Using bisection method to find p3 for f(x) = sqrt(x) - cos(x), I got the right answer on paper p3 = .625
I am having a trouble converting the problem and solving it using Jupyter notebook. Any tips?
Jupyter Notebook code
0.625 is the third guess in the approximation process so the for loop should be completed before the fourth iteration and the code should be as follows.
import math
##problem statement
#finding a root of f(x) = sqrt(x)-cos(x) between the interval [0,1]
#f(0) = -1 f(1) = 0.45969.. so there is at least one root between the interval [0,1]
a = 0
b = 1
maxIt = 100
negative_result = math.sqrt(a) - math.cos(a)
positive_result = math.sqrt(b) - math.cos(b)
for i in range(maxIt):
if i == 3:
print("answer is: ", guess)
break
guess = (a+b)/2
guess_result = math.sqrt(guess) - math.cos(guess)
if guess_result*negative_result >= 0:
a = guess
else:
b = guess
And the output is
0.625
Related
I'm trying to translate a C routine from an old sound synthesis program into R, but have indexing issues which I'm struggling to understand (I'm a beginner when it comes to using loops).
The routine creates an exponential lookup table - the vector exptab:
# Define parameters
sinetabsize <- 8192
prop <- 0.8
BP <- 10
BD <- -5
BA <- -1
# Create output vector
exptab <- vector("double", sinetabsize)
# Loop
while(abs(BD) > 0.00001){
BY = (exp(BP) -1) / (exp(BP*prop)-1)
if (BY > 2){
BS = -1
}
else{
BS = 1
}
if (BA != BS){
BD = BD * -0.5
BA = BS
BP = BP + BD
}
if (BP <= 0){
BP = 0.001
}
BQ = 1 / (exp(BP) - 1)
incr = 1 / sinetabsize
x = 0
stabsize = sinetabsize + 1
for (i in (1:(stabsize-1))){
x = x + incr
exptab [[sinetabsize-i]] = 1 - (BQ * (exp(BP * x) - 1))
}
}
Running the code gives the error:
Error in exptab[[sinetabsize - i]] <- 1 - (BQ * (exp(BP * x) - 1)) :
attempt to select less than one element in integerOneIndex
Which, I understand from looking at other posts, indicates an indexing problem. But, I'm finding it difficult to work out the exact issue.
I suspect the error may lie in my translation. The original C code for the last few lines is:
for (i=1; i < stabsize;i++){
x += incr;
exptab[sinetabsize-i] = 1.0 - (float) (BQ*(exp(BP*x) - 1.0));
}
I had thought the R code for (i in (1:(stabsize-1))) was equivalent to the C code for (i=1; i< stabsize;i++) (i.e. the initial value of i is i = 1, the test is whether i < stabsize, and the increment is +1). But now I'm not so sure.
Any suggestions as to where I'm going wrong would be greatly appreciated!
As you say, array indexing in R starts at 1. In C it starts at zero. I reckon that's your problem. Can sinetabsize-i ever get to zero?
Recently, I got started with Julia's (v1.0.3) DifferentialEquations.jl package. I tried solving a simple ODE system, with the same structure as my real model, but much smaller.
Depending on the solver which I use, the example either solves or throws an error. Consider this MWE, a Chemical Engineering model of a consecutive / parallel reaction in a CSTR:
using DifferentialEquations
using Plots
# Modeling a consecutive / parallel reaction in a CSTR
# A --> 2B --> C, C --> 2B, B --> D
# PETERSEN-Matrix
# No. A B C D Rate
# 1 -1 2 k1*A
# 2 -2 1 k2*B*B
# 3 2 -1 k3*C
# 4 -1 1 k4*B
function fpr(dx, x, params, t)
k_1, k_2, k_3, k_4, q_in, V_liq, A_in, B_in, C_in, D_in = params
# Rate equations
rate = Array{Float64}(undef, 4)
rate[1] = k_1*x[1]
rate[2] = k_2*x[2]*x[2]
rate[3] = k_3*x[3]
rate[4] = k_4*x[2]
dx[1] = -rate[1] + q_in/V_liq*(A_in - x[1])
dx[2] = 2*rate[1] - 2*rate[2] + 2*rate[3] - rate[4] + q_in/V_liq*(B_in - x[2])
dx[3] = rate[2] - rate[3] + q_in/V_liq*(C_in - x[3])
dx[4] = rate[4] + q_in/V_liq*(D_in - x[4])
end
u0 = [1.5, 0.1, 0, 0]
params = [1.0, 1.5, 0.75, 0.15, 3, 15, 0.5, 0, 0, 0]
tspan = (0.0, 15.0)
prob = ODEProblem(fpr, u0, tspan, params)
sol = solve(prob)
plot(sol)
This works perfectly.
