User:NorwegianBlue/refdesk/mathematics

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The area of a square on the surface of a sphere.
Moved to a separate subpage.

Cubic equations with three real, irrational roots
(Copied from the reference desk, maths section).

The equation 8x3 - 6x + 1 = 0 has the solutions sin 10º, sin 50º and -sin 70º, and is solved readily using trigonometric methods. However, attempting to solve the equation using Cardano's method (see Cubic equation) yields some rather nasty expressions, such as


 * $$\frac{1}{2}\sqrt[3]{-\frac{1}{2}+\frac{1}{2} i\sqrt{3}} + \frac{1}{2}\sqrt[3]{-\frac{1}{2}-\frac{1}{2} i\sqrt{3}}$$

Question 1: Is it possible to reduce this to an expression involving radicals and non-complex rational numbers only? Question 2: Can anybody give an example of a cubic equation with three different real irrational roots, where the roots can be expressed using radicals and non-complex rational numbers only? --NorwegianBlue 16:58, 21 May 2006 (UTC)
 * I don't know about the first question, but my guess is that no. About the second, the equation
 * $$x^3 - (\sqrt{2}+\sqrt{3}+\sqrt{5})x^2 + (\sqrt{6}+\sqrt{10}+\sqrt{15})x - \sqrt{30} = 0$$
 * Satisfies your conditions. If you want the coefficients to be integers as well, I guess this is equivalent to the first question. -- Meni Rosenfeld (talk) 17:50, 21 May 2006 (UTC)
 * Yes, I did want the coefficients to be integers. And I suspect you are right that the answer to both questions is no. If so, is anybody aware of a proof? --NorwegianBlue 17:56, 21 May 2006 (UTC)
 * I think that the answer to Question 1 in general is "no", because of the result refered to here:
 * "One of the great algebraists of the 20th century, B.L. van der Waerden observes in his book Algebra I, that the Casus Irreducibilis is unavoidable. There will never be an algebraic improvement of the cubic formula, which avoids the usage of complex numbers."
 * However, it may still be possible that for this particular equation, such a formula exist. Perhaps that reference is enough to get you started if you are really interested. You could also try asking User:Gene Ward Smith. -- Jitse Niesen (talk) 03:05, 22 May 2006 (UTC)
 * Thank you, Jitse, as far as I can see, the reference answers both questions. And I had no idea that it was this very problem that initiated the study of complex numbers! --NorwegianBlue 16:57, 22 May 2006 (UTC)


 * Solving the Casus Irreducibilis, the case of three real roots, of the cubic equation requires the use of complex numbers if you insist on doing it with radicals; this was a strong factor in the adoption of complex numbers by European mathematicians, which may be the first clear example of modern European mathematics getting somewhere which the rest of the world hadn't gotten to earlier and better. The article cubic equations goes into great deal on how you can solve the cubic equation algebraically, not just in terms of transcendental functions, without using complex numbers. In that case, the algebraic funcion C1/3(x) can be used. While this can be computed in terms of trig functions, that does not make it a transcedental function, any more than the fact that P1/3(x) = x1/3 can be computed in terms of logs and exponentials makes it a transcendental function. In general, solving a solvable algebraic equation of degree not a power of two, all of whose roots are real, requires complex numbers if you use radicals but no complex numbers if you use Chebychev radicals C1/n. The cubic equation article is long--would an article on the Casus Irreducibilis help, I wonder? Gene Ward Smith 19:38, 24 May 2006 (UTC)


 * Thank you Gene, for commenting my question. I do realize that sin 10º, sin 50º and -sin 70º are algebraic numbers, being solutions of the equation 8x3 - 6x + 1 = 0, but I did insist on expressing the solution with radicals. I'm grateful for the response that it cannot be done without using complex numbers. As to whether there should be a separate article on the Casus Irreducibilis, I think that would be overkill, but it wouldn't hurt if, at the end of the section on Cardano's method, you mentioned the fact that, when applied to an equation with three real, irrational roots, it will always give a solution which includes a sum of two conjugate complex numbers, such that the imaginary parts cancel out. --NorwegianBlue 21:47, 25 May 2006 (UTC)


 * Second thought, maybe for some special classes, it can be done by denesting Nested radicals? (See the reference section, through you may have to use an algebra system) --Lemontea 02:37, 24 May 2006 (UTC)


 * Well, the Cardano solution is expressed in nested radicals involving complex numbers. I did try to construct cubic equations using nested radicals of real numbers only, but always ended up with irrational coefficients. It is quite some time since this problem nearly drove me nuts (33 years, to be exact), and I didn't have access to a computer algebra system. --NorwegianBlue 14:25, 24 May 2006 (UTC)


 * Anyway, at least now I know that it is impossible for your example above - see Exact trigonometric constants, which said "No finite radical expressions involving real numbers for these triangle edge ratios are possible because of Casus Irreducibilis. 9×2X-sided 70°-20°-90° triangle - enneagon (9-sided) 80°-10°-90° triangle - octakaidecagon (18-sided)" PS:my fault, the reference I cited actually duels with nested square root more, and for general radicals, not all are denestable. PPS:If question 2 doesn't require there to be three real roots, then $$x^3+6x+2=0$$ satisfy the other requirement. --Lemontea 14:50, 24 May 2006 (UTC)
 * Thanks. The point was that there be three real roots. The real solution to your example ($$\sqrt[3]{2}-\sqrt[3]{4}$$) would have satisfied the conditions if only there were two more real, irrational roots that could be expressed in a similar manner. However, the example is only a combination of the equations $$y^3=2$$ and $$z^3=4$$. Subtract the equations, i.e. $$y^3-z^3 = -2$$, factorize the l.h.s., substitute with $$x = y-z$$, observe that $$yz = 2$$ and that $$x^2+6 = (y^2+yz+z^2)$$, and you have your equation. --NorwegianBlue 19:10, 25 May 2006 (UTC)

Algebra stumper
Years ago, a calculus professor gave me this one. The problem is to either solve the following set of equations by giving x, y and z; or to prove no such solution exists (among the complex numbers)



x + y + z = 1 \,\!$$



x^2 + y^2 + z^2 = 2 \,\!$$



x^3 + y^3 + z^3 = 3 \,\!$$ —The preceding unsigned comment was added by Plf515 (talk • contribs) 03:50, 2006 November 24.


