Rest of division (f(x) mod g(x) = h(x)) I mean for example (3.15 mod 3.14 = 0.01).
In General what are Derivatives of that kind of function ?
dh(x)/dx
dh(x)/df(x)
dh(x)/dg(x)
As
h(x) = f(x) - (f(x) div g(x)) * g(x)
you will need to define a derivative for the integer-valued function f(x) div g(x). Depending on what you really had in mind, this involves the Dirac delta distribution.
Related
To approximate the function with Cheboshev polynomials, it is necessary to operate on the interval [-1,1]. How can these constants be recalculated if I want to approximate on another interval?
specifically, I use maple and the following loop:
(https://i.stack.imgur.com/TWT74.png)
but I don't know how to modify the function to calculate in an interval, for example [-pi,pi]
If you have a function f(x) defined on [-pi, pi] then you can transform it to a function g(u) on [-1, 1] by a linear change of variable:
u = -1 + 2 * (x + pi) / (2*pi).
Then you can approximate g by a polynomial P(u), and then transform P(u) to the polynomial Q(x) by the inverse change of variables:
x = -pi + (2*pi) * (u + 1) / 2.
I'm working on an assignment for class and I'm a little lost with trying to write a function out which will calculate Newton's quotient.
This is what the questions is asking
The derivative of a function f(x) can be approximated by the Newton's quotient (f(x+h) - f(x))/h
where h is a small number. Write a function to calculate the Newton's quotient
for f(x) = exp(x). The function should take two scalar arguments, x and h.
Use a default value of h=1e-6.
Test your function at the point x=1 using the default value of h, and compare
to the true value of the derivative f'(1) = e^1.
So far I have written the code as so
x=1
newton = function(x, h = 1e-06){
quotiant = ((x+h) - x)/h
return(x = exp(x))
}
y = newton(1,h)
print(y)
I can see this is wrong, but I don't really understand how I can fix this, and what exactly I'm trying to calculate.
I have also tried this code
x=1
newton = function(x, h = 1e-06){
quotiant = ((x+h) - x)/h
}
y = newton(1,h)
print(y)
which I think gives me the right answer, but again I don't really understand what I'm calculating.
Your function doesn't evaluate the values of x and x+h using the exponential function. In your two examples you are either just returning the exponential of x, or not using the exponential function at all. What you want is something like this:
newton = function(x, h = 1e-06){
quotient = (exp(x+h) - exp(x))/h
quotient
}
newton(1)
I have a function which looks like:
g(x) = f(x) - a^b / f(x)^b
g(x) - known function, data vector provided.
f(x) - hidden process.
a,b - parameters of this function.
From the above we get the relation:
f(x) = inverse(g(x))
My goal is to optimize parameters a and b such that f(x) would be as close as possible
to a normal distribution. If we look on a f(x) Q-Q normal plot (attached), my purpose is to minimize the distance between f(x) to the straight line which represents the normal distribution, by optimizing parameters a and b.
I wrote the below code:
g_fun <- function(x) {x - a^b/x^b}
inverse = function (f, lower = 0, upper = 2000) {
function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}
f_func = inverse(function(x) g_fun(x))
enter code here
# let's made up an example
# g(x) values are known
g <- c(-0.016339, 0.029646, -0.0255258, 0.003352, -0.053258, -0.018971, 0.005172,
0.067114, 0.026415, 0.051062)
# Calculate f(x) by using the inverse of g(x), when a=a0 and b=b0
for (i in 1:10) {
f[i] <- f_fun(g[i])
}
I have two question:
How to pass parameters a and b to the functions?
How to perform this optimization task, meaning find a and b such that f(x) would approximate normal distribution.
Not sure how you were able to produce the Q-Q plot since your provided examples do not work. You are not specifying the values of a and b and you are defining f_func but calling f_fun. Anyway here is my answer to your questions:
How to pass parameters a and b to the functions? - Just pass them as
arguments to the functions.
How to perform this optimization task, meaning find a and b such that f(x) would approximate normal distribution? - The same way any optimization task is done. Define a cost function, then minimize it.
Here is the revised code: I have added a and b as parameters, removed the inverse function and incorporated it inside f_func, which can now take vector input so no need for a for loop.
g_fun <- function(x,a,b) {x - a^b/x^b}
f_func = function(y,a,b,lower = 0, upper = 2000){
sapply(y,function(z) { uniroot(function(x) g_fun(x,a,b) - z, lower = lower, upper = upper)$root})
}
# g(x) values are known
g <- c(-0.016339, 0.029646, -0.0255258, 0.003352, -0.053258, -0.018971, 0.005172,
0.067114, 0.026415, 0.051062)
f <- f_func(g,1,1) # using a = 1 and b = 1
#[1] 0.9918427 1.0149329 0.9873386 1.0016774 0.9737270 0.9905320 1.0025893
#[8] 1.0341199 1.0132947 1.0258569
f_func(g,2,10)
[1] 1.876408 1.880554 1.875578 1.878138 1.873094 1.876170 1.878304 1.884049
[9] 1.880256 1.882544
Now for the optimization part, it depends on what you mean by f(x) would approximate normal distribution. You can compare mean square error from the qq-line if you want. Also since you say approximate, how close is good enough? You can go with shapiro.test and keep searching till you find p-value below 0.05 (be ware that there may not be a solution)
shapiro.test(f_func(g,1,2))$p
[1] 0.9484821
cost <- function(x,y) shapiro.test(f_func(g,x,y))$p
Now that we have a cost function how do we go about minimizing it. There are many many different ways to do numerical optimization. Take a look at optim function http://stat.ethz.ch/R-manual/R-patched/library/stats/html/optim.html.
