Wikipedia:Reference desk/Mathematics

March 14

Messy integral

I was working on some probability stuff and reduced the problem to the integral

$\int _{0}^{\infty }{\frac {\exp(-x^{2})}{c+\exp(x)}}dx$

The c is a constant. Any ideas or thoughts on how to proceed? I tried mathematica's online integrator, and it finds no closed form indefinite integral. It looks convergent. Thanks in advance, --TeaDrinker 01:42, 14 March 2007 (UTC)

Are you supposed to find out whether it is convergent or divergent, or are you supposed to find a value?

because if you want to merely find out if it is convergent of divergent, you can use the comparison theorem and prove that a larger function is convergent, or a smaller function is divergent. In this case, exp(x) is always positive for all real numbers x (I'm assuming we are working in the reals here). So we can eliminate that from the denominator, and we have a larger function (which needs to be proven to be convergent):

$\int _{0}^{\infty }{\frac {\exp(-x^{2})}{c}}dx$

You can then take the 1/c factor outside of the integral:

${\frac {1}{c}}\int _{0}^{\infty }\exp(-x^{2})dx$

Now you need to prove that exp(-x^2) is convergent. You can either use the maclaurin series and integrate term by term, or you can use another application of the comparison theorem and find a bigger function. --Ķĩřβȳ♥ŤįɱéØ 03:51, 14 March 2007 (UTC)

If you need to find the exact value, then you can use integration by parts, using the fact that

$\int \exp(-x^{2})dx={\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x)+C$

where erf is the Error function. --Ķĩřβȳ♥ŤįɱéØ 03:59, 14 March 2007 (UTC)

Thanks for the replies; I was actually trying to find the exact value... Worst case is I can do a numerical integration, since as you point out, it is bounded by exp(-x^2) which convergest fairly quickly. The integration by parts seems like a great way to try, I'll give it a go. Thanks in any event. --TeaDrinker 20:15, 14 March 2007 (UTC)

I love math!

$\int _{0}^{\infty }{\frac {\exp(-x^{2})}{c+\exp(x)}}dx=\lim _{t\to \infty }\int _{0}^{t}{\frac {\exp(-x^{2})}{c+\exp(x)}}dx$

• $[u={\frac {1}{c+\exp(x)}}]$
• $[du=-(c+\exp(x))^{-2}(\exp(x))dx]\!$
• $[dv=\exp(-x^{2})dx]\!$
• $[v={\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x)]$

And of course, since: $\int u\,dv=uv-\int v\,du$

$=\lim _{t\to \infty }\int _{0}^{t}{\frac {1}{c+\exp(x)}}\,\exp(-x^{2})dx$

$=\lim _{t\to \infty }({\frac {1}{c+\exp(x)}}{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x))|_{0}^{t}-\lim _{t\to \infty }\int _{0}^{t}{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x)\,-(c+\exp(x))^{-2}(\exp(x))dx$

Simplifying:

$=0-0-\lim _{t\to \infty }\int _{0}^{t}{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x)dx+\,\lim _{t\to \infty }\int _{0}^{t}(c+\exp(x))^{-2}(\exp(x))dx$

The first integral can be simplified to:

$-\lim _{t\to \infty }{\frac {1}{2}}{\sqrt {\pi }}\int _{0}^{t}\operatorname {erf} (x)dx$

Noting that

$\int \operatorname {erf} (x)dx=x*\operatorname {erf} (x)+{\frac {exp(-x^{2})}{\sqrt {\pi }}}$

...to be continued. (Wow, I love this problem)

--Ķĩřβȳ♥ŤįɱéØ 21:55, 14 March 2007 (UTC)

So now we have:

$=-\lim _{t\to \infty }({\frac {1}{2}}{\sqrt {\pi }}(x*\operatorname {erf} (x)+{\frac {exp(-x^{2})}{\sqrt {\pi }}})|_{0}^{t})$

Evaluated at t and 0:

$=-\lim _{t\to \infty }({\frac {1}{2}}{\sqrt {\pi }}(t*\operatorname {erf} (t)+{\frac {exp(-t^{2})}{\sqrt {\pi }}})-(0*\operatorname {erf} (0)+{\frac {exp(-0^{2})}{\sqrt {\pi }}}))$

Simplify the right side:

$=-\lim _{t\to \infty }({\frac {1}{2}}{\sqrt {\pi }}(t*\operatorname {erf} (t)+{\frac {exp(-t^{2})}{\sqrt {\pi }}})-{\frac {1}{\sqrt {\pi }}})$

Further simplification:

$=-\lim _{t\to \infty }({\frac {{\sqrt {\pi }}t*\operatorname {erf} (t)}{2}}+{\frac {exp(-t^{2})}{2}}-{\frac {1}{\sqrt {\pi }}})$

Limit laws:

$=-\lim _{t\to \infty }({\frac {{\sqrt {\pi }}t*\operatorname {erf} (t)}{2}})-\lim _{t\to \infty }({\frac {exp(-t^{2})}{2}})+\lim _{t\to \infty }({\frac {1}{\sqrt {\pi }}})$

Simplify the last 2 terms:

$=-\lim _{t\to \infty }({\frac {{\sqrt {\pi }}t*\operatorname {erf} (t)}{2}})-0+{\frac {1}{\sqrt {\pi }}}$

Oh boy, I think I am confused now, as the first time seems to evaluate to infinity... someone else help? --Ķĩřβȳ♥ŤįɱéØ 22:31, 14 March 2007 (UTC)

