Wikipedia:Reference desk/Archives/Mathematics/2007 March 5

= March 5 =

Permutations problem
I need this for finding out how many simulations I need for my research. Suppose I have m numbers 1, 2, 3,..,m. I have to make a non-increasing sequence of size N out of these. I can use any number as many times as I want. How many such permutations are possible? (e.g., if m = 3, N = 2, we have these 6 possibilities: [{3,3}, {3,2}, {3,1}, {2,2}, {2,1}, {1,1}]). m could be less than, equal to or greater than N.

I could find this recurrence relation:



\pi (N,m) = \sum_{i=1}^m \pi (N-1,i) $$

Along with the boundary conditions π(N,1) = 1 and π(1,m) = m. π here is the required function that gives out the number of permutations.

First, is this correct, and second, how can I get a closed form solution for this? Thanks. deeptrivia (talk) 01:05, 5 March 2007 (UTC)


 * I think this is just Combinatorics, $${{m + N - 1} \choose N}$$; also see Multiset --Spoon! 02:25, 5 March 2007 (UTC)


 * But there is an additional restriction that going from left to right, the numbers cannot increase (non-increasing sequence).


 * In a combination, order doesn't matter. That's equivalent to what you are doing, imposing an arbitrary order (non-increasing). In other words, there is a simple bijection between multisets and non-increasing sequences. --Spoon! 02:25, 5 March 2007 (UTC)


 * Oh yeah. You're right! Thanks a ton! deeptrivia (talk) 02:47, 5 March 2007 (UTC)

I need (...want) a diagram drawn.
Yeah, I told myself I would quit with the frivolous Reference Desk question titles, but for the curious, it would have been 'A series of tube'. Anyway, I'm trying to visualise a tube that has one end inserted into the other, but I'm having difficulty. So I tried to draw myself a 2D diagram to help me, but it frankly sucks, so I need someone to draw it more better. The blue lines represents each end of the tube and the orange line represents... nothing, really. Hopefully you understand what I'm getting at, because I don't know how I would explain, as there is no 2D equivalent of the word 'tube'. They're not parallel lines because they are forced to get closer and closer. Hey, that's another thing: since mathematically lines are not meant to have thickness, does that mean my lines could actually be parallel? Vitriol 04:04, 5 March 2007 (UTC)


 * What has thickness got to do with lines being parrallel? The line still has its direction and the directions must be the same to be parrallel. Also what do you want the picture for? Why not just go get a peice of paper and make a tube that gets bigger and bigger, then you can put it inside it self.--210.49.152.114 06:26, 5 March 2007 (UTC)
 * Because I can't look inside the tube, and I don't think paper will bend that way. Also, I thought that lines could be parallel even if they weren't straight. Guess I was wrong? Vitriol 14:13, 5 March 2007 (UTC)


 * Rather than parallel lines your probably thinking of Parallel curves. Your question brings to mind Ouroboros the worm which ate itself. The various spirals might be of help. --Salix alba (talk) 17:49, 5 March 2007 (UTC)
 * [[Image:TORUSA-1 Torus mit variablem Ringdurchmesser.PNG|thumb|Yarr!]] This dealy seems to be what I was needing. Instead of driving me dotty it drove me to dots, apparently. Vitriol 19:52, 5 March 2007 (UTC) 
 * You may like to play a bit with the following Tex file. In fact, your figure is less symmetric than it seems at first glance and you have to make some aesthetical decision:

\documentclass{article} \pagestyle{empty} \usepackage{pst-eps} \usepackage{pst-plot} \pagestyle{empty} \begin{document} \begin{TeXtoEPS} \begin{pspicture}(-2.333,-2.333)(2.333,2.333) %900=360+360+180,990=900+90,1260=990+270,1350=1260+90  %2340=1260+360+360+360   \parametricplot[plotpoints=1000]{0}{900}{t dup sin 1000 div mul t    dup cos 1000 div mul} \parametricplot[linecolor=red,plotpoints=1000]{990}{1260}{t dup sin 1000 div mul t   dup cos 1000 div mul} \parametricplot[plotpoints=1000]{1350}{2340}{t dup cos 1000 div mul t   dup sin 1000 div mul} \psline[linecolor=green](0,0)(-2.333,0) \psline[linecolor=green](0,-.9)(0,-1.35) \end{pspicture} \end{TeXtoEPS}% \end{document} 88.73.127.20 21:42, 7 March 2007 (UTC)

Compound Interest, 2 loans, 2 rates, how to split payments?
I am trying to find a mathematical solutions to problems like this: A house loan is $100,000 at 6% PA, and a car loan is $20,000 at 10% PA, acculating monthly. And a person has say $1,000 per month to pay both. The minimum repayments are $6,000 per year on house and $2,000 per year on car. Where should the extra $4,000 per year be paid, on the house or the car, or a certain ratio house:car etc., such that total interest paid over both loans is minimised. Obviously the full $1,000 per month is avaialable until both are paid off. So after the car is paid off, all $1,000 would go onto the house.

