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April 18

Determinant solving

Please solve the determinant at http://demonstrations.wolfram.com/FivePointsDetermineAConicSection/ 171.226.88.246 (talk) 12:19, 18 April 2012 (UTC)

That's a quadratic equation, so you can solve (say) for y in terms of x using the quadratic formula. Expect a mess though. You can get a simpler explicit parametrization by directly applying the Braikenridge–Maclaurin construction. Sławomir Biały (talk) 12:24, 18 April 2012 (UTC)

the graph is Compact

Suppose $X$ is compact and $f:X\rightarrow Y$ is a function. Prove that $\{(x,f(x))\in X\times Y:x\in X\}$ is compact if and oNly if $f$ is continuous. The metric on $X\times Y$ is $d_{X}(x_{1},x_{2})+d_{Y}(y_{1},y_{2})$ Widener (talk) 15:55, 18 April 2012 (UTC)
First of all, I believe it is true that a sequence $\{(x,y)_{n}\}$ converges iff the sequences $\{x_{n}\}$ and $\{y_{n}\}$ converge if you use this metric. Therefore proving "f continuous => graph compact" is easy: X being compact implies that any sequence $\{x_{n}\}$ has a convergent subsequence $\{x_{n_{j}}\}$, and since $f$ is continuous, the sequence $\{f(x_{n_{j}})\}$ converges and so too does the sequence $\{(x_{n_{j}},f(x_{n_{j}}))\}$, and all subsequences in the graph are of that form. Now, the converse. :B Widener (talk) 15:55, 18 April 2012 (UTC)

Did you mean to say that the graph of f is compact iff f is continuous? — Fly by Night (talk) 17:14, 18 April 2012 (UTC)

Yes, that is what I meant. Oops! Widener (talk) 03:19, 19 April 2012 (UTC)

I'm going to similar encouragement as before for abstraction. I haven't thought in terms of sequences in a long time. A good exercise: Let X and Y be topological spaces with Y Hausdorff, and the graph of a function f (the set of (x,f(x))) be denoted by G. Then G is closed in X×Y if and only if f is continuous.

If X is also Hausdorff (which is the case when you are assuming both spaces are metric spaces) you can replace closed with compact with the following strategy:

• Assuming f is continuous, the continuous image of a compact set is compact. Argue that this makes the graph compact.
• Assuming the graph is compact, a compact subspace of the Hausdorff space X×Y is closed, hence the graph is closed

**Since the graph is closed, f is continuous, by the exercise.

Please excuse silly mistakes if I make any, I'm knocking rust off my topology. Eliminated one misstep already. Rschwieb (talk) 18:04, 18 April 2012 (UTC)

I don't think the exercise is right. For instance the function $f:\mathbb {R} \to \mathbb {R}$ given by

$f(x)={\begin{cases}0&x\leq 0\\1/x&x>0\end{cases}}$

has a closed graph but is discontinuous. (Note: the closed graph theorem is for linear functions.) Sławomir Biały (talk) 19:25, 18 April 2012 (UTC)

Doesn't the statement assume that the domain is compact? If I recall, the real line is not compact. — Fly by Night (talk) 21:39, 18 April 2012 (UTC)

I was replying to Rs post, which claimed that a fn is continuous iff its graph is closed. Sławomir Biały (talk) 22:13, 18 April 2012 (UTC)

Ah! Fair enough. Sorry. — Fly by Night (talk) 23:55, 18 April 2012 (UTC)

Thank Slawomir: that counterexample does look familiar, and now I see in my scribblings that I relied on the projected image of a closed set being closed. But then there is another question: isn't the example restricted to [0,1] a counterexample to the OP's problem about a compact graph implying the map is continuous?! Haha, I bet the funky distance function stipulated at the beginning is going to save the day, and I'm going to end up eating my words about abstraction :) Rschwieb (talk) 00:34, 19 April 2012 (UTC)

Yeah I think I will eat my words. Using that metric, it's very easy to check that f preserves convergent sequences. Rschwieb (talk) 02:03, 19 April 2012 (UTC)

I don't see how. Is my approach so far correct? Widener (talk) 13:19, 19 April 2012 (UTC)

Yes, I agree with everything you wrote in your first paragraph. Based on that, it looks like you're totally aware that continuity of f is equivalent to f preserving convergent sequences (by preserving I mean, if x_i converges to x, then f(x_i) converges to f(x) ). You used one direction of that successfully above, and now you can use it the other direction to show that if the graph G is compact, then f preserves all convergent sequences (hence is continuous). Sequences are more fun than I remember them, suddenly! Rschwieb (talk) 13:59, 19 April 2012 (UTC)

All I can derive is that all sequences $\{f(x_{n})\}$ have a convergent subsequence. Since the graph is compact, every sequence $\{(x_{n},f(x_{n})\}$ has a convergent subsequence $\{(x_{j_{n}},f(x_{j_{n}})\}$. It follows that $\{f(x_{j_{n}})\}$ converges. However, I see no reason it for it to converge to $f(\lim x_{j_{n}})$ or even if it did, that $\lim f(x_{n})=f(\lim x_{n})$ for all sequences in X. Widener (talk) 15:57, 19 April 2012 (UTC)

Write out what it means for $(x_{j_{n}},f(x_{j_{n}}))$ to converge to (a, f(a)) in that metric on X×Y. Not only does it say $f(x_{j_{n}})\rightarrow f(a)$, but it also says $x_{j_{n}}\rightarrow a$. Rschwieb (talk) 16:19, 19 April 2012 (UTC)

But there's no reason to expect $(x_{j_{n}},f(x_{j_{n}}))$ to converge to (a, f(a)) for some a. Could it not converge to (a,b) where $b\neq f(a)$? Widener (talk) 05:31, 20 April 2012 (UTC)

No, it could not converge to a point outside the graph: the graph is closed! Rschwieb (talk) 13:34, 20 April 2012 (UTC)

Oh of course; compact implies closed doesn't it. So every sequence $\{(x_{n},f(x_{n}))\}$ has a subsequence $\{(x_{j_{n}},f(x_{j_{n}}))\}$ which converges to $\{(a,f(a))\}$. But does this mean every sequence $\{(x_{n},f(x_{n}))\}$ converges? It's not immediately obvious - does it have something to do with the fact that $X$ is also compact? Widener (talk) 14:14, 20 April 2012 (UTC)

I had the following contradiction to prove uniqueness, but now that I look at it, we might want to back up and do the whole proof via this contradiction:

• Suppose the sequence x_i converges to x but lim f(x_i) does not converge to f(x).
• Deduce the existence of an ε>0 and a subsequence of the x_i, call them y_i, with images more than ε from f(x).
• Use compactness of the graph to find a subsequence of the y_i, call it z_i, such that (z_i,f(z_i)) converges to (a,f(a)).
• Use the given metric to conclude (as we did before) that x=a, so that f(z_i) converges to f(x).
• Derive a contradiction since no subsequence of the f(y_i) approaches f(x).

