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March 23

Cartesian form

How would I translate this from polar to cartesian form? r = sin θ + 1

Thanks for your help. -- Sturgeonman 00:08, 23 March 2007 (UTC)[reply]

Multiply through by r and think about each term. x42bn6 Talk 00:14, 23 March 2007 (UTC)[reply]

x = r cos θ

y = r sin θ

sin θ = y/r

r = sin θ + 1

r = (y/r) + 1

r*r = y + r

r*r - r - y = 0

Solve for r = f(y) using the quadratic equation. WARNING! beware of any false root(s)

Thus r^2 = f(y)*f(y)

Replace r^2 with x^2 + y^2

Giving x^2 + y^2 = f(y)*f(y)

202.168.50.40 02:44, 23 March 2007 (UTC)[reply]

Here is a shortcut in the above approach. From r² = y + r, eliminate r using r² = x² + y². --Lambiam ^Talk 07:27, 23 March 2007 (UTC)[reply]

Just substitute r=√(x²+y²), sin θ = y/r. Consider carefully the r=0 case. --CiaPan 07:24, 23 March 2007 (UTC)[reply]

It's a cardioid curve (although the article does not give a cartesian equation). Gandalf61 13:43, 23 March 2007 (UTC)[reply]

Philosophy logic/mathematical logic

I saw this question after doing my p-set. It seems fairly simple but I can't get it. Prove that if $\alpha$ is a wff in sentential logic and no sentence symbol appears twice in $\alpha$ then $\alpha$ and $\lnot \alpha$ are both satisfiable. Do I use the compactness thm?

I wish I could figure out how to type math formulas like everyone else, sorry. Simon24.123.234.125 00:16, 23 March 2007 (UTC)[reply]

You can prove the result using structural induction. Try proving this: for any wff

\alpha

with

n

logical connectives, if no propositional variables in

\alpha

occur more than once, there is at least one truth-value assignment that will cause

\alpha

to evaluate to true, and there is at least one truth-value assignment that will cause

\alpha

to evaluate to false. (Math symbols in the original question re-typed.) --71.175.24.57 04:31, 23 March 2007 (UTC)[reply]

Also, note that if

\alpha

is of the form

\beta \diamond \gamma

, where

\diamond

is a binary logical connective, the sets of propositional variables in

\beta

and

\gamma

are disjoint. --71.175.24.57 04:39, 23 March 2007 (UTC)[reply]

See Help:Formula for markup to typeset mathematics. We prefer not to use the TeX forms, like "<math>\alpha</math>" (yielding "

\alpha

"), inline; instead use "α" (yielding "α"). The difference is especially pronounced with superscripts and subscripts. A table of mathematical characters to copy and paste can be found here. Thus we may write "¬α" instead of "

\lnot \alpha

", and "β⋄γ" (or "β♢γ" or "β◇γ") instead of "

\beta \diamond \gamma

". However, depending on the reader's font and browser configuration, some symbols may appear as "missing character" glyphs. --KSmrq^T 13:42, 23 March 2007 (UTC)[reply]

"Theory": model-theoretic or proof-theoretic concept?

I've seen seemingly conflicting usage of the term "theory" in mathematical logic. Some authors use it to refer to a set of formulas closed under semantic entailment. The Wikipedia article on theory, on the other hand, says

A theory in this sense is a set of statements in a formal language, which is closed upon application of certain procedures called rules of inference.

In this usage, a theory is a concept relative to a proof system.

My question is: Is there a well-agreed-upon standard usage for the term? If so, which usage is standard? --64.236.170.228 14:04, 23 March 2007 (UTC)[reply]

Herbert B. Enderton, in A Mathematical Introduction to Logic, ISBN 978-0-12-238450-9, says (page 144):

We define a theory to be a set of sentences closed under logical implication. That is, T is a theory iff T is a set of sentences such that for any sentence σ of the language,
$T\models \sigma \Rightarrow \sigma \in T.\,\!$

On the following page is an example:

For a class ${\mathcal {K}}$ of structures (for the language), define the theory of ${\mathcal {K}}$ (Th ${\mathcal {K}}$ ) by the equation
$\mathrm {Th\ } {\mathcal {K}}=\{\sigma :\sigma {\text{ is true in every member of }}{\mathcal {K}}\}.\,\!$

A. S. Troelstra's Lectures on Linear Logic, ISBN 978-0-937073-77-3, a much more advanced work, says (page 70):

Definition. A theory T is a set of sequents, and a sequent S is derivable in T if S can be derived from sequents in T used arbitrarily often as axioms. Notation: T ⊢ S. ∎

This depends on (page 3):

A sequent is an expression Γ ⇒ A.

