Wikipedia:Reference desk/Archives/Mathematics/2008 April 22

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[edit] April 22

[edit] Most used math?

What is the most used math?--208.102.189.190 (talk) 02:38, 22 April 2008 (UTC)

It really depends on how you split it up categorically, but I would say arithmetic is the most used. And excepting that and basic algebra, I would say Statistics is the most used. That is just my opinion. A math-wiki (talk) 06:19, 22 April 2008 (UTC)

Counting. Bo Jacoby (talk) 09:37, 22 April 2008 (UTC).

Geometry, just because it's instinctive doesn't mean it's not math. So you use geometry whenever you take a step. Taemyr (talk) 12:36, 22 April 2008 (UTC)

Yeah, it's either basic arithmetic or geometry (some rather advanced geometry, in fact - it's far far easier to catch a ball than it is to calculate where to ball is going to land using pen and paper). --Tango (talk) 16:18, 22 April 2008 (UTC)

Identity. It's the first mathematical thought babies have and people hold on to it forever! —Preceding unsigned comment added by 90.198.200.119 (talk) 21:57, 24 April 2008 (UTC)

[edit] Real Analysis

Okay, so in my real analysis class our teacher gave us a function.. $f:(0,1)\rightarrow\mathbb{R}$ such that $f (x) = 0$ if $x$ is rational and $f (x) = [1 / x] - 1$ is $x$ is irrational where [x] is truncation function which returns the integer part of $x$ .

So, our teacher keeps saying that the Riemann integral $\int_0^1f(x)dx=\infty$ . But I don't understand how this can be. I remember reading somewhere that a function f is Riemann integrable if and only if the set of discontinuities on the interval which you are integrating has measure zero. Since in this example, the discontinuities are all at rational numbers (which are countable therefore have measure zero), this function is integrable. So if we plot this function, we simply get rectangles of width $\frac{1}{k}-\frac{1}{k+1}$ with height $\frac{1}{k}$ . So,
$\int_0^1f(x)dx=\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})\frac{1}{k}=\frac{\pi^2}{6}-1$ So, the integral does not only exist, we can find the exact value of the integral. Can somebody please elaborate on this? What is the error in my reasoning? Did the teacher make a mistake?A Real Kaiser (talk) 05:27, 22 April 2008 (UTC)

This function is not Riemann integrable (which is not the same as saying the integral is infinite).
~~The function's Lebesgue integral is infinite. Perhaps this is what the teacher meant?~~ The function is Lebesgue integrable.
The criterion you state is incorrect for both Riemann and Lebesgue integrals. Riemann integrals can fails to exist when the function only has countably many discontinuities, and Lebesgue integrals can exist when the function is discontinuous everywhere.
This function is discontinuous everywhere, not just at rational points.
See Dirichlet function which is very similar.
~~The rectangles each have height k, not $\tfrac1k$ . Thus the sum is $\sum_{k=1}^{\infty}\frac{1}{k+1}=\infty$ .~~
Remember that the Lebesgue integral is robust to changes of measure zero. Thus it is in this case equal to the integral of $\lfloor1/x\rfloor^{-1}$ , which is itself not that much different from $x - 1$ . Thus, we shouldn't be surprised that it is infinite.

-- Meni Rosenfeld (talk) 08:52, 22 April 2008 (UTC)

Meni, are you sure that you're not thinking about $\lfloor 1/x \rfloor$ ? The integral in the original post is finite, and Mathematica agrees (up to 5 decimal places) that it is $\frac{\pi^2}{6}-1$ . The teacher was probably thinking about $\lfloor 1/x \rfloor$ also. As for Riemann integrability, a function is (properly) Riemann integrable if and only if it is bounded and its discontinuity set is a zero set. Our article zero set refers to something else, but here zero set means that it can be covered by a union of open intervals while making the measure of the union arbitrarily small. Having measure zero is necessary, but not sufficient. 134.173.93.127 (talk) 09:37, 22 April 2008 (UTC)

You are of course correct. I'd like to emphasize that it is the Lebesgue integral which is finite, the function is still discontinuous everywhere and its Riemann integral doesn't exist. -- Meni Rosenfeld (talk) 10:08, 22 April 2008 (UTC)

It's certainly not Riemann-integrable. Even if you'd never heard that having a set of points of discontinuity of positive measure entails non-Riemann-integrability, you could see that's it's not Riemann-integrable just by thinking about the upper and lower sums. It is not Lebesgue-integrable since its Lebesgue integral over the interval (0, 1) is not finite. It is measurable and its Lebesgue integral over that interval is ∞. Michael Hardy (talk) 22:17, 22 April 2008 (UTC)

