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August 21

A function

What functions f(x) are there such that f[f(x)]=x? Obviously excluding f(x)=constant of course. 76.230.213.76 (talk) 00:13, 21 August 2010 (UTC)[reply]

Assuming you can graph these, have a look at functions which are symmetric about the line y=x: for example, f(x)=1/x, f(x)=1-x etc. Your example is incorrect - for example, f(x)=0 gives f(f(1))=f(0)=0 which does not equal 1. Maybe you were thinking of f(x)=x, which is trivially symmetric about y=x (since it 'is' y=x). 84.45.219.231 (talk) 00:31, 21 August 2010 (UTC)[reply]

Such a function is called an involution. Algebraist 00:36, 21 August 2010 (UTC)[reply]

(also called a self-inverse function) Dbfirs 11:51, 21 August 2010 (UTC)[reply]

Another interesting concept to think about is the functional square root. That is, given a function f, can we find or create a function g such that g(g(x))=f(x)? After that you might consider the idea fractional derivatives. Not all these avenues of thought have proved fruitful but they are interesting to think about nonetheless. Zunaid 07:01, 21 August 2010 (UTC)[reply]

Graphically plotting a flight course

What would be the best way to graphically display a flight course given data (per second) of elements such as airspeed, altitude, vertical acceleration, heading, pitch, etc?

Thank You —Preceding unsigned comment added by 74.113.218.132 (talk) 00:16, 21 August 2010 (UTC)[reply]

You mean on a graph or chart?Smallman12q (talk) 02:15, 22 August 2010 (UTC)[reply]

If your data consists only of speeds and accelerations then you need to integrate to be able to plot the course. Someone might point you towards some special software, but you could easily set up the approximate calculations on a spreadsheet, and then graph the resulting displacements to display a flight course. Errors would accumulate over a long course. Dbfirs 06:41, 22 August 2010 (UTC)[reply]

Hiya!

Does anyone know a proof that if A is a square matrix then there's a matrix B s.t. AB = BA = det(A) I. I know B is called the adjugate of A, but how do you find such a B or proof such a B exists/ Thanks. —Preceding unsigned comment added by 110.20.59.41 (talk) 12:29, 21 August 2010 (UTC)[reply]

You might want to take a look at the adjugate matrix article. -- The Anome (talk) 12:49, 21 August 2010 (UTC)[reply]

(edit conflict) Try the adjugate matrix article. It's mentioned in this subsection. It's a consequence of the Laplace expansion. — Fly by Night (talk) 12:51, 21 August 2010 (UTC)[reply]

I know but how do you PROVE that the adjugate exists? Sorry I wasn't clear and thanks for youre answer. —Preceding unsigned comment added by 110.20.9.174 (talk) 12:51, 21 August 2010 (UTC)[reply]

It has an explicit definition, and so must exist. The definition involves taking determinants of sub-matrices. That's just elementary arithmetic and algebra. Try this subsection of the adjugate matrix article. The only thing that needs to be proved is that a square matrix M, multiplied by its adjugate matrix adj(M), is the determinant of M multiplied by the identity matrix: i.e. M⋅adj(M) = det(M)⋅I. — Fly by Night (talk) 12:52, 21 August 2010 (UTC)[reply]

To be precise, note that the existence of such a

B

is fairly obvious, and the relation does not characterize the adjugate. Precisely, if

A

is not invertible,

B:=0

obviously verifies the relation, although in general it's not the adjugate of

A.

On the other hand, if

A

is invertible,

B:=\det(A)A^{-1}

obviously verifies the relation too (and in this case, it is the adjugate). Just to say that the existence of such a

B

is fairly obvious, even though it does not characterize the adjugate. --pm a 21:53, 21 August 2010 (UTC)[reply]

I'm not sure that's right, PMA. I've just done a simple example in the 2 × 2 case and it seems that the property M⋅A^T = det(M)⋅I₂ determines A uniquely. Let M = (m_p,q) be a 2 × 2 matrix. Assume that there exists a matrix, A = (a_p,q), with the property that M⋅A^T = det(M)⋅I₂. Multiply out and solve to give a_1,1 = m_2,2, a_1,2 = −m_2,1, a_1,2 = −m_2,1 and a_2,2 = m_1,1. Notice that A is exatly the adjugate matrix of M. — Fly by Night (talk) 20:50, 22 August 2010 (UTC)[reply]

Sorry if I wasn't clear. I mean: that property does not characterize the adjugate, in that for a matrix A there are in general several matrices B verifying the relation. Precisely, if A is invertible, then such a B is unique and it is the adjugate; if A is not invertible, then there are infinitely many such B -for instance, any scalar multiple of adj(A), or more generally any B in the algebra generated by adj(A) will do it.

Oh I see what you mean. Yeah, for any given matrix A₀ there might be a number of matrices with the property that M⋅A₀^T = det(M)⋅I₂. Is that because the space of n × n matrices has zero divisors and as such is not an integral domain? I was thinking of adj as an operator from the space of n × n to itself. Sorry for misunderstanding you. — Fly by Night (talk) 19:19, 24 August 2010 (UTC)[reply]

Thanks pm. Bt can someone give the explciit calculation? I can't find this in linear algebra books. In one book, it says the existence of the adjugate is another form of the Cayley-Hamilton theorem. Is this true? If so how? Thanks for you answer. —Preceding unsigned comment added by 114.72.245.87 (talk) 00:11, 22 August 2010 (UTC)[reply]

Try this subsection of the article on the Cayley-Hamilton theorem. — Fly by Night (talk) 20:24, 22 August 2010 (UTC)[reply]

Still I found a bit unclear your request for a proof of the existence of the adjugate, as already remarked by Fly by Night. It would help if you made explicit the definition of adjugate of a matrix you are referring to (note that "adjugate of A = the matrix B s.t. AB = BA = det(A) I" is not correct, as I was saying) --pm a 22:14, 22 August 2010 (UTC)[reply]

Thanks pm. But ... what I'm after is a proof that for A there's a B and AB = det(A)I = BA. I mean OK one exist. But when you prove it you have to go through the calc. with the matrix. But why is this obvious? Doesn't the matrix calc. get messy? There's a cleaner way to prove it or no? —Preceding unsigned comment added by 110.20.6.240 (talk) 23:01, 22 August 2010 (UTC)[reply]

Dude, try this subsection of the article on the Cayley-Hamilton theorem. It's there in black-and-white! It says that "These relations are a direct consequence of the basic properties of determinants." — Fly by Night (talk) 19:24, 24 August 2010 (UTC)[reply]