Wikipedia:Reference desk/Archives/Mathematics/2007 October 16

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October 16

Thomas and Finney 9th ED even answers

I am teaching myself calculus and am in dire need of the even answers to the 9th ed Calculus book by Thomas and Finney. I am a hard worker and like the challenges that the even problems pose, but I am unable to check to see if my answers are correct. I tried to find a teaches ed. on amazon and barnes and nobles, but they don't sell this. Where can I get these answers, does anyone have a link, or any advice. I'd really appreciate it. thanks, Rob —Preceding unsigned comment added by 71.56.231.40 (talk) 03:55, 16 October 2007 (UTC)

Contact the publisher of the textbook. Which, I believe, is Addison-Wesley in this case. Textbook publishers typically publish "Answers and Solutions Manuals" -- sometimes for students only (just the odd questions), sometimes for teachers only (both the odd and even questions), sometimes for both students and teachers. (Joseph A. Spadaro 04:19, 16 October 2007 (UTC))

Don't check your answers. If you are sure that you are right, then it's OK. If you are not sure, then recheck until you are convinced. Trust your own judgement rather than that of the author. Bo Jacoby 10:07, 16 October 2007 (UTC).

It is a very bad idea to never check your answers against an authoritative source. Students often have some misunderstandings about the material, and when this is the case you can be as convinced of your answer as you want - it will not make it correct. -- Meni Rosenfeld (talk) 18:06, 16 October 2007 (UTC)

It's common that textbooks present questions in matched pairs: The even and odd questions which are adjacent test similar skills. So if you can do problem #17, then you should be able to apply that to #18. If you're still unsure, you can post your work here and ask whether you have it right (or where you went wrong). Do you not have access to a math teacher? Donald Hosek 18:32, 16 October 2007 (UTC)

Hi Meni! When you check a correct answer you are likely to happily quit that problem instead of studying the problem until you are satisfied. This cost has to be compared to the cost of believing that an answer is correct when in fact it is not. The purpose of doing exercises is to provide understanding rather than to provide correct answers. Bo Jacoby 20:31, 16 October 2007 (UTC).

Oh, I do not deny the importance of struggling with a problem until you truly understand the question, your answer, and the way you reached it. I only say the occasional check is necessary to avoid going astray. -- Meni Rosenfeld (talk) 12:32, 17 October 2007 (UTC)

You have chosen a popular text, so surely instructors manuals are out there. But your quest raises some issues. A teacher will want to reserve the unanswered questions for homework and tests. Such a teacher will not sell the manual to a used-book shop. In class, we have "right answers" (though books do have mistakes!); outside class, we confront questions without any such assurance. We must learn strategies to cope with our uncertainties. Mistakes are most likely when our mental models are rough and under construction, but we are also less likely to trust our results then; mistakes are most costly when our confidence is misplaced. In the professional world, we rely on peer review, standard checklists, redundant calculation using a completely different method, and so on. Throughout our lives we continue to learn, and we continue to make mistakes. A teacher (or answers book) can be great if we have one, but most of the time we don't. You, however, are fortunate, especially compared to students 100 years ago; you have the Web, an incredible resource for learning calculus. You can find instructional materials (including video), people who will answer your specific questions, other students who can offer aid and comfort — and (of course) misinformation. :-) --KSmrq^T 07:47, 17 October 2007 (UTC)

Of course, there is always the option of checking answers yourself indirectly. For instance, if the problem is finding x for x = x^2 + 2x + 3, (a simple example) you can check your answer by putting it in the original equation. If you're supposed to find a derivative or an indefinite integral, you can plot the answer with a function plotter, and see if their features look like you'd expect. This isn't as exact as checking the correct answer in a list, and you may even think you got it right when you didn't (eg. you found one solution when there were two), but it does have the important bonus that figuring out how to test your answer can give you important insights into the matter you're studying. After all, when you start using calculus for real, there won't be an answer booklet either. risk 21:56, 17 October 2007 (UTC)