Wow, impressive work. I think the mistake is on the line

$=\lim _{t\to \infty }({\frac {1}{c+\exp(x)}}{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x))|_{0}^{t}-\lim _{t\to \infty }\int _{0}^{t}{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (x)\,-(c+\exp(x))^{-2}(\exp(x))dx$

The $\int vdu$ should be the product of the terms; I think a negative sign may have slipped in as a subtraction. I think the integral should be

$\lim _{t\to \infty }\int _{0}^{t}{\frac {{\frac {1}{2}}{\sqrt {\pi }}\operatorname {erf} (z)\exp(z)}{(c+\exp z)^{2}}}dz$

I was trying a substitution on this, but did not get anywhere. Thanks again, --TeaDrinker 23:24, 14 March 2007 (UTC)

Wow, all that work, for nothing, because of a mistake. I'm sorry about that. In any case, I think another application of parts is needed, with ${\frac {d}{dz}}\operatorname {erf} (z)\exp(z)$ being evaluated with the product rule. Denominator as dv. I have to eat lunch now, and then I have tutor someone in Calculus. Cheers! --Ķĩřβȳ♥ŤįɱéØ 23:40, 14 March 2007 (UTC)

what is the limit of this integral ?

f(x)=x^n now presum the arc lengh of f(x) on the interval(0,1) arclenght=integral{sqr(1+n^2x^2(n-1)}limits of integral is from 0to 1 what is, lim(n→∞)integral{sqr(1+n^2x^2(n-1)}?is it=2or what? 80.255.40.168 12:23, 14 March 2007 (UTC)ARTHER

Yes it is 2.  --Lambiam^Talk 15:12, 14 March 2007 (UTC)

is this right?

IF,integral(f(x)dx)=integral(dx),does this mean f(x)=1? 80.255.40.168 12:29, 14 March 2007 (UTC)NEIL

It depends on the precise definition of integral used. If you use the Riemann integral, the answer is simply yes. For the Lebesgue integral, f can be any function that is 1 almost everywhere, which means that exceptions are allowed on a set of Lebesgue measure 0. For example, f(x) = 0 if x is an integer, and f(x) = 1 otherwise.  --Lambiam^Talk 15:22, 14 March 2007 (UTC)

Umm - maybe I am missing something here, but isn't the function "f(x) = 0 if x is an integer, and f(x) = 1 otherwise" bounded and continuous almost everywhere, and so Riemann integrable ? (Unlike the Dirichlet function, which is nowhere continuous and not Riemann integrable). If so, it's still a good counterexample to the questioner's proposition, but Riemann integrability is not a sufficient condition to the make the proposition true. Gandalf61 15:34, 14 March 2007 (UTC)

I think you're right (meaning I was wrong).  --Lambiam^Talk 16:40, 14 March 2007 (UTC)

What we need is that f is continuous. Given that, the fundamental theorem of the calculus and the mean value theorem imply that f is everywhere 1. Algebraist 22:25, 14 March 2007 (UTC)

Sudoku grids

How many possible standard sudoku grids are there? (Reflections and rotations counting as distinct). Obviously cell (1,1) can be any of 9 values, cell (1,2) any of 8 etc to cell (1,9) which only has one option left. On the next row, cell (2,1) can be any of 6, so we get a number of options per cell grid as:

987|654|321
654|???|321
321|321|321
-----------
6??|???|321
5??|???|321
4??|???|321
-----------
333|333|321
222|222|2?1
111|111|111

But I can't work out what the options are for the ? cells. The final number of grids is presumably the product of the values in each cell. -- SGBailey 13:59, 14 March 2007 (UTC)

This problem is quite difficult. As stated on Sudoku, the answer is 6,670,903,752,021,072,936,960. You'll have to read the original paper for the proof. Algebraist 14:03, 14 March 2007 (UTC)

Note that the factorization of that number is $2^{20}\cdot 3^{8}\cdot 5\cdot 7\cdot 27704267971$, so it is not a product solely of small integers. --Tardis 15:39, 14 March 2007 (UTC)

BECAUSE for cell 4,2, there are a few amounts of possible numbers - if 4,1 = 1,2, 5,1 = 2,2, 6,1 = 3,2, then there are 6. If those are 6 different values, it's 3. It'd require you to use some sort of tree, I think. ST47Talk 20:17, 14 March 2007 (UTC)

The paper referenced above used computers to give an answer through brute force. It uses some damn good optimization techniques to do that. Using a rather naive approach on my computer suggests I'll have an answer in 211 billion years :D I did run it for a 4x4 sized grid rather than a 9x9 (it ran instantly), and there are 288 distinct combinations for that. --h2g2bob 15:59, 17 March 2007 (UTC)

Any Mathematics Tests?

I have learned advanced mathematics and I am applying for univeristy , I wanna ask is there any test for me to impress the people at the admission department? —The preceding unsigned comment was added by 76.199.98.111 (talk) 16:29, 14 March 2007 (UTC).

