Wikipedia:Reference desk/Archives/Mathematics/2010 August 17

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August 17

Starting values for Root-finding algorithms (polynomials)

Are there any algorithms for finding appropriate starting values for algorithms such as Brent's method? I think what would be needed would be 1. some approximation of the highest and lowest possible roots, and 2. what resolution to search at to make sure you aren't skipping any roots. —Preceding unsigned comment added by 70.162.15.58 (talk) 00:54, 17 August 2010 (UTC)

In general, there is no easy way to do this, but in case of polynomial roots, you can always use Sturm's theorem. (Igny (talk) 03:14, 17 August 2010 (UTC))

See Durand-Kerner method. Bo Jacoby (talk) 14:59, 17 August 2010 (UTC).

integral of sinx cosx

I am exposing my ignorance of trig here... When I do u substitution with u = sinx I get (1/2)(sinx)^2 but my ti89 gives me -(1/2)(cosx)^2. What am I missing here? are these two results equivalent? -- 99.20.118.197 (talk) 04:02, 17 August 2010 (UTC)

What happens when you do the substitution u = cosx? What is sin²x + cos²x (the most important trig identity)? How does C (the constant of integration) fit into this? -- 111.84.234.215 (talk) 05:05, 17 August 2010 (UTC)

The easiest check is to differentiate both of them and with both of them you end up with $\sin {x}\cos {x}$. We can therefore safely say they are both integrals of $\sin {x}\cos {x}$. Are they equivalent? Let's plug in some numbers. ${\frac {1}{2}}\sin ^{2}{0}=0$.   $-{\frac {1}{2}}\cos ^{2}{0}=-{\frac {1}{2}}$. So clearly they are not equivalent. Look at the coefficient of integration next and see if you spot anything. Readro (talk) 08:21, 17 August 2010 (UTC)

The integral can be looked at in a number of different ways.

• Substitute u = sin x, with (du/dx) dx = cos x dx:

$\int \sin x\cos xdx=\int udu={\frac {1}{2}}u^{2}+C={\frac {1}{2}}\sin ^{2}x+C.$

• Substitute u = cos x, with (du/dx) dx = -sin x dx:

$\int \sin x\cos xdx=-\int udu=-{\frac {1}{2}}u^{2}+C=-{\frac {1}{2}}\cos ^{2}x+C=-{\frac {1}{2}}(1-\sin ^{2}x)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{2}}+C.$

• Multiply by 2, take 0.5 outside the integral, and convert to sin 2x:

$\int \sin x\cos xdx={\frac {1}{2}}\int 2\sin x\cos xdx={\frac {1}{2}}\int \sin 2xdx={\frac {1}{2}}(-{\frac {1}{2}}\cos 2x)+C={\frac {1}{4}}(2\sin ^{2}x-1)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{4}}+C.$

However you look at it, though, the antiderivative is the same up to a constant – and this difference is accounted for by the constant of integration C. —Anonymous Dissident^Talk 10:39, 17 August 2010 (UTC)

Can't somebody mention the trigonometric identity that says

$\sin x\cos x={\frac {\sin(2x)}{2}}$?

Michael Hardy (talk) 14:57, 17 August 2010 (UTC)

OK, I see someone did mention that. But why wasn't it the first thing mentioned here? To anyone who knows trigonometry, that would be the first thing that comes to mind. Michael Hardy (talk) 14:59, 17 August 2010 (UTC)

Probably because this isn't what the OP asked. He asked how to reconcile his (correct) computation for the integral with the output of the calculator. -- Meni Rosenfeld (talk) 17:50, 17 August 2010 (UTC)

Thanks alot everyone. I worked out your examples on paper and read the article on C. Before I posted my question I did get the feeling that I could do some pythagorean substitution to change sine to cosine ... but the piece of the puzzle I was missing was an understanding of +C. Thanks again -- 99.20.118.197 (talk) 15:31, 17 August 2010 (UTC)

Simple math problem

Hi. The rate of my dorm for 38 weeks is 1900, so it gives 50 every week (it's even written on the page that it is 50). I have to pay each month 237,5. On the average, a month has 30,42 days, which gives a month 4,35 weeks. 4,35 weeks times 50 is 217,5. Where are those 20?! (I didn't give the currency, but I think it's ok). 83.31.113.33 (talk) 14:05, 17 August 2010 (UTC)

It looks like you are right. Either you're paying for some extra service, there's an additional tax, there's a mistake in their billing or published numbers, or they're just ripping you off. -- Meni Rosenfeld (talk) 14:31, 17 August 2010 (UTC)

237.5 is exactly one eighth of 1900. 38 weeks is just under 9 months. So maybe it is simpler for your bank and whoever is letting the dorm to process 8 equal monthly payments than it would be to process 8 equal payments and then an odd amount for the "short" month at the end. Sensible thing is to ask whoever is letting the dorm to clarify how many monthly payments you will be making. Gandalf61 (talk) 14:40, 17 August 2010 (UTC)

Hm, I think you are right. If anyone wants to check their information, it's here: [1] (Lyon St, twin). 83.31.113.33 (talk) 14:42, 17 August 2010 (UTC)

If you read the rates details [2] it says you get the 38 weeks for 8 monthly payemnts of £237.50. It says that the weekly amount is for info only. It all matches up because you aren't paying for 8.7 months. -- SGBailey (talk) 16:29, 18 August 2010 (UTC)