Heres a block of code that plots a function over a range, as well as the at a single input :
a = 1.0
f(x::Float64) = -x^2 - a
scatter(f, -3:.1:3)
scatter!([a], [f(a)])
i would like to plot the line, tangent to the point, like so:
Is there a pattern or simple tool for doing so?
That depends on what you mean by "pattern or simple tool" - the easiest way is to just derive the derivative by hand and then plot that as a function:
hand_gradient(x) = -2x
and then add plot!(hand_gradient, 0:0.01:3) to your plot.
Of course that can be a bit tedious with more complicated functions or when you want to plot lots of gradients, so another way would be to utilise Julia's excellent automatic differentiation capabilities. Comparing all the different options is a bit beyond the scope of this answer, but check out https://juliadiff.org/ if you're interested. Here, I will be using the widely used Zygote library:
julia> using Plots, Zygote
julia> a = 1.0;
julia> f(x) = -x^2 - a;
[NB I have slightly amended your f definition to be in line with the plot you posted, which is an inverted parabola]
note that here I am not restricting the type of input argument x to f - this is crucial for automatic differentiation to work, as it is implemented by runnning a different number type (a Dual) through your function. In general, restricting argument types in this way is an anti-pattern in Julia - it does not help performance, but makes your code less interoperable with other parts of the ecosystem, as you can see here if you try to automatically differentiate through f(x::Float64).
Now let's use Zygote to provide gradients for us:
julia> f'
#43 (generic function with 1 method)
as you can see, running f' now returns an anonymous function - this is the derivative of f, as you can verify by evaluating it at a specific point:
julia> f'(2)
-4.0
Now all we need to do is leverage this to construct a function that itself returns a function which traces out the line of the gradient:
julia> gradient_line(f, x₀) = (x -> f(x₀) + f'(x₀)*(x-x₀))
gradient_line (generic function with 1 method)
this function takes in a function f and a point x₀ for which we want to get the tangent, and then returns an anonymous function which returns the value of the tangent at each value of x. Putting this to use:
julia> default(markerstrokecolor = "white", linewidth = 2);
julia> scatter(f, -3:0.1:3, label = "f(x) = -x² - 1", xlabel = "x", ylabel = "f(x)");
julia> scatter!([1], [f(1)], label = "", markersize = 10);
julia> plot!(gradient_line(f, 1), 0:0.1:3, label = "f'(1)", color = 2);
julia> scatter!([-2], [f(-2)], label = "", markersize = 10, color = 3);
julia> plot!(gradient_line(f, -2), -3:0.1:0, label = "f'(-2)", color = 3)
It is overkill for this problem, but you could use the CalculusWithJulia package which wraps up a tangent operator (along with some other conveniences) similar to what is derived in the previous answers:
using CalculusWithJulia # ignore any warnings
using Plots
f(x) = sin(x)
a, b = 0, pi/2
c = pi/4
plot(f, a, b)
plot!(tangent(f,c), a, b)
Well, the tool is called high school math :)
You can simply calculate the slope (m) and intersect (b) of the tanget (mx + b) and then plot it. To determine the former, we need to compute the derivative of the function f in the point a, i.e. f'(a). The simplest possible estimator is the difference quotient (I assume that it would be cheating to just derive the parabola analytically):
m = (f(a+Δ) - f(a))/Δ
Having the slope, our tanget should better go through the point (a, f(a)). Hence we have to choose b accordingly as:
b = f(a) - m*a
Choosing a suitably small value for Δ, e.g. Δ = 0.01 we obtain:
Δ = 0.01
m = (f(a+Δ) - f(a))/Δ
b = f(a) - m*a
scatter(f, -3:.1:3)
scatter!([a], [f(a)])
plot!(x -> m*x + b, 0, 3)
Higher order estimators for the derivative can be found in FiniteDifferences.jl and FiniteDiff.jl for example. Alternatively, you could use automatic differentiation (AD) tools such as ForwardDiff.jl to obtain the local derivative (see Nils answer for an example).