If you really want to test yourself you can try a past paper from the International Mathematical Olympiad. A lot will depend on the university, some will make offers purely on exam grades, other like Oxford/Cambridge (UK) set a special maths paper for applicants. Others will use an interview, I don't remember getting any maths quiz during my interviews. The situation in the US may be different. --Salix alba (talk) 17:48, 14 March 2007 (UTC)

The IMO would certainly impress admissions departments, but it doesn't test "advanced mathematics" so much as a deeper knowledge and understanding of elementary and somewhat offbeat mathematics. For a more helpful response, you should tell us at least what country you're in, what qualifications/tests you have/will have already taken, what extra mathematics you have learned, and whether you're applying for a degree in mathematics or in some other scientific subject. —Blotwell 18:26, 14 March 2007 (UTC)

Somewhat personal interjection: If you have a propensity for math, are in high school in the states, and want to go to college, see if your school of choice accepts Advanced Placement credit, and take one of the AP exams for calculus. (their website is here: [1]) They are dirt cheap compared to college courses, I took 3 for $70 each, and saved myself a few thousand dollars by getting credit at my college for entry level courses. I'm not an admissions worker, but demonstrating that you can handle college level material (the AP exams) doesn't hurt. Atropos235 00:10, 15 March 2007 (UTC)

Check out our list of mathematics competitions. − Twas Now ( talk • contribs • e-mail ) 10:28, 17 March 2007 (UTC)

March 15

Solids of revolution

To begin, I apologize for asking a rather specific calculus homework question. But I'm a little stumped and I have about five problems left to finish, all of which involve the same twist in the normal solid of revolution process...the regions we are rotating contain the axis of revolution. For instance, one problem asks us to find the volume of the region bounded by $y=-x+3$ and $y=x^{2}-3x$ when it is rotated about the x-axis. The curves intersect at (−1, 4) and (3, 0), but I'm not sure where to move from here. None of the typical methods seems to yield the correct answer, 20.367π. If anyone has a suggestion or a prod in the right direction, I'd be grateful. Thanks! 12.226.88.48 22:16, 15 March 2007 (UTC)

Disk_integration may be of help. See the part about "hollow". --Ķĩřβȳ♥ŤįɱéØ 22:26, 15 March 2007 (UTC)

If you rotate by 180°, the effect is that of replacing x by −x. This means the region can also be described as bounded by y = x + 3 and y = x² + 3x. If you graph all four functions (which you need to do the understand the following), you see a few more points of intersection, including at (0, 0), (0, 3), and (−2 + √7, 5 − √7) – the latter where y = −x + 3 intersects y = x² + 3x. In rotating full circle, the triangle-like figure having these three points as corners sweeps out a volume that is also sweeped out by the part having positive x-values of the "blade" bounded by the original two functions. This means that standard methods for determining the volume will count this twice. You have to compensate for this, which you can do by separately determining the volume sweeped out by the triangle-like figure and subtracting it from the original answer, or by first splitting up the area into the "positive" part of the blade, and the smaller triangle-like figure with corners (0, 3), (−2 + √7, 5 − √7), and (1, 4), next determining for each the volume of the region sweeped out by a rotation, and finally adding the results.  --Lambiam^Talk 23:40, 15 March 2007 (UTC)

Perhaps I'm missing something, but it seems to me that Lambiam is describing a rotation around the y axis. As the original question was about a rotation about the x axis, the specific details of this solution may not be relevant --- though the overall tactic is an excellent one. Tesseran 07:08, 16 March 2007 (UTC)

Right, my specific dissection was for a rotation about the y-axis, but the same approach with the roles of x and y swapped should work for the y-axis. A new point of intersection is then (1, 2).  --Lambiam^Talk 08:58, 16 March 2007 (UTC)

Hey guys, are you sure you talk about same rotation? Original poster says the area is rotated about the x-axis, Labmiam talks about replacing x with -x, wwich suggests rotation about y-axis... —CiaPan 07:23, 16 March 2007 (UTC)

See above.  --Lambiam^Talk 08:58, 16 March 2007 (UTC)

Nevertheless Lambian has identified the key point of this problem. The region bounded by the curves has two parts, one below the x-axis and one above the x-axis. Each of these parts sweeps out a solid of revolution and these two solids have a common part. The volume of this common part needs to be subtracted from the sum of the volumes of the solids of revolution in order to get the answer. Geoffcobra 23:19, 16 March 2007 (UTC)

March 17

how to sketch hyperbolas without table?

hi,i've got no clue on how to sketch hyperbolas without using the table. i've searched through several year 10 textbooks (i'm in year 10) and found nothing. what are the methods for sketching hyperbola without using a table? eg. 1) y = 2/(3-x) eg. 2) y = x/(x-1) eg. 3) y = (x+1)/(x-2)

THANKS for any help--Lil devilz 09:35, 17 March 2007 (UTC)Dave

For the first, think about the sketch of the hyperbola 2/x. What does it look like? How does it compare to 2/(x-3)?

For the second and third, you will have to apply a little trick.

${\frac {x}{x-1}}={\frac {x}{x-1}}+{\frac {-1}{x-1}}+{\frac {1}{x-1}}$

This is okay since we are essentially adding zero. (-1/(x-1)+1/(x-1) = 0).

$=({\frac {x}{x-1}}+{\frac {-1}{x-1}})+{\frac {1}{x-1}}={\frac {x-1}{x-1}}+{\frac {1}{x-1}}$

This is okay since we can simplify the terms in any order. And we're just performing addition according to the laws of fractions, so everything's okay. Now,

$=1+{\frac {1}{x-1}}$

The 1/(x-1) is just a regular hyperbola that is shifted across. What does adding 1 do to it? You basically do this trick to the third question as well. —The preceding unsigned comment was added by 149.135.84.54 (talk) 11:05, 17 March 2007 (UTC).

Here is an alternative way of looking at the problem.

These three hyperbolas are all rectangular hyperbolas where the asymptotes are parallel to the x and y axes. You can recognise this because there are no $x^{2}$ or $y^{2}$ terms in the equations.

