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How can I fit this equation to a set of data (x,y)
y = a x ^ b + c
I tried least square error method (in Maple software) and power low but it doesn't work!
I solved it!
Y = a x ^ b ; Y = y-c
so 'a' and 'b' can be obtained from least square method, if 'c' is known. Therefor error can be calculate as a function of 'c'. (only 'c' and not 'a' or 'b')
No I should minimize the square error witch is a function of only 'c'.
It's possible to solve this by a numeric method.
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Generate a vector of 1000 Poisson random numbers with λ = 3. Make a histogram and a boxplot of the 1000 numbers. Find the expected value of the vector in Rstudio
Try with this:
set.seed(123)
#Code
v <- rpois(1000,lambda = 3)
#Hist
hist(v)
#Boxplot
boxplot(v)
#Mean
mean(v)
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How do I get the coefficient of a regression in R? In the equation of the Regression line, where y = a + bx, the Angular coefficient would be the "b". I would also like to know how to calculate the intercept - the "a" of the equation
I only speak enough spanish to understand a little bit, try:
Model <- lm(y ~ 1 + x)
summary(Model)$coefficients
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I can't figure out how to plot the F distribution in R, given two degrees of freedom using standard normal variates. Any suggestions?
You can use curve()
curve(df(x, df1=1, df2=2), from=0, to=5)
Here is the documentation of curve()
df is the density of the F distribution. This can be found in ?distributions and follows the standard naming conventions dnorm for normal distribution, dt for t distribution, etc. The F distribution has two degrees of freedom parameters. Use pf if you want the CDF.
x = seq(0, 5, length = 100)
plot(x, df(x = x, df1 = 1, df2 = 1))
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Whats is the fastest way to find the maximum root of a cubic function in R?
a x^3 + b x^2 + c x + d = 0
Is there anything wrong with the base function polyroot?
Description
Find zeros of a real or complex polynomial.
An example of a cubic
polyroot(c(1,3,3,1))
# [1] -1+0i -1+0i -1-0i
Here is a function to find the maximum non-complex root of a polynomial...
maxReal <- function(params){
x <- polyroot(params)
reals <- sapply(x, function(i) isTRUE(all.equal(Im(i),0)))
max(Re(x)[reals])
}
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A reviewer of a paper I submitted to a scientific journal insists that my function
f1[b_, c_, t_] := 1 - E^((c - t)/b)/2
is "mathematically equivalent" to the function
f2[b0_, b1_, t_] := 1 - b0 E^(-b1 t)
He insists
While the models might appear(superficially) to be different, the f1
model is merely a re-parameterisation of the f2 model, and this can be
seen easily using highschool mathematics.
I survived High School, but I don't see the equivalence, and FullSimplify does not yield the same results. Perhaps I am misunderstanding FullSimplify. Is there a way to authoritatively refute or confirm the assertion of the reviewer?
If c and b are constant, you can factor them out relatively easily given the property of the power operator:
e^(A + B) = e^A x e^B...
so
e^((c - t)/b) = e^(c/b - t/b) = e^(c/b) x e^(-t/b) = b0 x e^(-t/b)
The latter expression is commonly used to simplify linear differential equation.