Recent content by jjou

There you go. Can you do the rest of the problem now?

Thanks! This makes sense. So... \lim_{x\rightarrow\infty}\frac{2^x}{x^{12}}=\lim_{x\rightarrow\infty}\frac{(\ln2)2^x}{12x^{11}}=...=\lim_{x\rightarrow\infty}\frac{(\ln2)^{12}2^x}{12!}=\infty Which means that for very large x, 2^x does eventually exceed x^12, which gives us the third...

If that is indeed the problem in the book, fine. Then answer my question: What is A\cap A' for any set A? In other words, what is the intersection of a set and its complement?

I tried integration by parts too, and it was giving me trouble. So I did some research. The solution I found hopefully is not the only one, since it's a little complicated (in my opinion). Here are the basic steps: 1. Show that G(\frac{3}{2})=\frac{1}{2}G(\frac{1}{2}) where G is the gamma...

I just realized that I had a typo in my original response. I was using the variables k, m, n for most of it and then switched to k, l, m at the very end... I have since edited the post just to keep it consistent. Let's back up a bit: We know a is even, a=2k. so a^3=8k^3=2b^3+4c^3. Let's...

First of all, your solution will be an ordered triplet (a, b, c) - so you cannot apply the well ordering principle to your solution set. You CAN however apply it to the set of a's for which there exist corresponding b and c which form solution triples. In other words, your idea is perfectly...

Are you sure you typed the problem correctly? If so, all of Statdad's deductions are correct so far. Now, when we look at case 3: w\in A\cap X\cap A'. This says that w is an element of A and w is an element of A' (the complement of A). What is wrong with this statement?? Another way to...

(Problem from practice math subject GRE exam:) At how many points in the xy-plane do the graphs of y=x^{12} and y=2^x intersect? The answer I got was 2, but the answer key says 3. Intuitively, by the shape of their graphs, I would say two. I tried to calculate actual values for x...

Easier method: Since delta > 0, there exists some natural number N which satisfies 1/N < delta. Then \frac{(1-\epsilon)^{n+1}}{n+1}<\frac{(1-\epsilon)^n}{n}<\frac{1}{n}<\frac{1}{N}<\delta Thanks so much! :)

Ah, I dropped a "ln" in the numerator in that last inequality - have since changed it. Is it right now? n>\frac{\ln{\delta}}{\ln{(1-\epsilon)}} Or is something else off?

Aha, I changed my mind. \frac{(1-\epsilon)^{n+1}}{n+1}<(1-\epsilon)^{n+1}<(1-\epsilon)^n<\delta Which holds iff n>\frac{\ln{\delta}}{\ln{(1-\epsilon)}}. Yes?

Is this inequality supposed to hold for every x (in the set on which f_n converges uniformly)? Yup. \lim_{n\rightarrow\infty}\int_0^1\frac{x^n}{1+x^n}dx=\lim_{n\rightarrow\infty}\int_0^{1-\epsilon}\frac{x^n}{1+x^n}dx+\lim_{n\rightarrow\infty}\int_{1-\epsilon}^1\frac{x^n}{1+x^n}dx Then...

To simplify the entries of R_{-\theta}, think about a point on the unit circle. First, how do the x and y coordinates relate to the sine and cosine functions? Secondly, pick a \theta. Look at the x and y values corresponding to that \theta. How do they relate to the x and y values...

Just a note: unless the above is a typo, it seems that you have a pretty severe misconception about vectors. a, c, and d are elements of \mathbb{R}, i.e. scalars. Thus none of them can "be the zero vector." The zero vector in this case (since we are working in \mathbb{R}^3) is the vector (0...

To show that U\cap W\neq\{(0,0,0)\}, you need to find some other vector (x,y,z) which belongs in both U and W where at least one of x, y, or z is not zero. In other words, if (a, 0, a) = (c,d,c+2d), what equations can you set up to solve for a, c, and d? Do all three variables have to equal 0?