 * Suppose a is one of x, y, and z. Then $$(a - x)(a - y)(a - z) = 0$$, so
 * $$a^3 - (x+y+z)a^2 + (xy+xz+yz)a - (xyz)a = 0\,\!$$
 * We already know one of the coefficients, and we can find the other two:
 * $$(x + y + z)^2 = 1\,\!$$
 * $$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\,\!$$
 * $$2 + 2(xy + xz + yz) = 1\,\!$$
 * $$xy + xz + yz = -1/2\,\!$$
 * $$(x + y + z)(x^2 + y^2 + z^2) = 2\,\!$$
 * $$x^3 + y^3 + z^3 + x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2 = 2\,\!$$
 * $$(x + y + z)(xy + xz + yz) = -1/2\,\!$$
 * $$3xyz + x^2y + x^2z + xy^2 + xz^2 + y^2z + yz^2 = -1/2\,\!$$
 * $$x^3 + y^3 + z^3 - 3xyz = 5/2\,\!$$
 * $$3 - 3xyz = 5/2\,\!$$
 * $$xyz = 1/6\,\!$$
 * So, x, y, and z are the roots of the cubic polynomial $$a^3 - a^2 - (1/2)a - 1/6$$, which are about 1.43, 0.215 + 0.265i, and 0.215 - 0.265i. —Keenan Pepper 05:30, 24 November 2006 (UTC)


 * We might explicitly point out that all three equations lead to symmetric polynomials in x, y, and z. This has a double relevance. First, as shown, it assists in finding one solution. Second, it tells us that any permutation of one solution is also a solution.
 * A modern tool, that may not have been available when the problem was first posed, is the computation of a Gröbner basis for the three polynomials using Buchberger's algorithm. One such basis here is
 * $$\begin{align}

&6 z^3-6 z^2-3 z-1 ,\\ &2 y^2+2 z y-2 y+2 z^2-2 z-1 ,\\ &x+y+z-1. \end{align}$$
 * The cubic in z should look familiar. --KSmrqT 06:52, 24 November 2006 (UTC)


 * Expressed algebraically, the real root of the 3rd degree polynomial equals:
 * $$\frac{w}{w^2-w-1}, \mbox{ where } w=\sqrt[3]{5 + \sqrt{26}}.$$
 * This was found by first using the substitution $$z := 1/\zeta$$ and subjecting the result to Cardano's third degree technique of interrogation. --Lambiam Talk  09:14, 24 November 2006 (UTC)

Thanks~ Plf515 10:56, 24 November 2006 (UTC)plf515

Integration in Maxima
I'm experimenting with open-source computer algebra systems, and am trying out Maxima. I've encountered a problem when doing symbolic integration. The value returned from integrate, although displayed as an algebraic expression, behaves differently from the same expression entered by hand. Here is an example of what is going on:


 * f(x):=45^(-x^2);

$$ \, f\left( x\right) :={45}^{-{x}^{2}} $$


 * F(x):=integrate(f(x),x);

$$ \, F(x):=\operatorname{integrate}(f(x),x)$$


 * F(x);

$$ \frac{\sqrt{\pi }\,\operatorname{erf}\left( \sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\,x\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }} $$
 * // Seems reasonable...
 * // Now I'd like to evaluate F(x) for a specific value of x


 * F(5);

Attempt to integrate wrt a number: 5#0: F(x=5)
 * -- an error. To debug this try debugmode(true);
 * // Hmm, why doesn't it simply substitute x=5?
 * // Ok, lets try defining G(x) = F(x) from scratch, and see what happens...

$$ \frac{\sqrt{\pi }\,\operatorname{erf}\left( \sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\,x\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }} $$
 * G(x):=sqrt(%pi)*erf(sqrt(log(5)+2*log(3))*x)/(2*sqrt(log(5)+2*log(3)));

$$\frac{\sqrt{\pi }\,\operatorname{erf}\left( 5\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }\right) }{2\,\sqrt{\log\left( 5\right) +2\,\log\left( 3\right) }} $$
 * G(5);

$$ \, 0.45422679973746 $$
 * float(G(5));

$$ \, 1 $$
 * F(t)/G(t);


 * F(5)/G(5);

Attempt to integrate wrt a number: 5#0: F(x=5)
 * -- an error. To debug this try debugmode(true);

My question is this: How do I make F(x) behave as though it had been entered by hand. --NorwegianBluetalk 19:19, 14 January 2007 (UTC)


 * When you wrote F(x):=integrate(f(x),x) you intended the right side to be evaluated before the function was defined, but that is not what you actually said. So what is happening is that F(5) means integrate(f(5),5) and the error is exactly what you would expect. Try the quote-quote input:
 * F(x):= '' (integrate(f(x),x));
 * This forces the integral to be evaluated before the function is defined. --KSmrqT 23:36, 14 January 2007 (UTC)
 * Yes, that fixed it. Thanks! --NorwegianBluetalk 20:59, 15 January 2007 (UTC)

A couple of calculus questions
Hi, all, hope you can help me with a couple of things.

Firstly, how do I go about evaluating this?

$$\int_{-\infty}^{\infty} e^{-x^2}\,dx$$

I know what the answer is from looking at a page of integrals, but I'm more interested in the process.

Also, how do you differentiate something in the form of, say, $$3^x$$ ? I tried using the chain rule to give $$ \frac{d}{dx} 3^x = \frac{d(3^x)}{d3} \times \frac{d3}{dx} = x(3^{x-1}) \times 0 $$, but clearly the answer is not 0 for all x, as this implies. Neither of these are homework, but I don't mind working through them myself if you give me some hints. Thanks, 80.169.64.22 19:45, 29 June 2007 (UTC)


 * The first example has no antiderivative, except one artifically constructed just for the purpose, the error function. But the second example is easy once we rewrite it as exp(x log(3)), then apply the chain rule. Very creative, your derivative with respect to 3; but it has the fatal flaw that 3 is not a variable! --KSmrqT 20:18, 29 June 2007 (UTC)


 * Check out the limits on the first question; we want a specific definite integral. The standard technique for this one is to square the integral and then transform from cartesian co-ordinates to plane polars. Algebraist 20:32, 29 June 2007 (UTC)


 * I suspect the OP may not be familiar with the tools involved; I will therefore sketch a calculation based on Algebraist's suggestion, and invite requests for anything that needs to be explained further:
 * $$\left(\int_{-\infty}^{\infty} e^{-x^2}\ dx\right)^2 = \int_{-\infty}^{\infty} e^{-x^2}\ dx \int_{-\infty}^{\infty} e^{-y^2}\ dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}\ dx\ dy =$$
 * Changing to polar coordinates, $$x = r \cos{\theta}$$ and $$y = r \sin{\theta}$$, which has Jacobian $$r$$:
 * $$=\int_0^{2\pi}\int_0^{\infty}e^{-r^2}r\ dr\ d\theta = \int_{0}^{2\pi}\frac12\ d\theta = \pi$$
 * $$\int_{-\infty}^{\infty} e^{-x^2}\ dx = \sqrt{\pi}$$
 * -- Meni Rosenfeld (talk) 21:03, 29 June 2007 (UTC)


 * KSmrq: you can differentiate 3, or any other constant (you should know this), it can be considered a function f(x) = 3. The problem is that we are talking about exponentials, rather than the differentiability of a constant function.


