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The other symbol you can use is "-", e.g. A - B. I don't know which one is more standard though. Personally, I like "-" better.

I'm know the statement is true if a[k] > 0 for all k. I'm sure there is a conditionally convergent series that makes the statement false but I can't think of any.

[math]a_n^2 = \frac{1}{n^2}[/math] [math]\sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2)[/math] which does converge.

Your IP changes everytime you connect to the internet (or whatever network you are on that is running DHCP) or when the lease time of the IP you were given expires. No. Top level domains are given to you. If you want a DNS server, try BIND. It's very powerful. Running a DNS server for one website is overkill in my opinion. Just ask your ISP to do it for you.

I know "word" means 4 bytes, but this wasn't explicitly given. My point is that the example gives useless info. that I don't need and doesn't give info. that I do need to understand the example.

I hope this isn't to over-the-top for this forum. I have some comments on the following example from a book of mine on parallel computing: Consider a computation running on a machine with a 1 GHz clock, 4-word cache line, single cycle accesss to the cache, and 100 ns latency to DRAM. The computation has a cache hit ration of 1 KB of 25% and 32 KB of 90%. Consider two cases: first, a single threaded execution in which the entire cache is available to the serial context, and second, a multithreaded execution with 32 threads where each thread has a cache residency of 1 KB. If the computation makes one data request in every cycle of 1 ns, in the first case the bandwidth requirement to DRAM is one word every 10 ns since the other words come from the cache (90% cache hit ratio). This corresponds to a bandwidth of 400 MB/s. In the second case, the bandwidth requirement to DRAM increases to three words every four cycles of each thread (25% cache hit ratio). Assuming that all threads exhibit similar cache behavior, this corresponds to 0.75 words/ns, or 3 GB/s. My basic comment is on the initial information about the computer in the example. They include irrelevant information such as the cache line and the latency of memory but forgot to include what the size of each word is in bytes! 0.75 words/ns = 0.75 gigawords/s or 750 megawords/s. If 750 megawords is equivalent to 3 gigabytes, then each word is 4 bytes. Judging by this example, do you think this books is weird? I think it is. I will post another weird example from this book after I get some feedback here.

Both stack and heap are areas in memory. The difference between the two comes about from there usage (in terms of memory management). A stack is an area of memory that behaves like a stack ADT. It is used for storing information/objects that will not grow in size and will not be moved around in memory (in other words, the stack is used for storing static data objects). The stack requires very little janitor work from the OS/programmer. The heap on the other hand, is a dynamic area of memory. You can use the heap to create data objects that are likely to grow (like vectors, maps, sets, etc.). The heap tends to suffer from fragmentation issues and needs to be garbage collected every so often to maintain its usability. Hence, the heap require much intervention from the memory management system of the OS or the programmer.

I can't seem to formulate any proper delta-epsilon argument to prove the continuity of the arcsin function. Maybe using the inverse function theorem is the only viable path.

I think the inverse function theorem is overkill. To prove that a function of a single variable x is contiunous on at a if: 1. f(a) is defined, 2. [math]\lim_{x \to a} f(x)[/math] exists, and 3. [math]\lim_{x \to a} f(x) = f(a) [/math]. 1 is necessarily trivial. You can prove 2 and 3 using a delta-epsilon argument which is what I think is being sought for.

Just some comments on the nomenclature being employed. You might want to stick with the word derivative here (i.e. the derivative of a function...) since the word differential is reserved for something else. "d" isn't really an operator. The operator here is d/dx. Again, the word differential is misused here. I hope my critiques are helpful.

[math]-\log_a{x} = y \equiv \log_a{x} = -y[/math] Let u = -y. Then, [math]\log_a{x} = u \rightarrow a^u = x \equiv a^{-y} = x[/math]

Make the substitution [math]\log_x{3} = \log_3{3}/\log_3{x}[/math] Manipulate the equation until you get [math](\log_3{x})^2 - \log_3{x} - 2 = 0[/math] Then make the substitution [math]u = \log_3{x}[/math] Solve for u, then solve for x.