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February 15

Formula for partial derivative of A with respect to B given that f(A,B) is constant, applied to vectors

Suppose that we have a vector-valued function $f({\vec {a}},{\vec {b}})$ of the type $(\mathbb {R} ^{n},\mathbb {R} ^{m})\rightarrow \mathbb {R} ^{k}$ .

I am trying to find a formula for the value of

$\left({\frac {\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{f}$ .

This is equivalent to the Jacobian matrix of a function $g({\vec {x}})$ such that if $f({\vec {a}},{\vec {b}})={\vec {0}}$ , then $g({\vec {b}})={\vec {a}}$ .

There is a technique for finding this kind of partial derivative of scalar functions, and I tried to generalize it to vector-valued functions:

df({\vec {a}},{\vec {b}})=\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}d{\vec {a}}+\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{\vec {a}}d{\vec {b}}

Holding $f({\vec {a}},{\vec {b}})$ constant, we get:

\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}d{\vec {a}}=-\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}\right)_{\vec {a}}d{\vec {b}}

Multiplying both sides by $(d{\vec {b}})^{-1}$ and $\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}\right)_{\vec {b}}^{-1}$ :

(d{\vec {a}})(d{\vec {b}})^{-1}=-\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}}_{\vec {b}}\right)^{-1}\left({\frac {\partial ({\vec {f}}_{0}...{\vec {f}}_{k})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{m})}}_{\vec {a}}\right)

I am not sure, but I believe that $(d{\vec {a}})(d{\vec {b}})^{-1}$ is equivalent to $\left({\frac {\partial ({\vec {a}}_{0}...{\vec {a}}_{n})}{\partial ({\vec {b}}_{0}...{\vec {b}}_{n})}}\right)_{f}$ , which, if it is correct, gives me an answer to my original question. Is this valid, or did I make a mistake somewhere?

On a side note, I'm new to both Wikipedia's formatting and LaTeX; if you have any comments on my formatting, I'd like to hear them.

24.130.128.99 (talk) 02:14, 15 February 2009 (UTC)[reply]

 —Preceding unsigned comment added by 24.130.128.99 (talk) 02:12, 15 February 2009 (UTC)[reply]

I believe you are right, see implicit function theorem. Upd: k must be equal to n for the inverse matrix to make sense. (Igny (talk) 04:58, 15 February 2009 (UTC))[reply]

The Problem I am stuck with

Question

A car of mass 1200 kg, towing a caravan of mass 800 kg, is travelling along a motorway at a constant speed of 20 m/s. There are air resistance forces on the car and the caravan, of magnitude 100N and 400 N respectively. Calculate the magnitude of the force on the caravan from the towbar, and the driving force on the car.

The car brakes suddenly , and begins to decelerate at a rate of 1.5 m/s2. Calculate the force on the car fro the towbar. What effect will the driver notice?

I did get the first part which is quite easy. Given the objects travel at constant speed, pull on them must equal to the resistive forces to achieve equilibrium(constant speed).

Thus: Magnitude of the force on the caravan from towbar = 400N

     Driving force = 500N

MY PROBLEM

I am totally lost at the second part of the question:

The car brakes suddenly , and begins to decelerate at a rate of 1.5 m/s². Calculate the force on the car from the towbar. What effect will the driver notice?

The book says the answer to this part is: 800N forwards; it will appear that the car is being pushed from behind.

But it doesn't say anything about why this is so. My question or help required is that why this is so? —Preceding unsigned comment added by 202.72.235.208 (talk) 08:54, 15 February 2009 (UTC)[reply]

If the 800 kg caravan is decelerating at 1.5 m/s², what net force must be acting on it ? 400N of this force comes from air resistance - where does the rest come from ? The car exerts a force on the caravan through the towbar - what does Newton's third law then tell you about the force exerted by the caravan on the car ? Gandalf61 (talk) 09:24, 15 February 2009 (UTC)[reply]

First, a translation for US readers: "caravan" = "trailer". Next, there is an apparent assumption that only the car is braking. Next, we need a diagram:

         400N -> +------+
           __    |      |
         _/  \_  | 800kg|
100N -> |1200kg|-|      |
        +-O--O-+ +-O--O-+

Now calculate the total deceleration force needed on the trailer:

F = ma = (800kg)1.5m/s² = 1200kg•m/s² = 1200N

Now, if there's a 1200N deceleration force on the trailer, and 400N of that is initially provided by wind resistance, the additional 800N must be provided by the tow bar. Note that either the rate of deceleration will decrease, or the braking force and force transmitted by the tow bar must increase, as the speed (and therefore wind resistance) decreases. StuRat (talk) 16:17, 15 February 2009 (UTC)[reply]

