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March 17[edit]

how to sketch hyperbolas without table?[edit]

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[reply]

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).[reply]

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)[reply]

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)[reply]

Can u solve this?[edit]

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).[reply]

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)[reply]

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. Mart inp23 17:53, 17 March 2007 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

Basic Math Skills, using Chisum Block[edit]

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).[reply]

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)[reply]

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)[reply]

I believe you are seeking the Chisenbop method which can be found in Wikipedia at [[1]] . A visual tutorial can be found at [2] .

Powers[edit]

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

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)[reply]

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).[reply]

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

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

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

abstract algebra[edit]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]