 * Differentiating 3 is something else than differentiating with respect to 3.--P.wormer 07:32, 30 June 2007 (UTC)


 * On the second (differentiation) problem, you might consider implicit differentiation. Consider $$\ln y = x\ln 3$$. Thus
 * $$\frac{y'}{y} = \ln 3$$
 * Substitute y and solve for y-prime. All the best (and great questions!), --TeaDrinker 21:13, 29 June 2007 (UTC)

$$\frac{d}{dx}\ a^x = a^x ln(a)$$

for constant and positive a.

For the first question, check Meni Rosenfeld's rather elegant method for evaluating it.


 * It's not mine. I was in the middle of trying to asymptotically evaluate an integral in the complex plane when Algebraist gave a gentle reminder of the "right" way to do it. -- Meni Rosenfeld (talk) 10:38, 30 June 2007 (UTC)

Maths resources
I am delighted by your question. It takes uncommon wisdom and humility to acknowledge a need to learn more and to seek assistance. In this case, your need is a shared one, and there are quality sources on the web. One place to start is The Math Forum at Drexel, where you will find a page of resources. For online mathematics texts, the AMS has links, though many are on more advanced topics. Still, you can find jewels like Calculus, by Gilbert Strang, a Professor of Mathematics at MIT. You might also browse topics at The Mathematical Atlas. If I have a chance, I'll see if I can dig up a good elementary discussion of the binomial theorem. Meanwhile, these sources should keep you in good reading. --KSmrqT 22:07, 29 December 2006 (UTC)

Radicals
Just a quick question: is it correct to have a radical inside of another radical, or should I attempt to work the second out of the first radical? I guess "correct" isn't the best word. I guess I'm asking is it more formal format to have it out? 75.18.9.71 03:48, 30 January 2007 (UTC)


 * It depends on the context, but I would certainly say that there is nothing wrong with having square roots inside other square roots. Do you mean something like $$\sqrt{2 + \sqrt{2}}$$? –King Bee (T • C) 04:05, 30 January 2007 (UTC)
 * Yes, that's exactly what I mean. Thanks :-) -75.18.9.71 04:11, 30 January 2007 (UTC)


 * That's perfectly ok. In fact, there are some numbers defined by an infinite amount of nested radicals. Check out that article. -- Ķĩřβȳ ♥  Ťįɱé  Ø  07:13, 30 January 2007 (UTC)


 * If there is an unnested version, it will be considered simpler, so by the unwritten rule that (in the absence of other rules to the contrary) the simpler expression is preferred, it is the better form. For example, rather than
 * $$\sqrt[3]{-2+\sqrt 5}\,,$$
 * use
 * $$\frac{-1+\sqrt 5}{2}\,.$$
 * --Lambiam Talk 14:13, 30 January 2007 (UTC)


 * How do you go about to simplify such an expression? I tried the expression in Maxima (which I've only just begun to get acquainted with). It confirmed, of course, that both expressions evaluate to 0.6180339887499. But I was unable to make it simplify the expression (tried procedures ratsimp, radcan, trigsimp and trigreduce, i.e. all the buttons that had the words "simplify" or "reduce" on them). It was no more successful at simplifying
 * $$\frac{\sqrt{5}-1}{2\,{\left( \sqrt{5}-2\right) }^{\frac{1}{3}}}$$.
 * How does one approach the problem "by hand"? --NorwegianBluetalk 22:08, 30 January 2007 (UTC)


 * One possible approach, which can have varying degrees of success, it to conjecture that $$\sqrt[3]{-2+\sqrt 5}$$ will be of the form $$a + b \sqrt{5}$$ with a and b rational. This gives $$(a+b\sqrt{5})^3=-2+\sqrt{5}$$, or $$a^3+3a^2b\sqrt{5}+3ab^25+5b^3\sqrt{5}=-2+\sqrt{5}$$ which gives $$a^3+15ab^2=-2$$ and $$3a^2b+5b^3=1$$. Solving this (the substitution $$t=\tfrac{a}{b}$$ helps) should give the result. -- Meni Rosenfeld (talk) 22:26, 30 January 2007 (UTC)


 * Thanks for the quick reply! --NorwegianBluetalk 22:45, 30 January 2007 (UTC)


 * From Exact_trigonometric_constants
 * Main article: Nested radicals
 * In general nested radicals cannot be reduced.
 * But if for $$ \sqrt{a + b\sqrt c} $$,
 * $$ R = \sqrt{a^2 - b^2 c} $$ is rational,
 * and both $$ d = \pm \sqrt{ \frac{ a \pm R }{2}} $$
 * and $$ e = \pm \sqrt{ \frac{ a \pm R }{2c}} $$ are rational,
 * with the appropriate choice of the four $$\pm$$signs,
 * then $$ \sqrt{a + b\sqrt c} = d + e\sqrt c $$
 * Example:
 * $$ 4\sin 18^\circ = \sqrt{6 - 2 \sqrt 5} = \sqrt 5 - 1 $$

Solving cubic equations
In school, we learnt how to solve cubic equations of the form ax3 + bx2 + cx + d.

To factorise it into (x+r)(x+s)(x+t), we have to find a value for r by trial and error (with the factor theorem), then solve a quadratic equation to get the other two roots, s and t.

In the exams, the value of r will usually be 3 or less, so finding r by trial and error will not take a long time.