If the answer should be infinite, where has the OP gone wrong? The method looks good to me... --Tango (talk) 22:19, 22 April 2008 (UTC)

Michael, are you making the same mistake I did? -- Meni Rosenfeld (talk) 23:28, 22 April 2008 (UTC)

I wonder what it is about that function. I made that mistake too, and even posted it. I removed my post when I noticed it, but I probably should have just crossed it out as a warning... 134.173.93.127 (talk) 23:46, 22 April 2008 (UTC)

Maybe it's because the inversion is denoted once by

1 / x

and once by

- 1

. Instead of interpreting the function as having two distinct inversions, they coalesce in our mind to a single inversion represented in two different ways. -- Meni Rosenfeld (talk) 09:18, 23 April 2008 (UTC)

Thanks everyone, for their input but what is the verdict? Is this function Lebesgue integral? What is the Lebesgue integral's value? Is this function Riemann integral? What is the Riemann integral value? I say that this function is both Riemann and Lebesgue integrable and hence both integrals are equal and the value is $\frac{\pi^2}{6}-1$ .A Real Kaiser (talk) 19:05, 23 April 2008 (UTC)

The function is Lebesgue integrable, and the Lebesgue integral is $\frac{\pi^2}{6}-1$ . The function is nowhere continuous, and thus not Riemann integrable. For any sequence of partitions, you could select sample points that give you Riemann sums converging to 0 or converging to $\frac{\pi2}{6}-1$ , for example. It's also not hard to see that the upper and lower Darboux integrals are different. If instead you looked at the function which is $\lfloor 1/x \rfloor$ on irrationals and 0 on rationals, then it still wouldn't be Riemann integrable, but this time the Lebesgue integral would be $\infty$ 134.173.93.127 (talk) 19:25, 23 April 2008 (UTC)

I appreciate your gratitude for our input, but this post seems to indicate you didn't actually read it. -- Meni Rosenfeld (talk) 21:55, 23 April 2008 (UTC)

I think he read it and got confused by the fact that everyone else was getting confused about what the questions was actually asking. Everyone was contradicting each other - it was rather difficult to see what the final conclusion was. --Tango (talk) 22:14, 23 April 2008 (UTC)

There was unanimous agreement that the function is Lebesgue integrable and not Riemann integrable. The only confusion was about the value of the Lebesgue integral, and I think it was pretty clear what the correct answer was. Even if the correspondence was confusing (which I don't think it was), it's the OP's obligation to make an effort to understand it. -- Meni Rosenfeld (talk) 23:29, 23 April 2008 (UTC)

[edit] Dominated Convergence Theorem

On a similar note, how can one show that
$\int_0^1\sin (x) \ln (x) dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)(2n)!}$ . What I have done is that I tried to find the power series representation of $f (x) = sin(x)ln(x)$ and then simply integrate term by term. So in order to find the power series, I constructed a differential equation with initial conditions $x 2 y'' + x 2 y = 2 x cos(x) - sin(x)$ with $y(\pi)=0, y''(\pi)=\frac{-2}{\pi}$ (which is the simplest one I could come up with). My $f (x)$ is the solution to this differential equation and if I find the power series solution to this equation, I will have my power series (by the existence and uniqueness theorem). But when I tried to solve this non-homogeneous using the power series method, it got a bit ugly. Now, I think that another way of solving this is to use the dominated convergence theorem. But can anyone suggest some methods with which I can come up with my $g (x)$ and my sequence $f n (x)$ so that I can apply the DCT?A Real Kaiser (talk) 05:40, 22 April 2008 (UTC)

Wouldn't there be a much simpler way to find $\int_0^1\sin (x) \ln (x) dx$ ??? Like using integration by parts, or em I wrong in thinking that ln(x) has an elementary anti-derivative?? (going to check that) A math-wiki (talk) 06:25, 22 April 2008 (UTC)

Well integrating ln(x) isn't the problem, the problem is the anti-derivative of that product is not elementary, you need something akin to the Cosine Integral, Ci(x) (according to mathematica's free online integrator), to get the anti-derivative of that function. A math-wiki (talk) 06:40, 22 April 2008 (UTC)

Quick thought: if you could find a dominating function, how about doing something like

$f_n(x) = \sin(x) \ln(x) \chi_{[\frac{1}{n},1]}(x)$

and then integrating

f n

by parts? You'll get a nice power series expansion for the resulting integral, and the terms that are blowing up should cancel. Warning: haven't tried it. 134.173.93.127 (talk) 09:57, 22 April 2008 (UTC)