You can therefore compare them with XY= $c^{2}$. You have to find the centre of the hyperbola and its size. In this case the origin is X=0, Y=0 and there are points (c,c) and (-c,-c) on the curve.

In 2) (x-1)(y-1) = 1 and so the origin is x=1, y=1 and $c^{2}$ =1. If $c^{2}$ is negative as in question 1), you will have to compare it with XY=-1. Geoffcobra 11:17, 17 March 2007 (UTC)

In other words, you can re-arrange the equation of each of your three hyperbolas to put it in the form (x-a)(y-b)=c (where a or b may be 0). You then have a horizontal asymptote which is the line y=b and a vertical asymptote which is the line x=a. The asymptotes divide the x-y plane into four quadrants. If c is positive the two branches of the hyperbola are in the upper right and lower left quadrants; if c is negative the two branches of the hyperbola are in the lower right and upper left quadrants. Gandalf61 12:57, 17 March 2007 (UTC)

Can u solve this?

x-3y=1

2x-6y=2

must be solved by substitution..... —The preceding unsigned comment was added by 89.56.167.132 (talk) 17:21, 17 March 2007 (UTC).

If you multiply the first equation by 2 you get the second equation. So the solution is not unique - both equations have the same set of solutions, which is a line in the x-y plane. Gandalf61 17:35, 17 March 2007 (UTC)

um...hello its supposed to be solved by substitution not by elimination...we already knew that they all cancel out and equal 0....but thanks anyways....

Well, what are you trying to find then? As the two lines are exactly the same (try drawing them), solving them by subsitution or whatever will always fail to give a value for x and y, as it's impossible to get the point of intersection where the lines are the same (persuming this is what you are asking), as they lie on top of eachother. Please remember that the reference desk if not for homework help. Martinp23 17:53, 17 March 2007 (UTC)

um...hello too: can't be solved by substitution..... sorry about that on behalf of all linear equations.  --Lambiam^Talk 19:28, 17 March 2007 (UTC)

Suppose the question was correctly stated (a bold assumption); then perhaps we are meant to assign a value to y, say, and find a value for x. This would, quite literally, be a solution (one of many!) by substitution.

Otherwise, without further context or clarification, I think we're agreed this is a questionable question.

Also, I would like to remind questioners to please read and follow the directions at the top of the page. Especially, note we don't do homework (we only guide), and please sign your posts. Your cooperation benefits everyone. Thanks. --KSmrq^T 05:05, 18 March 2007 (UTC)

To solve this system by substitution, you can solve the first equation for x and substitute into the second equation. This simplifies to 2=2, which is true for any value of x.

The three possibllities of kinds of solutions for a 2 variable, 2 equation system are:

1. One value for x and one value of y, which is the "normal" solution; the graphs of the two equations intersect in one point.

2. An infinite number of values for x ,each with a specific value for y, which is what we get when we end up with a true numeric equation such as 0=0 or 2=2; the graphs of the two equations are both the same line.

3. No solution, which is what we get when we end up with a false numeric equation such as 0=7 or 2=9; the graphs of the two equations are parallel lines that never intersect.--MathMan64 05:49, 18 March 2007 (UTC)

Basic Math Skills, using Chisum Block

In the late 1970's, in a one-room elementary schoolhouse in Wyoming, our teacher taught us a basic math computation technique he called "Chisum Block."

I have been searching the internet, looking to find out more about this technique as I can only remember bits and pieces of it. Do you have any information or ideas of where to look that might help me??? —The preceding unsigned comment was added by AliceMaybelAnnie (talk • contribs) 22:17, 17 March 2007 (UTC).

What can you remember about the technique? What is it used for? If you can provide some details, we might be able to help you. Phils 02:13, 18 March 2007 (UTC)

Possibly Pattern blocks, more details would certainly help. Were they like any of the ones here? --Salix alba (talk) 08:51, 18 March 2007 (UTC)

I believe you are seeking the Chisenbop method which can be found in Wikipedia at [[2]] . A visual tutorial can be found at [3] .

Powers

Does zero to the power of zero equal one?Dudforreal 05:04, 18 March 2007 (UTC)

A natural question, given the conflict between x⁰ = 1 and 0^y = 0. Obviously both cannot be correct, so there must be limitations on x and y. For an extended discussion, see Zero to the zero power in our article on exponentiation. --KSmrq^T 05:20, 18 March 2007 (UTC)

Yes and no. It makes sense sometimes to treat 0^0 as being 1, and it sometimes makes sense to treat 0^0 as being 0. See the article just linked to. —The preceding unsigned comment was added by 149.135.99.162 (talk) 05:21, 18 March 2007 (UTC).

I would say it's undefined, just like 0/0. StuRat 00:07, 20 March 2007 (UTC)

It's far worse than "undefined"; it's indeterminate. –King Bee ^{(τ • γ)} 20:27, 22 March 2007 (UTC)

Thankyou all very much  :).Dudforreal 06:02, 18 March 2007 (UTC)

abstract algebra

ques1. every subgroup of an infinite group is infinite ? give reason also/

ques.2 every permutation in Sn is a product of disjoint transposition ,where n>1 ?

ques.3 if G is a group with respect to addtion , then it is also group with respect to subtraction.?

1) All you need is one example to disprove it, like "the group of whole numbers less than 100 is a subset of the group of whole numbers". StuRat 00:10, 20 March 2007 (UTC)

1) If G is infinite, and H is finite, then GxH is a group, with H a subgroup.

2) Nope.