Slope of a Line

In essence, the slope of a line (first degree linear equation) is defined as the change in y-values divided by the change in x-values (for any points on that line). This is sometimes (verbally) also referred to as "delta y over delta x" or "rise over run" or "y-two minus y-one over x-two minus x-one", etc. Written as a fraction, we would see (y2-y1) / (x2-x1) or some similar. Invariably, students often tend to mistakenly place the "x" values in the numerator and the "y" values in the denominator (in reverse of the correct formula). Does anyone know of any type of easy mnemonic device to help students remember that the y's go on top and the x's go on the bottom of the ratio/fraction/slope formula? Admittedly, the letters "x" and "y" do not lend themselves easily to every-day English words ... but I would like to find an easy mnemonic for this. Thanks. (Joseph A. Spadaro 04:28, 16 October 2007 (UTC))

I don' know if this is any help but perhaps just "rise over run"? Thought that is possibly just as tough to remember at y over x but it does have the visualisation aspect as well? --AMorris (talk)●(contribs) 04:35, 16 October 2007 (UTC)

My experience is that students don't quite understand what the words "rise over run" even mean. It is a string of mathematical words which they cannot intuit. I would like something to remember "y over x" ... like "Yolanda tripped over the X-ray machine" ... but perhaps something less contrived and more realistic. (Joseph A. Spadaro 04:44, 16 October 2007 (UTC))

"Rise over run" should work fine if the students are well versed in the Cartesian Plane. SO if your having trouble with that, then have them brush up on the Cartesian Plane. A math-wiki 06:16, 16 October 2007 (UTC)

Really? Cartesian plane (a redirect to Cartesian coordinate system) doesn't contain either one of those words anywhere! At Wiktionary (wikt:run) we have this definition of run: "The top of a step on a staircase, also called a tread, as opposed to the rise." That is presumably the definition referred to in the phrase "rise over run". It's way down the definition list at number 23, and that's only counting the ones in the Noun section, for a word which is more commonly used as a verb. The corresponding definition of rise referring to the vertical part of a step on a staircase does not appear in wikt:rise at all. I'd say this is pretty strong evidence against "rise over run" being understandable. Maybe there was a time when kids all practiced carpentry at home and knew all the staircase-building terminology. Now I think that phrase is just one more useless thing you memorize in high school Algebra because the textbooks for some reason expect you to already know those obscure meanings of "rise" and "run". --tcsetattr (talk / contribs) 07:44, 16 October 2007 (UTC)

Isn't it simple enough just to remember that the slope corresponds to the "steepness"? So if a line is really steep, the slope will be a large number, to achieve that you need to divide the large change in y by the small change in x. 130.88.47.48 12:26, 16 October 2007 (UTC)

Follow Up: Thanks for the replies, but they are not really addressing my question asked. Yes, I know what a slope is ... and I certainly understand all of the replies above. My question is: students often mix up the x and the y (in reverse). I'd like a mnemonic so that they can easily remember "y over x" ... example: Yolanda tripped over the X-ray machine. Students simply do not intuit "rise over run." Also, if they (beginning students) were indeed so clear and un-fuzzy on the mathematical concepts referred to above (steepness, Cartesian plane, dividing a large change in y by a small change in x, etc.) -- I would probably not need a mnemonic at all. Thanks. (Joseph A. Spadaro 13:12, 16 October 2007 (UTC))

(I have not thought of one as yet) One problem here is to ask "should students be associatinf y with vertical and x with horizontal" - In two dimensions it's what we always do I know - but when they get onto 3 dimensions x,y are often the horizontal plane and z the vertical - so a mnemonic might confuse them later on. At school I remember using analogies such as roads and railway tracks - I think using real world examples to demonstrate the method might work better than a mnenomic.87.102.12.235 15:00, 16 October 2007 (UTC)

A simple way to associate the usual 2D axes with their direction is "x goes aCROSS", hence y is the other one. If v/h is to be remembered, just think of "very hot".…81.154.108.36 16:12, 16 October 2007 (UTC)

Alright, since you insist: "Young seX". They'll remember it, especially if you refer to the function going up and down. It's in the fine tradition of memory aids like "Oh, be a fine girl/guy, kiss me", used to help remember the spectral sequence of stars. (Who would remember OBAFGKM otherwise? And never mind how we got that alphabet soup!) Or, if you prefer restraint, "Why seX?" is another possibility. The point is not insight or logic; the more outrageous and vivid the phrase, the more it will be remembered. If you want to work "over" in there, I'm sure you can. Of course, if your students are too young they might not think much about sex; a little older, and they'll think of it constantly. You decide.