However, if r is large, finding r by trial and error will take a long time. Are there any faster, more systematic ways of finding r, besides trial and error? — Preceding unsigned comment added by 218.186.9.3 (talk • contribs) 09:18, 2007 February 13


 * (Please sign your posts with four tildes, " ~ ".) The roots of a cubic polynomial can be expressed in closed form, so in theory no searching is required. However, the technique is more difficult than using the quadratic formula. When the coefficients, a, b, c, and d, are real numbers (with a nonzero), the cubic is guaranteed to have a real root; but it may have only one. For example,
 * $$ x^3-3 x^2+2 x-6 \,\!$$
 * has only one real root.
 * But we can say much more. The coefficients are normalized so that the x3 term has coefficient 1. If we list them all, (1,−3,2,−6), the −6 has the largest absolute value. The Cauchy bound says that all real roots must lie between −m and +m, where m is one plus the maximum absolute value. In this example, the roots will be between −7 and 7.
 * We can do better. The positive coefficients before the −6 are 1 and 2, summing to 3; and before the −3 we have just 1. For an upper bound we can choose the maximum of 6&frasl;3 and 3&frasl;1, which will be 3, and again add 1. That is, the maximum real root is 4. Negating every other term (so we are evaluating the polynomial with −x), we can similarly find a lower bound, which here will be −1.
 * There is also a method known as Descartes' rule of signs that may help us restrict the number of positive or negative real roots. In this example it is no help with the positive roots (we will have three or one), but it tells us we cannot have a negative root.
 * This example polynomial has integer coefficients, and may have integer roots. If r is such a root, then it must divide the constant term, so the only possibilities here are 1, 2, 3. We have used the known bounds; and we know that 0 cannot be a root unless the constant term is 0.
 * We also have ways to bracket roots. If the polynomial evaluates to positive for x = a and negative for x = b, then there must be at least one root between a and b. Using a Sturm sequence, we can learn even more.
 * This does not exhaust our inventory of tools, but perhaps this is enough for now. Computer programs for solving cubics typically use closed form solutions. Above degree four, however, a famous theorem states that we have no closed form solutions. --KSmrqT 11:48, 13 February 2007 (UTC)


 * You might want to read cubic equation. &#x2013; b_jonas 14:17, 13 February 2007 (UTC)

Best calculus textbook
I've heard very often of Tom Apostle's Calculus I and Calculus II written in the 40s and 50s, which are very proof-heavy, well, entirely proof-based. What are your opinions on the best calculus textbooks? ['Mαc Δαvιs '] X (How's my driving? ) ❖ 08:28, 19 February 2007 (UTC)


 * I can also recomend Leithold's book. Specially for beginners it is a good choice. Mr.K. (talk) 13:48, 19 February 2007 (UTC)


 * More information about your expectations would help. Do you want a graphic-rich easy introduction? Well-grounded proofs? Higher mathematics insight? Emphasis on engineering/physics applications? A modern revision? One of the non-standard analysis presentations?
 * Most texts I've seen do not impress me. Richard Courant and Fritz John do. Try a variety of books online, including those listed by Stef and at AMS. --KSmrqT 04:27, 20 February 2007 (UTC)

Bernoulli numbers
In the first paragraph of the Bernoulli number page it is written that there is no known elementary description for Bernoulli numbers .What is the meaning of "elementary description"? I thought that it meant that there is no mathematical rule to find the nth Bernoulli number that is not a sum of an infinite series because there was none on the page .but then I found somewhere on the internet a simple definition that is not a sum of an infinite series and also it does not depend on previous values of $$B_n$$ .So, my question is :what does "elementary description" mean ? --George 04:04, 4 November 2007 (UTC)


 * I suppose the meaning intended here is that there is no definition in the form of an expression using only elementary functions, in other words, a closed-form solution of a defining equation for the Bernoulli numbers. --Lambiam 06:38, 4 November 2007 (UTC)


 * George - did you say you have somewhere seen a finite and non-recursive expression for $$B_n$$ ? Do you remember where that was ? The simplest "rule" for finding the value of $$B_n$$ that I can think of is


 * $$B_0=1$$
 * $$B_n=\left(\frac{-1}{n+1}\right)\sum_{j=0}^{n-1}{n+1\choose{j}}B_j\mbox{ ; }n>0$$


 * but that is recursive, so it probably doesn't qualify as an "elementary description". Gandalf61 09:57, 4 November 2007 (UTC)


 * Might it have been a generating function you saw? They're easy to mistake for formulas. Black Carrot 06:42, 5 November 2007 (UTC)


 * The formula is $$B_n=\sum_{k=0}^n\sum_{j=0}^k j^n (-1)^j \frac{\binom{j}{k}}{k+1}$$

I'm not sure if this can be considered an elementary description ,if it's not one please explain the reason .Thanks --George 10:20, 5 November 2007 (UTC)


 * Since j ranges from 0 to k, should $$\binom{j}{k}$$ in the numerator actually be $$\binom{k}{j}$$ instead ? Gandalf61 10:55, 5 November 2007 (UTC)
 * Well, the Gamma function is non-elementary, so in some sense the binomial coefficient function is not elementary, either. This might have some relevance to the claim that no elementary formula exists. -- Meni Rosenfeld (talk) 13:11, 5 November 2007 (UTC)


 * yes, I'm sorry, it is $$B_n=\sum_{k=0}^n\sum_{j=0}^k j^n (-1)^j \frac{\binom{k}{j}}{k+1}$$, I just mistyped the formula. --George 14:59, 5 November 2007 (UTC)

Learning maths from scratch
I always want to learn maths all over again, but I don’t know how to start with. Now, I’m undergraduate, I “parted” with everything science since high school, and I did almost nothing more than very simple statistics (like poisson distribution, but I don’t really remember it) and calculus (I don’t remember how to differentiate). So, if I want to know “advanced maths”, like what is being taught at university, what sorts of books (maybe in English) may I use (as an adult learner)?