R. Goldblatt, in Topoi: The Categorial Analysis of Logic, ISBN 978-0-444-86711-7, essentially concurs (page 496):

A set $\mathbb {T}$ of sequents will be called a theory, just as for a set of formulae.

His definition of sequent (given just before) is:

By a sequent we mean an expression Γ ⊃ ψ, where Γ is a finite set of formulae, and ψ a single formula.

Likewise, Peter J. Freyd and Andre Scedrov, in Categories, Allegories, ISBN 978-0-444-70367-5, go with (Appendix B):

A THEORY is a set of assertions.
ASSERTIONS are expression A ⇀ B, where A, B are formulae. Such an assertion is read A TOLERATES B.

We cannot help but notice the variety of notation! Also, we see some differences in technical detail. Especially, we see that the advanced works say nothing about closure.

Going with the topos bias, we consult one last work. Michael Barr and Charles Wells, Toposes, Triples, and Theories, ISBN 978-0-387-96115-6, say in the Preface to the first edition:

Theories, which could be called categorical theories, have been around in one incarnation or another at least since Lawvere’s Ph.D. thesis. Lawvere’s original insight was that a mathematical theory—corresponding roughly to the definition of a class of mathematical objects—could be usefully regarded as a category with structure of a certain kind, and a model of that theory—one of those objects—as a set-valued functor from that category which preserves the structure. The structures involved are more or less elaborate, depending on the kind of objects involved. The most elaborate of these use categories which have all the structure of a topos.

I rather doubt this glut of data satisfies the question, but perhaps it gives some sense of the challenge in trying to write an article about the modest notion of "theory". --KSmrq^T 21:56, 23 March 2007 (UTC)[reply]

The above establishes unequivocally that the answer to the question "Is there a well-agreed-upon standard usage for the term?" is a resounding No, which makes the next question moot. By the way, the definition with closure under semantic entailment was new to me. Does it give rise to any interesting theory? --Lambiam ^Talk 23:01, 23 March 2007 (UTC)[reply]

(I am the original questioner) I don't know the answer to your last question. Like I said, I've seen conflicting use of the term. Of course, we can solve the problem by giving different names to differently-defined "theories". I am still curious about which of the definitions would give rise to a more useful and "robust" concept. --64.236.170.228 15:41, 27 March 2007 (UTC)[reply]

Abstract algebra

which of the folling statement are true ? Give reason also

Ques1:If H and K are disjoint normal subgroup of G such that |G|=|HK| than |G|=H (multipcation sign)K. Here multication sign indicate the direct product of groups
Ques2:(R\{0},.)≤(R,+)
Ans:since R\0 is a group wrt to multiplication and R+ is group wrt to addition the identity differ so it is false because a subgroup should have same identity

Please tell if i am going right

Ques3:For any two set H and K,H\K=H(intersection sign)(HŲK)but K is compliment in this bracket

—The preceding unsigned comment was added by 59.95.246.25 (talk) 15:43, 23 March 2007 (UTC).[reply]

I'm finding it hard to decipher your questions. For instance, in question one, there isn't really even a conclusion, and I assume that by "disjoint", you mean "intersect to the trivial subgroup", since any two subgroups necessarily share at least one element.

For question two, I think that (R\{0},.) is the group of real numbers (without 0) under multiplication, and (R, +) is the group of real numbers under addition. If this is the case, the former cannot be a subgroup of the latter because the operation isn't even the same. Perhaps you mean "isomorphic to" instead of "is a subgroup of".

For question three, again, if "\" means "set minus", then just use De Morgan's Laws to get the answer. –King Bee ^{(τ • γ)} 19:15, 23 March 2007 (UTC)[reply]

You're asked to determine (and presumably prove) whether the statement is true, regarding that first question.

What is the semantics of the relation "(multipcation sign)"? --Lambiam ^Talk 00:37, 24 March 2007 (UTC)[reply]

I think the OP means the direct product of groups; I thought that would be obvious, unless you're asking some form of pedagogical question to the OP, in which I think you should adjust your indenting :)

(please sign your posts with ~~~~) Hello. For Question 1, a sketch is as follows: since H and K are normal and disjoint (except for the identity), you can show the set of elements of the form h.k, for h in H and k in K is a subgroup isomorphic to H x K. Then you use the fact about their sizes (assuming they are finite) to see it is all of G.

