Math Is Fun Forum

Well, {0} is obviously an equivalence class unto itself.

In the integers modulo 12, there four units, namely 1, 5, 7, 11 (these are the integers that are coprime with 12).

Thus U = {1,5,7,11} (the set of units) is another equivalence class.

To find the other equivalence classes, just “multiply” each of the other nonzero elements with U.

2U = {2,10} = 10U
3U = {3,9} = 9U
4U = {4,8} = 8U
6U = {6}

Hence the equivalence classes are {0}, {1,5,7,11}, {2,10}, {3,9}, {4,8}, {6}.

Let’s take the numbers 1, 2, 3.

The root mean square is √[(1² + 2² + 3²)∕3] = √(14⁄3) ≈ 2.17.

The mean square root is (√1 + √2 + √3)∕3 ≈ 1.38.

The two are different.

Then Euler’s formula comes into play.

http://en.wikipedia.org/wiki/Planar_gra … 7s_formula

Simultaneous non-linear equations in x and y, I believe.

Here’s a hint: Notice that the second equation is exactly the same as the first with x and y interchanged? Therefore one possibility is x = y. Try putting that into one of the equations and see what happens.

When you’ve found a basis for the space, you will have “described” the vectors in the space (since every vector is a linear combination of the basis vectors).

[x] always rounds down, never up.

Also note that it’s not the same as taking the integer part of x for x < 0: [−0.5] = −1, [−2.3] = −3, etc.

Let the two columns be the ith and jth columns, where i ≠ j. Then the dot product of the columns is the product of the ith row of A[sup]T[/sup] with the jth column of A. This is the (i,j)th entry of the product A[sup]T[/sup]A = I, the identity matrix. Since  i ≠ j. this entry is 0.

The converse is certainly not true. Take A to be the zero matrix, for example.

You can’t do it that way – h[sub]p−1[/sub] is not an integer.

Hello and welcome to the site!

I have a very strong suspicion that you got one of the signs in your equation wrong. Instead of

I’m very sure it should be

with a minus sign for the tanx. Then instead of xsecx you would be integrating xcosx – which is really a piece of cake. The solution to the second equation above satisfying the given initial condition would then be

Frankly I don’t see how you can integrate xsecx without making so much of a mess that the question becomes not worth answering at all.