Recent content by Unco

The first 5 columns of your row-reduced matrix show that S forms a basis. This is enough. The last (sixth) column does give you the coefficients of the linear combination of p1, ..., p5 giving p. The entire 5x6 matrix is the augmented matrix for the system [\mathbf{p_1} \, \ldots \...

No; sketch the curve. Beginning at theta=0, it returns to itself at theta=pi. Hence your integral is going around the curve twice.

We can only go by what you wrote: Correcting the example I gave (I originally had it switched around, sorry): make that X=[0,2], then f is a discontinuous bijection with a continuous inverse.

Without certain conditions on X the statement to prove is false, anyway. A common counterexample is with X = [0,1) U (1,2], f(x)=x on [0,1] and f(x)=x+1 on (1,2]. EDIT: X should be [0,2].

If r(t) = x(t)i + y(t)j + z(t)k, then r''(t) = x''(t)i + y''(t)j + z''(t)k. How you would you normally solve x''(t)=6 for x(t), given x'(0) and x(0) ? What is troubling you?

I bet you are! (plus or minus a twist) Consider I = \int (u + a^2)^{-\frac{3}{2}} \, du. Let u = r^2, so du = 2r\, dr. Then I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}

First, we won't need to consider a basis... First direction: Suppose T^2(x) = T_0(x) = 0 for all x\in V. We want to show R(T) \subseteq N(T). To show that R(T) \subseteq N(T), since an element of R(T) is of the form T(x) for some x in V, we let T(x)\in R(T) and we wish to show that T(T(x)) =...

Right,though the last column of zeros is superfluous. This time taking vectors in row form (which is fine), the matrix you have here (albeit again with superfluous column of zeros) is what you would set up to see if {p, p1, p2, ..., p5} is linearly independent. You have shown they are not. But...

And I would agree with you... Using the substitution u=\ln{x}, the integral \int_e^\infty \frac{1}{x(\ln{x})^3} \, dx becomes -\frac{1}{2} \int_1^\infty \frac{1}{u^2} \, du.

I only wish to elaborate on Lanedance's post with a systematic spin. Think of the matrices in the set S as vectors in the vector space M_{23} (the set of all 2x3 matrices). A basis for M_{23} is \mathbf{e_1} = \begin{pmatrix}1 & 0 & 0\\ 0&0&0\end{pmatrix}, \mathbf{e_2} = \begin{pmatrix}0 & 1...

Are you given some restriction on x (i.e., the domain of f(x) = 3x - 1/(2x)) ? Otherwise, as you have discovered, f(x) does not have an inverse (an inverse function must be single-valued!).

Well, no, but it probably helps to replace x with u in what Jazznaz wrote. To derive \frac{d}{du} \ln{u} = \frac{1}{u} first let y = \ln{u}. Then e^y = u. Now use implicit differentiation to find \frac{dy}{du} (which is \frac{d}{du} \ln{u} ) in terms of u.

Denote f(x0)=a=5.8, f(x1)=b, f(x2)=c, ..., f(x6)=g. The formula you stated (let me replace a and b with x and y, respectively, to avoid confusion with your problem) requires values of f at x, y and their midpoint (x+y)/2, so to apply it to the problem one would group the intervals as {x0, x1...

If A is a linear map that takes basis elements of R3 (say) b1, b2, b3 to c1, c2, c3, respectively, then, by definition, the matrix representation of A with respect to the basis {b1, b2, b3} is given by [c1 c2 c3].

If what I wrote was unclear it is I who should be apologising! Well done!