3) We require the group operation to be associative - you might run into problems with that particular thing, depending on how, exactly, you define the group operation from subtraction. Michiexile 17:44, 22 March 2007 (UTC)

Either I'm very stupid or there is a problem with mathematica

In[1]:=ClearAll["Global`*"]

In[2]:= A = {{0.99,0.5},{0.01,0.5}}
Out[2]= {{0.99,0.5},{0.01,0.5}}

In[3]= Z = {100,0}
Out[3]= {100,0}

In[4]:= B = A * Z
Out[4]= {{99.,50.},{0,0}}

Now why do I not get B = {99,1} as

${\begin{bmatrix}99\\1\end{bmatrix}}={\begin{bmatrix}0.99&0.5\\0.01&0.5\end{bmatrix}}{\begin{bmatrix}100\\0\end{bmatrix}}$ 220.239.107.13 07:23, 18 March 2007 (UTC)

I don't know Mathematica, but it looks like the elements of A are column vectors. So then you get

$B=AZ={\begin{bmatrix}0.99&0.01\\0.5~&0.5~\end{bmatrix}}{\begin{bmatrix}100\\0\end{bmatrix}}={\begin{bmatrix}99\\50\end{bmatrix}}.$

I don't understand, though, why the output is a matrix and not a single vector. Presumably it has to do with the semantics of the binary operation "*", which apparently is not the usual matrix-times-vector product here.  --Lambiam^Talk 08:11, 18 March 2007 (UTC)

Matrix multiplication is Dot, not Times (ie., A.Z not A*Z). {{0.99, 0.5}, {0.01, 0.5}}.{100, 0} works as desired.

This is the correct diagnosis and fix. Mathematica declares Times, abbreviated *, to be Orderless, meaning commutative, so we know that won't work. The NonCommutativeMultiply, abbreviated **, can be used to avoid that assumption, but there are few simplification rules, especially none for matrices. Mathematica's Dot, abbreviated ., is not considered commutative (Orderless), unlike the mathematician's dot product, and is explicitly intended for matrix multiplication and tensor contraction. (But see also DotProduct.) Moral: Think twice before you let a physicist design a system for mathematics. --KSmrq^T 10:06, 18 March 2007 (UTC)

Huh? What do you mean by Think twice before you let a physicist design a system for mathematics? —Preceding unsigned comment added by 129.78.64.102 (talk • contribs) 01:16, 2007 March 19

Stephen Wolfram, who began Mathematica and is responsible for most of its basic design, came to it from a background in theoretical physics.

Reviews such as this by Richard Fateman note deep flaws in the design of Mathematica. Talking with Wolfram, one is struck by his intelligence; but some of his other qualities (examples here) are less appealing. Mathematica shows side effects.

Perhaps I painted with too broad a brush when I extrapolated to all physicists; however, mathematicians have long chided physicists for their cavalier disregard of rigor. --KSmrq^T 12:42, 19 March 2007 (UTC)

Standard multiplication is designed to be threadable, so multiplying a matrix (a vector of vectors really) by a vector gives a threaded vector of the products of the respective vectors of the left matrix with the respective elements (numbers) of the right vector; matrix multiplication ("Dot") is not standard multiplication at all but rather a special case of an inner product (you could use "Inner" for this computation as well). To make the indices work right, you need to denote your column vector by {{100},{0}} (note the extra braces). Then all is well. Baccyak4H (Yak!) 19:51, 19 March 2007 (UTC)

Sorry it still does NOT work

In[1]:=ClearAll["Global`*"]

In[2]:= A = {{0.99,0.5},{0.01,0.5}}

Out[2]= {{0.99,0.5},{0.01,0.5}}

In[3]= Z = {{100},{0}}

Out[3]= {{100},{0}}

In[4]:= B = A * Z

Out[4]= {{100} {0.99, 0.5}, {0} {0.01, 0.5}}

You missed the point contrasting "threadable" multiplication (which is what you computed) and "inner product" multiplication (which is what you want).

Try B = Dot[A, Z], or even B = A.Z Baccyak4H (Yak!) 02:24, 20 March 2007 (UTC)

You need to enter the following: A = {{0.99, 0.5}, {0.01, 0.5}}; Z = {100, 0}; B = A . Z for it to work. Please follow what others have told you!

(to Ksmrq) I think your attribution to the perceived flaws of Mathematica to that of a physicist's mind is a bit strange -- I know Wolfram is a physicist at heart, but if your criticism of the dot product functionality in Mathematica is that one of the properties of Dot is that it is Orderless is a bit bizarre. Would you rather have one function for matrix multiplication where the matrices commute and one function for matrix multiplication where they don't? Recall that the Orderless attribute means that the arguments of the function can be sorted in order. This is obviously not permissible for matrix-matrix and matrix-vector multiplication.

Most of the criticism in a cursory examination of the review in which you referred to me seems to have been present in early versions of Mathematica and have been resolved, or reflects a misunderstanding with how the user is to use Mathematica's functionality. I could do a careful analysis, and will probably find that some points in the review have merit, but overall Mathematica in my opinion is a superior system over all the computer algebra systems that I have used. —The preceding unsigned comment was added by 149.135.29.252 (talk) 10:10, 20 March 2007 (UTC).

March 18

Generic plot diagrams

I'm wanting to produce some clear, simple time-domain and frequency domain plots of audio signals for a audio processing assignment I'm working on. Audacity does both, but I want to produce more "generic" plots where the particular GUI of the audio editor is not visible. For instance, if I want to display the time-axis information in Audacity, I have to show other parts of the program which is most undesirable.

http://img.waffleimages.com/2125764ae49011cfd7de2c41f1d106b708c6cb7e/img24.png

The picture above is an example of what I'm looking for. In particular, there are horizontal and vertical lines at the marked values for the axes. Naturally, I'll want control over how "zoomed in" I am to a specific plot, vertically and horizontally, and make it simple to load in any wave file.

