Wikipedia:Reference desk/Archives/Mathematics/2008 January 21

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January 21

Differential Equation

This is not a homework problem - I thought of it while I was playing with my cat. If you dangle a string and spin it the right way, it forms a sort of standing wave. There are points along it, usually at most 3, that are almost stationary, and the rest looks like a squished sine wave spinning around its x-axis. In trying to figure out its shape, I got stuck with an equation I couldn't solve. To start, I assume no air resistance, that the string is the same mass all the way along and completely flexible, and that it has attained its ideal shape - a stationary planar wave spinning around a vertical axis. Call the axis of rotation the x-axis, and a perpendicular line the y-axis, with the y-axis spinning with the string. Since I wasn't sure how to solve it directly, I made the string massless with point masses along it at equal intervals, and drew a free body diagram for each mass. The first mass, at the bottom end of the string, has only two forces on it: mg downwards, and F₁ along the string. Every other mass, say the nth along from the bottom, has three forces: mg downwards, and F_n-1 and F_n along the string. Equating X-components, F_n,x = nmg, because no mass is moving up or down. The Y-components, then, must add up to the actual acceleration (a_n) of the point. Since the point moves in a circle at constant angular velocity, acceleration is towards the x-axis, and since all points move with the same angular velocity, their accelerations are proportional. That is, there's some constant a such that a_n = ay_n for all n, where y_n is the y-coordinate of the point mass. Equating Y-components, then, $F_{n,y}=ma\sum _{k=1}^{n}y_{k}$. In other words, each bit of the string must support the weight of everything below it, and keep everything below it moving. Since the force on each mass must be in the direction of the next mass, that is along the string, F_n,y/F_n,x = (y_n+1-y_n)/(x_n+1-x_n). Since the masses are at even intervals along the string, there's some constant d such that d² = (y_n+1-y_n)²+(x_n+1-x_n)² for all n. Substituting the first two equations into the third, I get ${\frac {a\sum _{k=1}^{n}y_{k}}{ng}}={\frac {y_{n+1}-y_{n}}{x_{n+1}-x_{n}}}$ for all n. Once I had the shape of the problem, I shifted it to differentials. n becomes a continuous independant variable measuring distance along the string, x and y the dependant variables. If f(n)=y, ${\frac {a\int _{0}^{n}f(s)\,ds}{g}}={\frac {y'}{x'}}$ and 1 = (y')²+(x')². I differentiate the first, so (a/g)y = (x'y"-y'x")/(x')². Plugging in the other equation and its derivative to get rid of x, (a/g)y²(1-(y')²)³ = (y")². Choosing b=a/g, I'm left with by²(1-(y')²)³ = (y")² Unfortunately, I'm not sure how to solve this for y. Any ideas? Black Carrot (talk) 00:09, 21 January 2008 (UTC)

Decide whether y' shall mean dy/dx or dy/ds. The differential equation must be supplemented by boundary conditions: the length of the chain (S=integral ds), the fixed end of the chain (x(0)=y(0)=0), and the loose end of the chain (x"(S)=y"(S)=0). Note the trivial (unstable) solution that the chain is straight and vertical, (x(s)=s, y(s)=0). The other solutions are characterized by the number of crossings of the x-axis. It's a nice problem. You may substitute some power series of s (or is it x?) for y in the differential equation and thus transform it into a difference equation in the coefficients. Have fun! Bo Jacoby (talk) 02:19, 21 January 2008 (UTC).

y' is dy/dn, and x' is dx/dn. s is a dummy variable in the integration, like k was in the summation. I've started pushing a power series through it, but it's a bit messy with a degree-six polynomial. Black Carrot (talk) 02:49, 21 January 2008 (UTC)

I've hit a bit of a difficulty with the power series. When I expand it around some points, it doesn't form a wave - all the odd-degree terms are zero. Specifically, if the x term is zero, corresponding to one of the peaks of the wave, all the odd-degree terms are as well. This means that if y is analytic anywhere, it isn't for very far. Is there a way to pull some bounds on that out of the equation? Black Carrot (talk)05:22, 21 January 2008 (UTC)

Odd-degree terms being zero does not prevent a power series from being a wave - cosine is a wave with odd-degree terms zero. But cos(x) written parametric as (x(s),cos(x(s))) where (dx/ds)^2+(dy/ds)^2=1 is not handy. You might prefer considering the differential equation for the function y(x) rather than for (x(s),y(s)). And you might want to study the catenary article for a similar, but somewhat simpler, problem. Observe that your problem is related to the quantum mechanics of hydrogen-like atoms where boundary conditions also give raise to discrete quantum numbers labelling the solutions of the differential equation, but quantum mechanics is easier because the differental operator of the Schrödinger equation is linear. Bo Jacoby (talk) 08:14, 21 January 2008 (UTC).

