Wikipedia:Reference desk/Archives/Mathematics/2010 August 26

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August 26

Volume of revolution

How do you find the volume of revolution generated by the area enclosed between $y=x^{n}$, $x=0$, $y=a^{n}$, $y=b^{n}$ rotated around the x axis?--115.178.29.142 (talk) 03:24, 26 August 2010 (UTC)

It would be easier to help if you show us how much you've already done. The obvious first step here is to draw a sketch, however crude, of the curves and find out what species the region in question belongs to. Have you done that? 85.226.206.72 (talk) 06:18, 26 August 2010 (UTC)

... and tell us if you don't understand the methods given in the article: Solid of revolution. Dbfirs 06:50, 26 August 2010 (UTC)

Inclined sit-up

OK, tonight I did 3 sets of 5 inclined sit-ups. The incline was at approximately 25-30 degrees. I placed a 25 lb weigh on my chest, and sat straight up. I also tried to maintain myself parallel to the ground for a few seconds, which was surprisingly difficult.

Can you help me come up with a formula, based on my body weight of 185 lbs, for how many calories I actually burned tonight. I do not know the breakdown of how weigh is distributed in the body (sorry). I would like to have an accurate representation, so if we have to add just a little bit for calories wasted, that's fine with me (I imagine most exercises have this type of fudge factor, though I could be wrong). Magog the Ogre (talk) 04:14, 26 August 2010 (UTC)

The amount of calories burned is pretty small, but you are building muscle tone. If you are trying to lose weight through mathematics, you might like the downloadable book The Hacker's Diet. 67.117.146.38 (talk) 08:41, 26 August 2010 (UTC)

I did a very rough calculation and it turned out about 5 kcal (it could easily be 10 or something).

Assuming that by "formula" you meant just that rather than a number, then I'd say the energy burned per sit-up is

${\frac {mgh(\sin(\alpha _{2})-\sin(\alpha _{1}))}{\eta }}$

Where

• m is the moving mass (not including, say, your legs)
• g is the acceleration due to gravity
• h is the average (weighted by mass) distance of the moving mass from the fulcrum
• $\alpha _{1},\ \alpha _{2}$ are the angles at the start and finish (negative angle for the mass being below the fulcrum)
• $\eta$ is the efficiency of the muscles involved.

If you're using metric units you'll get a result in Joules; you'll need to convert them using $1kcal=4184J$.

-- Meni Rosenfeld (talk) 15:10, 27 August 2010 (UTC)

Optimal decision making under uncertainty.

100 cups are placed on a desk. Underneath one of them is $100, and the rest contain nothing. At any time, you may pick a cup and claim the contents of it. You may only do this once. You may also pick a cup and identify the contents of it, but doing so costs $1. You may do this as many times as you like. What is the optimal strategy to maximise the expected profit?--Alphador (talk) 09:04, 26 August 2010 (UTC)

Keep spending $1 until you find the $100. On average you'll have looked at ~50 cups before you find it and expect to earn ~$50. Dragons flight (talk) 09:26, 26 August 2010 (UTC)

With the obvious, very minor caveat that if you investigate 99 and don't find it, take the 100th without investigating. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:55, 26 August 2010 (UTC)

This would be much harder if, as you wait idly trying to make up your mind, other players are allowed to participate in the same game with the same set of cups. The mathematics for deciding when to act are described in Nash equilibrium (to maximize your own profit), and the mathematics to maximize the total profit for all participants (i.e., somebody gets the $100, and the least amount of money is spent) is described in Pareto optimality. Nimur (talk) 21:41, 26 August 2010 (UTC)

It's customary to say "thank you" when people answer one's question, along with any followup questions. I don't know about others, but for me, knowing that my answer was helpful gives me closure which allows me to stop thinking about the question. -- Meni Rosenfeld (talk) 14:50, 27 August 2010 (UTC)

Assigning probabilities to curve fittings

Let's suppose you're given the following data points: (0,1) (1,2) (2,4) (3,8) (4,16) (5,32) (6,64) (7,128) (8,256) (9,512)
There are infinitely many curves that will fit these data points, however if you saw this as a natural phenomenon (for instance, the population of bacteria with respect to time), intuition would tell you that the most likely equation that would fit these is $y=2^{x}$. Is there any mathematical justification for believing that $y=2^{x}$ is the equation which most probably fits these data points out of the infinitely many that could potentially fit them?--Alphador (talk) 11:10, 25 August 2010 (UTC)

The general principle for this is Occam's razor. One concretization for it is that your prior probability for a sequence (or anything else) should be smaller the higher its Kolmogorov complexity. Because the sequence $y=2^{x}$ has such a short description, it will have a relatively high prior probability, and its posterior probability will be overwhelming after observing these data points. -- Meni Rosenfeld (talk) 11:01, 26 August 2010 (UTC)

The exponential function assumes fractional values while the population at any time is an integer. So the population is not an exponential function.

The exponential function grows without limit while the population is limited. So the population is not an exponential function.

So the best you can get is an approximate fit in a finite interval of time.