My “junior high school” maths is about factorizing, polynomials, simple geometry (calculating angles) and so on; which areas do high school maths (I mean, for learning science subjects) and university maths cover? I’m looking for books for learning more advanced things (compared to my present level) and I don’t just concentrate on calculus or statistics (or something like that). Any suggestions? --Fitzwilliam (talk) 14:10, 29 February 2008 (UTC)


 * From my experience, university maths tends to recap all you learnt at school in the first year - try just going along to the first year maths lectures. -mattbuck (Talk) 14:39, 29 February 2008 (UTC)
 * Especially maths lectures targeted at science students, rather than maths students. My Maths department (at a UK uni, I know it's a little different elsewhere) offers a module called "Mathematics for Scientists and Engineers" which covers a wide range of mathematical topics at a fairly basic level (by Uni standards) without assuming much (if any) prior knowledge. If there are similar modules are your uni, those would be the ones to go to. --Tango (talk) 16:11, 29 February 2008 (UTC)


 * I'd go to a Maths Department. They love to find new students. Imagine Reason (talk) 23:15, 29 February 2008 (UTC)


 * What is your motivation? Do you want to learn advanced maths primarily because you feel it will be useful, or is it more for fun? If you have forgotten how to differentiate, maybe you should take a course in calculus (actually analysis) anyway, using a text that does not just give the rules and formulas but also precise definitions of concepts like the real numbers, limit, and continuity, and rigorous proofs. Other topics you may study that don't immediately require much prior knowledge are linear algebra and projective geometry. Also consider elementary number theory and combinatorics. A nice book is Concrete Mathematics; although aiming at hopeful computer scientists, it is also quite valuable for mathematicians. I'd also advice you not to go immediately very deep into one field of maths, but to first build up a fairly broad basic knowledge of various fields. Much of the more advanced stuff in maths requires some knowledge of other fields. --Lambiam 00:26, 1 March 2008 (UTC)

cookbook curve fit
I need something from a Numerical Methods book that I gave away twenty years ago! Given a set of points $$x_i, y_i$$, what's a compact expression for the polynomial (of minimal degree) that fits exactly? —Tamfang (talk) 20:36, 28 September 2008 (UTC)


 * $$\sum_i y_i \prod_{j \ne i} \frac{x - x_j}{x_i - x_j}$$? -- BenRG (talk) 21:23, 28 September 2008 (UTC)


 * That's the one, thanks. —Tamfang (talk) 21:49, 28 September 2008 (UTC)


 * Here's how to remember it so you'll never have to look it up again: by linearity the polynomial you want is the sum of $$y_i$$ times a polynomial that's 1 at $$x_i$$ and zero at the other points. To be zero at the other points it must have factors of $$(x - x_j)$$ for every $$j \ne i$$, i.e. it must be a multiple of $$\prod_{j \ne i}(x - x_j)\,$$. To make that equal one at $$x_i$$ you divide by the result of plugging $$x_i$$ into it, that is, $$\prod_{j \ne i}(x_i - x_j)\,$$. -- BenRG (talk) 22:33, 28 September 2008 (UTC)

Dumb question about infinity
Are there an infinite number of infinities? Sorry if this is a dumb question. Jooler (talk) 20:21, 13 October 2008 (UTC)


 * There are no dumb questions about infinity, it's a very confusing topic! The answer is "it depends". There are different things "infinity" can mean. It can be used to mean "going on forever" (when doing limits and things), so if you're talking about a sequence indexed by natural numbers, there's just one infinity. If you're talking about a function on the real numbers, there are two, positive and negative, if you're talking about a function on the plane or some high dimensional space, then there are infinitely many, one in each direction (the infinity being the same as the number of real numbers or the "cardinality of the continuum"). On the other hand, there are infinite ordinals (numbers used for ordering things) and cardinals (numbers used for counting things), there are infinitely many of each of those too (I'm not sure which infinity - I think there are more ordinals than any infinity you can think of, so I guess that's a kind of super-infinity!). --Tango (talk) 18:07, 13 October 2008 (UTC)


 * So, as the set of real numbers is larger than the set of integers, and the set of complex numbers is larger still, can you logically say that the infinity represents the count of one set is larger than than the infinity that represents the count of another set?


 * $$\infin_r > \infin_i $$


 * Jooler (talk) 20:22, 13 October 2008 (UTC)


 * The infinity which is the cardinality ("size") of the real numbers is, indeed, large than the cardinality of the integers. The complex numbers are actually the same size as the real numbers in terms of cardinality. For example, you can pair them up by taking a complex number with its real and imaginary parts written as decimal expansions and then form a single real number by interspersing the digits, so 0.25+0.36i would become 0.2356 (there are details to be worked out, but the basic idea works). Real numbers can obviously be mapped to complex numbers by simple inclusion - a real number is already a complex number. --Tango (talk) 20:30, 13 October 2008 (UTC)


 * Most definitely. Consider the natural numbers - it's obvious that there are an infinite number of these, and that infinity is $$\aleph_0$$, and it's called a countable infinity. However, by Cantor's diagonal argument, we know that there is no bijection between the set of natural numbers and the set of real numbers (though there is between the naturals and rationals). It's clear that there are more real numbers than rational numbers, so the Cardinality of the continuum is $$> \aleph_0$$, and could therefore be said to be more infinite. We call this an uncountable infinity. In fact, $$\vert\mathbb{R}\vert = 2^{\aleph_0}$$, though opinion is divided as to whether this is the smallest uncountable infinity (which is $$\aleph_1$$). For more information, look at the article on the Continuum Hypothesis. -mattbuck (Talk) 20:32, 13 October 2008 (UTC)

Ok, so as $$\infin_r > \infin_i $$, can it be concluded that there must be an infinite number of infinities between those two infinities? Or is this becoming absurd? Jooler (talk) 20:57, 13 October 2008 (UTC)


 * As for whether there can be an infinity between those two, read what mattbuck had to say above (with more at Continuum Hypothesis, which is exactly the question you asked). As for an infinite number of infinities, if we are talking about cardinalities, then for any set S, the power set of S is necessarily larger, so we can always get a larger and larger infinity by taking the power set over and over again.  Eric.  131.215.45.106 (talk) 21:13, 13 October 2008 (UTC)
 * Right, but what he literally asked was whether there were infinitely many cardinals between $$\aleph_0$$ and $$2^{\aleph_0}$$. That's not exactly the same question as CH, although it has a similar answer (namely, the question can't be settled from the accepted axioms of set theory alone, and whether that's a defect of the question, or just indicates the weakness of the accepted axioms, is a bone of contention). --Trovatore (talk) 21:17, 13 October 2008 (UTC)
 * Indeed, it's closely related to CH - it's a stronger version of the negation of CH. --Tango (talk) 21:26, 13 October 2008 (UTC)
 * Well, you can look at it that way if you want, although I think that these days a significant current of thought suggests $$\aleph_2$$ for the cardinality of the continuum (for example, this is the value that follows from the proper forcing axiom, and it's the value that shows up in certain models related to Woodin's Ω-logic, though I've never quite got it straight whether Woodin's anti-CH argument specifically supports the value $$\aleph_2$$). --Trovatore (talk) 22:12, 13 October 2008 (UTC)
 * My bad, I misread what he wrote. I seem to be in the habit of that lately (a sign I should be sleeping instead of being on WP?).  My apologies to Jooler for the mistake.  Eric.  131.215.159.210 (talk) 06:59, 14 October 2008 (UTC)