As I look at it, I don't immediately see that the compactness of X is needed in this direction. You'll remember though that it was necessary for the other direction, though. Rschwieb (talk) 18:09, 20 April 2012 (UTC)

I got this by the way. Thanks for your help . Widener (talk) 08:43, 22 April 2012 (UTC)

I should be thanking you for helping me knock rust off my analysis/topology. It was a fun problem! Rschwieb (talk) 15:09, 23 April 2012 (UTC)

April 19

Mercator and stereographic projection

Hi. If you consider the Mercator projection as a C-valued function f of lattitude $\theta$ and longitude $\phi$, and rotate the map 90 degrees to the right so the meridian $\phi =0$ is the real axis, you get

$f(\phi ,\theta )=\ln(\sec \theta +\tan \theta )-i\phi$.

Similarly if you consider the stereographic projection from the North pole as a C-valued function g of lattitude $\theta$ and longitude $\phi$, you get that

$g(\phi ,\theta )=(\sec \theta +\tan \theta )\cos \phi +i(\sec \theta +\tan \theta )\sin \phi =(\sec \theta +\tan \theta )\exp(i\phi )$.

Now consider $F=g\circ f^{-1}$. Were F = exp (which is what I'm supposed to prove), you would expect that $F\circ f=g$, but in fact $F_{0}\circ f=g$ for $F_{0}(z)=\exp({\bar {z}})$. Can someone give me a hint about where I've gone wrong in my definitions of f and g? Thanks for any help. —Anonymous Dissident^Talk 04:52, 19 April 2012 (UTC)

there is a kind of fuction named L-fuction,what meanings the "L"?

p-adic number what meanings the "adic",there is not in EN dictionary — Preceding unsigned comment added by Cjsh716 (talk • contribs) 07:26, 19 April 2012 (UTC)

"-adic" comes from the common ending in words like "dyadic" - pertaining to two, twofold; "triadic" - pertaining to three, threefold; "pentadic" - pertaining to five, fivefold etc. In a similar way, "n-ary" is used as a generalisation of "binary", "trinary" etc. Gandalf61 (talk) 10:31, 19 April 2012 (UTC)

I'm also curious where the "L" in L-function comes from. Rckrone (talk) 13:55, 20 April 2012 (UTC)

I believe it is an allusion by Dirichlet to the Legendre symbol. That seems to be pure OR on my part as I can't find anything about it. Dmcq (talk) 12:07, 22 April 2012 (UTC)

Interpretation of Beta hyperparameters

Our article Conjugate prior states that for Bernoulli and Binomial distributions the alpha and beta hyperparameters of the Beta conjugate prior may be interpreted as "a-1 successes, b-1 failures". I don't understand this however. Previously I have always interpreted the parameters as corresponding to "a successes, b failures", and I can't work out the motivation for this alternative interpretation. Certainly in the Bernoulli case, the 'most likely' value of the probability parameter is a / (a + b) (i.e. the ratio of successes to trials), and this seems to favour the latter interpretation? Thanks --Iae (talk) 20:22, 19 April 2012 (UTC)

Also I have had practical applications before where I define one or both of my prior Beta hyperparameters to be less than one, to show that my prior knowledge should have very low weight. The interpretation of this as meaning a negative number of successes (or failures) seems unnatural. --Iae (talk) 20:25, 19 April 2012 (UTC)

The expression ${\binom {n}{i}}p^{i}(1-p)^{n-i}$ is the probability that i is the number of successes in n Bernoulli trials each having the success probability p. The same expression is the likelihood that the success probability has the value p when n trials has produced i successes. This defines a beta distribution with mean value ${\frac {i+1}{n+2}}$ and standard deviation ${\sqrt {\frac {(i+1)(n-i+1)}{(n+2)^{2}(n+3)}}}$. Before any observation we have n=i=0 and the prior distribution is uniform on 0≤p≤1. Then p=0.5(3). (concise notation). Bo Jacoby (talk) 10:07, 22 April 2012 (UTC).

If we assume that the absolute prior is uniform (which is maximum entropy), then a,b correspond to a-1/b-1 success/failure (since a=1, b=1 is uniform). Alternatively, we could say that even a uniform distribution represents some information - that we have supposedly seen a success and a failure, and thus understand that either is possible and we can't guess the exact probability - and then it would correspond to a/b success/failure. -- Meni Rosenfeld (talk) 20:22, 23 April 2012 (UTC)

I don't understand why "even a uniform distribution represents some information". Before any observation is made we only know how to distinguish success from failure, but that does not give us any information about the success probability. p_deduction=i/n is the probability that a random one of the n already observed trials was a success. p_induction=(i+1)/(n+2) is the probability that the next trial will be a success. Bo Jacoby (talk) 12:07, 24 April 2012 (UTC).

square footage

How do I figure the square footage of a room? — Preceding unsigned comment added by 71.171.25.31 (talk) 20:59, 19 April 2012 (UTC)

I believe that would be a reference to the square footage, or the area, of the floor of the room? Would that be correct? Bus stop (talk) 21:07, 19 April 2012 (UTC)

If the room can be approximated by a polygon, you can use the generalized polygon area formulae for any room shape. Nimur (talk) 21:54, 19 April 2012 (UTC)

As most rooms are rectangular, multiplying the length by the width, both measured in feet, normally does the trick. --Tagishsimon (talk) 22:54, 19 April 2012 (UTC)

And some are a number of rectangles. You can add and subtract individual rectangles to get the total:

□ □ □
□ □ □
□ □ □ □
□ □ □ □
□ □ □ □

If each box is 3.5 feet square, that would make the room 5×3.5, or 17.5 feet long, and 4×3.5, or 14 feet wide. This gives us 17.5×14 or 245 square feet total. From this we must subtract the missing rectangle, which is 1 box by 2 boxes, or 3.5×7 feet, for a total of 24.5 square feet. 245 - 24.5 = 220.5 square feet.