Related
I got an apparently quite common Julia error when trying to use AD with forward.diff. The error messages vary a bit (sometimes matching function name sometimes Float64)
MethodError: no method matching logL_multinom(::Vector{ForwardDiff.Dual{ForwardDiff.Tag{typeof(logL_multinom), Real}, Real, 7}})
My goal: Transform a probability vector to be unbounded (θ -> y), do some stuff (namely HMC sampling) and transform back to the simplex space whenever the unnormalized posterior (logL_multinom()) is evaluated. DA should be used to overome problems for later, more complex, models than this.
Unfortunately, neither the Julia documentation, not the solutions from other questions helped me figure the particular problem out. Especially, it seems to work when I do the first transformation (y -> z) outside of the function, but the first transformation is a 1-to-1 mapping via logistic and should not cause any harm to differentiation.
Here is an MWE:
using LinearAlgebra
using ForwardDiff
using Base
function logL_multinom(y)
# transform to constrained
K = length(y)+1
k = collect(1:(K-1))
# inverse logit:
z = 1 ./ (1 .+ exp.(-y .- log.(K .- k))) # if this is outside, it works
θ = zeros(eltype(y),K) ; x_cumsum = zeros(eltype(y),K-1)
typeof(θ)
for i in k
x_cumsum[i] = 1-sum(θ)
θ[i] = (x_cumsum[i]) * z[i]
end
θ[K] = x_cumsum[K-1] - θ[K-1]
#log_dens_correction = sum( log(z*(1-z)*x_cumsum) )
dot(colSums, log.(θ))
end
colSums = [835, 52, 1634, 3469, 3053, 2507, 2279, 1115]
y0 = [-0.8904013824298864, -0.8196709647741431, -0.2676845405543302, 0.31688184351556026, -0.870860684394019,0.15187821053559714,0.39888119498547964]
logL_multinom(y0)
∇L = y -> ForwardDiff.gradient(logL_multinom,y)
∇L(y0)
Thanks a lot and especially some further readings/ explanations for the problem are appreciated since I'll be working with it moreoften :D
Edit: I tried to convert the input and any intermediate variable into Real / arrays of these, but nothing helped so far.
I am a new user of Scilab and I am not a mathematician.
As my end goal, I want to calculate (and plot) the derivative of a piece-wise defined function, see here.
I tried to start small and just use a simple (continuous) function: f(x) = 3*x.
My Google-Fu lead me to the numderivative function.
Problem: It seems that I do not understand how the argument x works since the result is not a 1D-array, instead, it is a matrix.
Update 1: Maybe I use the wrong function and diff is the way to go. But what is then the purpose of numderivative?
PS: Is this the right place to ask Scilab-related questions? It seems that there are several StackOverflow communities where Scilab-related questions are asked.
// Define limits
x0 = 0;
x1 = 2;
// Define array x for which the derivative will be calculated.
n = 100;
x = linspace (x0, x1, n);
// Define function f(x)
deff('y=f(x)','y=3*x');
// Calculate derivative of f(x) at the positions x
myDiff = numderivative(f,x)
(I expect the result 3 3 and not a matrix.)
numderivative(f,x) will give you the approximated derivative/Jacobian of f at the single vector x. For your example it yields 3 times the identity matrix, which is the expected result since f(x)=3*x. If you rather need the derivative of f considered as a function of a single scalar variable at x=1 and x=2, then numderivative is not convenient as you would have to make an explicit loop. Just code the formula yourself (here first order formula) :
// Define limits
x0 = 0;
x1 = 2;
// Define array x for which the derivative will be calculated.
n = 100;
x = linspace (x0, x1, n);
// Define function f(x)
deff('y=f(x)','y=3*x');x = [1 2];
h = sqrt(%eps);
d = (f(x+h)-f(x))/h;
The formula can be improved (second order or complex step formula).
I accidentally (but fortunately) found the elegant solution to your end goal (and also to mine) of calculating and plotting the derivatives through the SciLab documentation itself via the splin (https://help.scilab.org/docs/6.1.1/en_US/splin.html) and interp (https://help.scilab.org/docs/6.1.1/en_US/interp.html) functions. This elegantly works not only for piece-wise defined functions but also for functions defined numerically by data.