Thanks for all the answers guys. Unfortunately my maths education stopped at A-Level, and that was 25 years ago, so some of the above is over my head, but its all helpful! Jooler (talk) 22:21, 13 October 2008 (UTC)


 * I've seen arguments for variants of a new axiom which would imply that there was exactly one new infinity between $$\aleph_0$$ and $$2^{\aleph_0}$$. So the subject is by no means dead with saying one could assume this either way. Dmcq (talk) 09:45, 14 October 2008 (UTC)
 * Yes, that's what Trovatore was talking about above. Algebraist 10:26, 14 October 2008 (UTC)
 * Sorry, true - I should read through all the answers more carefully Dmcq (talk) 18:16, 14 October 2008 (UTC)

What does the "d" stand for? (in dx/dy)
In article Kinematics, there's a formula:


 * " Instantaneous velocity (the velocity at an instant of time) is defined as
 * $$ \boldsymbol v = \frac {d \boldsymbol r}{d t} \, $$ "

May I ask, what do the "d" stand for? Thank you. 60.0.162.131 (talk) 09:55, 26 May 2009 (UTC) (Matthew 百家姓之四 without signing in)
 * "Change in". So you essentially have the change in radius over the change in time. How much the radius is changing as time changes. See derivative and differential (infinitesimal), among others. &mdash; Anonymous Dissident  Talk 10:11, 26 May 2009 (UTC)
 * In the context of Kinematics, r is a displacement vector, not a radius. Gandalf61 (talk) 10:24, 26 May 2009 (UTC)
 * You're correct. Sorry, slip of the mind. &mdash; Anonymous Dissident  Talk 10:39, 26 May 2009 (UTC)
 * The small d is Leibniz notation in differential calculus that means an infinitessimally small change in the variable that follows. Cuddlyable3 (talk) 10:14, 26 May 2009 (UTC)
 * The 'd' probably comes from the word "difference" (or its equivalent in whatever language Leibniz liked to work in). --Tango (talk) 10:49, 26 May 2009 (UTC)

While the section does try to explain the notation, it seems to do this while going to great lengths not to ever mention the word derivative or link to further reading on the subject; instead the reader is presumed to understand what an "infinitesimally small displacement" is, what an "infinitesimally small length of time" is, and how the one can be divided by the other. I'm not sure this is really productive. — JAO • T • C 11:29, 26 May 2009 (UTC)


 * Thank you all, Now I clearly understand what does "d" means. I feels Wikipedia is really a warm community. Thanks! Matthew 百家姓之四 Discussion 討論  11:33, 26 May 2009 (UTC)


 * Oh, may I ask another question? In article Jerk (physics), it says:
 * "Jerk is defined by the following equation:$$\vec j=\frac {\mathrm{d} \vec a} {\mathrm{d}t}=\frac {\mathrm{d}^2 \vec v} {\mathrm{d}t^2}=\frac {\mathrm{d}^3 \vec r} {\mathrm{d}t^3}$$"
 * I can understand t2, but what is a "d2"? I actually understand "d" as a "Δ", so I get difficulty here. Thank you! Matthew 百家姓之四 Discussion 討論  12:35, 26 May 2009 (UTC)


 * $$\frac {\mathrm{d}^2 \vec v} {\mathrm{d}t^2}$$ may be a little counter-intuitive, but it's Leibniz's notation for the second derivative: it actually means $$\frac {\mathrm{d} \frac {\mathrm{d} \vec v} {\mathrm{d} t}} {\mathrm{d} t}$$, that is the rate of change in the rate of change in velocity. It does not actually have anything to do with $$t^2$$. — JAO • T • C 12:44, 26 May 2009 (UTC)


 * Thank you for your patient explanation. Matthew 百家姓之四 Discussion 討論  13:16, 26 May 2009 (UTC)
 * It can also be thought of as $$\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)^2\vec v$$, which may make the choice of notation a little clearer. --Tango (talk) 14:21, 26 May 2009 (UTC)

Coin flipping
If a coin has 50% chance of landing "tails" when flipped, and I flip it 25 times in a row, what are the chances that I never get 4 (or more) consecutive "tails"? What is the formula by which I can calculate this? --Theurgist (talk) 14:53, 17 December 2014 (UTC)
 * The number of sequences is a(29)=14564533 where a(n) is the nth Tetranacci number, so the probability is 14564533/225 or about 43%. See for formulas. --RDBury (talk) 00:10, 18 December 2014 (UTC)
 * Why 29? From the OESIS page: "a(n) = number of compositions of n-3 with no part greater than 4". 25+3=28. Is 29 used to correct for heads/tails symmetry? --46.9.44.66 (talk) 21:52, 19 December 2014 (UTC)
 * Look a bit further down. "a(n+4) = number of 0-1 sequences of length n that avoid 1111." --RDBury (talk) 01:47, 20 December 2014 (UTC)
 * Ah, thanks! --NorwegianBluetalk 11:40, 20 December 2014 (UTC)

Polynacci numbers, followup question

 * I attempted to find the answer to 's question using elementary probability calculations before 's answer. I tried to separate the problem into individual cases (a sequence of heads that begins at position 1, a sequence of heads that begins at position 2, etc), but soon realized that I was unable to make the "separate" cases non-overlapping. I also realized that the question could easily be answered by brute force calculation, and have now written a small program that does so, and which confirms 's answer. Experimenting a little with variations of the program, I find that an analogous approach holds true for sequences of 2, 3, 5 and 6 tails. The number of sequences of 25 coin tosses with no runs of more than 1 tail is fibonacci_number(25+2), the sequences with no runs of more than 2 tails is tribonacci_number(25+3), the sequences with no runs of more than 3 tails was the original question, the number of sequences with no runs of more than 4 tails is pentanacci_number(25+5), the number of sequences with no runs of more than 5 tails is hexanacci_number(25+6). I assume that this holds true in general. My question: Is there some intuitive or easily proven reason why this is so? --NorwegianBluetalk 19:39, 20 December 2014 (UTC)