If all else fails, you can draw the room on graph paper, where one square equals one square foot, and then total up the squares. StuRat (talk) 06:04, 22 April 2012 (UTC)

April 20

Limit question

Here's a question for our able analysts. No differentiation is allowed, so no chain rule, no l'Hôpital's rule, and no Taylor series please.

$\lim _{h\to 0}\left({\frac {\ln(x+h)-\ln x}{h}}\right)=\lim _{h\to 0}\left({\frac {1}{h}}\cdot \ln \left({\frac {x+h}{x}}\right)\right)=\ ?$

We all know that the answer is 1/x, but what are the absolute bare minimum assumptions needed to prove it? Assume that

$\ln x=\int _{1}^{x}{\frac {\operatorname {d} \!\alpha }{\alpha }}\,.$

This implies that ln(a) + ln(b) = ln(ab), ln(a) – ln(b) = ln(a/b) and ln(a^k) = k⋅ln(a). Is it possible to evaluate the limits using these and only these properties? — Fly by Night (talk) 01:31, 20 April 2012 (UTC)

Please excuse me for modifying your post by swapping that h for an x. I guess "no differentiation" rules out the fundamental theorem of calculus.

Do you still allow the limit (1+x)^1/x definition of e? If so then

$\lim _{h\to 0}\left({\frac {1}{h}}\cdot \ln \left({\frac {x+h}{x}}\right)\right)=\ln \left(\lim _{h\to 0}\left({\frac {x+h}{x}}\right)^{1/h}\right)=\ln \left(\lim _{k\to 0}\left(1+k\right)^{1/k}\right)^{1/x}=\ln(e)^{1/x}=1/x$

Sorry in advance if this is not what you're looking for. I thought by the time I asked and you confirmed you'd like to see it I would have had to type it anyway. Rschwieb (talk) 01:58, 20 April 2012 (UTC)

I've got a very ugly way to do this, find f and g functions of h and x so both limit to 1 / x and for all x ln's difference quotient is bound above and below by them for all h in a neighborhood of 0. I'd use f(x,h) = ln (exp(1/x) (x + h)/(x - h)) for the lower and g(x,h) = (1/h) [ln (x / x) + (h / x)] = 1 / x for the upper bound (This should follow from showing that exp(b) / b limits to infinity as b goes to 0). I worked this out at work on a post it note, so I apologize if there is any error;even if so, the general idea is sound. PS, I know you didn't introduce exp, however, there is x and y so ln x < 1 < ln y and ln is continous, so there is some z with ln z = 1; it follows that ln (z ** a) = a, so just define exp(x) as z ** x that z so ln z = 1, you don't need any properties specific to e, just that it exists and is > 1. Phoenixia1177 (talk) 04:36, 20 April 2012 (UTC)

If you're looking for a non-rigorous, but more intuitive argument, notice that your difference quotient can be written ${\frac {1}{h}}\cdot \int _{x}^{x+h}{\frac {\operatorname {d} \!\alpha }{\alpha }}\,.$, which is the average of 1 / t over the interval [x, x + h]. It makes sense that when h is 0 that this average would be 1 / x since the interval, at that point, is just the point x; the average of a single value is just that value. Like I said, not rigorous, but I think it makes more sense than weird limit arguments; not sure if that's useful to you or not, though :-) Phoenixia1177 (talk) 04:50, 20 April 2012 (UTC)

Sorry, to blather on, but I was thinking that you could make the above a real argument if you really wanted to. You could show that for the integral of step function so the interval divides up evenly for each value that the integral form gives the actual average. Then, from there, argue that as the interval shrinks to a point you can bound it above and below by a sequence of step functions that must limit to x; or some such. Really, it'd just be bounding the limit like above mixed with a real round about use of the definition of the integral as the limit of a sum over step functions; not pretty, but it could make the above a legitimate proof. Sorry if that is not easy to read, by the way, I don't get time to use wiki out of work, but am usually doing ten things at once at work...:-)Phoenixia1177 (talk) 04:57, 20 April 2012 (UTC)

To me, this seems to be the right approach given the constraints. For what it's worth, it's the mean value theorem for integrals (which some might say is "more fundamental" than the fundamental theorem—it is at least more intuitive). Sławomir Biały (talk) 12:13, 20 April 2012 (UTC)

Proof of order of operations

How can we show using the field axioms of say, the real numbers that operations should be performed in a certain order (i.e. multiplication before addition)? I tried manipulating but it didn't really give me any insight...

$1+ab=a({\frac {1}{a}}+b)=a({\frac {1}{a}}+{\frac {ab}{a}})=a({\frac {1+ab}{a}})$

We can't. It's just a matter of convention whether $a+b\cdot c=a+(b\cdot c)$ or $a+b\cdot c=(a+b)\cdot c.$ It is not 'provable'. --CiaPan (talk) 12:54, 20 April 2012 (UTC)

Gotcha, but I'm just surprised it's not a consequence of the axioms. I mean, what gives multiplication the right over addition, for instance? Suppose if write it out like $a+b\cdot 3=(a+b)+(a+b)+(a+b)$, instead of $a+b+b+b$? Why? — Preceding unsigned comment added by 77.100.75.68 (talk) 13:13, 20 April 2012 (UTC)

It's just the way we decided to write things. We could have all decided that all statements must be fully parenthesized to be well formed, but that would be more writing so mathematicians decided to have the conventions we use. They are merely conventions. Consider this: if we had a completely different notation would we need different axioms even though we are expressing the same idea? Or consider this statement a b 3 * +, which is a+b*3 in postfix. RJFJR (talk) 13:29, 20 April 2012 (UTC)

