Wikipedia:Reference desk/Archives/Mathematics/2008 April 6

= April 6 =

Arc-length and area
This question's been bothering me for a while: find a continuous function f such that $$f(0)=f(1)=0\,\!$$, $$f(x) \ge 0$$ for $$0 \le x \le 1$$, the area under the curve f(x) between x=0 and x=1 is 1, and that the fuction which will have the smallest possible arclength. I know that a circle has the smallest perimeter to area ratio, but obviously a circle with a radius of 1/2 can't have an area of 1. Any ideas? Thanks. —Preceding unsigned comment added by 65.92.4.130 (talk) 01:55, 6 April 2008 (UTC)


 * Try $$\sin{\pi x}$$ (I'm only guessing on that one but it might be what your looking for) A math-wiki (talk) 07:43, 6 April 2008 (UTC)
 * The area under $$f(x)=\sin(\pi x)\,\!$$ is $$\frac{2}{\pi}\,\!$$. —Preceding unsigned comment added by 65.92.4.130 (talk) 16:03, 6 April 2008 (UTC)
 * There is no solution, and I will tell you why. The semicircular arc has area &pi;/4<1. Put the arc on pillars so that f(x)=1&minus;&pi;/4+(x(1&minus;x))1/2. This function has minimum arclength = &pi;/2 and area = 1 but it does not satisfy the condition f(0)=f(1)=0. If you measure the arclength from (0,0) to (1,0) you still get minimum arclength = 2, but now the function is not continuous. A continuous approximation to this function with area = 1 will have greater arclength. So your problem has no solution. Bo Jacoby (talk) 08:22, 6 April 2008 (UTC).
 * Try the following function: Every x not bigger than 1/2 satisties f(x)=4x, and every x not smaller than 1/2 satisfies f(x)=4-4x. Eliko (talk) 09:26, 6 April 2008 (UTC)
 * Area=1 is OK, but arclength=171/2~4.12 is far from minimum, as a continuous approximation to f(x)=1&minus;&pi;/4+(x(1&minus;x))1/2 has arclength arbitrarily close to 2. Bo Jacoby (talk) 11:32, 6 April 2008 (UTC).
 * Isn't the area under a semicircle with radius 1/2 not &pi;/4 but &pi;/8 ? So to make the area under the curve equal to 1, the two "pillars" need to be of height 1-&pi;/8, and the arc length is 2(1-&pi;/8)+&pi;/2 = 2+&pi;/4 = 2.78 approx. Even with this correction, I can't think of a unit area curve with a shorter arc length - I tried the half-ellipse with semi-axes 1/2 and 4/&pi;, but its arclength is a little over 2.9. Gandalf61 (talk) 12:34, 6 April 2008 (UTC)
 * Thank you Gandalf61, you are right. Note that the circular arc is better than an arc of any other shape, such as an ellipse. If the problem was not to find a function, but to find a curve connecting (0,0) with (1,0) such that the area above the x-axis inside the curve is =1 and the curve length is minimum, then a circular arc, a little more than a semicircle, is the answer. Bo Jacoby (talk) 14:29, 6 April 2008 (UTC).
 * I believe that people are misinterpreting my question. I is not possible that the question has no solutions, for I can think of at least one function that has 0 and 1 as its zeroes, is greater than zero over the region between x=0 and x=1, and has an area of 1 between x=0 and x=1:$$f(x)=6x-6x^{2}\,\!$$. This function clearly obeys the first restriction. Additonally, $$\int_0^1 6x-6x^{2}\, dx = 1$$. The arclength of this function over x=0 and x=1 is $$\frac{\sqrt{37}}{2}-\frac{\ln(\sqrt{37}-6)}{2}$$. However, this function may not have smallest arclength of functions that satisfy the first three restrictions. My question is: which function does? I suspect it would be an ellipse.
 * I think they've interpreted the question correctly, but haven't done a great job of explaining why there's no answer. Bo and Gandalf are trying to tell you that the set of all possible arc lengths for your functions is an open interval with a lower bound of 2+pi/4. It's as if you'd asked: of all positive numbers, which one is lowest? There isn't an answer because no matter how small a positive number is, it can still be made smaller and remain positive. You can't say "of course there must be a smallest positive number, I can name 3 positive numbers off the top of my head!" Zero is the lower bound, but it's not an answer because it's not positive. Assuming Bo and Gandalf's answer is correct, it would be possible to give a function with arc length 2+pi/4+0.01, or 2+pi/4+0.001, or 2+pi/4+0.0001, etc. but not 2+pi/4 itself because that requires a couple of vertical line segments on the sides, which can't be described in the form of a function y=f(x). --tcsetattr (talk / contribs) 21:36, 6 April 2008 (UTC)
 * Yes, of course. I've must of misinterpreted what the others have said. Thanks. —Preceding unsigned comment added by 65.92.4.130 (talk) 21:48, 6 April 2008 (UTC)

