similar to: numerical integrals

Greetings, Sorry, the last message was sent by mistake! Here it is again: I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix

Dear ALL I have this function in R: func_LN <- function(data){ med_ge <- matrix(c(rep(NA,nrow(data)*ncol(data))), nrow = nrow(data), ncol=ncol(data), byrow=TRUE) T <- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) Tdiff<- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) T1<- c(rep(NA,ncol(data)))

Hello, I'm trying to create a line graph with a legend, but have no success controlling the legend. Since nothing I've tried seems to work, I must be doing something systematically wrong. Can anyone point this out to me? Here's my data: > weights # A tibble: 1,246 ? 3 Date J K <date> <dbl> <dbl> 1 2000-02-13 133 188 2

Rui, thanks so much for your help. Your explanation and example were clear and concise. Thanks for taking the time and effort to help me. -Kevin On 5/12/23 16:06, Rui Barradas wrote: > ?s 14:24 de 12/05/2023, Kevin Zembower via R-help escreveu: >> Hello, I'm trying to create a line graph with a legend, but have no >> success controlling the legend. Since nothing I've

Suppose I have data organized in the following way: (P_i, M_j, S_k) where i, j and k and indexes for sets. I would like to analyze the data to get for example the following information: what is the average over k for (P_i, M_j) or what is the average over j and k for P_i. My question is what would be the way of doing this in R. Specifically how should I write the data in a csv file and how do I

See below. On 5/16/23 10:52, Christopher Ryan wrote: > I"m more of a lattice guy than a ggplot guy, but perhaps this is part of > the problem: > > ..... > ? ? ?geom_point(aes(y = m_K, color = "red")) +? ##### >> you've > associated "K" with the color red > ? ? ?geom_smooth(aes(y = m_K, color = "red")) + >

I cannot understand the following error printed out when I try to get the extrema of my time series. I would appreciate some suggestion as I really cannot interpret the error. I might not be using a proper set of parameters in calling such functions. I am learning by doing ... > aa.peak <- wavCWTPeaks (aa.tree) Error in `row.names<-.data.frame`(`*tmp*`, value = c("1",

I have a P3-800 with two IDE drives in a software RAID1 configuration. Each drive is on a separate IDE channel. Now anytime there is HD activity, I hear "beeps" and "cutting out" on a call using the X100P card. I ran the zttest program, and discovered HD activity would drop the accuracy down to between 2% and 50%. However I noticed if I disabled one drive in the RAID1

Based on what follows, the most favourable construction is that the documentation, by failing to say that the function should be used only for completely balanced designs, is deficient. Even for such designs, there is a bug that needs correction. The discussion is lengthy because I think it important to document, at least wrt effects and means, exactly what model.tables() does. My preference is

Hi R users, I’d like to get an expected value for a minimum value from order statistics of sample size, say, 5 for standard normal distribution, and the formula would be 5* Integral (from -inf to Inf) x*f(x)* (1-F(x))^4,where f(x) and F(x) are a standard normal density and a cumulative distribution function respectively. After searching R posts, I applied the following double integral

Hi folks, I am having a question about efficiently finding the integrals of a list of functions. To be specific, here is a simple example showing my question. Suppose we have a function f defined by f<-function(x,y,z) c(x,y^2,z^3) Thus, f is actually corresponding to three uni-dimensional functions f_1(x)=x, f_2(y)=y^2 and f_3(z)=z^3. What I am looking for are the integrals of these three

reorder() is probably the best way to order the levels in a vector without manually specifying the order. But reorder() orders by default in an increasing order: "The levels are ordered such that the values returned by ?FUN? are in increasing order." Is there a way to do what reorder() does, but order the levels according to a _decreasing_ order of the values? Sverre

Hi, New to R, using version 2.0.1 (2004-11-15) on debian Linux (sid), kernel 2.6.8-2-686. I have data in files with separate vectors on each row of the file, e.g., $ cat /tmp/stats freq,0,1,2,3,4,5,6,7,8,9,16,17,18,19,20,... noise,49,47,48,48,50,47,48,47,46,50,48,54,49,47,49,... signal,99,0,100,0,0,100,0,100,100,0,100,101,100,0,0,... pctrcv,5,0,5,0,0,5,0,5,11,0,5,5,5,0,0,...

Scot, Thanks for the info. I will try your code out to verify the result, but before I do that, will your code (SAS and R) work with variable names that are longer than 8 characters long without truncating the variable name in R? Also, I wonder about using your method or the PROC EXPORT method with larger data sets. The data sets I will be working with for the most part will not be that large,

Hi R experts, I have the following timeseries data: #example data structure a <- c(NA,1,NA,5,NA,NA,NA,10,NA,NA) c <- c(1:10) df <- data.frame(timestamp=a, sequence=c) print(df) where i would like to linearly interpolate between the points 1,5, and 10 in 'timestamp'. Original timestamps should not be modified. Here the code I use to run the interpolation (so far): # linear

Hi all, newbie question: I am trying to set up a very simple loop without succeeding. Let's say I have monthly observation of two variables for a year - Sales_2012_01, Sales_2012_02, Sales_2012_03, .... (total sales for jan 2012,feb 2012, etc.) - Customers_2012_01, Customers_2012_02, .... (total number of customers for jan 2012, etc.) and I want to create new monthly variables in

Signed-off-by: Scott Seago <sseago at redhat.com> --- AUTHORS | 17 ++++++ README | 10 +++ conf/ovirt-agent | 12 ++++ conf/ovirt-db-omatic | 12 ++++ conf/ovirt-host-browser | 12 ++++