You're right, it doesn't. That was silly. I've made more progress with differentials, though. I can reduce it to a first-order equation: 1=(-(b^1/2/2)y²+c)(1-(y')²)^1/2, where c is an arbitrary constant. This is seperable, giving the hideous (but hopefully accurate) integral $\int _{}^{}{\sqrt {\frac {hy^{2}+c}{hy^{2}+c-1}}}\,dy=\int _{}^{}\,dn$, where h=-b^1/2/2 and c is an arbitrary constant. Another sticking point. Black Carrot (talk) 19:26, 21 January 2008 (UTC)

Exponential naming

"Square" and "cube" are apparently used to describe the exponents 2 & 3 because of their relationship to geometrical forms. Are there any forms that could lend their names to higher powers? Retarius | Talk 02:47, 21 January 2008 (UTC)

I read a book a bit ago that called four-dimensional figures "planoplanes". I think it was the Arithmetica Infinitorum. Black Carrot (talk) 02:53, 21 January 2008 (UTC)

Actually, in that direction, many n-dimensional objects are named after 2 or 3-dimensional objects they resemble, with some modifier (hyperplane, hypersphere, hypercube, n-ball, etc.) Black Carrot (talk) 02:55, 21 January 2008 (UTC)

Apparently quartic functions can be called "biquadratic", so maybe the fourth power could be a bisquare. Black Carrot (talk) 03:03, 21 January 2008 (UTC)

I was attempting to explain powers of numbers to someone I was tutoring when the question arose: "If three is a cube, what does four make?" (Golden Rule: To discover how little you know about something try explaining it to someone who knows nothing about it.) I tried to imagine a four-dimensional form to use as a name-source but, of course, that fourth dimension is Time in conventional discourse. Retarius | Talk 03:12, 21 January 2008 (UTC)

That would be a better opportunity to teach the theory of language than math. It shows up in all sequences that people started naming before they realized they couldn't stop. First, second, third, then nth. Eleven, twelve, then n-teen. Linear, quadratic, cubic, quartic, quintic, n-degree. Ten, hundred, thousand, million, n-illion. Black Carrot (talk) 03:27, 21 January 2008 (UTC)

Now I've had a chance to look at some of the concepts you've referred to I think a valid terminology could be derived from the table shown in the article on the hypercube. Although "Two hexeracted" might lose in a cost/benefit analysis against "Two to the sixth". Many thanks for your guidance, Black Carrot. Retarius | Talk 04:00, 21 January 2008 (UTC)

That's a good word. I wish I had a copy of that book still, I think it had some other odd words for powers. That was back when algebra was some newfangled toy, and I think he made up a lot of stuff. For instance, according to his article he was the first person to use the infinity symbol in print. Black Carrot (talk) 05:26, 21 January 2008 (UTC)

I usually refer to fourth powers as Quarts, (for Quartic) and Quints for 5ths and so forth. This is mainly due to my having studied Polynomials a fair amount on my own. A math-wiki (talk) 08:04, 21 January 2008 (UTC)

See zenzizenzic and zenzizenzizenzic. Gandalf61 (talk) 14:52, 21 January 2008 (UTC)

Apparently a question of long standing...Zenzizenzic? Now that's a word for quiz nights! Retarius | Talk 01:44, 22 January 2008 (UTC)

free shell account for math?

I have only a very slow computer with very slow dialup access, any chance I could do mathematical computation (matlab, mathematica, etc) on a normal free shell account somewhere? I don't know where to begin.

