Recent content by MrBailey

my mind is slipping...sorry the convolution of f(x) and g(x) from -inf to inf

Hi, all. Just a little confused over this one (okay...a lot confused): convolution of f(x) and g(x) from -inf to inf where f(x) = e^-x and g(x) = x I would really appreciate some pointers on this one. thanks, Bailey (edit) forgot the range

Okay, is this on the right track? So, S_n = a_1 + a_2 + \ldots + a_n and S'_n is a rearrangement of the series where the terms move no more than 2 places. Then |S'_n - S_n| \leq |a_{n-1}| + |a_n| + |a_{n_1}| + |a_{n_2}| where n < n_1 < n_2 From that we get: S_n - (|a_{n-1}| + |a_n| +...

Hello, I could use a big helping hand in trying to understand an example from a text. Let's say I have a convergent series: S=\sum_{n=1}^{\infty}{a_n} Okay, so now: \sum_{n=1}^{\infty}{a'_n} is a rearrangement of the series where no term has been moved more than 2 places. So, the exercise...

Got it... u_n=\frac{k}{(n-1)^2} where k is a constant. This is just the series: k \sum{\frac{1}{n^2}} which we know convergesWhew! Bailey

Hi all! Here's something I'm having difficulty seeing: Suppose u_n > 0 and \frac{u_{n+1}}{u_n} \leq 1-\frac{2}{n} + \frac{1}{n^2} if n \geq 2 Show that \sum{u_n} is convergent. I'm not sure how to apply the ratio test to this. It looks like I would just take the limit. I get: lim_{n...

Ugggh! I'm so blind sometimes...must be all the turkey I ate yesterday. I see the geometric series. Thanks, PM. Bailey

Hello all! I have the following infinite series: \frac{10}{x}+\frac{10}{x^2}+\frac{10}{x^3}+\ldots How would I find a function, f(x), of this series? I know the series converges for \vert x \vert > 1 I think the function is: f(x) = \frac{10}{x-1} but I'm not sure how to get it. Thanks, Bailey

Thanks...I wanted to make sure I was on the right track. Best wishes, bailey

I edited the above...hopefully it looks clearer. Isn't the derivative of the natural log 1/x and you keep on doing it "n" times...so you always end up with an "x" in the denominator when applying L'Hopital's rule...until you get 1/x with all of the "n" terms in the numerator. Please correct me...

Hi all. I'm slightly confused with the following limit prob: \lim_{x\rightarrow \infty} \frac{(ln (x))^n}{x} which I know = 0. (n is a positive integer) It looks like it's of indeterminate form, that is \frac{\infty}{\infty} Using L'Hopital's, it looks like you get another indeterminate form...

thanks! I see it now. Bailey

Hello out there. I hope everyone is doing well. I could use a little guidance on this: suppose f is continuous for all x, and \lim_{x\rightarrow -\infty}f(x) = -1 and \lim_{x\rightarrow +\infty}f(x) = 10 Show that f(x) = 0 for at least one x I know I need to use the Intermediate Value...

thanks...much clearer now Bailey

Hi, all. I'm working on some proof by induction problems. While I understand the concept, this one threw me for a loop. Let x_1=\sqrt{2} and x_{n+1}=\sqrt{2+x_n} Show that x_n < x_{n+1} I'd greatly appreciate help with this. Thanks, bailey