logarithm

I have plotted to vectors against each other, and they are already logarithmic and everything is fine with that. But now that I have my plot i want the grid to be logarithmic. I write "grid on" in my code, and I think there should be a way to do this in the plot, but I can't remember how. How do I make the grid logarithmic? If you use loglog , semilogx or semilogy instead of plot , the grid will automatically be on a log scale for the corresponding axes when using grid on . If you have already plotted the axes, you can execute the following on the command line: set(gca,'yscale','log') %# to

I have the following data: someFactor = 500 x = c(1:250) y = x^-.25 * someFactor which I show in a double logarithmic plot: plot(x, y, log="xy") Now I "find out" the slope of the data using a linear model: model = lm(log(y) ~ log(x)) model which gives: Call: lm(formula = log(y) ~ log(x)) Coefficients: (Intercept) log(x) 6.215 -0.250 Now I'd like to plot the linear regression as a red line, but abline does not work: abline(model, col="red") What is the easiest way to add a regression line to my plot? lines(log(x), exp(predict(model, newdata=list(x=log(x)))) ,col="red") The range of values for x

can someone please tell me How to find antilog for a number using java program? i am new to this java Math.log(10) gives the log value. now I want to take this output and verify using antilog that program is giving right value.please help me. mathematically: e^(ln x) = x in java: Math.pow(Math.E, (Math.log(x)) == x; //equals true You can use Logarithm class in Java to calculate log and antilog. More on this is provided here . 来源： https://stackoverflow.com/questions/18209676/how-to-find-antilog-for-a-number-using-java-programm

This question already has an answer here: Is there any fast algorithm to compute log2 for numbers that are all power of 2? 2 answers Fastest way of computing the power that a “power of 2” number used? 5 answers Is there an efficient way to find the log2 of a number assuming it's a power of 2. I know the obvious ways like having a table or for (log2=0;x!=1;x>>=1,log2++); But I am wondering if there is a more efficient/elegant way. You can just count the number of leading or trailing zero bits, because any exact power-of-two is represented as a single 1 bit, with all other bits 0. Many CPUs have

I have a short program here: Given any n: i = 0; while (i < n) { k = 2; while (k < n) { sum += a[j] * b[k] k = k * k; } i++; } The asymptotic running time of this is O(n log log n). Why is this the case? I get that the entire program will at least run n times. But I'm not sure how to find log log n. The inner loop is depending on k * k, so it's obviously going to be less than n. And it would just be n log n if it was k / 2 each time. But how would you figure out the answer to be log log n? For mathematical proof, inner loop can be written as: T(n) = T(sqrt(n)) + 1 w.l.o.g assume 2 ^ 2 ^ (t-1)<

I'm trying to write a Java program that can take values and put them into a formula involving log base 10. How can I calculate log 10 in Java? Mysticial It looks like Java actually has a log10 function: Math.log10(x) Otherwise, just use math: Math.log(x) / Math.log(10) http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html Louis Wasserman Math.log10(x) suffices for floating-point numbers. If you need integer log base 10, and you can use third-party libraries, Guava provides IntMath.log10(int, RoundingMode) . (Disclosure: I contribute to Guava.) sri You can do this too class log { public