You might think of it as being a consequence of "notational" (or syntactic) axioms: the rules that specify (in a context-free grammar, say) how to interpret mathematical notation. Notation is just a convenience; Frege's Begriffsschrift does without notation (in some sense), by simply drawing trees. Similarly, programmers in Lisp-like languages, like RJFJR suggests, fully parenthesize their expressions, so they say (+ a (* b c)). But mathematicians have found it convenient to come up with a notation that is terser in the common case. Paul (Stansifer) 05:31, 22 April 2012 (UTC)

Recursively enumerable theory

There are theories that are complete, consistent, and include elementary arithmetic, but no such theory can be effective. (quoted) Could you give any examples? --151.75.56.185 (talk) 15:27, 20 April 2012 (UTC)

True arithmetic. Widener (talk) 17:13, 20 April 2012 (UTC)

April 21

Integration - Calculus

As part of an assignment, I've derived this expression (dx/dt=-0.0004286x) by following my course notes. Now I'm meant to rearrange and integrate to show that it takes 1.49hours for x to go from 0.2 to 0.02. The coefficient -0.0004286 was arrived at using time in units of seconds. The notes said to remember that ʃdx/d=ln x. I haven't done calc for years, so I'd appreciate a little bit more help than that just to get me started. 203.27.72.5 (talk) 07:58, 21 April 2012 (UTC)

See separation of variables. Widener (talk) 10:07, 21 April 2012 (UTC)

I think the notes would have said $\int {\frac {dx}{x}}=\ln x+C$ Widener (talk) 10:09, 21 April 2012 (UTC)

No, they definitely leave out the "+ C". Maybe my lecturer is no better at calc than I am :P. 203.27.72.5 (talk) 20:37, 21 April 2012 (UTC)

Maybe they said $\int _{1}^{x}{\frac {dt}{t}}=\ln x$ then. Widener (talk) 06:34, 22 April 2012 (UTC)

Ok, I read the article on separation of variables. I followed it as best I could and I think I'm still wrong. Here's what I did: $Lety=x$
${\frac {dy}{dt}}=-0.0004286y$
${\frac {dy}{y}}=-0.0004286dt$
$\int {\frac {1}{y}}dy=\int -0.0004286dt$
$\ln |y|=-0.0004286t$
$e^{-0.0004286t}=y$
If that's wrong, what am I doing wrong? If it's right, how do I use that to show that it takes 1.49 hours for x to go from 0.2 to 0.02? 203.27.72.5 (talk) 21:33, 21 April 2012 (UTC)

Oh, wait...it is right and I worked the rest out for myself! Thanks so much for your pointers! 203.27.72.5 (talk) 21:38, 21 April 2012 (UTC)

Resolved

~ Need Solid Judgement On Mathematical Tools ~

Which of these is the best in this category of Mathematical Tools? The best would have

• comprehensive range of tools
• good usability of website, and UI of tools that anyone are able to use them

Calculators

• stattrek.com/online-calculator/
• easycalculation.com/
• wessa.net/
• vassarstats.net/
• alcula.com/calculators
• danielsoper.com/statcalc3/default.aspx

Feel free to suggest the best single math tools site if you already know from experience.

Test example:

If c = a/b

but you do not know a

The calculator will output a. Does this with c and b being unknown also. Same formula.

Thingstofollow (talk) 11:08, 21 April 2012 (UTC)

There is likely to be no "best single math tools site"- it'll depend on what exactly you want to do. I use http://www.wolframalpha.com. Try it on your example: http://www.wolframalpha.com/input/?i=c%3Da%2Fb and I think it does what you want. (a=bc) Staecker (talk) 12:54, 21 April 2012 (UTC)

Matrix Decompsotion

I wish to decompose the matrix ${\begin{bmatrix}1&0\\0&-1\end{bmatrix}}\,$ into some product of the matrices ${\begin{bmatrix}0&-1\\1&0\end{bmatrix}}\,$ and ${\begin{bmatrix}1&1\\0&1\end{bmatrix}}\,$, as well as their respective inverses. How might I go about this? Hints preferable to answers. Thanks. meromorphic [talk to me] 16:53, 21 April 2012 (UTC)

On second thoughts, I think we need -I in there as well, else the negative determinant cannot be achieved. meromorphic [talk to me] 19:34, 21 April 2012 (UTC)

I don't think this question has an affirmative answer given that the determinant of my matrix is negative and the determinant of the other two is positive. Thanks anyway. meromorphic [talk to me] 19:58, 21 April 2012 (UTC)

Most popular math (grades 1-9) materials?

What math textbooks/ebooks/instructional tools, for grades 1-9, are the most popular or common being used in classrooms today? --Agentundertables (talk) 18:33, 21 April 2012 (UTC)

The ones in China or India I'd guess ;-) Dmcq (talk) 10:51, 22 April 2012 (UTC)

April 22

Natural logarithms, limits and sequences of functions

Hello, I was inspired by a previous question about reconciling the integral definition of the natural logarithm

$\ln(x):=\int _{1}^{x}{\frac {\mathrm {d} t}{t}},$

with the limit expression for Euler's number

$e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.$

I thought I might try and do this by considering the sequence of functions

$f_{n}(x):=\int _{1}^{x}t^{{\frac {1}{n}}-1}\mathrm {d} t.$

If we further define $a_{n}$ as the solution to $f_{n}(a_{n})=1$, we can show that

$a_{n}=\left(1+{\frac {1}{n}}\right)^{n}.$

Now I'd like to show that in the limit $n\to \infty$ that $f_{n}(x)\to \ln(x)$ and therefore, by defining Euler's number by $\ln(e)=1$, that

$e=\lim _{n\to \infty }a_{n}$.