Integrals
Ok, the question I really want to ask is "How do integrals work" but I fear the answer to that would be less of a paragraph and more of a full calculus course. So, I'll try to narrow my question a little. As you can probably tell, I'm a beginner at this calculus type stuff, but as I understand it,

$$\int_{a}^{b} f(x)dx$$

Can be calculated by finding $$F(x)$$, an anti derivative of $$f(x)$$, and then doing $$F(b)-F(a)$$. So my question is, how would one go about calculating an integral if b is, say, infinity? Digger3000 (talk) 02:29, 6 April 2008 (UTC)


 * The usual physical representation of an integral is to find the area under a curve from point A to B, so changing that to find the area under the curve from A to ∞ would typically only yield ∞, unless the height of the curve happens to approach zero. StuRat (talk) 02:55, 6 April 2008 (UTC)


 * Calculus handles infinity and infinitesimals via limits, and the same idea works here. We can find the integral where the upper limit is infinite by approximating it with integrals with larger and larger upper limits -- if we can take the limit (that is, those increasingly better approximations converge towards a single finite value) then that's out answer. Thus:


 * $$\int_a^\infty f(x)\,dx = \lim_{l\to\infty}\int_a^l f(x)\,dx$$


 * Now, presuming that $$F(x)$$ is the appropriate antiderivative, and applying the fundamental theorem of calculus we get $$\lim_{l\to \infty} F(l)-F(a)$$, and the question is whether that limit exists. Depending on the function F it may. If it is, that's your answer. The keyword you may be after is improper integral; that should give you a much more solid grounding than the general intuitive handwaving I've just supplied. -- Leland McInnes (talk) 02:58, 6 April 2008 (UTC)

So, if the question is whether the limit exists, how do you answer that? How do you know if the limit exists and, if so, what it is? I'm sure it's pretty simple but oh well, I'm asking anyway. Digger3000 (talk) 03:16, 6 April 2008 (UTC)


 * The definition in Limit of a function should answer any questions you have about what makes a limit defined. A math-wiki (talk) 07:33, 6 April 2008 (UTC)
 * Perhaps an example will help: Suppose we wish to find $$\int_1^{\infty}\frac{1}{x^2}\ dx$$. An antiderivative is $$F(x)=\frac{-1}{x}$$. We have $$F(1)=-1$$ and $$\lim_{x \to \infty}F(x)=\lim_{x \to \infty}\frac{-1}{x}=0$$, thus the integral is equal to $$0-(-1)=1$$. -- Meni Rosenfeld (talk) 14:22, 6 April 2008 (UTC)

Antidifferentiation
How would you anti-differentiate (X+1)^.5? I figure it there is an X^1.5 but what about the inside function? 71.218.28.218 (talk) 04:25, 6 April 2008 (UTC)
 * This is essentially a u-substitution problem, but as a shortcut, ask yourself: what is d/dx of the inside function? -- Kinu t /c  04:47, 6 April 2008 (UTC)

d/dx of the inside function is 1 71.218.28.218 (talk) 04:49, 6 April 2008 (UTC)
 * So based on that, you have: $$u=x+1 \to du=dx$$. Set up your substitution and go from there: $$\int{u^{1/2}du}$$. -- Kinu t /c  04:52, 6 April 2008 (UTC)

im not good at substitution, but would it be {(1+x)^1.5}/1.5? —Preceding unsigned comment added by 71.218.28.218 (talk) 05:16, 6 April 2008 (UTC)


 * I believe it would be (2/3)*((X+1)^(3/2)). If that turns out to be wrong, please feel free to yell at me. Digger3000 (talk) 05:26, 6 April 2008 (UTC)

after looking at the wikibooks for calculus, i think your answer is right digger71.218.28.218 (talk) 06:06, 6 April 2008 (UTC)


 * You are both right: {(1+X)^1.5}/1.5 = (2/3)*((X+1)^(3/2)). --Lambiam 09:39, 6 April 2008 (UTC)

questions about radical expressions
Why is it important to simplify radical expressions before adding or subtracting?How is ading expressions similar to adding polynomial expressions? How is it different? What would a radical expression look like to simplify it? thank you —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 11:49, 6 April 2008 (UTC)


