You could use a spline to interpolate the points.
(And I'd consider increasing the number of points if possible, say to 200).
Then use a root finder, such as uniroot(), to solve for
f(i) - k
Where, k (a constant), would be 1e6, based on your example.
There are a number of variations on this
If I run this:
```
Y = c(1, 2, 4, 8, 16, 32, 64, 128)
X = 0:7
plot(Y~X, log='y')
model <- lm(log10(Y) ~ X)
abline(model)
predict(model, data.frame(Y=log10(100)))
```
I get a funny answer:
```
1 2 3 4 5 6 7 8
Thanks, I'll check it out. I ran the simulation and I got:
```
t = 1, N = 20,000
t = 2, N = 40,000
t = 3, N = 80,000
t = 4, N = 160,000
t = 5, N = 320,000
t = 6, N = 640,000
t = 7, N = 1,280,000
```
Hence the answer is t=6.{...} but the problem is to get that
fractional value. Would be possible
On 24/01/2021 2:57 p.m., Luigi Marongiu wrote:
Hello
I am trying to simulate a PCR by running a logistic equation. So I set
the function:
```
PCR <- function(initCopy, dupRate, Carry) {
ROI_T = initCopy
A = array()
for (i in 1:45) {
ROI_TplusOne <- ROI_T * dupRate * (1 -
Hello
I am trying to simulate a PCR by running a logistic equation. So I set
the function:
```
PCR <- function(initCopy, dupRate, Carry) {
ROI_T = initCopy
A = array()
for (i in 1:45) {
ROI_TplusOne <- ROI_T * dupRate * (1 - ROI_T/Carry)
A[i] <- ROI_TplusOne
ROI_T <- ROI_TplusOne
5 matches
Mail list logo