My problem is in making these statements rigorous. I tried to show that the sequence of functions converges uniformly to $\ln(x)$, but I don't think it does. Is it possible to say that this sequence converges in some other way? If it does converge, is it possible to extend the convergence of a function sequence to some convergence of its roots? Can anybody help me complete this argument? Thank you in advance. 129.234.53.19 (talk) 12:16, 22 April 2012 (UTC)

$f_{n}$ doesn't converge uniformly to ln on $(0,\infty )$, but it does on $[1,3]$, which is all you need once you've established that $a_{n}$ and $e$ are all in [1,3], which is technical (and can be made easier by taking a looser bound). ln is also uniformly continuous in this domain. From the fact that ln is increasing (and thus one-to-one) it follows that to show that $e=\lim _{n\to \infty }a_{n}$ it is sufficient to show that $1=\ln(e)=\ln(\lim _{n\to \infty }a_{n})$. I think it is true (and easy to prove) that if $f_{n}\to f$ uniformly, $f$ is uniformly continuous, and $a_{n}\to a$ then $f(a)=\lim _{n\to \infty }f_{n}(a_{n})$. Apply this and you're done. -- Meni Rosenfeld (talk) 20:13, 23 April 2012 (UTC)

Thank you, Meni. I'll have to wait until I have time to work through it, but your comments look like they'll carry me most of the way. 129.234.53.19 (talk) 12:58, 25 April 2012 (UTC)

real integrals by residue theorem

I am stuck wid this prblm as my answer isnt matching wid d book. integration of (sinsqarex divided by 5+4cosx) from 0 to 2pi. sorry 4 poor scripting. im using the substitution 4 sinx nd then sqaring it, i also tried sinx= 1-cos2x/2 but in both cases im not getting the correct answer. im getting poles to be 0 nd -1/2 and the numerator to be a biquadratic in z. in one solution one used, 1-cos2x/2= 1-zsqare/2 and i am still not getting past it.please highlight where im missin.59.165.108.89 (talk) 12:24, 22 April 2012 (UTC)

I just want to note that writing in this twitter style means making it easier for you to write while making it harder for other people to read. Why should we work to answer the question when you're not willing to do the minimal work of asking the question properly? Regards, Looie496 (talk) 18:30, 22 April 2012 (UTC)

lolz n00b Der iz pol @ $x=\pi +i\log 2$!!1! Clclate by using rect contour arnd pol, then snd top 2 &infinity;. Sławomir Biały (talk) 20:20, 22 April 2012 (UTC)

LMAO — Fly by Night (talk) 22:19, 23 April 2012 (UTC)

Polynomials in Zp

Hi,

Are there criteria for determining a given polynomial with no constant term is a zero polynomial in Zp?

Are there criteria for determining a given polynomial with a constant term is reducible in Zp (not irreducible)?

I'm not necessarily concerned with all cases. I'm more interested in special cases. Nkot (talk) 17:51, 22 April 2012 (UTC)

Are you interested in polynomials in one variable or several variables? By "a zero polynomial" do you mean the actual zero element of the polynomial ring, or maybe a polynomial which evaluates to zero at every point? Rckrone (talk) 19:15, 22 April 2012 (UTC)

Initially, I was envisioning one variable, but a polynomial with more variables might also be interesting if it had special properties relating to above. Also, by "zero polynomial," I did mean a polynomial that evaluates to zero at every point. Nkot (talk) 19:25, 22 April 2012 (UTC)

For question 1, Euler's theorem tells us a^p = a for all a in Z_p, so if you have a polynomial f with terms of degree p or higher, you can reduce those terms to get an equivalent (as a function) polynomial with degree less than p. The only polynomial with degree less than p which always evaluates to zero is the zero polynomial. (You can show that polynomials with degree less than p are in bijection with the functions Z_p → Z_p.) I think this works the same way for multiple variables.

I don't know a good answer for the second question. Rckrone (talk) 20:56, 22 April 2012 (UTC)

An algorithm similar to the Sieve of Eratosthenes can be used to find the irreducible polynomials in $\mathbb {Z} _{p}[x]$. Since there are finitely many polynomials below a given degree, if you want to determine whether $f\in \mathbb {Z} _{p}[x]$ is irreducible, you can just try all possible factors below, and you can optimize such an algorithm to only try the candidates which must show up in a factorization of $f$.

I don't know if there's a better way. --COVIZAPIBETEFOKY (talk) 02:02, 23 April 2012 (UTC)

See our articles on factorization of polynomials over a finite field and irreducibility tests, Berlekamp's algorithm, and Cantor–Zassenhaus algorithm. Gandalf61 (talk) 08:50, 23 April 2012 (UTC)

Name/Identity of trig conversion function?

Where

$\tan(V)={\frac {D}{N}}\tan(Q),\,\!$,

if

${\mbox{F}}(Q)={\sqrt {(D\sin(Q))^{2}+(N\cos(Q))^{2}}}\,\!$,

then

$\sin(V)={\frac {D\sin(Q)}{{\mbox{F}}(Q)}};\qquad \cos(V)={\frac {N\cos(Q)}{{\mbox{F}}(Q)}};\,\!$.

Does ${\mbox{F}}(Q)\,\!$ have a particular name/identity? ~Kaimbridge~ (talk) 18:03, 22 April 2012 (UTC)

April 23

Pi

What would you give a proper definition of Pi?How do every circle's C is always about just more than three times of its D?

Pi is defined as the ratio of a circle's circumference to its diameter. Are you asking for a proof that the ratio is constant for all circles? Or are you looking for an algebraic definition? KyuubiSeal (talk) 01:45, 23 April 2012 (UTC)

(ec) You just gave the definition, the ratio of a circle's circumference to it's diameter. Why it's an irrational number is a question for philosophers, not mathematicians. StuRat (talk) 01:46, 23 April 2012 (UTC)

Modulus

Is the set of all ordered k-tuples mod n Z^k/n or (Z/n)^k? --108.206.4.199 (talk) 03:04, 23 April 2012 (UTC)

I think the second one is more correct or at least more common, but either one would probably get the point across. If you're considering (Z/n)^k as a group, it's a quotient group of Z^k, but not by n since n isn't in Z^k. But (Z/n)^k is also a Z-module and is the quotient module of Z^k by nZ^k so Z^k/n works in that sense. Rckrone (talk) 16:15, 23 April 2012 (UTC)

integrals

Using the concept of differentials, is there a meaning to things like $\int x^{2}d(cosx)$? --85.250.73.51 (talk) 10:03, 23 April 2012 (UTC)

$\int x^{2}d(\cos x)=-\int x^{2}\sin xdx$

What is the problem? Bo Jacoby (talk) 10:11, 23 April 2012 (UTC).