 * I wouldn't say it is important to simplify before adding and subtracting, it just makes things easier. You can simplify later on if you want, it's just more work. It's similar to adding polynomials because you can treat things like $$\sqrt{2}$$ as a variable and collect like terms. I'm not sure it is significantly different for addition - for multiplication, you have to multiply and simplify the radicals rather than just writing x2 as you would with a polynomial, but that doesn't come up for addition. I don't understand the last question, could you rephrase? --Tango (talk) 14:08, 6 April 2008 (UTC)


 * This sounds just a tiny little bit like a homework question, but here's a hint as to why you should simplify before adding or subtracting: take the expression $$\sqrt{8} - 2\sqrt{2}$$ and try doing any addition or subtraction on it without simplifying things. Now try simplifying first, and then see if it's any easier or harder to work with. Confusing Manifestation (Say hi!) 22:56, 6 April 2008 (UTC)

question about Pythagorean theorem
What is the Pythagorean theorem?How is it used? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 11:51, 6 April 2008 (UTC)
 * Enter Pythagorean theorem in the search box to the left. PrimeHunter (talk) 12:16, 6 April 2008 (UTC)

Solutions to Tarski's Introduction to Logic
I have been reading Tarski's Intro to Logic, and trying the exercises at the end of the chapters. The solutions are not provided in the book. Does anyone know where I might find them ? —Preceding unsigned comment added by 70.251.246.119 (talk) 17:12, 6 April 2008 (UTC)
 * I doubt they are available in published form. We might be able to help with specific exercises you're having problems with. Algebraist 18:14, 6 April 2008 (UTC)

number theory problem
Hi. Can someone help me with the following problem: If a,b are integers and 9 divides $$a^2+ab+b^2$$ then show that 3 divides both a and b.--Shahab (talk) 17:35, 6 April 2008 (UTC)


 * This sounds like homework, so just some pointers: the equation divided by nine looks like this:
 * $$\frac{a^2 + ab + b^2}{9}$$


 * $$= \frac{a^2}{9} + \frac{ab}{9} + \frac{b^2}{9}$$


 * $$= \frac{a^2}{3^2} + \frac{ab}{3^2} + \frac{b^2}{3^2}$$
 * Now, do you see where to go? Can you "incorporate" those 3's along with the a's and b's? --Oskar 19:29, 6 April 2008 (UTC)
 * Thanks. (BTW it's not Homework.) I presume you want me to conclude that (a/3)^2+(a/3)(b/3)+(b/3)^2 is an integer and so a/3 and b/3 are integers. But non integers can also sum up to an integer.--Shahab (talk) 07:11, 7 April 2008 (UTC)
 * I think there's more to it than that - you can have 3 numbers add up to a multiple of 9 without each of them being a multiple of 9. --Tango (talk) 19:35, 6 April 2008 (UTC)
 * Interesting fact that I think is helpful here. Every square not a multiple of three, modulo 9, is 1 modulo 3. I’d break it into the cases that a is a multiple of 3 and b is not, and that both are not, and find a contradiction in either case. GromXXVII (talk) 19:50, 6 April 2008 (UTC)
 * Can you explain the fact more clearly. As far as I can understand it if $$n^2=3p+1,3p+2$$ then (n^2 mod 9) is congruent to 1 modulo 3. How do I prove this.--Shahab (talk) 07:11, 7 April 2008 (UTC)
 * What GromXXVII is saying is:
 * If n is a multiple of 3 then n2 is a multiple of 9.
 * If n is not a multiple of 3 then n2 is 1 modulo 3 (to prove this, just assume that n=3k+1 or n=3k+2 and square the expression on the right hand side) so n2 is 1, 4 or 7 modulo 9. Gandalf61 (talk) 10:01, 7 April 2008 (UTC)


 * Thanks for clearing that up. I still can't figure out the contradiction that I should reach though. Suppose a is a multiple of 3 and b isn't then 9 divides a^2 and (let) 9 divide b^2-1. What should I do now (The only thing I can conclude is 9 divides ab+1 but it doesn't seem useful.)? --Shahab (talk) 15:35, 7 April 2008 (UTC)


 * You could start with the identity


 * $$a^2+ab+b^2=(a-b)^2+3ab$$


 * so if the left hand side is a multiple of 9 (or even just a multiple of 3) then (a-b)2 is a multiple of 3, and so a-b is a multiple of 3. This just leaves you with just two cases to eliminate: a and b both equal 1 mod 3; and a and b both equal 2 mod 3. Gandalf61 (talk) 16:01, 7 April 2008 (UTC)


