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For example, I try to solve for x: 12*exp(-2x/4)=0.5, or x^2+3x^4+9x^5=10, is there any method in R to solve such equations?
Try with uniroot function
r1 <- uniroot(function(x) 12*exp(-x/2) - 0.5, c(-1000, 1000))
r2 <- uniroot(function(x) x^2 + 3*x^4 + 9*x^5 - 10, c(-1000, 1000))
r1$root
[1] 6.356108
r2$root
[1] 0.943561
You have to define:
a function of the type f(x) = 0 (and remove the second member of equality)
an interval within which to search for the solution
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What is a more elegant way to write the below function. I am trying to practice my function development skills, and I am simply trying to manually recreate the CIs of a linear model. I am very well aware of the confint(model) function, but still...
jad<-function(x, y) {
model<-lm(y~x)
std.err<-coef(summary(model))[, 2]
coef.model1<-coef(summary(model))[, 1]
upper.ci<-coef.model1+1.96*std.err
lower.ci<-coef.model1-1.96*std.err
print(upper.ci)
print(lower.ci)
}
What about the code below?
jad <- function(x, y) {
`colnames<-`(
coef(summary(lm(y ~ x)))[,1:2] %*% matrix(c(1, 1.96, 1, -1.96), nrow = 2),
c("upper.ci", "lower.ci")
)
}
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I'm having trouble solving a normal distribution problem in R. I'm unfamiliar with the syntax and would like some help.
If X~N(2,9), compute
a. P(X>=2)
b. P(1<=X<7)
c. P(-2.5<=X<-1)
d. P(-3<=X-2<3)
You are looking for the pnorm function. This is the normal CDF. So you want to do something like:
# A
1 - pnorm(2, mean = 2, sd = 9) # = 0.5
# B
pnorm(7, mean = 2, sd = 9) - pnorm(1, mean = 2, sd = 9) # = 0.255
I think you can figure out the last two yourself.
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Is there a way to solve a general recurrence relation of the form
a(n)=a(n-1) * a(n-2)....
I mean I can use the matrix method to solve a relation of the form
F(n)=a1*F(n-1) + a2*F(n-2).......+ ak*F(n-k)
but what to do when there is a '*' sign instead of '+'
Use logarithms:
a(n) = a(n-1) * a(n-2) * a(n-3) * ....
Take log of both sides:
log(a(n)) = log(a(n-1) * a(n-2) * a(n-3) * ...)
Use the fact that log(a * b) = log(a) + log(b) to split up the factors:
log(a(n)) = log(a(n-1)) + log(a(n-2)) + log(a(n-3)) + ...
Now, if you just say that F(n) = log(a(n)) then this equation looks just like your second equation. Use the matrix method to solve for log(a(n)):
log(a(n)) = X
Which leaves:
a(n) = e ^ X
(Assuming you take natural logarithms)
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T(n)=T(n-1) + lgn
My approach is:
Substituting n-1,n-2,n-3
Finally we get,
T(n)=T(1) + lg 2 +lg 3 and so lg n
=> T(n) = lg(2*3*4*5 n)
Hence T(n)=lg(n!).
But they give the answer as nlgn.
Is this a problem for computing complexity? If so then both you and "they" are correct.
O(lg(n!)) = O(lg(n^n)) = O(n lg(n))
More rigorously, from Stirling formula:
lg(n!) = n lg(n) - n + O(ln(n))
Therefore
O(lg(n!)) = O(n lg(n)) + O(n) + O(ln(n)) = O(n lg(n))
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the question is:
x dy/dx = 2y ; y(0)=0
because when i solve this problem the integration constant 'c' gets zero... and i have to find its value in order to calculate a solution to given IVP
Unless I'm mistaken, this question gives c = 0 for y(0) = 0
x*dy/dx = 2y
x*dy = 2y*dx
dy / 2y = dx / x
ln(2y) = ln(x) + c
e^(ln(2y)) = e^(ln(x) + c) = e^ln(x)*e(c)
2y = x + c
solving for y(0) = 0 gives c = 0, as you stated.
Why do you think c must not be 0?