Consider the function g(t)=f(t)−2^t. Then g(t)=0 for t=0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Consider only a finite number, N, of points of time in the finite interval, because you cannot measure with unlimited frequency. So this is like a lottery with N tickets, and K of these tickets are marked 'zero', and N−K of these tickets are marked 'nonzero', and you have opened n=10 tickets, out of which k=10 are all marked 'zero' and n−k=0 are marked 'nonzero'. The likelihood function L(K)=$\scriptstyle {\binom {K}{k}}{\binom {N-K}{n-k}}$=$\scriptstyle {\binom {K}{10}}$ for 0≤K≤N. The maximum likelihood estimate is K=N. So YES, there is a mathematical justification for believing that y = 2^x is the equation which most probably fits these data points out of the infinitely many that could potentially fit them. Bo Jacoby (talk) 18:30, 25 August 2010 (UTC). The question and the answer has been removed, and the question reinstalled without the answer. Don't know why. Bo Jacoby (talk) 11:07, 26 August 2010 (UTC).

Sorry, the question was removed several times, I restored the question as it appeared the first time while I should have looked for the last time. -- Meni Rosenfeld (talk) 11:22, 26 August 2010 (UTC)

how to integrate 1/(1+cosx)?

tried making denominator (1^2+(sqrt(cosx)^2)) but stuck after that. the answer supposedly has a tan(0.5x). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:34, 26 August 2010 (UTC)

Divide and multiple by 1-cos x -Shahab (talk) 19:11, 26 August 2010 (UTC)

You're right that the answer is tan(½x). Why not differentiate tan(½x) and then try to retrace the steps to prove the integration? For example:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {\text{d}}{{\text{d}}x}}\left({\frac {\sin \left({\frac {x}{2}}\right)}{\cos \left({\frac {x}{2}}\right)}}\right)={\frac {{\frac {1}{2}}\cos ^{2}\!\left({\frac {x}{2}}\right)+{\frac {1}{2}}\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}={\frac {1}{2}}\left({\frac {\cos ^{2}\!\left({\frac {x}{2}}\right)+\sin ^{2}\!\left({\frac {x}{2}}\right)}{\cos ^{2}\!\left({\frac {x}{2}}\right)}}\right)={\frac {1}{2\cos ^{2}\!\left({\frac {x}{2}}\right)}}.$

If you use the cosine angle formulas, which say that cos(α ± β) = cos(α)cos(β) ∓ sin(α)sin(β), then you will be able to show that 2⋅cos²(½x) = 1 + cos(x). Notice that cos(x) = cos(½x + ½x) = cos²(½x) − sin²(½x), and so 1 + cos(x) = 1 + cos(½x + ½x) = 1 + cos²(½x) − sin²(½x). Since cos²(½x) + sin²(½x) = 1 it follows that 1 − sin²(½x) = cos²(½x), and so 1 + cos(x) = 2⋅cos²(½x) as required. Thus:

${\frac {\text{d}}{{\text{d}}x}}\tan \!\left({\frac {x}{2}}\right)={\frac {1}{1+\cos x}}\ .$

If you run these trigonometric identities backwards then you'll have the proof. Don't forget the constant of integration! — Fly by Night (talk) 19:38, 26 August 2010 (UTC)

There's some material on this in the article titled tangent half-angle formula. Michael Hardy (talk) 20:43, 26 August 2010 (UTC)

differential equation

80(dx/dt) = 1 + k/(x+1) where k is a constant.

i can see that if k/(x+1) is combined with the dx/dt term, can integrate by ln. also tried multiplying whole equation by x+1 but then the 1 will become an x+1 and I can't transfer the x over. also tried making RHS (x+1+k)/(x+1) but after transferring reciprocal over to LHS, cant think of a formula for (x+1)/(x+1+k). —Preceding unsigned comment added by 218.186.8.247 (talk) 18:40, 26 August 2010 (UTC)

I'm going out on a limb, but you might have to solve this approximately or numerically - I'm pretty sure it's nonlinear. So: taylor series! k/(x+1) is exactly k(x+1)^-1, which can be expanded and approximated as k*(1-x). This linearizes the equation for small x. For large x, dx/dt approximately equals 1/80. So, you have a suitable first-order approximation. The function diverges, so if this were a real-world problem, it would be of little value to solve it exactly or approximate it very accurately anyway. Maybe a pure mathematician can come through and clean up my messy engineer's work. Nimur (talk) 21:34, 26 August 2010 (UTC)

Separation of variables is the way to go, try writing (x+1)/(x+1+k)=1-k/(x+1+k).--RDBury (talk) 23:25, 26 August 2010 (UTC)

Not nonlinear! Just ugly. Nimur (talk) 23:51, 26 August 2010 (UTC)

Separate variables:

$80{dx \over dt}={x+1+k \over x+1}$

and then

${80(x+1)\,dx \over x+1+k}=dt.$

Then integrate both sides. Michael Hardy (talk) 00:58, 27 August 2010 (UTC)

Substitute t=80s, dt=80ds, and get rid of the common factor 80:

${x+1 \over x+1+k}dx-ds=0.$

Substitute x=y-1-k, dx=dy:

$dy-k{dy \over y}-ds=0.$

Integrate:

$y-{k\log y}-s\,$ is constant.

Substitute back, y=x+1+k, s=t/80:

$x+1+k-k\log(x+1+k)-t/80\,$ is constant.

Discard the constants:

$x-k\log(x+1+k)-t/80\,$ is constant.

Exponentiate:

$(x+1+k)^{-k}e^{x-t/80}\,$ is constant.

Bo Jacoby (talk) 20:34, 27 August 2010 (UTC).