ is uniformly continuous, and that the fact that E is bounded and the function is bounded on each subinterval implies that the function is bounded on the whole of E. Widener (talk) 17:25, 23 April 2012 (UTC)
 * It's the right idea but I think you need to frame it a bit differently. If you want to prove the function is bounded, you need to prove that no sequence of points x1,x2,... exists such that f(xn) goes to infinity.  So assume that there is such a sequence.  Since E is contained in some compact set C, the sequence has a convergent subsequence (though the limit x is not necessarily in E).  Eventually these points are all very close to one another yet some pairs have |f(xi)-f(xj)| very large. Rckrone (talk) 23:18, 23 April 2012 (UTC)
 * It's a long time since I did this kind of stuff, but doesn't the following argument do it? Uniform continuity means that for all ε > 0 there exists a δ such that, for all x1 and x2 in E, |x1 - x2| < δ => |f(x1) - f(x2)| < ε. Take ε = 1 and subdivide E into intervals of width δ/2 - as E is bounded there are finitely many of these, say N. Then the value taken by f can't vary by more than Nε = N over E, and so f is bounded. Actually I think this is pretty much what Widener was saying, but in a simplified form. AndrewWTaylor (talk) 11:54, 24 April 2012 (UTC)
 * The problem is that E may have gaps. If E has a gap in it larger than δ, the values on either side can differ by an arbitrarily large amount, so the function can vary more than Nε.  It's probably possible to patch this up by arguing that there can't be more than a finite number of these gaps, and then showing that the argument really does work on each piece with no large gaps.  But this seems pretty messy already. Rckrone (talk) 12:39, 24 April 2012 (UTC)
 * @Widener: Your original post kind of reflects a trick I always keep in mind for uniformly continuous functions. The trick the definition permits you is: given an epsilon, you are guaranteed there is a delta such that the whole graph can be caught in nonoverlapping boxes of dimensions 2ε tall by δ wide (any positive number less than 2δ would be OK for a width, too). Thus if E is bounded, you can cover it with finitely many nonoverlapping closed intervals of width δ, and determine which horizontal edges of the finitely many boxes go highest/lowest to determine your bounds on f(E). Rschwieb (talk) 13:12, 25 April 2012 (UTC)

How to find a cubic root of a number in regular calculator models?
Max Viwe &#124;  Wanna chat with me?  19:32, 23 April 2012 (UTC)
 * Am I correct in assuming that by a "regular" calculator you mean something along the lines of a four-function one without an exponentiation button? -- Kinu  t/c 19:35, 23 April 2012 (UTC)
 * Ironically, this is trivial to do using a slide rule, and quite difficult to do with a 4 function calculator. To calculate an arbitrary root, you need a calculator that can perform exponentiation; else, you will need to do some work on your own, (possibly with the aid of the 4-function calculator to do the arithmetic for you).  There are numerous iterative methods to estimate roots; there are several logarithmic identities that you can use to simplify the math; but there's no "quick" solution.  In the general case, you are solving the cubic function x3 - k = 0, and you can use any numerical or algebraic technique you like to solve for x.  Nimur (talk) 19:40, 23 April 2012 (UTC)
 * You can do this easily enough with Newton's method. -- Meni Rosenfeld (talk) 19:46, 23 April 2012 (UTC)
 * Were I hypothetically to be stranded on a desert island with just a 4-function calculator, and I needed to calculate the cube-root of k (only one time), I would run Newton's method by hand to solve for x^3 - k = 0, . Were I to require calculation of many cubic roots, I would use the four-function calculator to help me write out a precomputed table of logarithms so that when its battery eventually died, I'd still be able to do fast and accurate arithmetic to a high number of decimal places.  Nimur (talk) 19:46, 23 April 2012 (UTC)
 * If you were stranded with "just a ... calculator", what would you use for paper or a writing implement? --  ♬  Jack of Oz  ♬  [your turn]  19:52, 23 April 2012 (UTC)
 * Obviously, I would create clay tablets and use a suitable cuneiform numeral glyph system. Though I would not be able to remember Babylonian numerals off the top of my head, I would at least know that they invented a system that was easy to carve into semisoft clay, unlike our modern arabic numerals, which are better-suited for figuring on paper, rather than carving into sand or mud.  When you are stranded on a desert island, you have to make effective use of your limited natural resources.  Nimur (talk) 23:23, 23 April 2012 (UTC)
 * (slaps forehead) Well, of course! Cuneiform glyphs, what else.  Sorry for the dumb question.  --  ♬  Jack of Oz  ♬  [your turn]  01:48, 24 April 2012 (UTC)
 * Jack, you're a crack-up Benyoch...Don't panic! Don't panic!... (talk) 08:22, 25 April 2012 (UTC)
 * Of course, this raises the question of how you would best go about calculating those log tables with only your four-function calculator. -- The Anome (talk) 11:44, 24 April 2012 (UTC)
 * The answer is simple, you make a PCuneiform Tablet. It will compute anything if you have a proper stylus. Benyoch...Don't panic! Don't panic!... (talk) 08:22, 25 April 2012 (UTC)
 * Taylor series expansion of the exponential function. I won't pretend to assert that this is the best or fastest way to reduce logarithm generation to simple arithmetic; but it's the method I could remember and re-derive if stranded on a desert island.  Nimur (talk) 18:11, 24 April 2012 (UTC)
 * That makes sense to me. And once you've got the coefficients computed you can use something like Briggs' finite difference method to compute the polynomial at a large number of evenly-spaced values to create the table. -- The Anome (talk) 22:36, 24 April 2012 (UTC)
 * Just by the way, the usual term in English is cube root (parallel with square root), not cubic root. A "cubic root" sounds to me like the root of a general cubic equation. --Trovatore (talk) 01:55, 24 April 2012 (UTC)
 * You can try and guess it, just guess a number with one significant figure and multiply it with itself twice. Adjust your guess using swedish rounding depending on the answer. For instance, find the cube root of 675654.