The former will give you the corresponding first derivative of your array values while the latter will give you up to the third derivative. Both are actually meant to be used in conjunction because the interp function requires the corresponding derivatives for your y values, which you can easily get through the splin function. Using this method is better than using the diff function because the output array is of the same size as the original one.
Illustrating this through your code example:
// Define limits
x0 = 0;
x1 = 2;
// Define array x for which the derivatives will be calculated.
n = 100;
x = linspace (x0, x1, n);
// Define function f(x)
deff('y=f(x)','y=3*x');
// Calculate derivative of f(x) at the positions x
myDiff = splin(x, f(x));
// Calculate up to the third derivative of f(x) at the positions x
// The first derivatives obtained earlier are required by this function
[yp, yp1, yp2, yp3]=interp(x, x, f(x), myDiff);
// Plot and label all the values in one figure with gridlines
plot(x', [yp', yp1', yp2', yp3'], 'linewidth', 2)
xgrid();
// Put the legend on the upper-left by specifying 2 after the list of legend names
legend(['yp', 'yp1', 'yp2', 'yp3'], 2);
To demonstrate, let's start with a simple multi-variable function f(x,y) = xy^2.
I am trying to find a command that would allow me to numerically integrate f(2, y) = 2y^2 from y = 0 to y = 2. (i.e. the original function is multi-variable, but only one variable remains when actually doing the integration)
I needed to define the function that way as I need to obtain the results using different values of x. (probably going to involve for-loop, but that is another story)
I have tried to walk through Cubature's user guide but apparently did not find anything useful. Maybe I have missed it
Can anyone help?
In such case it is simplest to use an anonymous function wrapper:
using QuadGK
f(x,y) = x*y^2
intf(x) = quadgk(y -> f(x, y), 0, 2)
if the anonymous function would be longer you could write:
intf(x) = quadgk(0, 2) do y
f(x, y)
end
This is an exact equivalent of the latter but do syntax allows you to write longer bodies of an anonymous function.
Now you can write e.g.:
julia> intf(1)
(2.6666666666666665, 4.440892098500626e-16)
julia> intf(2)
(5.333333333333333, 8.881784197001252e-16)
julia> intf(3)
(8.0, 0.0)
Background
I read here that newton method fails on function x^(1/3) when it's inital step is 1. I am tring to test it in julia jupyter notebook.
I want to print a plot of function x^(1/3)
then I want to run code
f = x->x^(1/3)
D(f) = x->ForwardDiff.derivative(f, float(x))
x = find_zero((f, D(f)),1, Roots.Newton(),verbose=true)
Problem:
How to print chart of function x^(1/3) in range eg.(-1,1)
I tried
f = x->x^(1/3)
plot(f,-1,1)
I got
I changed code to
f = x->(x+0im)^(1/3)
plot(f,-1,1)
I got
I want my plot to look like a plot of x^(1/3) in google
However I can not print more than a half of it
That's because x^(1/3) does not always return a real (as in numbers) result or the real cube root of x. For negative numbers, the exponentiation function with some powers like (1/3 or 1.254 and I suppose all non-integers) will return a Complex. For type-stability requirements in Julia, this operation applied to a negative Real gives a DomainError. This behavior is also noted in Frequently Asked Questions section of Julia manual.
julia> (-1)^(1/3)
ERROR: DomainError with -1.0:
Exponentiation yielding a complex result requires a complex argument.
Replace x^y with (x+0im)^y, Complex(x)^y, or similar.
julia> Complex(-1)^(1/3)
0.5 + 0.8660254037844386im
Note that The behavior of returning a complex number for exponentiation of negative values is not really different than, say, MATLAB's behavior
>>> (-1)^(1/3)
ans =
0.5000 + 0.8660i
What you want, however, is to plot the real cube root.