I may start a math degree this fall. —Preceding unsigned comment added by 212.51.122.8 (talk) 03:19, 21 January 2008 (UTC)

What OS do you have? Why not download mathematica and run it? --wj32 ^t/c 03:53, 21 January 2008 (UTC)

Well, Mathematica isn't free. If you're going to download something, you might try SAGE. —Bkell (talk) 05:44, 21 January 2008 (UTC)

Oops, I was thinking of Maxima, not Mathematica. Stupid names. --wj32 ^t/c 06:18, 21 January 2008 (UTC)

You definitely don't need Matlab for a math degree. Mathematica can be useful but is far from being necessary (and can even be harmful if you over-rely on it). -- Meni Rosenfeld (talk) 09:28, 21 January 2008 (UTC)

If you start a degree this fall, try asking at the collage about any shell account they provide. – b_jonas 10:01, 22 January 2008 (UTC)

Numerical methods

The measured heights are 2950cm and 35 cm, while the true values are 2945cm and 30 cm respectively. compare the absolute and relative errors —Preceding unsigned comment added by 210.212.161.105 (talk) 06:43, 21 January 2008 (UTC)

Think about this, those two numbers are vastly different in size, so for example being 5cm off on the larger measurement is really not a big deal for practical purposes, but on the smaller one it could be a real issue, this is where absolute error and relative error differ. One of these would directly compair the difference between (expected and observed) or (true and measured) in our case the other would account for the fact that the two true values are different in size, and obtain a percentage for the error. Can you guess which one is which?? A math-wiki (talk) 08:02, 21 January 2008 (UTC)

Okay, I've compared them. -- Meni Rosenfeld (talk) 09:25, 21 January 2008 (UTC)

Also see Approximation error. GromXXVII (talk) 12:55, 21 January 2008 (UTC)

Limit question

I'm trying to find the limit as x approaches 0 of (1 - cos 4x)/x. I rewrote it as (cos 4(0) - cos 4x)/(0 - x) and took the derivative of the subtrahend in the numerator to get -4 sin 4x. Substituting 0 in for x, that gives me 0, which was indeed given as the right answer. But my question is, was I justified in rewriting
lim as x approaches 0 of (1 - cos 4x)/x

as
lim as x approaches 0 of (cos 4(0) - cos 4x)/(0 - x)

given that that seems to change the sign of the denominator? Is it acceptable in this case because x is approaching 0 and the sign before 0 doesn't matter? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 (talk) 19:30, 21 January 2008 (UTC)

No, that rewriting is not justified. However, you could have written lim(1 - cos 4x)/x=lim(-(cos 4(0) - cos 4x)/(0 - x))=-lim(cos 4(0) - cos 4x)/(0 - x)=-0=0. Algebraist 20:16, 21 January 2008 (UTC)

Or use l'Hôpital's Rule, which would give you $\lim _{x\to 0}{\dfrac {4\sin {4x}}{1}}={\dfrac {0}{1}}=0$. --Kinu ^t/_c 20:28, 21 January 2008 (UTC)

Gotcha. Thanks! —anon —Preceding unsigned comment added by 70.19.20.251 (talk) 20:38, 21 January 2008 (UTC)

Evaluating a definite integral

I'm trying to evaluate ∫₀²(2x³ + 3)dx. Here's what I have so far:

∆x = 1/n

c = a + i∆x = 0 + i(1/n) = i/n

lim [as n --> infinity] of ∑ [from 1 to n] of (2(i³/n³) + 3)(1/n)

lim [as n --> infinity] of ∑ [from 1 to n] of (2(i³/n⁴) + 3/n)

lim [as n --> infinity] of ((2/n⁴)∑ [from 1 to n] of i³ + (1/n)∑ [from 1 to n] 3)

lim [as n --> infinity] of ((2/n⁴)((n²)(n + 1)²/4) + (1/n)3n)

lim [as n --> infinity] of (1/2 + 3)

1/2 + 3

7/2

But that's not one of the choices given; it says here that the answer is 14. Why is this? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 (talk) 20:47, 21 January 2008 (UTC)

Because you have computed (correctly) $\int _{0}^{1}2x^{3}+3dx$. You'll have to change things a bit to take account of the correct limits of integration. Algebraist 21:01, 21 January 2008 (UTC)

How do I do that? Should my ∆x be (b - a)/n? —Preceding unsigned comment added by 70.19.20.251 (talk) 21:13, 21 January 2008 (UTC)

Yes. Note that this correct ∆x is used both in the formula for c, and you multiply every term by it. I'll get you started:

$\Delta x={\frac {b-a}{n}}={\frac {2}{n}}$

$c_{i}=a+i\Delta x={\frac {2i}{n}}$

$\int _{0}^{2}(2x^{3}+3)dx=\lim _{n\to \infty }\sum _{i=1}^{n}(2c_{i}^{3}+3)\Delta x=\cdots$

-- Meni Rosenfeld (talk) 21:36, 21 January 2008 (UTC)

Thanks so much! I got it right now. —Preceding unsigned comment added by 70.19.20.251 (talk) 22:00, 21 January 2008 (UTC)

Intersection of two cones

There are two cones with intersecting axes. They don't necessarily have the same slope. That is, one may be pointier than the other (I don't know the way you're supposed to say it). I assume the intersection between them is a conic section. How do I find the plane the intersection is on? Specifically, I need a vector at right angles to it. — Daniel 21:49, 21 January 2008 (UTC)

I forgot to add: the cones are the same width where their axes intersect.