Couldn't you perhaps use a Taylor series expansion of the function:

$f(t) = \int_t^1\sin (x) \ln (x) dx$

around the point t=1? That should be pretty easy if you use the fundamental theorem of calculus and the fact that when g(a) is defined, $\int_a^a g(x) dx = 0$ , though you might have to look carefully at Taylor series convergence. --Prestidigitator (talk) 20:09, 22 April 2008 (UTC)

Erm, the obvious way to go about this is to write the sine as a power series. You can then exchange the series and the integral by dominated convergence, and you end up with integrals of the form $\int_0^1 x^{2k+1} \ln(x) dx$ which are easy to compute using partial integration. Bikasuishin (talk) 19:46, 23 April 2008 (UTC)

[edit] Anti - derivative

Hi, This may be simple, but how can I find the integral of, square root of (r² - x² ) dx ??? (r is a constant) ie., integral of (r² - x²) ^1/2—Preceding unsigned comment added by 116.68.71.178 (talk) 06:59, 22 April 2008 (UTC)

Use trigonometric substitution. –Pomte 08:19, 22 April 2008 (UTC)

... or think visually and geometrically. $\int_0^t \sqrt{r^2-x^2}\, dx$ is the area under the curve $y=\sqrt{r^2-x^2}$ between x=0 and x=t. And the curve $y=\sqrt{r^2-x^2}$ is a well-known geometric shape. Even better - use both approaches and check that they give the same answer. Gandalf61 (talk) 09:19, 22 April 2008 (UTC)

Actually, I wanted to know the integral , to see how I can find the area in a circle cut off by a chord (ie., a segment).............Thanks for the answer! (I'm only starting to learn differentiation and integration and don't know much. This helps a lot) 116.68.70.17 (talk) 12:33, 22 April 2008 (UTC) A 15-year old.

If you want an exercise in integration, go for it. But if you just want the answer, you don't even really need calculus; just a little trig. You know the area of a circle and the angle subtended by the whole shape (2π), so you should be able to figure out the area of a circular arc subtended by a given angle. Then figure out the area of the triangle formed by the center of the circle and the ends of the chord by determining the dimensions of the triangle. --Prestidigitator (talk) 20:19, 22 April 2008 (UTC)

If you want to do it to practice some calculus (and why not - it's useful to have an exercise where you can compare your answer to what it should be) you will probably want to introduce a change of variables via substitution, as Pomte said. In this case, I might suggest trying something like

x = r sinθ

. Confusing Manifestation(Say hi!) 22:36, 22 April 2008 (UTC)

x= r sin \theta ?? I'm sorry I didn't get that. Can you clarify what \theta stands for?? (Am I being dense here?? ) 116.68.70.86 (talk) 12:29, 23 April 2008 (UTC)A 15-year old

θ is an arbitrary angle - although in this problem it happens to be the angle that the radius of the circle makes with the y axis. The substitution works as follows:

x = r sinθ

$\sqrt{r^2-x^2}=\sqrt{r^2-r^2 \sin^2 \theta}= r\sqrt{1-\sin^2 \theta}=r\cos\theta$

$\frac{dx}{d\theta} = r\frac{d \sin \theta}{d \theta}=r \cos \theta$

$\Rightarrow \int \sqrt{r^2-x^2}\, dx = \int r^2 \cos^2 \theta\, d \theta$

See our article on trigonometric substitution. Gandalf61 (talk) 12:59, 23 April 2008 (UTC)

Thanks a lot Mr. Gandalf! I understand now, and this is certainly a much better method than trying to integrate $\sqrt (r^2- x^2)$ as it is, with my little knowledge of calculus...... —Preceding unsigned comment added by 116.68.70.44 (talk) 09:05, 24 April 2008 (UTC)

[edit] anew technique to solve some equations.

Iam not sure but I think I have developed asimple technique to solve equations like,e^x=sin(x).I need your opinions if this technique works.Now,if,f(x)=x,at x=x0,then, f(f(f(…….f(x0))=x0.This technique depends on choosing apoint around the interval where, g(x)=h(x).It doesnot matter which point we choose as long as it is inside the interval.for example, Cos(x)=x,if we start with any point within the interval (0,pi\2),say,0,then we will get, Cos(cos(cos(…….cos(0))=0.73908….it does not matter wich point we choose within (0,pi\2),we will always get the same result.As for the equation,e^x=sin(x),we put, [Inv(sin(e^x)=x],as long as sin(x)has aperiod of(pi).If we start with(-pi),then we should get for the first approximation,[ Inv(sin(e^-pi)=0.04322…=x1].and for the second approximation,[Inv(sin(e^-(pi+x1)=x2]….and so on until we be close enough and the point would be(-pi-xn),where sin(-pi-xn)=e^(-pi-xn).We also could apply the same way at the zeros of sin(x) and find all of the other solutions.thank youHusseinshimaljasimdini (talk) 11:30, 22 April 2008 (UTC)