Things like this, with nontrivial stuff inside the d, are basically what you are doing when you do Integration by substitution. Staecker (talk) 10:29, 23 April 2012 (UTC)

Indeed ...

$u=\cos x$

$\Rightarrow {\frac {{\text{d}}u}{{\text{d}}x}}=-\sin x$

$\Rightarrow {\text{d}}u=-\sin x\,{\text{d}}x$

(pause while pure mathematicians in the room are revived with smelling salts) Gandalf61 (talk) 11:09, 23 April 2012 (UTC)

We're not fainting, we're napping through the boring parts! Rschwieb (talk) 15:07, 23 April 2012 (UTC)

Minor details

Let $f$ be a uniformly continuous real valued function defined on a bounded set $E\subset \mathbb {R}$. Prove that $f$ is bounded on $E$.

I think I understand intuitively why this is true, and this is the proof that I have written down: Fix some $a\in E$ and some $\epsilon >0$. Since $f$ is continuous at $a$, there exists some $\delta$ such that $f(x)\in B_{\epsilon }(f(a))$ for all $x$ in the set of points $\{x\in E:d(x,a)<\delta \}$. Note therefore that $f$ is bounded on this set of points. Now consider an increasing sequence $(x_{n})_{n=1}^{m}$ of points in $\mathbb {R}$ such that $\forall x\in E:x_{0}\leq x\leq x_{m}$ and that $d(x_{j},x_{j+1})<\delta$ for $j=1,...,m-1$. Since $f$ is uniformly continuous, such a sequence exists and it follows that each on each interval $[x_{j},x_{j+1}]$, $f([x_{j},x_{j+1}])\subset B_{\epsilon }(f(x_{j}))$. Therefore the function is bounded on its whole domain.

I think I may be making a few unjustified assertions which intuitively I know are true but that I have not formally proven, such as the fact that a such a sequence exists if $f$ is uniformly continuous, and that the fact that E is bounded and the function is bounded on each subinterval implies that the function is bounded on the whole of E. Widener (talk) 17:25, 23 April 2012 (UTC)

It's the right idea but I think you need to frame it a bit differently. If you want to prove the function is bounded, you need to prove that no sequence of points x₁,x₂,... exists such that f(x_n) goes to infinity. So assume that there is such a sequence. Since E is contained in some compact set C, the sequence has a convergent subsequence (though the limit x is not necessarily in E). Eventually these points are all very close to one another yet some pairs have |f(x_i)-f(x_j)| very large. Rckrone (talk) 23:18, 23 April 2012 (UTC)

It's a long time since I did this kind of stuff, but doesn't the following argument do it? Uniform continuity means that for all ε > 0 there exists a δ such that, for all x₁ and x₂ in E, |x₁ - x₂| < δ => |f(x₁) - f(x₂)| < ε. Take ε = 1 and subdivide E into intervals of width δ/2 - as E is bounded there are finitely many of these, say N. Then the value taken by f can't vary by more than Nε = N over E, and so f is bounded. Actually I think this is pretty much what Widener was saying, but in a simplified form. AndrewWTaylor (talk) 11:54, 24 April 2012 (UTC)

The problem is that E may have gaps. If E has a gap in it larger than δ, the values on either side can differ by an arbitrarily large amount, so the function can vary more than Nε. It's probably possible to patch this up by arguing that there can't be more than a finite number of these gaps, and then showing that the argument really does work on each piece with no large gaps. But this seems pretty messy already. Rckrone (talk) 12:39, 24 April 2012 (UTC)

@Widener: Your original post kind of reflects a trick I always keep in mind for uniformly continuous functions. The trick the definition permits you is: given an epsilon, you are guaranteed there is a delta such that the whole graph can be caught in nonoverlapping boxes of dimensions 2ε tall by δ wide (any positive number less than 2δ would be OK for a width, too). Thus if E is bounded, you can cover it with finitely many nonoverlapping closed intervals of width δ, and determine which of the finitely many boxes goes highest/lowest to choose your bounds. Rschwieb (talk) 14:01, 24 April 2012 (UTC)

How to find a cubic root of a number in regular calculator models?

Max Viwe | Wanna chat with me? 19:32, 23 April 2012 (UTC)

Am I correct in assuming that by a "regular" calculator you mean something along the lines of a four-function one without an exponentiation button? --Kinu ^t/_c 19:35, 23 April 2012 (UTC)

Ironically, this is trivial to do using a slide rule, and quite difficult to do with a 4 function calculator. To calculate an arbitrary root, you need a calculator that can perform exponentiation; else, you will need to do some work on your own, (possibly with the aid of the 4-function calculator to do the arithmetic for you). There are numerous iterative methods to estimate roots; there are several logarithmic identities that you can use to simplify the math; but there's no "quick" solution. In the general case, you are solving the cubic function x³ - k = 0, and you can use any numerical or algebraic technique you like to solve for x. Nimur (talk) 19:40, 23 April 2012 (UTC)

You can do this easily enough with Newton's method. -- Meni Rosenfeld (talk) 19:46, 23 April 2012 (UTC)

Were I hypothetically to be stranded on a desert island with just a 4-function calculator, and I needed to calculate the cube-root of k (only one time), I would run Newton's method by hand to solve for x^3 - k = 0, . Were I to require calculation of many cubic roots, I would use the four-function calculator to help me write out a precomputed table of logarithms so that when its battery eventually died, I'd still be able to do fast and accurate arithmetic to a high number of decimal places. Nimur (talk) 19:46, 23 April 2012 (UTC)

If you were stranded with "just a ... calculator", what would you use for paper or a writing implement? -- ♬ Jack of Oz ♬ ^{[your turn]} 19:52, 23 April 2012 (UTC)

Obviously, I would create clay tablets and use a suitable cuneiform numeral glyph system. Though I would not be able to remember Babylonian numerals off the top of my head, I would at least know that they invented a system that was easy to carve into semisoft clay, unlike our modern arabic numerals, which are better-suited for figuring on paper, rather than carving into sand or mud. When you are stranded on a desert island, you have to make effective use of your limited natural resources. Nimur (talk) 23:23, 23 April 2012 (UTC)