I don't object to mnemonics, but I'm sure you agree that in a case like this the long-term objective is to replace rote memorization with true insight. When I hike in the mountains, my legs and lungs tell me about vertical distance; the word "slope" has a tangible physical meaning! When your students get that vivid a mental model, they should have little trouble reconstructing the Δy/Δx formula. --KSmrq^T 07:04, 17 October 2007 (UTC)

I'm a math teacher, and I teach slope as "rise over run". However, as you note, students don't just intuit "rise" and "run". Therefore, I spend a few minutes talking about why a change in the y-coordinate is called a "rise", and why a change in the x-coordinate is called a "run". Then, when I write down the slope formula, I write all three: ${\frac {rise}{run}}={\frac {\Delta y}{\Delta x}}={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$, and I draw arrows from the words "rise" and "run" to a picture of the appropriate right triangle joining the two points. After that, it's just a matter of repetition...

Actually, I just went into the next room and asked the students there how they remember, and one suggested that she remembers, "y is high". That also sounds good. Auditory learners, visual learners... it takes all kinds, right? -GTBacchus^(talk) 20:48, 16 October 2007 (UTC)

Thanks for all of the input ... it was very helpful ... much appreciated ... (Joseph A. Spadaro 20:03, 21 October 2007 (UTC))

Clarification

"For $m,n\in \mathbb {N}$. If m is a multiple of n then any m-th power is also an n-th power."

I assume this means:

$(m=nx)\to (a^{m}=a^{nx})$?

But I don't understand the sentence "any m-th power is also an n-th power" so it's a guess. Could someone put this out in notation correctly, I have trouble visualising verbose statements. —Preceding unsigned comment added by 91.84.143.82 (talk) 08:08, 16 October 2007 (UTC)

The quoted sentence could be translated like this:

$\forall m(\forall n((\exists x(m=nx))\rightarrow (\forall y((\exists z(y=z^{m}))\rightarrow (\exists u(y=u^{n}))))))$

Morana 09:44, 16 October 2007 (UTC)

Perhaps loosely, with less quantifiers, the statement could be expressed as $(m=nx)\to (a^{m}=(a^{x})^{n})$, . That is, if m is x times n, then the m-th power of any a is the n-th power of a^x. Thus, for example, every fourth power of an integer is a square number. —Ilmari Karonen (talk) 10:37, 16 October 2007 (UTC)

Perhaps it should be clarified that a number is called "an mth power" if it is the mth power of some (in this case, integer) quantity.

With that behind us, I'll try to combine the best from Morana's and Ilmari's replies: Assume m is a multiple of n. Then, every number which is the mth power of some quantity, is also the nth power of some (other) quantity. Hence "any mth power is also an nth power". -- Meni Rosenfeld (talk) 15:40, 16 October 2007 (UTC)

What no one has said, but what we all draw upon, is the law of exponents that says a^pq =(a^p)^q. We can see the truth of this as a consequence of the meaning of an exponent and the definition of multiplication. Exponentiation is repeated multiplication:

$a^{p}=\underbrace {a\times \cdots \times a} _{p}\,\!$

The product of p by q repeats p:

$p\times q=\underbrace {p+\cdots +p} _{q}\,\!$

Putting the two together, we have

${\begin{aligned}a^{pq}&=\underbrace {a\times \cdots \times a} _{pq}\\&=\underbrace {(\underbrace {a\times \cdots \times a} _{p})\times \cdots \times (\underbrace {a\times \cdots \times a} _{p})} _{q}\\&=\underbrace {a^{p}\times \cdots \times a^{p}} _{q}\\&=(a^{p})^{q}.\end{aligned}}$

This is what the question seems designed to explore. --KSmrq^T 06:45, 17 October 2007 (UTC)

chirality

I'm assuming the universe is chiral (please correct if wrong)

Given that the universe is chiral - and having no external reference for chirality - I don't know that it is possible to tell which chirality it is (left or right so to speak).

