Recent content by Johnson04

You may also integrate in the following way: I = \int\limits_{0}^{\infty}\frac{1}{\sqrt{1+y^2}}\int\limits_{0}^{\infty}\frac{1}{1+(\frac{x}{\sqrt{1+y^2}})^2}\frac{1}{\sqrt{1+y^2}}dxdy = \int\limits_{0}^{\infty}\frac{1}{\sqrt{1+y^2}}[\arctan(\frac{x}{\sqrt{1+y^2}})]|_{0}^{\infty}dy =...

This may not be true. Think about this: P_k=\{a, b-\sum\limits_{n=1}^{k-1}\frac{b-a}{2^n}}, b-\sum\limits_{n=1}^{k-2}\frac{b-a}{2^n}}, \cdots, b-\frac{b-a}{4}-\frac{b-a}{2}, b-\frac{b-a}{2}, b\}

Well, I believe I totally understand the related properties of the Cantor set now. First, not all points in Cantor set are endpoints of the removed open sets. Second, the complement of the Cantor set only contains countably infinitely many open sets, that means the Cantor set does contain only...

I think you just need to check their respective definitions. A \subseteq B just means that \forall x\in A, x\in B. This definition does not say anything about the elements in B. In other words, \forall x\in B, it could be either in A or not in A. If \forall x\in B, implies x\in A, then A=B; if...

Thanks Dick! I think I see what you mean. Each interval contains a unique rational number, because the set of all rational numbers is dense on real line. On real line there are only at most countably infinite many open intervals. Actually this conclusion make the proof of the Cantor set is...

I think that was my question: if the complement of the Cantor set on [0,1] is the set of uncountably infinitely many disjoint open sets. I think it is, because the end points of the open sets form the Cantor set. It can be proved that the Cantor set is uncountable set by using the Cantor's...

Thanks a lot. I see, since any points in a open set are interior points, the points in the any union of the open sets are still interior points, accordingly, the union of the open sets is still open, no matter if the union is finite, countably infinite or uncountably infinite. Well, I was...

This is not a homework problem, just a question from a discussion with my classmates about the Cantor set. The original goal is to prove Cantor set is closed. My earlier attempt is to show the complement of the Cantor set is open. Since when construct the Cantor set each time the sets removed...

For this problem, in order for the group G to be cyclic, is the abelian condition necessary? In other words, if the problem is restated as: "if a finite group of order pq, where p and q are distinct primes, the the group is cyclic", is it still true? The reason I asked this question is that...

Homework Statement Let G be a finite group which possesses an automorphism \sigma such that \sigma(g)=g if and only if g=1. If \sigma^2 is the identity map from G to G, prove that G is abelian. Homework Equations Show that every element of G can be written in the form x^{-1}\sigma(x) and...

Homework Statement Let E be the set of all x\in [0,1] whose decimal expansion contains only the digits 4 and 7. Is E compact? Is E perfect? Homework Equations The Attempt at a Solution My answer is: E is compact and perfect. By Heine-Borel theorem (E is compact equivalent to E...

Homework Statement u''(x) - (x^6 + 3*x^2)*u(x) = 0, u(-1) = u(1) = 1, -1 <= x <= 1 The Attempt at a Solution Consider v''(x) - s^2 * v(x) = 0, the auxiliary equation is: r^2 - s^2 = 0. Since (x^6 + 3*x^2) >= 0, s >= 0. Thus, I got r = s or -s. Suppose v(x) = a*exp(s*x)+b*exp(-s*x)...

Thanks for replying! \lim \limits_{\vec h \rightarrow 0} \frac{|\sum \limits_{i=1}^n h_i^2|}{|\vec h|} was obtained from \lim \limits_{\vec h \rightarrow 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h|}{|\vec h|} . Since I defined the linear transform at \vec x =...

Homework Statement \vec f_1 : \bf R^n \rightarrow R^1 , \vec f_1(\vec x) = \|\vec x\|^2, calculate \vec f_1'(\left[ \begin{array}{c} 1 \\\vdots\\ 1 \end{array} \right]) Homework Equations N/A The Attempt at a Solution I calculated the derivative as follow: Since \vec f_1(\vec x) =\|\vec x...

Once you find the general solution, which is in the form: x(t) = constant * something or something + constant, by using x(0) = 1, you can determine the specific values of the constants. In other words, you can use the initial condition to obtain the specific solutions.