...and the driver may notice that he must apply increasing force on the brake pedal to keep the deceleration constant, and that the wind, engine and tire noises decrease. Cuddlyable3 (talk) 21:01, 15 February 2009 (UTC)[reply]

StuRat, that's a simply phenomenal piece of AsciiArt. Kudos! --DaHorsesMouth (talk) 21:28, 15 February 2009 (UTC)[reply]

Thanks ! StuRat (talk) 17:41, 16 February 2009 (UTC)[reply]

Thanks people for your help and specially thankful to StuRat for his visual description, really appreciate that kind of helping hand. By the way, air resistance was told to be kept constant in this problem unless otherwise stated in this one, it was written at the top of the chapter, SORRY FORGOT TO MENTION. The answer is that the car had to provide extra 800N braking force despite its own 1700N needed braking for such deceleration.

                                  400N ->+------+
           __                            |      |
         _/  \_                          | 800kg|
100N -> |1200kg|<-800N-------------800N->|      |
        +-O--O-+                         +-O--O-+

Braking(1700 + 800)N ->

Towbar acts as a couple, meaning 800N forces bothways parallely. Thus the reaction force is 800N. —Preceding unsigned comment added by 202.72.235.202 (talk) 18:34, 16 February 2009 (UTC)[reply]

Collatz-like sequence

Does anyone know of any results concerning the Collatz-like sequences generated by

f(n)={\begin{cases}n/2&{\mbox{if }}n\equiv 0{\pmod {2}}\\3n-1&{\mbox{if }}n\equiv 1{\pmod {2}}\end{cases}}

I have found papers on various generalisations of the Collatz conjecture, but I haven't found any results on this specific case.

As far as I can tell, there are three loops:

1\rightarrow 2\rightarrow 1

5\rightarrow 14\rightarrow 7\rightarrow 20\rightarrow 10\rightarrow 5

17\rightarrow 50\rightarrow 25\rightarrow \cdots \rightarrow 68\rightarrow 34\rightarrow 17

and every sequence I have tested eventually enters one of these loops. Having found three loops, I was surprised not to find more - why just three ? Gandalf61 (talk) 09:14, 15 February 2009 (UTC)[reply]

Never mind - I have just realised that these are essentially the same as Collatz sequences if we replace n with −n. Gandalf61 (talk) 11:37, 15 February 2009 (UTC)[reply]

Resolved

StuRat (talk) 15:31, 15 February 2009 (UTC)[reply]

Name the curve

What is the name of the curve(of four cusps)described by the enclosure of a moving straight line of length A, wherein the end points of line A move along their respective X and Y axis?

The equation given for this curve is: x^2/3 + y^2/3 = A^2/3. The curve is similar to the hypocycloid of four cusps,(the astroid), however the line generation appears to be uniquely different.

Vaughnadams (talk) 19:26, 15 February 2009 (UTC)[reply]

According to our article, that is an astroid. Algebraist 19:44, 15 February 2009 (UTC)[reply]

It looks a bit like this. Cuddlyable3 (talk) 20:51, 15 February 2009 (UTC)[reply]

Maths: discovery or invention?

Are mathematical developments discoveries or inventions, and does someone's answer to this question effect the conclusions they can draw? Thanks in advance. 86.8.176.85 (talk) 19:46, 15 February 2009 (UTC)[reply]

This depends on your Philosophy of mathematics. That article has some positions that various people have held. Algebraist 19:53, 15 February 2009 (UTC)[reply]

This is a perennial source of discussion for philosophers; there is no clearly correct answer. It's analogous to solving a crossword puzzle – would you say that you created the solution, or that you discovered it? Even usage among mathematicians is varied. I typically say that I discover a new mathematical object but I invent a new technique. — Carl (CBM · talk) 19:58, 15 February 2009 (UTC)[reply]

One often speaks of constructing a new object also. Algebraist 20:45, 15 February 2009 (UTC)[reply]

Mathematics is discovery in an abstract universe that does not exist but is useful to invent. I think the questioner means affect not effect the conclusions one can draw. The answer to the first question does not affect the conclusions one can draw, only the mathematics can prove or disprove a mathematical conclusion. Cuddlyable3 (talk) 20:46, 15 February 2009 (UTC)[reply]