After a little experimentation, it looks like if the cones have the same slope, the normal of the plane they intersect on is the cross product of their axes. The plane doesn't go through the intersecting axes as I thought it would, so I need to find a point on the plane. If the cones don't have the same slope, they appear to intersect on a hyperbola stretched infinitely in the third dimension. For what I'm doing, it would probably be best just to approximate it as a plane by averaging the slopes and taking the first case. Can anyone tell me how to find the point I need for the first case? — Daniel 00:39, 22 January 2008 (UTC)

I don't think I understand your question. Could you be more specific, or sketch a picture or include a formula or something? Do you, for instance, include "oblique" cones? Black Carrot (talk) 01:21, 22 January 2008 (UTC)

I include only the curved surface of the cone, as described by y*y=x*x+z*z, but rotated and stretched. I've managed to get the problem down to rotating a line by a certain amount around the origin in each direction and finding the y value where the two lines produced intersect. I got y=1/(a(cos(ɵ)/a²+cos(ɵ)/b²)), but that doesn't seem to work. If you want to double-check my work, just ask and I'll put it on here. — Daniel 05:34, 22 January 2008 (UTC)

Yeah, let's see it. Do I understand correctly that the tips of the cones coincide? Black Carrot (talk) 09:05, 22 January 2008 (UTC)

Never mind. I solved it myself. I'm apparently better at solving the problem then explaining it. — Daniel 18:27, 22 January 2008 (UTC)

Have I found the absolute minimum correctly?

I need to find the absolute minimum value of f(x) = 2x³ - 3x² - 12x, which is defined on the interval [-3, 2]. Relative extrema are marked by sign changes in the derivative, which is f'(x) = 6x² - 6x - 12 = 6(x² + x + 2) = 6(x - 2)(x + 1). So the possibilities are -3, -1, and 2. f(-3) = -45, f(-1) = 7, and f(2) = -20, so the answer is -45, isn't it? —Preceding unsigned comment added by 70.19.20.251 (talk) 22:07, 21 January 2008 (UTC)

You're fine, apart from a couple of typos (x² + x + 2 should be x² - x - 2). Algebraist 22:14, 21 January 2008 (UTC).

Whoops. Thanks!

Least-powers question

I'm looking for the critical numbers of f(x)= x√(16 - x²). I took the derivative: x•(1/2)(16 - x²)^1/2 + (16 - x²)^1/2. I factored out 16 - x², giving me (16 - x²)[(x/2) + 16 - x²]. So my critical numbers are ±4 and ... what? I tried running the contents of the other parenthesis through the quadratic formula, but I got the gobbledygook 1/4 ± √(257)/8 ... —Preceding unsigned comment added by 70.19.20.251 (talk) 22:33, 21 January 2008 (UTC)

You have an error in your calculation of the derivative. -- Meni Rosenfeld (talk) 22:38, 21 January 2008 (UTC)

You neglected the chain rule and you left 1/2 as the exponent where you needed (1/2) − 1. Michael Hardy (talk) 23:23, 21 January 2008 (UTC)

I see, I had typed it up wrong. Corrected: I'm looking for the critical numbers of f(x)= x√(16 - x²). I took the derivative: x•(1/2)(16 - x²)^-1/2 + (16 - x²)^1/2. I factored out (16 - x²)^-1/2, giving me (16 - x²)^-1/2[(x/2) + 16 - x²]. But my question still remains ... —Preceding unsigned comment added by 70.19.20.251 (talk) 23:46, 21 January 2008 (UTC)

It's still not correct. You still haven't applied the chain rule, which means that when differentiating ${\sqrt {16-x^{2}}}$ you need to multiply by the derivative of $16-x^{2}$. If this is unclear, take a look at the article. -- Meni Rosenfeld (talk) 23:51, 21 January 2008 (UTC)

Got it. Thanks so much! —Preceding unsigned comment added by 70.19.20.251 (talk) 00:09, 22 January 2008 (UTC)