(slaps forehead) Well, of course! Cuneiform glyphs, what else. Sorry for the dumb question. -- ♬ Jack of Oz ♬ ^{[your turn]} 01:48, 24 April 2012 (UTC)

Jack, you're a crack-up Benyoch^{...Don't panic! Don't panic!...} (talk) 08:22, 25 April 2012 (UTC)

Of course, this raises the question of how you would best go about calculating those log tables with only your four-function calculator. -- The Anome (talk) 11:44, 24 April 2012 (UTC)

The answer is simple, you make a PCuneiform Tablet. It will compute anything if you have a proper stylus. Benyoch^{...Don't panic! Don't panic!...} (talk) 08:22, 25 April 2012 (UTC)

Taylor series expansion of the exponential function. I won't pretend to assert that this is the best or fastest way to reduce logarithm generation to simple arithmetic; but it's the method I could remember and re-derive if stranded on a desert island. Nimur (talk) 18:11, 24 April 2012 (UTC)

That makes sense to me. And once you've got the coefficients computed you can use something like Briggs' finite difference method to compute the polynomial at a large number of evenly-spaced values to create the table. -- The Anome (talk) 22:36, 24 April 2012 (UTC)

Just by the way, the usual term in English is cube root (parallel with square root), not cubic root. A "cubic root" sounds to me like the root of a general cubic equation. --Trovatore (talk) 01:55, 24 April 2012 (UTC)

You can try and guess it, just guess a number with one significant figure and multiply it with itself twice. Adjust your guess using swedish rounding depending on the answer. For instance, find the cube root of 675654.

A first guess could be 100:

100³ = 1000000: over

50³ ≈ 100000: under

80³ ≈ 500000: under

90³ ≈ 700000: over

85³ ≈ 610000: under

88³ ≈ 680000: over

87³ ≈ 660000: under

87.5³ ≈ 670000: under

87.8³ ≈ 677000: over

87.7³ ≈ 675000: under

87.75³ ≈ 675700: over

87.73³ ≈ 675200: under

87.74³ ≈ 675400: under

87.745³ ≈ 675600: under

87.748³ ≈ 675630: under

87.749³ ≈ 675660: over

...

The answer lies between 87.748 and 87.749. Plasmic Physics (talk) 02:58, 24 April 2012 (UTC)

That's almost Newton's method, but you aren't making full use of all the information available to you. You know the error in each guess, which allows you to select a better next-guess. In your method, you're not taking advantage of this information, so you converge slower. If you had a calculator, you could just compute the gradient, and you'd get closer to the correct answer much faster. However, it's fair to say that guess-and-check might be faster if working with cuneiform wedges on sand tablets: calculating the gradient requires long division, so while guess-and-check requires more iterations, it may actually compute faster (more iterations, but less wall-clock time), if multiplication can be performed much more rapidly than division. This is called algorithmic analysis and makes up a large chunk of modern computer science. In fact, if we were designing a calculator to solve cube roots, we'd have to think long and hard about this exact tradeoff. Nimur (talk) 05:29, 24 April 2012 (UTC)

What do you mean by I have the error in each guess? Assuming I didn't calculate the exact answer using a scientific calculator, I cannot compare each guess with the answer to find the error in each guess. Plasmic Physics (talk) 08:32, 24 April 2012 (UTC)

Do you mean to compare the cube of the guess with the known cube? Plasmic Physics (talk) 08:34, 24 April 2012 (UTC)

Yes. Suppose our guess is x and the true cube root is x plus an error ε where ε is small compared to x. Then we know that

$k=(x+\epsilon )^{3}=x^{3}+3x^{2}\epsilon +3x\epsilon ^{2}+\epsilon ^{3}\approx x^{3}+3x^{2}\epsilon$

where in the last step we have used the fact that ε is small compared to x. So an approximate value for our error term ε is

$\epsilon _{\text{approx}}={\frac {k-x^{3}}{3x^{2}}}$

and we add this approximate error to our guess x and repeat. Worked example:

100^3 = 1000000

(675654 - 1000000) / (3*100^2) = -10.8115

Next guess = 100 - 10.8115 = 89.1885

89.1885^3 = 709457.8193

(675654 - 709457.8193) / (3*89.1885^2) = -1.4165

Next guess = 89.1885 - 1.4165 = 87.7720

87.7720^3 = 676188.8159

(675654 - 676188.8159) / (3*87.7720^2) = -0.0231

Next guess = 87.7720 - 0.0231 = 87.7489

etc. Gandalf61 (talk) 09:00, 24 April 2012 (UTC)

As always there is a bit in WIkipedia about this Cube root#Cube root on standard calculator Dmcq (talk) 13:38, 24 April 2012 (UTC)

Like we said, it depends on what functions his 'normal' calculator can do. If it had a power function, it could be as simple as using the power of a third to calculate the cubee root. Plasmic Physics (talk) 14:06, 24 April 2012 (UTC)

April 24

Primality question

I just gave a test to my students. One of the questions basically came down to "suppose x is prime and y is the least number greater than x such that: 1) x does not divide y; and 2) every number less than x which divides y also divides x. Show that y is prime." I realize that (1) is unnecessary, but this was intended to be an exercise in induction, not number theory. My question is, do we actually need that x is prime? I can't seem to come up with any examples showing that it's necessary, but I haven't looked out too far.--130.195.2.100 (talk) 02:28, 24 April 2012 (UTC)

After a little thought, my question can be rephrased as: is there a prime p and positive integers a and b with p greater than a and no primes between $p^{b}a$ and $p^{b+1}$?--130.195.2.100 (talk) 03:52, 24 April 2012 (UTC)