You can go with
plot(x -> x < 0 ? -(-x)^(1//3) : x^(1//3), -1, 1)
to enforce real cube root or use the built-in cbrt function for that instead.
plot(cbrt, -1, 1)
It also has an alias ∛.
plot(∛, -1, 1)
F(x) is an odd function, you just use [0 1] as input variable.
The plot on [-1 0] is deducted as follow
The code is below
import numpy as np
import matplotlib.pyplot as plt
# Function f
f = lambda x: x**(1/3)
fig, ax = plt.subplots()
x1 = np.linspace(0, 1, num = 100)
x2 = np.linspace(-1, 0, num = 100)
ax.plot(x1, f(x1))
ax.plot(x2, -f(x1[::-1]))
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
plt.show()
Plot
That Google plot makes no sense to me. For x > 0 it's ok, but for negative values of x the correct result is complex, and the Google plot appears to be showing the negative of the absolute value, which is strange.
Below you can see the output from Matlab, which is less fussy about types than Julia. As you can see it does not agree with your plot.
From the plot you can see that positive x values give a real-valued answer, while negative x give a complex-valued answer. The reason Julia errors for negative inputs, is that they are very concerned with type stability. Having the output type of a function depend on the input value would cause a type instability, which harms performance. This is less of a concern for Matlab or Python, etc.
If you want a plot similar the above in Julia, you can define your function like this:
f = x -> sign(x) * abs(complex(x)^(1/3))
Edit: Actually, a better and faster version is
f = x -> sign(x) * abs(x)^(1/3)
Yeah, it looks awkward, but that's because you want a really strange plot, which imho makes no sense for the function x^(1/3).
Does anyone know how to minimize a function containing an integral in MATLAB? The function looks like this:
L = Int(t=0,t=T)[(AR-x)dt], A is a system parameter and R and x are related through:
dR/dt = axRY - bR, where a and b are constants.
dY/dt = -xRY
I read somewhere that I can use fminbnd and quad in combination but I am not able to make it work. Any suggestions?
Perhaps you could give more details of your integral, e.g. where is the missing bracket in [AR-x)dt]? Is there any dependence of x on t, or can we integrate dR/dt = axR - bR to give R=C*exp((a*x-b)*t)? In any case, to answer your question on fminbnd and quad, you could set A,C,T,a,b,xmin and xmax (the last two are the range you want to look for the min over) and use:
[x fval] = fminbnd(#(x) quad(#(t)A*C*exp((a*x-b)*t)-x,0,T),xmin,xmax)
This finds x that minimizes the integral.
If i didn't get it wrong you are trying to minimize respect to t:
\int_0^t{(AR-x) dt}
well then you just need to find the zeros of:
AR-x
This is just math, not matlab ;)
Here's some manipulation of your equations that might help.
Combining the second and third equations you gave gives
dR/dt = -a*(dY/dt)-bR
Now if we solve for R on the righthand side and plug it into the first equation you gave we get
L = Int(t=0,t=T)[(-A/b*(dR/dt + a*dY/dt) - x)dt]
Now we can integrate the first term to get:
L = -A/b*[R(T) - R(0) + Y(T) - Y(0)] - Int(t=0,t=T)[(x)dt]
So now all that matters with regards to R and Y are the endpoints. In fact, you may as well define a new function, Z which equals Y + R. Then you get
L = -A/b*[Z(T) - Z(0)] - Int(t=0,t=T)[(x)dt]
This next part I'm not as confident in. The integral of x with respect to t will give some function which is evaluated at t = 0 and t = T. This function we will call X to give:
L = -A/b*[Z(T) - Z(0)] - X(T) + X(0)
This equation holds true for all T, so we can set T to t if we want to.
L = -A/b*[Z(t) - Z(0)] - X(t) + X(0)
Also, we can group a lot of the constants together and call them C to give
X(t) = -A/b*Z(t) + C
where
C = A/b*Z(0) + X(0) - L
So I'm not sure what else to do with this, but I've shown that the integral of x(t) is linearly related to Z(t) = R(t) + Y(t). It seems to me that there are many equations that solve this. Anyone else see where to go from here? Any problems with my math?