My assumption is that more customisable, generic plots can be produced using some kind of maths package , but if there's a nice GUI program for outputting custom generic plots that would be preferred. Thanks. 164.11.204.51 18:34, 18 March 2007 (UTC)

If you're looking for free software, you could check out the tools listed in Category:Free plotting software. If you are using Linux, you may be specifically be interested in Category:Linux graph plotting software (all of which is listed as free). See also Wikipedia:How to create graphs for Wikipedia articles.  --Lambiam^Talk 20:40, 18 March 2007 (UTC)

In particular, I need to plot information from audio files (wave e.t.c) in domains and ranges of my choice. The wikipedia tutorials don't seem to point towards anything like this, although I'm sure there must be a function in a free windows plotting software to do this? —The preceding unsigned comment was added by 164.11.204.51 (talk) 22:31, 18 March 2007 (UTC).

You probably want the computing reference desk for this. Think of this as a two stage process 1) extract sutible numeric data from a wav, 2) plot it. I'd guess your more likely to find two different bits of software for the different tasks, and the CS desk may be able to help with 1). It might also be worth looking at some of the visulisation effects for winamp and similar, many of these are based around processing the audio data and turning them into visuals. Maybe theres some code you can hack out of one of these. --Salix alba (talk) 23:07, 18 March 2007 (UTC)

Differentiation of Trig

Why is the derivative of sin(x) only cos(x) when x is measured in radians? Algebra man 18:56, 18 March 2007 (UTC)

To avoid all ambiguity, let sinrad denote the sine function whose argument is in radians, and likewise for cosrad, so ^d/_dxsinrad(x) = cosrad(x). Define

sinscaled_a(x) = sinrad(2πx/a),

and again likewise for cosscaled. So, for example, sinscaled₃₆₀(x) gives the sine of angle x measured in degrees. Now, by the chain rule,

^d/_dx sinscaled_a(x) = ^d/_dx sinrad(2πx/a) = cosrad(2πx/a) ^d/_dx(2πx/a) = (2π/a) cosrad(2πx/a) = (2π/a) cosscaled_a(x),

This only equals cosscaled_a(x) if a = 2π. I hope this qualifies as a "reason".  --Lambiam^Talk 19:25, 18 March 2007 (UTC)

The working out of the derivative of sin(x) requires the sine of small angles rule which states that sin(x)~x only when x is in radians. Alexs letterbox 06:41, 19 March 2007 (UTC)

Which is quick to show informally. Define a radian as 1/(2π) of a circle. Start from a point x = r on the positive x axis, and draw a very short arc, moving through θ radians counter-clockwise. Then draw a line from the top of that arc to the origin. This gives an almost-triangle, except that the short side is an arc, not a line. For a short arc that's negligible, though, so we can pretend that the arc is a line, and that our shape is a triangle. (It's actually a sector of a circle, like a slice of cake.)

The long sides are both the same length r, because the shape is a sector of a circle. The arc has a length of θr. (If it were the entire circle, then θ would be 2π, so we would have the familiar formula for the circumference of a circle, 2πr.) The angle θ was very small, and the angles of a triangle should sum to 180 degrees; this means that the other two angles are almost ninety each, so we have a right triangle. The sine of θ is equal to opposite/hypoteneuse = θr/r = θ. If we were using different angle units, and not radians, then the formula for the arc length would have been different.

The first answer sort of begs the question, I think. I hope this is more convincing. —The preceding unsigned comment was added by 71.192.58.216 (talk) 10:03, 19 March 2007 (UTC).

All the information is in the preceeding answers; I shall attempt a synthesis.

• The derivative of a function f at a point x is the ratio of a small displacement in input from x divided into the resulting small displacement in output from f(x).
• Therefore sin(2x) will have a derivative twice as large as sin(x), for example; and in general the derivative will depend on the units of input. (This is true for any non-constant function.)
• At x = 0, cos(x) is exactly 1 no matter what input units we use.
• Therefore we can have only one possible choice of input unit for the sine function if its derivative is to match.
• Radian measure (arc length) works, as shown by a geometric argument.
• Therefore radian measure is the unique measure permitting the derivative of sin(x) to be cos(x').

This line of reasoning does not prove that the correspondence holds for all values of x, but it does show that if it is to hold at all, we must use radians. (I assume that anyone who could handle the general proof wouldn't be asking this question.) --KSmrq^T 23:36, 19 March 2007 (UTC)

Trend and Pattern

I am analysing some biological data, and am having to comment on the trend and pattern in the data. I was wondering if any statisticians could tell me what exactly is the difference between a trend and pattern in data ??? 89.241.4.129 19:40, 18 March 2007 (UTC)

If you plot some data as dependent on other data, and the plotted points seem equally scattered around a steadily increasing or decreasing curve, this reveals what is usually called a "trend". Often but not necessarily the independent variable is time. The concept of "pattern" is much more general; it can be applied to any recognizable regularity, for example that some variable usually registers lower during weekends than on workdays. See also Trend estimation.  --Lambiam^Talk 20:55, 18 March 2007 (UTC)

Please help really hard

please help me with these questions there reallyy hard the answers can only be between 1-31

1. A number n where if the sum of the digits of the number is divisible by n, then the number itself is divisible by n.
2. A number that is neither prime nor the sum of two distinct primes.
3. The only perfect number of the form x n + y n
4. The largest integer that is not the sum of two or more different primes.
5. The maximum number of pieces into which a pizza can be cut by making 6 cuts

if you could just answer these for me id be so greatful you have no idea thanks to anyone who can answer them for me remember the answer can only be between 1-31.

thank you, anon —The preceding unsigned comment was added by 68.193.39.27 (talk) 22:53, 18 March 2007 (UTC).