I was under the impression the answer would be it doesn't matter. Since the interest rates don't change, loan amounts don't change and total repayment remains constant, any extra paid on one loan just results in more interest on the other loan, and the total amount of interest paid will always be constant no which which way the extra payments are split. But other people say you pay the highest interest rate first, and others say it depends on ratio of amounts owed to their interest etc.--Dacium 04:39, 5 March 2007 (UTC)


 * Answering your question literally, it's always the higher interest rate that you should attack. The reason is that interest is linear: reducing the house's principal to $96k causes it to accrue exactly $4k•6% less interest (per year), while reducing the car's principal to $16k causes it to accrue exactly $4k•10% less interest per year.  (Note that the change in interest has nothing to do with the remaining amount of principal.)  Obviously the latter reduction in interest is better.  However, there is (at least) one important complication in practical planning: suppose that the car was actually the better loan: $20k at 4%.  While the preceding argument would now suggest paying on the house, if you can finish paying on the car, you will no longer have to make its minimum payments, which may be quite helpful if you have a variable income or highly variable other expenses.  It also could be the case that the loans are compounded differently, but no compounding can increase an interest rate r to more than $$e^r$$, where r is expressed as a decimal: for your 10%, that's $$e^{.1}=10.517\%$$.  As such, it's unlikely that compounding switches which loan really accrues more interest &mdash; but note that for something like a hideous credit card's 29%, compounding can yield up to 33.64%!.  Does that help?  --Tardis 15:41, 5 March 2007 (UTC)
 * Not really :-). You see while the higher rate does have more interest, paying more on it just lengths the term of hte other loan. I did spread sheets to work out each period and made the extra payments go on the car, then next time onto the house, then half on each. Every answer came out to have exactally the same interest paid and exactally the same amount of time before both were repaid! So I believe if you have two loans it does not matter which way you split payments if your total money to repay loans remains the same. I guess the only way to know for sure is to solve by formula: (ln(R)-ln(R-Ar))/ln(1+r)=n, and nR-A=interest paid. So assuming two loans, you would have n1R1-A1 + n2R2-A2 = total interest paid. Find the minimum of that equation considering that n1 = (ln(R1)-ln(R1-A1r1))/ln(1+r1) and n2 = (ln(R2)-ln(R2-A2r2))/ln(1+r2), where R is payment size, r is interest rate, A is amount borrowed, r= rate/12, since its monthly. So r1=0.06/12 r2=0.1/12, A1=100000, A2=20000. The conditions to solve against are R1+R2=1,000 and also R1 >= 6000/12 and R2 >= 2000/12.    Not sure how to work this out o_O--Dacium 21:40, 5 March 2007 (UTC)
 * No Tardis is right, you should pay the higher interest loan off first. As an extreme example, say there are two loans each for 100,000 one with 10% interest PA and one with 0% interest PA. Let's say you always have to pay the interest each year and you have 10,000 to pay off the captial. The extreme ends of the spectrum of possibilities are 1) paying first the 0% interest loan completely 2) paying first the 10% interest loan completely.
 * If you take the first option, in the first ten years you will pay 10,000 of the captial of the 0% interest loan and 10,000 as an interest on the 10% loan but without reducing the captial of the 10% loan at all. Then in the next 10 years you pay off the 10% interest loan with total interest 10,000+9,000+...+1,000=55,000.
 * If you take the second option, in the first ten years you will pay 10,000 of the capital of the 10% interest loan each year and the interest payments are again 10,000, then 9,000, and so on, with total interest payed 55,000. Then in the last ten years you pay off the 0% interest loan.
 * The difference here, is that in the first example you spend 10,000 in each of the first ten years for paying interest on the 10% loan but in the second example you save this money, a total of 100,000.
 * In general you can see that paying the high-interest loan first will save interest payments. Stefán 19:04, 6 March 2007 (UTC)

This may be homework, but I hope it's OK...
The question is:

The function y = 5x2 - 4x + 12. Write down its derivative and show that when x = 6, a small increase in x of p% causes an increase in y of approximately 2p%.

I know how to differentiate the function, and I understand the formula delta y = dy/dx * delta x (sorry, I don't know how to insert a delta sign on the computer).

If the question asked for "a small increase in x of p", I would know how to do the question: dy/dx = 10x - 4, delta y = (10x - 4) * p.

However, p is a percentage value, so I don't have the value of delta x to submit into the formula. I know the formula delta x / x * 100 = p, but I still need delta x in that equation.

How do I obtain a value for delta x? Or can I do the question without knowing delta x?

Please answer quickly - I have to submit this quickly and my teacher will kill me if I don't.


 * One percent of some value is 1/100 of that value. So, if x increases by 1%, you get (x + 1/100x) instead of initial value of x. That means the increment is (1/100)x. Hope that helps... I can't do you homework. [[image:smile eye.png]] CiaPan 14:54, 5 March 2007 (UTC)


 * If you substitute x=6 in the expressions for y and dy/dx, you'll find that (delta y)/y = (56/168)*delta x, where delta x is p% of 6. You want 100*(delta y)/y, which I'm sure you can now get...86.132.239.217 16:37, 5 March 2007 (UTC)


 * You are being asked to show that (&Delta;y/y)/(&Delta;x/x) is approximately 2 when x is 6. Re-arranging the ratio gives you (&Delta;y/&Delta;x)(x/y), which, for small &Delta;s, is approximately (dy/dx)(x/y). Find y and dy/dx when x is 6, and then show that (dy/dx)(x/y) is approximately 2. Gandalf61 18:04, 5 March 2007 (UTC)

An Indeterminate form to an indeterminate form power
The question asks to find the limit as x approaches 0 from the positive side for x x x (That is x^x^x). I tried taking the natural log of both sides so ln(y)= lim x to 0 of xxln(x). But then there is still an indeterminate form, and I can't figure out how to make it a fraction. What should I try?


 * This question was already addressed above, and a hint was given to you. Concentrate first on finding $$\lim_{x \to 0^+}x^x$$ Use that result to compute the limit $$\lim_{x \to 0^+}x^x\ln(x)$$. Lastly, please read the directions above and remember to sign your posts using four tildes (like this: ~ ) –King Bee (&tau; • &gamma;) 16:36, 5 March 2007 (UTC)