Not all Fixed points are attractors. Taemyr (talk) 12:11, 22 April 2008 (UTC)

As to your technique. If you can transform your equation to the form x=f(x), for some function f, then the solutions of the equation must be exactly equal to the fixed points of f. So the set of fixed points for ln(sin(x)) is equal to the solution set for e^x=sin(x). So if we assume attractive fixed points and that you start "close enough" to a fixed point, the procedure described will approximate a solution. It's important to note that you will only ever get approximations, because the value of x that we are after is the limit of the sequence x0,x1,x2... so if you want the exact answer you would have to prove the limit, not just iterate. This means that your technique is a numerical approach to solving equations. Taemyr (talk) 12:29, 22 April 2008 (UTC)

(After edit conflict) What you have discovered is called an iterative method, and is part of the area of mathematics called numerical analysis. There are two difficulties with this type of solution method (which have both been pointed out by Taemyr).

The first is that you only get an approximate numerical solution to an equation, rather than an exact algebraic solution. For practical purposes, that may not be a problem as long as you can place an upper bound on the difference between the approximation and the actual solution, and as long as this error bound decreases rapidly so that you don't have to do hundreds of iterations to get a reasonably accurate answer.

The second, more fundamental, problem is that iterative methods do not always converge. For example, if you try to use your simple iterative method to solve the equation x=2cos(x) you will find that it does not "settle" to a fixed value, even though we can tell graphically that there is a solution quite close to x=1.03. Gandalf61 (talk) 12:47, 22 April 2008 (UTC).OK.I got it.Thank you.Husseinshimaljasimdini (talk) 07:31, 23 April 2008 (UTC)

[edit] Name of a shape

reposted from Language desk as there is probably a mathematical name also. -- Q Chris (talk) 13:04, 22 April 2008 (UTC)

I'm asking this here not at maths as I am seeking the name commonly used rather the maths name fo a shape. EG Diamond not Rhombus. The shape I want a name for is the shape made as an arc of a circle (under 180 degrees) with its reflection. A bit like the current Doctor Who logo. -- SGBailey (talk) 13:52, 21 April 2008 (UTC)

In America, we might call it a football shape. I was curious about the technical name for it, so I looked it up and I guess it is a kind of Lens (geometry). Recury (talk) 14:14, 21 April 2008 (UTC)

Under some circumstances is is also called a mandorla, Italian for almond. SaundersW (talk) 15:43, 21 April 2008 (UTC)

There's also lenticular, although it's not exactly an everyday word. —Steve Summit (talk) 22:28, 21 April 2008 (UTC)

Almond-shaped is common in English, though not for something as narrow as the Dr Who logo. kwami (talk) 23:07, 21 April 2008 (UTC)

I will repost this in the mathematics desk, as there is probably a mathematical name also. -- Q Chris (talk) 13:04, 22 April 2008 (UTC)

Lens is the mathematical name for this shape. If you are looking for a more impressive name, you could call it a vesica piscis, as long as the arcs are specifically arcs of a circle. Gandalf61 (talk) 13:31, 22 April 2008 (UTC)

[edit] question about quadratic equation

How do you know is a quadratic equation will have one, two, or no solutions? How do you find a quadratric equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 14:20, 22 April 2008 (UTC)

All quadratic equations have two solutions, just those solutions may not be distinct or real. For instance the equation x² + 16 has solutions of $\pm 4i$ . As for your second and third questions, try factorising any quadratic you like, and thus find its two roots. Then see if you can make a different quadratic with the same roots. -mattbuck (Talk) 14:43, 22 April 2008 (UTC)

To tell how many real roots a quadratic has, you need to calculate the discriminant, (the bit inside the square root in the quadratic formula: b²-4ac). If that's positive, it has two square roots, giving two roots to the quadratic. If it's 0, it has just one square root (0), so you have one root to the quadratic. If it's negative, it has no real square roots, so no real roots to the quadratic. And, as mattbuck says, factorisation should answer your last two questions - you just have to do it in reverse (the roots will give you factors, which you can then expand). --Tango (talk) 16:27, 22 April 2008 (UTC)

And just in case it isn't obvious from the above two answers, if the discriminant