Unfortunatly (for you anyway) we cannot do your homework for you. And these are (for me) difficult questions to give clues for without giving you the answer. - Akamad 04:38, 19 March 2007 (UTC)

These questions are weird (and some of them are ridiculously easy). I'll get you started. For number 1, take any one-digit number. For question 2, how about the number 4? See what I mean? Just think about them for a second. –King Bee ^{(τ • γ)} 20:02, 19 March 2007 (UTC)

you guys suck. i better have not failed my project because of you.

March 19

Liouville-Roth constant of a logarithm

I was reading the article Liouville number, and it mentioned the Liouville-Roth constant associated with an irrational number as being a sort of irrationality measure. Does anyone know if this has been calculated for logarithms of integers to integer bases, e.g., ln(5)/ln(3)? GTBacchus^(talk) 00:33, 19 March 2007 (UTC)

derivative of the inverse of a logit function

I have gotten stuck on a derivative that I did not think would be too hard. Please help me out if you have the chance. The function is simply:

$f(y)=e^{-x}\left(e^{x}+1\right)\,$

I know that if f(z) = e^x / (1 + e^x) then:

$f'(z)={\frac {e^{x}}{\left(e^{x}+1\right)^{2}}}\,$

and if f(a) = (1 / z) then f'(a) = (-1 / z^2).

So for the initial equation of:

f(y) = 1 / [e^x / (1 + e^x)]

I thought that maybe:

f'(y) = -1 / [e^x / (1 + e^x)^2]^2

But that does not seem to be the case.

I believe I solved for the correct answer using the d method and a do-loop on a computer.

When f(y) = 1 / [e^x / (1 + e^x)] and x = -0.40 I believe that

f'(y) = -1.49182.

But I do not know the equation for f'(y). None that I've tried give -1.49182 when x = -0.40.

Thanks for any thoughts. Sorry I did not take the time to format this post. I am using ^ to mean "raised to the power of" and e = 2.71828. Also I posted this question on March 19, but it seems to be in the March 13 section. Sorry about that.

Mark W. Miller 05:38, 19 March 2007 (UTC)

There is something a bit strange about your notation. The value of an expression like, for instance, 1 – cos(x) depends on the value of the variable x. If, in mathematics, something depends like that on something else, it is known as a function. Let us give a name to the function in this example, say Jack. To express that Jack is the function of the example, you can write this:

Jack(x) = 1 – cos(x).

This then implies that Jack(0) = 1 – cos(0), Jack(π/3) = 1 – cos(π/3), and so on. The definition applies to all possible values of x, so if you substitute the same expression for x in the left-hand side an all occurrences of x in the right-hand side, the equation remains true. The x here is known as a bound variable. Just like the choice of the name of the function (Jack) is arbitrary, the name of a bound variable can be chosen arbitrarily. Without change of meaning, we could have defined

Jack(u) = 1 – cos(u).

In your equations you use different variable names in the two sides of the equations that define the functions. I assume that you want to find the derivative of the function h defined by

h(x) = 1 / (e^x / (1 + e^x))

The chain rule is the rule to use when determining the derivative of a composite function. Putting f(z) = 1/z and g(x) = e^x / (1 + e^x), we have h(x) = f(g(x)). Then

h'(x) = f'(g(x)) g'(x) = – 1 / (e^x / (1 + e^x))² · e^x / (1 + e^x)²

This can be simplified dramatically. Instead of doing that, here is a much easier way to determine h' for this specific case:

h(x) = 1 / (e^x / (1 + e^x)) = (1 + e^x) / e^x = (1 + e^x)e^–x = e^–x + 1.

Then

h'(x) = –e^–x.

By the way, the functional inverse of the logit function is not this function h, but the sigmoid function. Function h is the multiplicative inverse of the latter.  --Lambiam^Talk 06:45, 19 March 2007 (UTC)

Thanks much. This morning, after sleeping on it, I came up with the equation:

h'(x) = ((e^x)² - (1+e^x)e^x) / ((e^x)²).

It gives the same answer as the equation you provided. I'm sure the equation I typed reduces to the one you gave.

Thank you again. Thanks also for the information about the sigmoid function.

I feel I should mention for completeness that I really want the partial derivative of a function:

f(x,y,z) = x / (y*z)

where:

x = e^a / (1 + e^a),

y = e^b / (1 + e^b), and

z = e^c / (1 + e^c).

However, now I am pretty sure:

f'b(a,b,c) = –e^–b ((e^a / (1 + e^a)) / (e^c / (1 + e^c))) or

more simply:

f'a(a,b,c) = (e^a / (1+e^a)²) / (y z)

f'b(a,b,c) = –e^–b (x / z), and

f'c(a,b,c) = –e^–c (x / y).

I would like to copy much of this to my home page for future reference if that is not against policy.

Thanks again.

