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I know it should be trivial, but tonight I'm not finding a solution.
Suppose I have a series of float in a given range such [0.25, 1.0]. For example:
{0.25, 0.625, 1.0}
What's the correct way to transform them in order to map the [0.25,1.0] interval to [0.0,1.0]?
The example sequence should become:
{0.0, 0.5, 1.0}
Second question, how to generalize that? How is the correct way to map a given interval [a,b] to [0,1]?
fun lin-map(list) :=
mx = max(list), mn = min(list)
return [ (x - mn)/(mx-mn) | x <- list ]
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I want to solve for x:
1 = a b e^(b(x-c))
That's where I'm stuck. I want to solve for x, and all the other letters are constants. I've forgotten how to solve that equation with the embedded "b(x-c)" for x.
Use logarithm
e^(b(x-c)) = 1/(ab)
ln(e^(b(x-c))) = ln(1/(ab))
b*(x-c) = -ln(a*b)
x-c = -ln(a*b)/b
x = c - ln(a*b)/b
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I have a number of the vector with the numbers.
test <- 0.495
vector <- c(0.5715122, 2.2860487, 5.1436096, 9.1441949)
This vector is the need to take an approximate number to the number 0.495.
Help me.
If I've understood correctly, you want to extract the value from a vector that is closest to your test value.
vector[which.min(abs(vector - test))]
#[1] 0.5715122
If two different values could be closest, you could do this:
vector <- c(0.5715122, 2.2860487, 5.1436096, 9.1441949, 0.4184878)
tol <- sqrt(.Machine$double.eps)
vector[which(abs(vector - test) - min(abs(vector - test)) < tol)]
#[1] 0.5715122 0.4184878
tol is a tolerance accounting for floating point accuracy and usually chosen based on help(".Machine").
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Lets say I have a vector c(1,2,3,4,5,6,7,8,9) how can I get a print of just 1,3,and 5 ?
For a sample of n random elements from vector X you can use sample(x = X, size = 3, replace = FALSE). To get the ith element of X you simply use X[i].
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I have googled for the meaning of ◁ without avail.
Can anyone explain to me what x ◁ y means? For example:
z ⊢ x ◁ y
could it mean that we can derive an agreement between x and y from z?
Thanks.
I had a look at the Wikipedia List of Mathematical Symbols and it seems that this symbol defines a normal subgroup.
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Is there a simplification for x^(log base (1/x) of N)?
Yes, it's 1 / N (the reciprocal of N), provided of course that both N and x are positive and x != 1.
x^(log[1/x](n)) = e^(log[1/x](n)*ln(x)) = e^((ln(n)/ln(1/x))*ln(x)) = e^(ln(n)*ln(x)/(-ln(x)) = e^(-ln(n)) = 1/n