Since there is a prime between x and 2x, y must be less than 2x; so y can't have any factors larger than x; y prime if x is is an obvious consequence of this. Suppose $x=p(1)^{a(1)}p(2)^{a(2)}...p(n)^{a(n)}$, then $y=p(1)^{b(1)}p(2)^{b(2)}...p(n)^{b(n)}$ with b(k) > a(k) some k or y is prime. If the former, then $c=p(k)^{a(k)+1}|y$, obviously c does not divide x, so c > x must occur; thus, $p(k)>p(1)^{a(1)}...p(k-1)^{a(k-1)}p(k+1)^{a(k+1)}...p(n)^{a(n)}$ must hold, so (assume p is increasing) k = n. This gets us to where you are. So, we want to show that there is a, b and prime p with p > a and no primes in $(p^{b}a,p^{b+1})$ We can minimize the interval by taking a = p-1, which gives us $(p^{b}(p-1),p^{b+1})$. If you look at our article on Bertrand's Postulate, you'll see that in 2010 Pierre Dusart showed that there is always a prime in the interval $(x,(1+k^{-1})x)$ with $k(x)=25ln^{2}x$ for large enough x. We can write our interval in the form $(x,(1+(p-1)^{-1})x)$ In this case, $k(x)=k(p^{b}a)$, which diverges to infinity with b, so Dusart's interval will be smaller than ours for large enough p, b, and a. Obviously, this doesn't show that y must be prime, but it does show that the size of b's that give prime y is limited by p. (there may be a better way to go about this, but it's the best I've got, I'm not really a Number Theorist.) Phoenixia1177 (talk) 08:18, 24 April 2012 (UTC)

Taking $n=sqrt(p^{b}(p-1))$ and assuming Legendre's Conjecture would give a prime q so $p^{b}(p-1)=n^{2}<q<(n+1)^{2}<p^{b+1}$, which would show that y must be prime. Though, the conjecture remains unproven as of yet. Phoenixia1177 (talk) 08:31, 24 April 2012 (UTC)

To answer the original question, I believe if you remove (1) then you do need that x is prime. Consider, for instance, x=4 and y=8. GromXXVII (talk) 14:51, 24 April 2012 (UTC)

Removing 1.) doesn't change anything, there is a z satisfying 2.) that is less than 2x, so since x < y <= z, y is to small to be divisible by x. Also, x = 4 has y = 5, not 8; 1 | 5 and 1 | 4. I think you may have misread the question, it is not required that if z | x, then z | y; only that z | y and z < x, then z | x. Phoenixia1177 (talk) 15:42, 24 April 2012 (UTC)

Your result would also follow from there is always a prime in (kx, (k+1)x) for all x and k; here k = p - 1 and x = (p-1)p ** b, but that's not proven either. Sorry to just keep saying if this conjecture holds than yes, just trying to be informative.Phoenixia1177 (talk) 19:25, 24 April 2012 (UTC)

Hungarian Algorithm using matrices

Hi All,

The Hungarian algorithm for the assignment problem, as it is commonly described involves steps like crossing out rows and columns. Is there anyway in which this algorithm can be expressed in terms of canonical matrix operations.

.Gulielmus estavius (talk) 18:22, 24 April 2012 (UTC)

Interpolation in 2D

So let us say that I have 9 points on a 2D mesh, making four units squares with x-y coordinates like this

(0,2),(1,2),(2,2)
(0,1),(1,1),(2,1)
(0,0),(1,0),(2,0)

and I also know the value of a function f(x,y) at these nine points as well. Now I need to do some interpolation (because I want the interpolated function to go through those nine known values). I need to know values at a bunch of points within those four squares. I am looking for something analogous to quadratic splines in 1D. At these nine points I only know the function values. I don't know any of the derivatives/partials. So the interpolating function could be a piecewise function made up of

f3(x,y), f4(x,y)
f1(x,y), f2(x,y)

for each of the four squares and at the boundaries the function values and the gradient matches so that the x,y partials at least are continuous. Another problem is that I will be interpolating at a LOT of points and I won't know any of them in advance so I need to compute the interpolating function before hand (preprocessing) and then just evaluate it at the needed points as they come up. So my question is how does this work in dimensions more than one? Should I assume f1,f2,f3,f4 to be a general quadratic in two dimensions like $f1(x,y)=ax^{2}+bx^{2}+cxy+dx+ey+g$
because I only need the first partials continuous? But then what should be the constraints? I am a little confused about the partials? How do I write the constraints so that the partials are equal? Would I just have a gigantic matrix at the end, solve the system, and have the interpolating coefficients? Something like Bilinear interpolation is too slow because it needs to be done everytime and my program spends a HUGE amount of time in it. I want to try doing all the work before hand and then just evaluate a polynomial when I need to. If it helps, I will eventually be doing this on a 4D mesh f(x,y,z,t) with maybe a 100 points in each direction so 100000000 points total. I can guess the matrix will grow horrendously large and just to solve THAT system will take some ingenuity. On that thought, is there ny way interpolation can be done independently on a square (so that the resulting linear systems stay small) but still have the gradient continuous for example? I know constant and linear interpolation do this but I just need the continuity of the gradient as well. I also saw the article on Bicubic interpolation but it seems a but more than what I need. So to keep the system as small as possible, I am trying to see something like a biquadratic interpolation. Any help/comments/suggestions will be very helpful. Thank you! 184.96.137.243 (talk) 05:14, 25 April 2012 (UTC)

A polynomial function of the form $f_{1}(x,y)=Dxy+Cx+By+A$ has 4 adjustable coefficients. The 4 equations f₁(0,0)=A, f₁(1,0)=C+A, f₁(0,1)=B+A, f₁(1,1)=D+C+B+A are easily solved. Filling each square with such a function provides a continuous - but not differentiable - interpolation. This is not what you ask for. The Discrete Fourier transform#Multidimensional DFT provides an interpolating trigonometric polynomial. It is differentiable everywhere. And the computation is feasible due the to FFT algorithm. This may be what you are looking for. Bo Jacoby (talk) 07:43, 25 April 2012 (UTC).

The common way of doing this is by using some sort of Spline. Spline interpolation may be the most useful reference. There is a vast literature on the subject and some efficient algorithms for solving your problem.--Salix (talk): 08:23, 25 April 2012 (UTC)

April 25

Prime <--> Fibonacci

Is there any intelligable, meaningful result from the concurrence of any prime number/s and the matching number/s in the Fibonacci series? Benyoch^{...Don't panic! Don't panic!...} (talk) 06:55, 25 April 2012 (UTC)

Take a look at our Fibonacci prime article. Gandalf61 (talk) 07:53, 25 April 2012 (UTC)