 * I found a nice explanation: . --NorwegianBluetalk 10:53, 21 December 2014 (UTC)
 * Applied to the specific problem:
 * To avoid 4 or more:
 * Case 1: The first is a head, the rest is satisfactory.
 * Case 2: the first is tail, the second is head, the reast is satisfactory
 * Case 3: the first and seconds are heads, the third is head, the rest is satisfactory
 * Case 4: the first, second and third are heads, the fourth is head, the rest is satisfactory.
 * These are mutually exclusive, and include all possible satisfactory sequences.
 * The number of sequences of 3 or fewer in a run are thus:
 * First case: F(n-1)
 * plus Second case: F(n-2)
 * plus Third case: F(n-3)
 * plus Fourth case: F(n-4)
 * F(4) = F(3) + F(2) + F(1)
 * F(n) = F(n-1) + F(n-2) + ... + F(n-p)
 * --NorwegianBluetalk 23:58, 26 December 2014 (UTC)
 * F(n) = F(n-1) + F(n-2) + ... + F(n-p)
 * --NorwegianBluetalk 23:58, 26 December 2014 (UTC)

Metaquestion: Why did my question receive no answers
Previous Question: I would appreciate your feedback about whether my understanding of ROC-curves is correct. I also have a couple of questions at the end. As this post clearly shows, my mathematical capabilities are limited, so please be gentle, and cautious about introducing terminology that I might have trouble understanding. If necessary, please translate my statements into more conventional terminology.

In my understanding, a ROC curve is a plot of true positive rates (TPR, Y-axis) vs false positive rates (FPR, X-axis), when the cutoff (CO) between what is considered a positive and a negative observation is varied such that it covers all reasonable values. The tangent at a given point (CO), can be estimated as


 * ΔTPR(CO)/ΔFPR(CO),

hence TPR(CO)/FPR(CO) is the derived function of the ROC curve, and the ROC-curve the antiderivative of the function TPR(CO)/FPR(CO).

In the following, I'll use the ROC curve in a medical context, and let 'm' represent the observed value of a diagnostic test, for which we have a ROC curve availabe. Then


 * TPR(CO)/FPR(CO) = p(m = CO ± eps|Disease)/p(m = CO ± eps|No disease) = the likelihood ratio function.

Q1: Am I right in thinking that the probability ratio in the previous line should be the probablity of 'm' being close to CO (in my notation ± eps), and not greater than CO?

Q2: I've read a couple of places (such as here: Choi BC (1998) Am J Epidemiol 148:1127–32. ) that it is valid to draw lines between several points on the ROC curve, say corresponding to "negative", "weak positive" and "strong positive", and that the slope of each line is a valid estimate of the likelihood ratio for test results that fall within the corresponding interval. Sounds reasonable, but exactly why is that so?

Q3: I've read many places (including in our article) that the area under the curve corresponds to the probability of a randomly chosen diseased individual getting a higher test result than a randomly chosen individual without the disease. Again, this sounds reasonable, but exactly why is it so?

Please see this question that I asked a couple of weeks ago. It received no answers. I know that the reference desk contributors are eager to help, and will answer questions that are answerable. Therefore, I suspect that there is a problem with the way the question was asked, that made it either difficult to understand, or difficult to answer. There is also the possibility that it was ignored because it was perceived as lazy, but my impression is that laziness on the part of the questioner tends to be commented. My question now is about the question itself. I am not requesting answers to the original question. I'm pretty confident that I have figured out the answers to Q1 and Q2 myself, some doubt about Q3 remains. I may re-ask the question if I learn how to ask it in a way that is more likely to attract answers. --NorwegianBluetalk 22:39, 18 September 2015 (UTC)


 * Hi,I remember your question, and I have some interest in your topic. I thought I knew a bit of the concepts, and I even looked in to it for a few minutes. But I quickly realized that it was fairly technical and specialized question, that general refs wouldn't help you much, and there wasn't much I could to in <10 min to help you. It may have helped if you included more refs and links in your question, so that people who might be able to help but aren't already as familiar as you with the topic could get up to speed quickly. There's a lot of "luck of the draw" in how many people see a question, what they know, how much time they have to spare, how much interest, etc. But there's also a sense I got of this being an area where there might be considerable variation in how terms are used, rigor of proof, ways of justifying conclusions, and all sorts of gritty details of research that just take a lot of time and attention to sort through. You may have better luck at math.stackexchange, quora, or other similar (sub)sites that can sometimes deal a little better with this kind of thing. SemanticMantis (talk) 01:42, 19 September 2015 (UTC)


 * My comments are similar to those above, but I would add that, as a visual thinker, it would have helped me if some graphics had been included. Otherwise it looks like a well posed Q. StuRat (talk) 01:45, 19 September 2015 (UTC)
 * You might have better luck with stats.stackexchange.com for something like this; add the roc tag to grab the attention of people who specialize in that area. In general, at least as far as I've seen, Stack Exchanges are more geared to professionals helping other professionals, while the WP reference desks are more for general questions. --RDBury (talk) 01:52, 19 September 2015 (UTC)


 * Thanks, everyone! --NorwegianBluetalk 14:50, 20 September 2015 (UTC)


 * , late answer, but anyway...
 * I've taken a look at it (don't visit RD/MA often ), the question was a bit confusing:


 * The tangent at a given point (CO), can be estimated as ΔTPR(CO)/ΔFPR(CO),
 * hence TPR(CO)/FPR(CO) is the derived function of the ROC curve
 * Disclaimer: not familiar with the topic, just my interpretation, which could be way off..
 * Is it about the continuous probability functions? These are turned into binary tests by integrating them (page 2260), and the ROC curve plots those values. (See note 1)
 * The tangent at a given point equals the ratio of the density functions (those in fig 1 in the link), and is also (by definition) ΔTPR(CO)/ΔFPR(CO), but not TPR(CO)/FPR(CO). (See note 2)
 * Each point on the curve gives you the fraction of the sick population and the fraction of the healthy population that would test positive at that CO. So if (TPR=0.9, FPR=0.1) is on the curve, 90% of all sick and 10% of healthy would test positive.


 * Context would make it easier to understand, preferably a link to the specific material and formulas you're asking about, not many people with a medical background here and most maths techniques have numerous applications, so people won't know the conventions/notations, etc. Most sources I found were like this.


 * TPR(CO)/FPR(CO) = p(m = CO ± eps|Disease)/p(m = CO ± eps|No disease)
 * Q1: Am I right in thinking that the probability ratio in the previous line should be the probability of 'm' being close to CO (in my notation ± eps), and not greater than CO?