Does this mean that it (the universe) is both chiralities at the same time (since there is no external reference)?

(Please redirect to correct desk if this is the wrong place - I assumed a theoretical treatment would be best - we have no philosophy desk)87.102.12.235 18:09, 16 October 2007 (UTC)

Perhaps it means the answer to the question is Mu. Without external reference, there is no distinction between left and right chirality, which does not mean we have both. There is no distinction in the cyclic group Z/2 between +1 and −1, which does not mean that 1 is both positive and negative in Z/2, but just that the distinction between these two makes no sense in that context.  --Lambiam 18:57, 16 October 2007 (UTC)

I think you're right - there is no distinction - so to suggest that there is/are "two types" is meaningless - that must be the bit of my thinking that is wrong. Thanks (I still can't get over this) There is no paradox as such then - more an example of applying concepts to which they do not apply. If anyone can tell me what's going on I'd appreciate it.87.102.12.235 19:43, 16 October 2007 (UTC)

..I got that the example of Z/2 is similar to what I was asking - Question does the 'universe' fall into a Z/2 group - I can see it has the same properties but don't know enough to know if it was an example or not.. I know very little about this topic.87.102.12.235 19:52, 16 October 2007 (UTC)

Sorry Lambiam, but there is a distinction between +1 and −1 in Z/2, because the equation xx=x is satisfied by x=+1 but not by x=−1. A better example: there is no distinction between +i and −i in the ring Z[i]/(i²+1). Bo Jacoby 20:51, 16 October 2007 (UTC).

Lambiam was, I believe, referring to Z/2 in its 'natural' form as the quotient of Z by 2 (with elements 0 and 1 and operation addition mod 2), not as ({+1,-1},*), in which there is trivially a distinction. Algebraist 21:03, 16 October 2007 (UTC)

It's worth nothing that in $\mathbb {Z} /2$, -1 and 1 are both shorthands for the coset {...-3, -1, 1, 3, ...} so yes, $xx=x$ is, in fact true for 1 and -1 in $\mathbb {Z} /2$. Algebraist's group, with elements 0 and 1 is typically notated as $Z_{2}$ and while its isomorphic to $\mathbb {Z} /2$, it's not exactly the same thing (some others reserve $Z_{2}$ for the group and $\mathbb {Z} /2$ for the ring). Donald Hosek 22:07, 16 October 2007 (UTC)

The human body is chiral as it is not quite identical to it's mirror image because the heart is not in the middle of the plane of approximate symmetry of the body. But "to tell which chirality it is (left or right so to speak)" is merely a matter of convention. In Lambiam's example of Z/2=({0,1},+), (which I misunderstood, thanks for explaining), the two symbols "+1" and "−1" refer to the same object, and any object is trivially indistinguishable from itself. Can different objects be indistinguishable too? If you swap the two electrons in a helium atom you get exactly the same helium atom, so the electrons are indistinguishable, yet there are two rather than one. One of the two electrons has "spin up", but it makes no sense to ask which one. You can talk about a multiset of electrons, not of a set. Bo Jacoby 07:22, 17 October 2007 (UTC).

Don't forget the liver that's way off centre..87.102.12.215 15:34, 17 October 2007 (UTC)

A more intreging possibility is that the universe is non-orientable, i.e. something like the projective plane. --Salix alba (talk) 08:25, 17 October 2007 (UTC)

Assume for a moment that the universe is R². We can change the chiariality by reflecting in the line x=0. Now embed R² in R³, and rotate through 180° around the y-axis, this has the same effect geometrically as the reflection without any change to the universe. If the universe can be represented as some sort of n-maniflod, the same trick can be performed by rotating in a suitabily higher dimensional state.