optim(c(1,1),cost)
This final line does not work, but without proper data and context this is as far as I can go. Hope this helps.
For an ocean shader, I need a fast function that computes a very approximate value for sin(x). The only requirements are that it is periodic, and roughly resembles a sine wave.
The taylor series of sin is too slow, since I'd need to compute up to the 9th power of x just to get a full period.
Any suggestions?
EDIT: Sorry I didn't mention, I can't use a lookup table since this is on the vertex shader. A lookup table would involve a texture sample, which on the vertex shader is slower than the built in sin function.
It doesn't have to be in any way accurate, it just has to look nice.
Use a Chebyshev approximation for as many terms as you need. This is particularly easy if your input angles are constrained to be well behaved (-π .. +π or 0 .. 2π) so you do not have to reduce the argument to a sensible value first. You might use 2 or 3 terms instead of 9.
You can make a look-up table with sin values for some values and use linear interpolation between that values.
A rational algebraic function approximation to sin(x), valid from zero to π/2 is:
f = (C1 * x) / (C2 * x^2 + 1.)
with the constants:
c1 = 1.043406062
c2 = .2508691922
These constants were found by least-squares curve fitting. (Using subroutine DHFTI, by Lawson & Hanson).
If the input is outside [0, 2π], you'll need to take x mod 2 π.
To handle negative numbers, you'll need to write something like:
t = MOD(t, twopi)
IF (t < 0.) t = t + twopi
Then, to extend the range to 0 to 2π, reduce the input with something like:
IF (t < pi) THEN
IF (t < pi/2) THEN
x = t
ELSE
x = pi - t
END IF
ELSE
IF (t < 1.5 * pi) THEN
x = t - pi
ELSE
x = twopi - t
END IF
END IF
Then calculate:
f = (C1 * x) / (C2 * x*x + 1.0)
IF (t > pi) f = -f
The results should be within about 5% of the real sine.
Well, you don't say how accurate you need it to be. The sine can be approximated by straight lines of slopes 2/pi and -2/pi on intervals [0, pi/2], [pi/2, 3*pi/2], [3*pi/2, 2*pi]. This approximation can be had for the cost of a multiplication and an addition after reducing the angle mod 2*pi.
Using a lookup table is probably the best way to control the tradeoff between speed and accuracy.
I have differential equations derived from epidemic spreading. I want to obtain the numerical solutions. Here's the equations,
t is a independent variable and ranges from [0,100].
The initial value is
y1 = 0.99; y2 = 0.01; y3 = 0;
At first, I planned to deal these with ode45 function in matlab, however, I don't know how to express the series and the combination. So I'm asking for help here.
**
The problem is how to express the right side of the equations as the odefun, which is a parameter in the ode45 function.
**
Matlab has functions to calculate binomial coefficients (number of combinations) and the finite series can be expressed just as matrix multiplication. I'll demonstrate how that works for the sum in the first equation. Note the use of the element-wise "dotted" forms of the arithmetic operators.
Calculate a row vector coefs with the constant coefficients in the sum as:
octave-3.0.0:33> a = 0:20;
octave-3.0.0:34> coefs = log2(a * 0.05 + 1) .* bincoeff(20, a);
The variables get combined into another vector:
octave-3.0.0:35> y1 = 0.99;
octave-3.0.0:36> y2 = 0.01;
octave-3.0.0:37> z = (y2 .^ a) .* ((1 - y2) .^ a) .* (y1 .^ a);
And the sum is then just evaluated as the inner product:
octave-3.0.0:38> coefs * z'
The other sums are similar.
function demo(a_in)
X = [0;0;0];
T = [0:.1:100];
a = a_in; % for nested scope
[Xout, Tout ]= ode45( #myFunc, T, X );
function [dxdt] = myFunc( t, x )
% nested function accesses "a"
dxdt = 0*x + a;
% Todo: real value of dxdt.
end
end
What about this, and you simply need to fill in the dxdt from your math above? It remains to be seen if the numerical roundoff matters...
Edit: there's a serious issue due to the 1=y1+y2+y3 constraint. Is that even allowed, since you have an IVP with 3 initial values given and 3 first order ODE's? If that constraint is a natural consequence of the equations, it may not be needed.