Recent content by dalle

for any other point x the radius of convergence is |x|. exp(-1/x^2) is holomorphic on C\{0} and the radius of convergence is the distance to the next singularity (which the function has only one at the point z=0)

dear wdlang, your function needs to be real analytic, being smooth is not enough. There is an excelent discussion of the difference by Dave Renfro at http://mathforum.org/kb/thread.jspa?forumID=13&threadID=81514&messageID=387148#387148 if the radius of convergence is infinite, than your...

dear ceasius, you are given a sequence of functions s_1 = \frac{4}{\pi} \sin x , s_2 =\frac{4}{\pi} (\sin x + \frac{\sin (3 x}{3})), s_3 = \frac{4}{\pi} ( \sin x + \frac{\sin (3x)}{3} + \frac{\sin(5x)}{5}),.. your task is to prove that \frac{d s_n(x)}{dx} =\frac{sin(2nx)}{\pi \sin x} ,x...

\sqrt{n+1}+\sqrt{n}> 2 \sqrt{n}. taking the inverse on both sides yields \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2 \sqrt{n}} your professor is just using a smaller upper bound for s_n. professors like to use bounds that are as small as possible:smile:

don't panic. all you have is a finite sum. the formula (f+g)' = f' + g' does work for finite sums! the only problem left is finding the right trigonometric identity to use!

you can find the formula in chapter 4.5 The Cotangent of "On Euler footsteps. Uppsala Lectures on Calculus" by Evgeny Shchepin. you can download it at http://www.pdmi.ras.ru/~olegviro/Shchepin/index.html or http://at.yorku.ca/i/a/a/z/20.htm

dear Haywood I don't know what kind of asypmtotic expasion you are looking for, so a short hint on how to solve the equation approximatly must suffice. For the root near 0 substitute the left hand side with a taylor polynomial. For the other root take logarithms on both sides.

dear foxjwill your solution is correct( more on this at the end) . Why can't we stop there? we know that that if a_1=a then we will get a constant sequence. we do not know if it will converge to a if we start at any other number! we will have to prove it! Let's consider an example: a_{n+1}...

an outline of your prove. a and b are different real numbers. d=abs(a-b) is a fixed real number depending on a and b \lim_{n \rightarrow \infty} a_n = a \in \mathbf{R} means that \forall \epsilon >0 \exists N(\epsilon) such that if n > N(\epsilon) \Rightarrow abs(a- a_n )< \epsilon ...

dear braindead101 i gave you an example of an function where U(s) holds for all s because you seemed unsure about what the meaning of U(s) is. Of course you are looking after a function which has a unique minimum (so that p(x) and q(x) will be satisfied).

dear sculptured there are many ways in topology to prove that the set A={ |z - 4| >= |z| : z is complex} is closed. I will show you 2 ways that don't require you to think of boundarys, interior points or exterior points. 1. you can prove that complement of A is an open set in C. Let's call...

\mathbf{R} \backslash \{0\} is the set R with the element 0 removed. well i will give you an example of what i had in mind. consider the function f(x)=x. U(x) will hold for all x because f is a bijective map, so the inverse image of each y \in R will consist of a set with one element.

dear foxjwill supose the limit a_n does exist and is a . then because the square root is a contious function , the limit a must be a solution of a = \sqrt{x+a} which you should be able to solve. Next you will have to show that a_n is constant if a_0 =a monoton increasing when -x<a_0 <...

dear braindead101 let us consider a function f such that f(0) =0 and f(x) \neq 0 \forall x \in \mathbf{R}\backslash\{0\} . Does U(0) hold?

expressing the summand as \frac{ \left( 1 + \frac{1}{n} \right)^n }{n} does help, you just have to give up finding a test but consider finding a divergent minorante. \frac{1}{n} < \frac{ \left( 1 + \frac{1}{n} \right)^n }{n} and we know that \sum_{n=1}^{\infty} \frac{1}{n} = \infty