It can, actually. A realist (about mathematical objects) is forced to conclude that the continuum hypothesis must be either true or false, and may be able to convince himself one way or the other. Some types of antirealist, on the other hand, are able to conclude that CH is without truth value. Algebraist 21:36, 15 February 2009 (UTC)[reply]

A related question would be whether mathematical concepts or techniques are copyrightable or patentable? It's a relevant question when you consider cryptology and compression technologies? I wonder what would would be the effect of someone having a patent on the Pythagorean theorem? -- Tcncv (talk) 07:58, 16 February 2009 (UTC)[reply]

I believe there is prior art. 76.126.116.54 (talk) 08:11, 16 February 2009 (UTC)[reply]

Theorems are just statements of fact, I don't think they are copyrightable or patentable. Mathematical algorithms can be patented, but are not copyrightable, as I recall (a given software implementation of the algorithm is copyrightable, though). --Tango (talk) 11:59, 16 February 2009 (UTC)[reply]

Mathematical algorithms are really no different from other mathematical facts and therefore should not be patentable. Indeed, generally they are not patentable outside the United States, see software patent. — Emil J. 13:48, 16 February 2009 (UTC)[reply]

Any copyrighted image or text that can be digitised becomes just a big binary number that may be judged to be an Illegal number. Psssst the secret number is 42 but you didn't hear that from me. Cuddlyable3 (talk) 15:11, 16 February 2009 (UTC)[reply]

Not serious series

1. ~~Counters of beans~~Fine mathematicians with too much free time, can you supply the last term of this series:

3 , 3 , 5 , 4 , 4 , 3 , 5 , ?

2. Riddle me this: why does six fear seven ? Cuddlyable3 (talk) 21:19, 15 February 2009 (UTC)[reply]

The answer to 1. is 5. Algebraist 21:23, 15 February 2009 (UTC)[reply]

2: Because 7 8 9. As for 1, I say the answer is pi. -mattbuck (Talk) 21:24, 15 February 2009 (UTC)[reply]

1. is A005589 at the OEIS. Algebraist 21:26, 15 February 2009 (UTC)[reply]

Interesting how they stopped at "one hundred", nicely sidestepping the issue of one hundred one versus one hundred and one (though still vulnerable to the challenge from a hundred). --Trovatore (talk) 23:02, 15 February 2009 (UTC)[reply]

The value of 4 for the noughth entry is also arguable. Algebraist 00:01, 16 February 2009 (UTC)[reply]

Mattbuck, the mission I give you is to take a LARGE piece of paper and write down the exact answer in very small print. Cuddlyable3 (talk) 14:29, 16 February 2009 (UTC)[reply]

3. I gave a junior class an introduction to using x as a variable (which some of them found baffling) by getting them to evaluate these expressions:

a)

2x+7-8/4

when

x=3

b)

0.5+5x+2

when

x=5

One student evaluated a) as 28 which is wrong. He got b) wrong too but at least he was consistent in his method. What was his answer to b) ? Cuddlyable3 (talk) 14:29, 16 February 2009 (UTC)[reply]

57.5 --Tango (talk) 14:46, 16 February 2009 (UTC)[reply]

That's not really wrong, is it? Concatenation can be denoted by juxtaposition just like multiplication. — Emil J. 15:02, 16 February 2009 (UTC)[reply]

You can denote anything you like however you like, but it's not standard. If you want to concatenate digits you would usually use some kind of symbol to denote it (perhaps

2\circ x

). --Tango (talk) 15:11, 16 February 2009 (UTC)[reply]

But the standard notation for concatenation of strings over a finite alphabet in combinatorics, theoretical computer science, and related areas is simple juxtaposition, that's the point. People also use stuff like

x\frown y

when they want to be extra explicit, but then again, there is explicit notation

x\cdot y

for multiplication as well. I've never seen

\circ

used for concatenation. — Emil J. 15:35, 16 February 2009 (UTC)[reply]

In those areas the symbols usually just represent an arbitrary alphabet, not actual digits (they may use the same symbols, but they aren't actually representing numbers). If you are doing addition and subtraction with them then it is clear that they are numbers. Juxtaposition of numbers (with at least one represented by a variable - obviously the juxtaposition of '2' and '3' means twenty-three) almost universally denotes multiplication. --Tango (talk) 16:15, 16 February 2009 (UTC)[reply]

I have seen || used as well when there are other operations floating around using juxtaposition, as in

x||y

GromXXVII (talk) 19:56, 16 February 2009 (UTC)[reply]

Mattbuck, almost finished are you? Cuddlyable3 (talk) 20:02, 16 February 2009 (UTC)[reply]