Mark W. Miller 16:56, 19 March 2007 (UTC)

I don't think it is against policy if you add where it is copied from. Instead of a copy you could also use a link to Wikipedia:Reference desk/Archives/Mathematics/2007 March 19#derivative of the inverse of a logit function, which right now is still a redlink but in a few days will become a link to this section.  --Lambiam^Talk 20:31, 19 March 2007 (UTC)

Fermat

That other day i was bored so i jotted Fermat's last theorem on a paper:$X^{n}+Y^{n}=Z^{n}$ Then i took the partial derivative first w.r.t to x....since y and z are treated as constants ,i got :$nX^{n-1}=0.$.

I did the same to y:$nY^{n-1}=0.$.

and to z:$nZ^{n-1}=0.$.

Knowing that n is different from zero (since n>2) it cancels in the three equations: $X^{n-1}=0.$ $Y^{n-1}=0.$ $Z^{n-1}=0.$

which gives us X=0, Y=0,or Z=0, the only solution in integers for the formula.

Ofcourse i couldn't have possibly proved Fermat's Last theorem .....of course not .....i am sure there's an error somewhere that i'm not finding. Could someone point it to me please? PS. Sorry if this is trivial or there's a principal error somewhere. I'm not a world class mathematician :) Hisham1987 17:47, 19 March 2007 (UTC)

The problem is in concluding that $nX^{n-1}=0$ etc. Suppose $x_{0},\ y_{0},\ z_{0}$ is a solution to the equation, so $x_{0}^{n}+y_{0}^{n}=z_{0}^{n}$. If you take $f(X)=X^{n}+y_{0}^{n}-z_{0}^{n}$, you will get $f(x_{0})=0$ for this particular $x_{0}$; you do not get $f(X)=0$ for every X, which has to hold if you want $f'(X)=0\,\!$. -- Meni Rosenfeld (talk) 19:41, 19 March 2007 (UTC)

If your (Hisham's) approach was valid, you would have succeeded in proving that the equation X¹ + Y¹ = Z¹ has no solutions except (0, 0, 0). Actually, also if n > 2, the equation has some other solutions in integers: (k, 0, k), (0, k, k), if n is even also (–k, 0, k) and (0, –k, k), and if n is odd also (k, –k, 0).  --Lambiam^Talk 20:21, 19 March 2007 (UTC)

Fermat's Last Theorem:

If an integer $n$ is greater than 2, then $a^{n}+b^{n}=c^{n}$ has no solutions in non-zero integers $a$, $b$, and $c$.

What you have shown is irrelevant to the theorem because the theorem already says that a, b, and c, must be nonzero. --Ķĩřβȳ♥♥♥ŤįɱéØ 23:36, 19 March 2007 (UTC)

Latex program for beginners

Is there an easy program to work with LaTex? Ideally a program where we can just click icons to insert a mathematical symbol. 132.231.54.1 18:30, 19 March 2007 (UTC)

I think Lyx (official page) may be what you are looking for. -- Meni Rosenfeld (talk) 19:35, 19 March 2007 (UTC)

I thought perhaps something smaller would be nicier.132.231.54.1 21:11, 19 March 2007 (UTC)

I started with Texmaker, which has buttons, but there's still a learning curve. If you can get some sample documents, they're a great way to get started, because you can see how things are already done. -GTBacchus^(talk) 02:40, 20 March 2007 (UTC)

I Don't Understand This!!

I have neven been able to solve questions with loans. These are the two problems I can't solve. I don't want the answer, but I need to understand how to do them.

1. Bob borrowed $9,000 for ten years at an annual rate of 18%. What is his monthly payment?

2. Simon obtained a 2 year loan at a simple interest rate of 13% annually. At the end of two years, he has payed $1454.44. What amount of money was he loaned?

What does annually mean in this anyway?

69.37.180.53 22:57, 19 March 2007 (UTC)

Just let me answer the part about what "annually" means. Obviously, in general it means once a year. When dealing with interest rates it could mean the payment is made once a year, or the interest compounds once a year, or that they are giving you either the nominal (no compounding) or effective (with compounding) annual rate. If they say there is an annual rate of 18%, and don't specify any other compounding, I would take that to mean the interest is compounded annually. I was a bit confused by the second question which states there is a 2 year loan with simple interest at a rate of 13% annually. The term "simple interest" means no compounding, they just give you the total interest over the entire period. But that evidently isn't what they really meant, due to the "13% annually" part. StuRat 23:30, 19 March 2007 (UTC)

If I understand the second question correctly, they mean that P dollars were borrowed, then, after 1 year, 13% interest was added, then, after 2 years, 13% interest was added again, including interest on the interest. At this point, the total (both principal and interest) was paid off. If my interpretation is correct, that means P(1.13)^2 = $1454.44, see if you can solve it from there. StuRat 23:37, 19 March 2007 (UTC)

P * 1.13^2 = 1454.44

sqrt(P) * 1.13 = sqrt(1454.44)

sqrt(P) = sqrt(1454.44) / 1.13

sqrt(P) = sqrt(1454.44) / sqrt(1.13*1.13)

sqrt(P) = sqrt(1454.44/(1.13*1.13))

P = 1454.44/(1.13*1.13)

P = 1454.44/1.13^2

202.168.50.40 01:58, 20 March 2007 (UTC)

Looks like you did that the long way. And aren't you going to solve for P ? StuRat 02:35, 20 March 2007 (UTC)

If you are interested in learning how to do <<compound interest rate calculations>>, you can find an very very old USENET message that explains how to do so.

• Goto http://groups.google.com
• Search for the string "surprise to me to learn that there are people who" . Make sure you type in the double quotes.

202.168.50.40 03:39, 20 March 2007 (UTC)

That would require me to use my calculator. I'm way too lazy for that! 202.168.50.40 03:28, 20 March 2007 (UTC)