For question 2, the above argument does not show one isn't a subgroup of the other, as there could be a morphism sending 1 to 0. The problem is that in (R\0,.) there is an element x not the identity such that x^2 is the identity. In (R,+), no such element exists (i.e. twice any non-zero number is not zero) so (R\0,.) can not be isomorphic to a subgroup. 131.111.8.96 07:31, 27 March 2007 (UTC)[reply]

Alternatively, use properties of cyclic groups: (R, +) is cyclic, and all subgroups of (R, +) are cyclic, but (R \ 0, *) is not. —The preceding unsigned comment was added by 149.135.55.229 (talk) 11:00, 27 March 2007 (UTC).[reply]

(R, +) is certainly not cyclic in any sense that I know. It contains many subgroups that are not cyclic as well: Q, for example, or Z[1/2] (the dyadic rationals). In any group, the subgroup <x> generated by a single element is cyclic; however, as 131.111.8.96 noted, in (R\0, *) the element -1 generates a cyclic subgroup of order 2. In (R, +) it is clear that any non-identity element generates an infinite cyclic group (isomorphic to Z) and so the former cannot be embedded in the latter.

The converse is true, however: (R, +) is isomorphic to a subgroup of (R\0, *). By examining the embedding you get, you can see that this trouble with -1 having multiplicative order 2 is, in a sense, the only obstruction to an isomorphism between these groups. (This can be formalized -- try it!) Tesseran 14:19, 27 March 2007 (UTC)[reply]

To 131.111.8.96 - "Subgroup of" demands that the operation be the same. If the operation is different, a subset cannot be a subgroup of its superset. If the OP meant "isomorphic to a subgroup of", then you have to say more. –King Bee ^{(τ • γ)} 17:26, 27 March 2007 (UTC)[reply]

Ugh, of course. I haven't thought about this stuff in a while, and I'm not thinking clearly enough when I do. I really need to keep this stuff up.

account

If you live in new york city, and you want to SELL a product to someone in utica new york, would you use the utica county sales tax of 9%, or that of new york city of 8.375%, or of new york state of 4%. Mab1700 18:08, 23 March 2007 (UTC)[reply]

If you're shipping the product to the customer, such that the customer takes possession of the product in Utica, then you unfortunately need to charge the Utica sales tax rate. (Your ST-100 forms will be messier than what one might hope for.) If the customer comes to your location in NYC, and takes possession of the product in NYC, then you charge the NYC tax rate. Supposedly Publication 750[1] tells you this, at the bottom of page 20. However, that paragraph uses the phrase "point of delivery", which to me anyway is unclear as to whether it refers to where the product is delivered from, or where the product is delivered to. The reason I know for sure that it means where the product is delivered to is because I own a small business in NY state, and I called the Dept. of Taxation and Finance to ask them essentially the same question when I was starting my business. If you want to call them and verify my answer, their sales tax information number is 1-800-698-2909.[2] MrRedact 21:39, 23 March 2007 (UTC)[reply]

And this is a mathematical question!!! How? 210.49.121.183 19:30, 23 March 2007 (UTC)[reply]

Accounting is a math-related discipline. StuRat 00:50, 24 March 2007 (UTC)[reply]

Still, this is a question from the legal side of accounting, and so should go at the Humanities desk. Neon Merlin 04:07, 26 March 2007 (UTC)[reply]

I would say legal questions belong in Misc., not Humanities, which is for arts, religion, and culture questions. And this question is close enough to math to go here, at least for me. StuRat 16:10, 26 March 2007 (UTC)[reply]

The question is not a math question at all. If you delete all occurrences of "of X%" in the question so that no numbers or any implied multiplications are mentioned, the question still makes sense and the answer remains the same. --64.236.170.228 15:54, 27 March 2007 (UTC)[reply]

It's a gray area. For example, would the question "Who was Einstein's first wife ?" belong on the Science Desk ? It's not strictly a science question, but it does relate to a scientist (two, in fact, since she was a scientist, as well), so many other people interested in science may be able to answer. Similarly, while this isn't strictly a math question, it does deal with an accounting issue, and accountants may read this board and be able to supply the answer. StuRat 16:06, 27 March 2007 (UTC)[reply]

Product rule

Is there any way of expanding the product rule to work for more than two functions of x? Algebra man 20:51, 23 March 2007 (UTC)[reply]

Sure, just try it. Write down [f(x)*g(x)]*h(x), and just do the product in the square brackets first. You'll probably see the pattern; if not, try expanding {[f*g]*h}*j, and see what it looks like with four functions. -GTBacchus^(talk) 20:58, 23 March 2007 (UTC)[reply]