It may also be worth asking on the science reference desk as there are some very strange lack of symmetry in the fundamental forces, and between the amount of matter and anti-matter in the universe. --Salix alba (talk) 19:42, 17 October 2007 (UTC)

See CP-violation and Parity violation. Discoveries that win a Nobel Prize should be better known.

In mathematics, any time we have an ordered list of two or more elements we can divide its possible rearrangements (permutations) into two categories: those that require an even number of element swaps (even parity), and those that require an odd number (odd parity). A particular arrangement tells us nothing about parity; it is the comparison, the rearrangement, that has parity. Chirality (from the Greek for "hand") concerns violation of symmetry, things that can change when a coordinate basis is rearranged by reflection. --KSmrq^T 00:04, 18 October 2007 (UTC)

I've never 'got' Parity violation (and probably never will) - if a 'left handed particle' has a different set of properties to a 'right handed particle' then they are not actually left and right handed versions - but different particles (that for some reason a scientist has decided to label 'left' and 'right' - these scientists think they can get away with murder (smile) - ..87.102.3.9 12:48, 18 October 2007 (UTC)

ADF Critical Values

At university I learnt how to do an ADF test. I don't need to have that explained.

But does anyone know where I can find the table of critical values for the ADF test?

The ADF page in wikipedia does not have the critical values. The only paper refered to doesn't appear to either.

I've already purchased one article from JStor and I'd prefer not to buy any more until I know for sure that it actually has the correct values. —Preceding unsigned comment added by 192.193.1.5 (talk) 23:40, 16 October 2007 (UTC)

You can find them in:

Fuller, W.A., (1976), Introduction to Statistical Time Series., John Wiley, New York.

Morana 23:53, 16 October 2007 (UTC)

Thank you very much. —Preceding unsigned comment added by 192.193.1.5 (talk) 23:58, 16 October 2007 (UTC)

Paths

Answering this question by clicking the link on the right will take you to a subpage of my userspace. - Mgm|^(talk) 22:22, 18 October 2007 (UTC)

I've developed an addiction for the game at http://www.blackflip.org Some of those puzzles are particularly devilish and I wondered what math applies to puzzles like this. Maybe reading up on those theories and rules makes solving some of them easier. - Mgm|^(talk) 08:48, 17 October 2007 (UTC)

Well I took a look at those puzzles, very addictive indeed. I think graph theory would help but It seems the most difficult part is deciding what color each horizontal row should be, based on the layout. In general, the study of problems like these falls under game theory. I also think analysis (mathematics) and perhaps Galois theory may help, but those are inherently more advanced and thus less easily understood. A math-wiki 11:27, 17 October 2007 (UTC)

I've a feeling a similar sort of logic to the Seven Bridges of Königsberg may prove to be of some help. To start workout what are illegal moves, for example you can't move horizontally from black to white. They you could try exaustive enumeration of simple 2 by 2 and 3 by 3 sub-grids, for example if you have a t-shape will imply something or other. game theory may be a readherring as the sort of games studied there are more like poker than strictly logical one, all though some ideas like the Decision tree may be of help. --Salix alba (talk) 14:22, 17 October 2007 (UTC)

This comes under combinatorial game theory in some sense, but I doubt that's a very useful observation. math-wiki: how on earth do you intend to apply Galois theory to this situation? Algebraist 15:46, 17 October 2007 (UTC)

Never mind Galois theory. Analysis, which I presume to mean mathematical analysis? Isn't that the polar opposite of what we have here? Salix has already commented on game theory, but I would like to emphasize that the study of problems like this certainly does not fall under what is usually called "game theory".