However, if a choose a different solver, say Rosenbrock23() or Rodas4(), the ODE is not solved and I get the following error:
ERROR: LoadError: TypeError: in setindex!, in typeassert, expected Float64,
got ForwardDiff.Dual{Nothing,Float64,4}
I won't paste the whole stacktrace here, since it is very long, but you can easily reproduce this by changing sol = solve(prob) into sol = solve(prob, Rosenbrock23()). It seems to me that the error occurs when the solver tries to derive Jacobians, but I have no clue why. And why does the default solver work, but others don't?
Please, could anyone tell me why this error occurs and how it can be fixed?
Automatic differentiation works by passing Dual types through your function, instead of the floats you would normally use it with. So the problem arises because you fix the internal value rate to be of type Vector{Float64} (see the third point here, and this advice). Fortunately, that's easy to fix (and even better looking, IMHO):
julia> function fpr(dx, x, params, t)
k_1, k_2, k_3, k_4, q_in, V_liq, A_in, B_in, C_in, D_in = params
# Rate equations
# should actually be rate = [k_1*x[1], k_2*x[2]*x[2], k_3*x[3], k_4*x[2]], as per #LutzL's comment
rate = [k_1*x[1], k_2*x[2], k_3*x[3], k_4*x[2]]
dx[1] = -rate[1] + q_in/V_liq*(A_in - x[1])
dx[2] = 2*rate[1] - 2*rate[2] + 2*rate[3] - rate[4] + q_in/V_liq*(B_in - x[2])
dx[3] = rate[2] - rate[3] + q_in/V_liq*(C_in - x[3])
dx[4] = rate[4] + q_in/V_liq*(D_in - x[4])
end
That works with both Rosenbrock23 and Rodas4.
Alternatively, you can turn off AD with Rosenbrock23(autodiff=false) (which, I think, will use finite differences instead), or supply a Jacobian.
I'm trying to implement the spiking neuron of the Izhikevich model. The formula for this type of neuron is really simple:
v[n+1] = 0.04*v[n]^2 + 5*v[n] + 140 - u[n] + I
u[n+1] = a*(b*v[n] - u[n])
where v is the membrane potential and u is a recovery variable.
If v gets above 30, it is reset to c and u is reset to u + d.
Given such a simple equation I wouldn't expect any problems. But while the graph should look like , all I'm getting is this:
I'm completely at loss what I'm doing wrong exactly because there's so little to do wrong. I've looked for other implementations but the code I'm looking for is always hidden in a dll somewhere. However I'm pretty sure I'm doing exactly what the Matlab code of the author (2) is doing. Here is my full R code:
v = -70
u = 0
a = 0.02
b = 0.2
c = -65
d = 6
history <- c()
for (i in 1:100) {
if (v >= 30) {
v = c
u = u + d
}
v = 0.04*v^2 + 5*v + 140 - u + 0
u=a*(b*v-u);
history <- c(history, v)
}
plot(history, type = "l")
To anyone who's ever implemented an Izhikevich model, what am I missing?
usefull links:
(1) http://www.opensourcebrain.org/projects/izhikevichmodel/wiki
(2) http://www.izhikevich.org/publications/spikes.pdf
Answer
So it turns out I read the formula wrong. Apparently v' means new v = v + 0.04*v^2 + 5*v + 140 - u + I. My teachers would have written this as v' = 0.04*v^2 + 6*v + 140 - u + I. I'm very grateful for your help in pointing this out to me.