 * But 3 divides a-b doesn't imply that 3 divides a and b. a=4, b=1. Am I missing ur point entirely here?--Shahab (talk) 16:43, 7 April 2008 (UTC)


 * 3 divides a-b implies that a and b are congruent to each other mod 3. So you need only check each of the three possible values that both of them might be congruent to. —David Eppstein (talk) 16:54, 7 April 2008 (UTC)


 * I need to show that 3 divides a and b. As far as I can understand Gandalf61 he wanted me eliminate the case a,b is 1mod3 on the basis of 3|(a-b) which is impossible. I will be highly obliged if someone could give a more explicit description of the method of proof. May I repeat that this is not homeowrk. (I am not a student of any formal school).--Shahab (talk) 17:14, 7 April 2008 (UTC)
 * Assume a and b are both 1 mod 3, then find a contradiction. Then assume a and b are both 2 mod 3 and find a contradiction. That leaves only the case that they are both 0 mod 3, ie 3 divides them both, as required. --Tango (talk) 17:22, 7 April 2008 (UTC)
 * I had got all that. My point is that if we start with the proposed method of proof (that 9|a^2-ab+b^2 implies 3|a-b imples 3|a and 3|b) there can be no contradiction since 3|a-b doesn't necessarily imply 3|a and 3|b. For example a=4 and b=1. 3|a-b and yet 3 doesn't divide 4 and 1. --Shahab (talk) 17:38, 7 April 2008 (UTC)
 * But $$4^2+4\cdot1+1^2=21$$ and $$9\not|21$$. You need to use the original assumption to find the contradiction. --Tango (talk) 18:41, 7 April 2008 (UTC)
 * OK. Now I got it. Thanks everyone.--Shahab (talk) 13:55, 8 April 2008 (UTC)

Maximing volume of cylinder in a sphere
Suppose I'm inscribing a cylinder within a sphere of radius 100m. I am trying to maximize the volume of the cylinder that is within that sphere, what do I want the radius and height of that cylinder to be? Am I trying to make both height and radius the same? For example, if I was given a fixed perimeter and was trying to construct a rectangle of maximum area, I want both length and width to be the same. Does the same hold true when finding the volume of a cylinder in a sphere? Thanks 99.240.177.206 (talk) 18:03, 6 April 2008 (UTC)


 * Once you've determined either the height, or the radius, the other is determined, so you can write the volume as a function of either one of them. Pick one, write down the formula, and then try and maximise it (try differentiating it). --Tango (talk) 19:39, 6 April 2008 (UTC)


 * By differenting it, I got both the height and radius to be roughly 7.62 m. Is this correct?

My beginning formula was: Volume = (pi)(r^2)(h) = (pi)(10-x)^2(20-2x)

99.240.177.206 (talk) 21:12, 6 April 2008 (UTC)


 * I don't know the answer, I haven't worked it out. What is x? --Tango (talk) 21:54, 6 April 2008 (UTC)


 * Based on some clever reasoning, I would guess that the answers your looking for are h=100m and r=50m for the Cylinder. This is a Related rates problem, if that's any help. A math-wiki (talk) 22:04, 6 April 2008 (UTC)
 * How does that cylinder fit in the sphere? To 99: first you used 100m, but your second posting suggests 10m. I'll just put the radius of the sphere at 1; then it can easily be scaled to anything.
 * A cylinder with maximum volume will fit snugly in the sphere and be in a symmetric position, with its axis intersecting the sphere in two diametrically opposite poles. Let h and r be the height and radius of the cylinder. The circular rims at each end of the cylinder lie entirely on the sphere, and the maximum distance within the sphere, from one rim point to the opposite point on the opposite rim, is two. This distance is also the hypotenuse of a right triangle with sides h and 2r, giving us a constraint:
 * h2 + 4r2 = 4.
 * The volume of the cylinder is πr2h. From the constraint we have r2 = 1 − $1/undefined$h2, giving us that the volume equals:
 * π(1 − $1/undefined$h2)h.
 * This reaches its maximum when h2 = $4/3$, and then r2 = $2/3$. --Lambiam 22:51, 6 April 2008 (UTC)

Oh thank you so much! I want to have your babies! 99.240.177.206 (talk) 23:47, 6 April 2008 (UTC)

Golden rectangle
if a rectangle has width to length ratio of 200mm: 323.6mm is it a golden rectangle? i need to show work--Dian2 (talk) 18:31, 6 April 2008 (UTC)
 * We will not do your homework for you. Look at Golden ratio. --Oskar 19:30, 6 April 2008 (UTC)


 * It's quite close. --Lambiam 22:21, 6 April 2008 (UTC)

Confidence interval mystery
Hi all:

While looking at the output of a multiple regression software I was mystified by how it calculated the confidence intervals for the x variables.