A math-wiki, your enthusiasm to provide help is appreciated, but perhaps some more reflection about whether your suggestions are truly relevent is in order? -- Meni Rosenfeld (talk) 21:28, 17 October 2007 (UTC)

After playing for a bit, I suspect a logic based way to solve these puzzles can be very useful. You can look at small subgrids of the whole puzzle and make certain logical conclusions. "If I'm ever going to make this square black, I'll have to come from there, so that row will need to be white. If I go over this square, then that becomes unreachable, etc". You can probably formulate a basic set of rules like this, and based on that you can determine for some subgrid a small set of possible paths to take along it. Once you run out of logical rules to apply, you can search the possible paths that are left exhaustively. If you have some paths left for each 3x3 grid, there will only be very few ways to combine those into one full path, if your set of rules is good enough, you will only have a few paths left to check. risk 18:34, 17 October 2007 (UTC)

(after edit conflict) Heres a partial method. Consider a game with n-rows. For the final state each row can be either all black or all white hence there are 2ⁿ possible final state. For each such final state mark the differences between the final state and the starting state with an X. A few things are imediatly obvious, each X must be connected by X's to the edges of the grid (or all X's are connected, but don't touch the edge). Next count the number of dead-ends, that is X's with only one imediate neighbour, here the edge of the board can be counted as X's. There can be at most 2 of these and if they exist the dead ends much be starting or ending points. Similarly T shapes and + shapes count as 1 and 2 end points. Once you eliminate the impossible final states you would have a much smaller set of possibilities. --Salix alba (talk) 19:11, 17 October 2007 (UTC)

Ok, so i only understand Galois Theory vaguely, my reference to analysis is mainly to constructivist analysis (again only a vague understanding), but from the description it would seem to have something to do with constructing the possible solutions from allowable moves.

After having played a lot more, here are a few nifty shortcuts. If your puzzle has one color in such a pattern that their are at most two dead ends (that is say a black square with three white ones adjacent to it) then it is basically analogus to a Hamiltonian circuit. the solution is generally very easy. If you have a one or more diagonals, try pathing right next to it, and then next to the path you created (a stair-step pattern). And if you have one vertical the is opposite colors it neighbor, you can usually solve these by pathing down the vertical (e.g. if you have a checkerboard pattern somwhere) With these few tricks and some logical thought, I can already solve most easy ones in under a minute and most moderately difficults ones in a few minutes tops. A math-wiki 22:52, 17 October 2007 (UTC)

The game is Polarium on the Nintendo DS, by the way. Don't know if the flash game is a rip-off or a proper version of it, but that's where I recognised it from.--PaulTaylor 22:57, 17 October 2007 (UTC)

I think it is a rip of Polarium, there is an acknowledgement of it at the bottom of the page. --Salix alba (talk) 07:29, 18 October 2007 (UTC)

I think I have a preliminary technique for finding general solutions. Consider the flipable rectangle, and m*n rectangle, that means your graph is an (m+2)x(n+2) set of vertices (remember that you can use the edge). All the vertices are linked in a "square lattice." There are $2^{n}$ possible combinations of colors, meaning that their are that many graphs to analize in total, but it is often possible to cut down on that number significantly. What your do to create these graphs is select one of the color combinations and mark all the vertices that need to be flipped. Then look to see if the graph (including the extra edges) has a path sure that the degree of all but 2 of the vertice is 2, the two should be degree 1 "Hamiltonian Case"(unless the path starts and stops next to itself "Euler Case"). This may not be a general solution but it definately comes close. The one thing I have not throughly investigated is how the edge squares which are used "at will" affect possible paths, but it would seem to allow a few different choices for the pather, so they all should be checked. A math-wiki 23:57, 17 October 2007 (UTC)

Upon putting it to the test, the restriction that the degree need be 2 is not actually the case, just as long as those with degree 3 are adjancent and have their 3rd edge on the same side so that your can parallel your earlier path. All in all, logical thought is still the most efficient. A math-wiki 00:39, 18 October 2007 (UTC)

Hamiltonian Case should be called Hamiltonian path Case, and Euler Case should be called Hamiltonian circuit Case. A math-wiki 00:54, 18 October 2007 (UTC)

It seems likely to be NP-complete, as it's closely related to finding Hamiltonian paths in a grid graph (A. Itai, C.H. Papadimitriou, J.L. Szwarcfiter, Hamiltonian paths in grid graphs, SIAM J. Comput. 11 (1982) 676–686.) If so, there wouldn't be a simple strategy that would be guaranteed to always work: you'd have to actually think to solve it. —David Eppstein 07:47, 18 October 2007 (UTC)