Take a look at the code that implements the Izhikevich model in R below. It results in the following R plots:
Regular Spiking Cell:
Chattering Cell:
And the R code:
# Simulation parameters
dt = 0.01 # ms
simtime = 500 # ms
t = 0
# Injection current
I = 15
delay = 100 # ms
# Model parameters (RS)
a = 0.02
b = 0.2
c = -65
d = 8
# Params for chattering cell (CH)
# c = -50
# d = 2
# Initial conditions
v = -80 # mv
u = 0
# Input current equation
current = function()
{
if(t >= delay)
{
return(I)
}
return (0)
}
# Model state equations
deltaV = function()
{
return (0.04*v*v+5*v+140-u+current())
}
deltaU = function()
{
return (a*(b*v-u))
}
updateState = function()
{
v <<- v + deltaV()*dt
u <<- u + deltaU()*dt
if(v >= 30)
{
v <<- c
u <<- u + d
}
}
# Simulation code
runsim = function()
{
steps = simtime / dt
resultT = rep(NA, steps)
resultV = rep(NA, steps)
for (i in seq(steps))
{
updateState()
t <<- dt*(i-1)
resultT[i] = t
resultV[i] = v
}
plot(resultT, resultV,
type="l", xlab = "Time (ms)", ylab = "Membrane Potential (mV)")
}
runsim()
Some notes:
I've picked the parameters for the "Regular Spiking (RS)" cell from Izhikevich's site. You can pick other parameters from the two upper-right plots on that page. Uncomment the CH parameters to get a plot for the "Chattering" type cell.
As commenters have suggested, the first two equations in the question are incorrectly implemented differential equations. The correct way to implement the first one would be something like: "v[n+1] = v[n] + (0.04*v[n]^2 + 5*v[n] + 140 - u[n] + I) * dt". See the code above for example. dt refers to the user specified time step integration variable and usually dt << 1 ms.
In the for loop in the question, the state variables u and v should be updated first, then the condition checked after.
As noted by others, a current source is needed for both of these cell types. I've used 15 (I believe these are pico amps) from this page on the author's site (bottom value for I in the screenshot). I've also implemented a delay for the current onset (100 ms parameter).
The simulation code should implement some kind of time tracking so it's easier to know when the spikes are occurring in resulting plot. The above code implements this, and runs the simulation for 500 ms.
I my main objectif is to obtain the same results on SAS and on R. Somethimes and depending on the case, it is very easy. Otherwise it is difficult, specially when we want to compute something more complicated than the usual.
So, in ored to understand my case, I have the following differential equation system :
y' = z
z' = b* y'+c*y
Let :
b = - 2 , c = - 4, y(0) = 0 and z(0) = 1
In order to resolve this system, in SAS we use the command PROC MODEL :
data t;
do time=0 to 40;
output;
end;
run;
proc model data=t ;
dependent y 0 z 1;
parm b -2 c -4;
dert.y = z;
dert.z = b * dert.y + c * y;
solve y z / dynamic solveprint out=out1;
run;
In R, we could write the following solution using the lsoda function of the deSolve package:
library(deSolve)
b <- -2;
c <- -4;
rigidode <- function(t, y, parms) {
with(as.list(y), {
dert.y <- z
dert.z <- b * dert.y + c * y
list(c(dert.y, dert.z))
})
}
yini <- c(y = 0, z = 1)
times <- seq(from=0,to=40,by=1)
out_ode <- ode (times = times, y = yini, func = rigidode, parms = NULL)
out_lsoda <- lsoda (times = times, y = yini, func = rigidode, parms = NULL)
Here are the results :
SAS
R
For time t=0,..,10 , we obtain similar results. But for t=10,...,40, we start to have differences. For me, these differences are important.
In order to correct these differences, I fixed on R the error truncation term on 1E-9 in stead of 1E-6. I also verified if the numerical integration methods and the hypothesis used by default are the same.
Do you have any idea how to deal with this problem?
Sincerely yours,
Mily
I am looking for a solution:
A= {0,1,2,3,4};
F(x) = 3x - 1 (mod5)
Could you help me to find the inverse. I am struggling with this as it seems to be not to be onto or 1to1.
Thank you for your help.
x = 2y + 2, where y = F(x)
-> 3x - 1 = 3(2y+2) - 1 = 6y + 5 = y (mod 5)
edit: if you want this to be evaluated for the list of principal values mod 5 [0,1,2,3,4], just evaluate 2y+2 for each of these, and what you get is [2,4,1,3,0]. Which, if you plug back into 3x-1, you get [0,1,2,3,4] as expected.