 * A first guess could be 100:
 * 1003 = 1000000: over
 * 503 ≈ 100000: under
 * 803 ≈ 500000: under
 * 903 ≈ 700000: over
 * 853 ≈ 610000: under
 * 883 ≈ 680000: over
 * 873 ≈ 660000: under
 * 87.53 ≈ 670000: under
 * 87.83 ≈ 677000: over
 * 87.73 ≈ 675000: under
 * 87.753 ≈ 675700: over
 * 87.733 ≈ 675200: under
 * 87.743 ≈ 675400: under
 * 87.7453 ≈ 675600: under
 * 87.7483 ≈ 675630: under
 * 87.7493 ≈ 675660: over
 * The answer lies between 87.748 and 87.749. Plasmic Physics (talk) 02:58, 24 April 2012 (UTC)
 * That's almost Newton's method, but you aren't making full use of all the information available to you. You know the error in each guess, which allows you to select a better next-guess.  In your method, you're not taking advantage of this information, so you converge slower.  If you had a calculator, you could just compute the gradient, and you'd get closer to the correct answer much faster.  However, it's fair to say that guess-and-check might be faster if working with cuneiform wedges on sand tablets: calculating the gradient requires long division, so while guess-and-check requires more iterations, it may actually compute faster (more iterations, but less wall-clock time), if multiplication can be performed much more rapidly than division.  This is called algorithmic analysis and makes up a large chunk of modern computer science.  In fact, if we were designing a calculator to solve cube roots, we'd have to think long and hard about this exact tradeoff.  Nimur (talk) 05:29, 24 April 2012 (UTC)
 * What do you mean by I have the error in each guess? Assuming I didn't calculate the exact answer using a scientific calculator, I cannot compare each guess with the answer to find the error in each guess. Plasmic Physics (talk) 08:32, 24 April 2012 (UTC)
 * Do you mean to compare the cube of the guess with the known cube? Plasmic Physics (talk) 08:34, 24 April 2012 (UTC)
 * Yes. Suppose our guess is x and the true cube root is x plus an error ε where ε is small compared to x. Then we know that
 * $$k = (x + \epsilon)^3 = x^3 + 3x^2 \epsilon + 3x \epsilon^2 + \epsilon^3 \approx x^3 + 3x^2 \epsilon$$
 * where in the last step we have used the fact that ε is small compared to x. So an approximate value for our error term ε is
 * $$ \epsilon_{\text{approx}} = \frac{k-x^3}{3x^2}$$
 * and we add this approximate error to our guess x and repeat. Worked example:
 * 100^3 = 1000000
 * (675654 - 1000000) / (3*100^2) = -10.8115
 * Next guess = 100 - 10.8115 = 89.1885
 * 89.1885^3 = 709457.8193
 * (675654 - 709457.8193) / (3*89.1885^2) = -1.4165
 * Next guess = 89.1885 - 1.4165 = 87.7720
 * 87.7720^3 = 676188.8159
 * (675654 - 676188.8159) / (3*87.7720^2) = -0.0231
 * Next guess = 87.7720 - 0.0231 = 87.7489
 * etc. Gandalf61 (talk) 09:00, 24 April 2012 (UTC)
 * Specifically this is the bisection method. With the bisection method you add a digit of precision every few iterations; with Newton's method you double the number of correct digits each iteration. -- Meni Rosenfeld (talk) 15:01, 27 April 2012 (UTC)
 * As always there is a bit in WIkipedia about this Cube root Dmcq (talk) 13:38, 24 April 2012 (UTC)
 * Like we said, it depends on what functions his 'normal' calculator can do. If it had a power function, it could be as simple as using the power of a third to calculate the cubee root. Plasmic Physics (talk) 14:06, 24 April 2012 (UTC)
 * Like we said, it depends on what functions his 'normal' calculator can do. If it had a power function, it could be as simple as using the power of a third to calculate the cubee root. Plasmic Physics (talk) 14:06, 24 April 2012 (UTC)