Just another math problem

What is 16,777,216 to the 2,079,600th power? Tamashiihiroka (talk) 23:33, 21 January 2008 (UTC)

I don't know about calling that a math problem, but this is roughly $3.130311574455724557\cdot 10^{15024527}$, give or take a gazillion or two. -- Meni Rosenfeld (talk) 23:39, 21 January 2008 (UTC)

Hmm, interesting talk is it not? hydnjo talk 02:16, 22 January 2008 (UTC)

Warning. Visiting that talk page, which is 1,660,001 bytes, crashed my browser.  --Lambiam 04:15, 22 January 2008 (UTC)

I just get a completely blank page. Apparently the server refuses to serve any of it to me. —Keenan Pepper 04:19, 22 January 2008 (UTC)

I just reverted silly edits — 20493 empty sections ('==PLEASE SPAM ME !!!==' or sth), which apparently cause DoS on TOC–formatting machine. --CiaPan (talk) 12:49, 22 January 2008 (UTC)

Errr, sorry about that. I was a little mad at myself at the time, and my computer crashed whenever I tried to revert the edits. Anyways, I'm over it now... Wait, IT WAS OVER A GIGABYTE AND A HALF???!!!!????!!!! Holy crap! I guess I win longest page on Wikipedia. lol. —Preceding unsigned comment added by Tamashiihiroka (talk • contribs) 13:00, 22 January 2008 (UTC)

No way. That was only as little as 1.6 megabyte.... --CiaPan (talk) 13:04, 22 January 2008 (UTC)

Oh, right. sorry. Just me being stupid. lol. duh. But if it WAS over a Gigabyte, that would be just plain weird. lol. Tamashiihiroka (talk) 14:17, 22 January 2008 (UTC)

...and according to Special:Longpages, the longest page in en-wiki is currently ‎Line of succession to the British Throne, 'only' 322,647 bytes. --CiaPan (talk) 12:32, 24 January 2008 (UTC)

It can also be expressed exactly as $2^{49910400}$, if that's any help. —Bkell (talk) 04:36, 22 January 2008 (UTC)

Are you by chance trying to compute the number of possible images that can be displayed on a 1080p screen? Such a screen is 1920×1080 pixels, which gives 2073600 pixels in all (not 2079600), and each pixel can display any of 16777216 colors, so the number of possible images that can be displayed is $16777216^{2073600}$, which is almost what you asked for. —Bkell (talk) 04:42, 22 January 2008 (UTC)

Err, yeah. I am. Would anyone be willing to calculate this new number? (I want to make a Youtube video of it using the voice of Cats in the AYBABTU video.) Tamashiihiroka (talk) 12:55, 22 January 2008 (UTC)

Do you realise that this number has about 15 million digits ? Reading at a rate of one digit per second, and without any breaks, it would take you about 6 months to read out the whole number. I guess you could read it out in hexadecimal - that would only take about 5 months, but reading "one" followed by about twelve and a half million zeros would be really boring. By the way, its last (least significant) decimal digit is a 6. Gandalf61 (talk) 13:33, 22 January 2008 (UTC)

.... Holy... could I have an approximate answer in scientific notation? Tamashiihiroka (talk) 14:17, 22 January 2008 (UTC)

Learn to use a decimal logarithm — then you'll get
$16777216^{2073600}=10^{\log(16777216^{2073600})}$ $=10^{2073600\times \log 16777216}\approx 10^{2073600\times 7.22472}$ $\approx 10^{14981179}\approx 10^{15\ million}$
CiaPan (talk) 14:28, 22 January 2008 (UTC)

Thanks. I'm still in middle school so I don't know this stuff yet.  :(. Tamashiihiroka (talk) 14:33, 22 January 2008 (UTC)

Scientific notation, using a base of 10, is useful because of our cultural background, but mathematically it's not much different from any other base. Since $16777216=2^{24}$, we can also use the rules of exponents to say

$16777216^{2073600}=(2^{24})^{2073600}=2^{24\times 2073600}=2^{49766400}$.

This is an exact expression for the answer, as opposed to the approximation you necessarily get if you use base 10. So if you wrote this number in binary, it would be a 1 followed by 49766400 (almost 50 million) zeros. Gandalf61 mentioned using hexadecimal, which is base 16; using the fact that $16777216=2^{24}=2^{4\times 6}=(2^{4})^{6}=16^{6}$, we can write $16777216^{2073600}$ as a power of 16, too. —Bkell (talk) 15:08, 22 January 2008 (UTC)