The output is as follows:



I have marked, in red, quantities that are necessary for calculating the confidence interval such as the mean (μ) and the number of observations (n). It is the standard deviation quantity, σ, that is eluding me right now. Using the confidence interval outputs, and the confidence interval formula:

$$\mu \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$$

I have back-traced the two sigmas for x1 and x2 to be:

$$\sigma_1 = 2.15975$$ $$\sigma_2 = -1.201709$$

However, I do not see these values anywhere in the software's output. so I am a bit mystified by how the software was able to calculate these confidence intervals?

Thanks.

L33th4x0r (talk) 19:40, 6 April 2008 (UTC)


 * It’s been awhile since I looked at regression output, and I’m really not a statistics person so I'm probably wrong: but it looks to me like it used a t test and thus got confidence intervals with the values from the t density function, not the normal distribution. That is, you’re using $$z_{\alpha / 2}$$ instead of whatever you call it if you get the values from the t distribution (or table).
 * Also, isn’t the Std. Err. The sigma? GromXXVII (talk) 19:58, 6 April 2008 (UTC)


 * If my memory from AP Statistics serves Grom is right. A math-wiki (talk) 22:06, 6 April 2008 (UTC)


 * As you appear to have calculated a negative standard deviation value (your σ2), I think you have probably made a slip somewhere. Gandalf61 (talk) 10:08, 7 April 2008 (UTC)


 * Grom is right. Coefficient estimate confidence intervals would be calculated using the t-values. --PalaceGuard008 (Talk) 10:35, 7 April 2008 (UTC)

Simplifying a radical expression: Problem type 2
Math is one of my most hated subjects I hate math more then anything in the world. I'm kinda understanding this radical expression stuff, but I am still not fully understanding it. Remember that I am horrible at math so I like simple explanations.

Notes: ^= Exponent
 * = Square Root symbol

Here is the question:
 * 98 x^11w

So 98 goes like this 49 . 2 .x^10. w. x

Then it goes

7 . 7 . 2 . x^10. w. x

Then the answer would be:

7x^5*2wx

I believe that is correct. But I am not sure. I'm still don't get it. I'm not exactly sure what I am doing here anyway. What am I doing? Can someone help me and clear this up to be more specefic with this kind of question?

Thank You

Always

Cardinal Raven

Cardinal Raven (talk) 22:58, 6 April 2008 (UTC)Cardinal Raven


 * You’re taking out squares. That is, $$49=7^2$$, so $$\sqrt{49}=7$$
 * Likewise, if you look at the Square Root article, it also says that $$\sqrt{xy}=\sqrt{x}\sqrt{y}$$, so that using $$x=49$$ and $$y=2$$, you get $$\sqrt{98}=\sqrt{49}\sqrt{2}=7\sqrt{2}$$
 * The variables work in the exact same way, so that $$\sqrt{x^2}=|x|$$, and $$x^{10}=x^2 x^2 x^2 x^2 x^2$$ so that $$\sqrt{x^{10}}=|x^5|$$. GromXXVII (talk) 00:09, 7 April 2008 (UTC)


 * But the question is, does your teacher consider $$|x^5|\sqrt{x}$$ to be simpler than $$\sqrt{x^{11}}$$ ? I certainly don't. StuRat (talk) 03:37, 7 April 2008 (UTC)
 * I thought that too. I would leave it as x11/2, personally (although that's not ideal, since it ignores that fact that we're looking for the positive square root). --Tango (talk) 14:34, 7 April 2008 (UTC)


 * What you are doing is:
 * Factor the expression:
 * 98·x11·w = 2·72·x11·w.
 * Separate out the even powers, which are squares:
 * 2·72·x11·w = 72·x10·2·x·w = (7·x5)2·2·x·w.
 * Take the square root, using √(a2·b) = |a|·√b:
 * √((7·x5)2·2·x·w) = |7·x5|·√(2·x·w) = 7·|x|5·√(2·x·w).
 * As another example, x3·y4·z5 = (x·y2·z2)2·x·z, so the square root is |x|·y2·z2·√(x·z).
 * The rule √(x·y) = √x·√y given by Grom above can only be safely used of you know that x and y are not both negative. --Lambiam 09:16, 7 April 2008 (UTC)