Haven't seen that formula in links I checked. Seems a casual way of saying that both become equal when eps goes to zero. Seems valid to me, no reason to limit me to one side of CO, it's not like this has influence on anything, or I'm missing context. (See note 3)


 * Q2: it is valid to draw lines between several points on the ROC curve, say corresponding to "negative", "weak positive" and "strong positive", and that the slope of each line is a valid estimate of the likelihood ratio for test results that fall within the corresponding interval.
 * It's the definition of ROC curve: Y-coordinate = TPR(CO) x-coordinate=FPR(CO), the slope of a line from origin (0,0) to a point is by definition y/x. For a line between two points on the curve you calculate the slope by subtracting the coordinates. For example: Two points TPR=0.5 FPR=0.3 and TPR=0.8 FPR=0.7, segment between: TPR=0.8-0.5=0.3 and FPR=0.7-0.3=0.4; between those points, the positive likelihood ratio is 0.3/0.4=0.75, so more False positives than true positives. (See note 4)


 * Q3: I've read many places (including in our article) that the area under the curve corresponds to the probability of a randomly chosen diseased individual getting a higher test result than a randomly chosen individual without the disease. Again, this sounds reasonable, but exactly why is it so?
 * The X and Y coordinates of a ROC are proportional to the number of healthy and infected people. so an increase of 0.10 in either direction represents 10 % of the related group. Movement at 45° represents the same % of people in both groups.
 * Suppose the curve starts by going up to 0.4, that means 40% of the infected (lets assume low values indicate infection) will test lower than all the rest. Then 0.2 to the right; 0.3 up; 0.7 right, 0.2 up, finally along the diagonal.
 * Calculate the odds of infected people scoring better (lower), than healthy: 40% score better than all 0.4*1; 30% score better than 80%; 0.3*0.8; 20% better than 10%: 0.2*0.1 and the last 10% have a chance of one in two to do better than the other 10%, so: 0.1*0.1*0.5. In total:  0.665.
 * The other group: 0.2 *0.6; 0.7*0.3 and 0.1*0.1*0.5: total: 0.335. You can work it out on paper, when you change the curve, the change in area will match the change in odds. Here, the area is 66.5%. (See note 5)


 * Oeps, I now see that the link you provided had not just the abstract, but the whole article... (9850136?)  <font style="color:DarkGreen;background-color:#FAFAFF;">Ssscienccce  (talk) 02:30, 21 September 2015 (UTC)

Thank you,, those were very helpful answers and links. I took the liberty of inserting yellow labels in your answer, in order to reference your replies more easily. Yes, I see that the question was confusing, reflecting my own confusion. Your reference to Johnson 2005, was very helpful indeed! --NorwegianBluetalk 20:38, 21 September 2015 (UTC)
 * Note 1: I was thinking about a smoothened ROC curve, and see that that would imply needing continuous probability distributions/densities.
 * Note 2: Duh! Confused, wrong thinking on my part here. Of course it isn't. I'm forgetting the Quotient rule, and being unclear about exactly what variable I'm differentiating with respect to.
 * Note 3: Q1 is very confused. I think I'd rather strike it than try to reformulate it.
 * Note 4: Thanks! I understand it now.
 * Note 5: Thanks! I'm still struggling a bit with this one, but it's getting late. I'll re-read your reply carefully in the morning, and I'm optimistic that I'll understand it then.


 * Wasn't very happy with that last answer myself. This may be a better way:
 * For every FPR value there exists a corresponding Cutoff value. Hundred FPR values from 0.01 to 1 (with 0.01 increments), have 100 corresponding cutoff values, CO0.01, CO0.02... CO1.00
 * This way you could divide the healthy people in 100 groups, based on their test result. Each group would contain 1% of the total healthy population. (follows from the definition of FPR).
 * The same for the diseased people. They too are split in 100 groups, corresponding to the TPR values 0.01, 0.02, 0.03 and so on... (note: the hundred cutoff values differ from those for the healthy group).
 * Randomly choose one healthy person. He would belong to one of the hundred groups, and since the groups are of equal size, they are all equally likely to be chosen. The same is true for a randomly selected diseased person.
 * As an example, we'll say that the healthy person is in healthy group 20 (FPR 0.20 to 0.21) and the other one in diseased group 40 (TPR 0.4 to 0.41). those values define a small square of 0.01*0.01 size on the ROC diagram.
 * If you draw vertical lines at FPR=0.2 and 0.21, they will cross the ROC curve at two points, (the Cutoff values that defined the group he belongs to) and the persons test value must be between those two.
 * Do the same for the horizontal lines at TPR =0.4 and 0.41, they cross the ROC curve at 2 points, the test value of the person is between the two.
 * If the small square where the horizontal and vertical lines cross is below the curve, the vertical lines will cross the curve further from the origin than the horizontal lines (on fig 1b at page 2261, the TPR=0.4 line crosses at about 0.1 FPR, FPR=0.2 line crosses at about TPR=0.57). Values further on the curve are lower, since more people test positive the further you go, so the healthy person has a lower test result than the sick person. If the point (little square) is above the curve, the reverse is true: the vertical FPR line intersects the curve closer to the origin, so the healthy person has a higher test value.
 * The number of 0.01*0.01 squares below the curve are proportional to the area below the curve. If the area is 90% then 9000 squares are below, 1000 above the curve. Picking one healthy and one sick person at random has a 90% probability of corresponding to a square below the curve, which corresponds to the healthy person having a lower test value than the diseased one. <font style="color:DarkGreen;background-color:#FAFAFF;">Ssscienccce (talk) 02:55, 22 September 2015 (UTC)
 * Thanks, I followed your argument carefully, with the help of a spreadsheet, pen and paper. I understand why the number of healthy people in each of the 100 boxes of healthy people is the same, and ditto for the number of diseased people in each of the 100 boxes of diseased people. I drew a diagram (only 10 categories of each), which turned out to look a lot like the example on page 2261 in the reference. Since any intersection point with the ROC curve, whether by a vertical or horizontal line, corresponds to a cutoff value (or in this argument, measurement), a point above the curve for a random pair of healthy and diseased individuals, implies that the healthy person (vertical line that has crossed the curve) has a measurement that is higher (closer to the origin) than that of the diseased person (horizontal line that will intersect the curve at a point further from the origin, i.e. has a measurement that is lower). I think I got it! If there were any errors in my restatement of your explanation, please warn me about it! Thanks a million. --NorwegianBluetalk 21:44, 22 September 2015 (UTC)