r,transformation,interaction,anova,sqrt

Just to close the question: interaction.plot(Gastropods$Zone, Gastropods$Species, Gastropods$sqrtAbundance, main= "Gastropod Interaction Plot", xlab = "Gastropod Zone", ylab= "Mean of SQRT Gastropod Abundance", legend = TRUE) ...

You need to statically import the method to use it as an unqualified method import static java.lang.Math.sqrt; or use double tosser = Math.sqrt(...); (importing from java.lang is unnecessary since those classes are imported by default)...

java,algorithm,numbers,primes,sqrt

1. Why is faster to compute i*i, sqrt (n) times. than sqrt (n) just one time ? Look at the complexities below. The additional cost of computing square root. 2. Why Math.sqrt() is faster than my sqrt() method ? Math.sqrt() delegates call to StrictMath.sqrt which is done in hardware or...

It's likely because bsqr is negative and taking the sqrt of a negative number doesn't work too well. >>> import math >>> math.sqrt(-1) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: math domain error Check your algebra/inputs. c (the hypoteneuse) should always be bigger than either...

The compiler and linker options with arguments must be separate, you can't compile them like you do. Make them separate, like e.g. gcc -Wall -o "%e" "%f" -lm ...

Most floating point numbers can't be stored precisely in Lua's number type (C's double by default). math.sqrt(2) is one of them. If you try: print(2 - math.sqrt(2) ^ 2) Output: -4.4408920985006e-016 which is a very small number, but still, making the two numbers not exactly equivalent. This floating point precision...

Try arrayfun: B = arrayfun(@(i) sqrt(A{i}), 1:length(A), 'Uni', 0); ...

The time complexity itself isn't different between the two loops (unless the complexity of sqrt itself is dependent on the number) but what is different is how many times you're computing the square root. Without optimisation like the compiler automatically moving loop invariant stuff outside of the loop (assuming that's...