I am trying to calculate some integrals, for example:
a = 1/sqrt(2);
b = -5;
c = 62;
d = 1;
f = exp(-x^2-y^2)*(erfc((sym(a) + 1/(x^2+y^2)*(sym(b)*x+sym(d)*y))*sqrt((x^2+y^2)*sym(10.^(c/10))))...
+ erfc((sym(a) - 1/(x^2+y^2)*(sym(b)*x+sym(d)*y))*sqrt((x^2+y^2)*sym(10.^(c/10)))));
h = int(int(f,x,-Inf,Inf),y,-Inf,Inf);
It will occur error like this:
Warning: Explicit integral could not be found.
Then, I try to use vpato calculate that integral,and get the result like this
vpa(int(int(f,x,-Inf,Inf),y,-Inf,Inf),5)
numeric::int(numeric::int(exp(- x^2 - y^2)*(erfc(((6807064429273519*x^2)/4294967296 + (6807064429273519*y^2)/4294967296)^(1/2)*(2^(1/2)/2 + (5*x - y)/(x^2 + y^2))) + erfc(((6807064429273519*x^2)/4294967296 + (6807064429273519*y^2)/4294967296)^(1/2)*(2^(1/2)/2 - (5*x - y)/(x^2 + y^2)))), x == -Inf..Inf), y == -Inf..Inf)
I already tried to change the interval [-Inf,Inf] to [-100,100], and get the same above result:
numeric::int(numeric::int(exp(- x^2 - y^2)*(erfc(((6807064429273519*x^2)/4294967296 + (6807064429273519*y^2)/4294967296)^(1/2)*(2^(1/2)/2 + (5*x - y)/(x^2 + y^2))) + erfc(((6807064429273519*x^2)/4294967296 + (6807064429273519*y^2)/4294967296)^(1/2)*(2^(1/2)/2 - (5*x - y)/(x^2 + y^2)))), x == -100..100), y == -100..100)
My question is why vpa in this case could not return to a real value?
There are something wrong in above Matlab code? (I, myself, could not find the bug so far)
Thank you in advance for your help.
It is unlikely that there is an analytic solution to this integral so using int may not be a good choice. In some cases vpa can be used to for a numeric solution. When this fails (by returning a call to itself) it may be for several reason: the integral may not exist, the integral may converge too slowly, singularities may cause issues, the integrand may be highly oscillatory or non-smooth, etc. Mathematica also struggles with this integral.
You can try calculating the integral numerically using integral2:
a = 1/sqrt(2);
b = -5;
c = 62;
d = 1;
f = #(x,y)exp(-x.^2-y.^2).*(erfc((a + 1./(x.^2+y.^2).*(b*x+d*y)).*sqrt((x.^2+y.^2)*10^(c/10)))...
+erfc((a - 1./(x.^2+y.^2).*(b*x+d*y)).*sqrt((x.^2+y.^2)*10^(c/10))));
h = integral2(f,-Inf,Inf,-Inf,Inf)
which returns 5.790631184403967. This compares well with Mathematica's numerical integration using NIntegrate. You can try specifying smaller absolute and relative tolerances for integral2 to get more accurate values, but this will result in much slower compute times.
Related
Firstly, I'm quite new to Matlab.
I am currently trying to do a definite integral with respect to y of a particular function. The function that I want to integrate is
(note that the big parenthesis is multiplying with the first factor - I can't get the latex to not make it look like power)
I have tried plugging the above integral into Desmos and it worked as intended. My plan was to vary the value of x and y and will be using for loop via matlab.
However, after trying to use the int function to calculate the definite integral with the code as follow:
h = 5;
a = 2;
syms y
x = 3.8;
p = 2.*x.^2+2.*a.*y;
q = x.^2+y.^2;
r = x.^2+a.^2;
f = (-1./sqrt(1-(p.^2./(4.*q.*r)))).*(2.*sqrt(q).*sqrt(r).*2.*a-p.*2.*y.*sqrt(r)./sqrt(q))./(4.*q.*r);
theta = int(f,y,a+0.01,h) %the integral is undefined at y=2, hence the +0.01
the result is not quite as expected
theta =
int(-((8*461^(1/2)*(y^2 + 361/25)^(1/2))/5 - (461^(1/2)*y*(8*y + 1444/25))/(5*(y^2 + 361/25)^(1/2)))/((1 - (4*y + 722/25)^2/((1844*y^2)/25 + 665684/625))^(1/2)*((1844*y^2)/25 + 665684/625)), y, 21/10, 5)
After browsing through various posts, the common mistake is the undefined interval but the +0.01 should have fixed it. Any guidance on what went wrong is much appreciated.
The Definite Integrals example in the docs shows exactly this type of output when a closed form cannot be computed. You can approximate it numerically using vpa, i.e.
F = int(f,y,a,h);
theta = vpa(F);
Or you can do a numerical computation directly
theta = vpaintegral(f,y,a,h);
From the docs:
The vpaintegral function is faster and provides control over integration tolerances.
I am solving a problem in my Macroeconomics class. Consider the following equation:
Here, k is fixed and c(k) was defined through the ```interp1''' function in Matlab. Here is my code:
beta = 0.98;
delta = 0.13;
A = 2;
alpha = 1/3;
n_grid = 1000; % Number of points for capital
k_grid = linspace(5, 15, n_grid)';
tol = 1e-5;
max_it = 1000;
c0 = ones(n_grid, 1);
new_k = zeros(n_grid, 1);
dist_c = tol + 1;
it_c = 0;
while dist_c > tol && it_c < max_it
c_handle = #(k_tomorrow) interp1(k_grid, c0, k_tomorrow, 'linear', 'extrap');
for i=1:n_grid
% Solve for k'
euler = #(k_tomorrow) (1/((1-delta)* k_grid(i) + A * k_grid(i)^alpha - k_tomorrow)) - beta*(1-delta + alpha*A*k_tomorrow^(alpha - 1))/c_handle(k_prime);
new_k(i) = fzero(euler, k_grid(i)); % What's a good guess for fzero?
end
% Compute new values for consumption
new_c = A*k_grid.^alpha + (1-delta)*k_grid - new_k;
% Check convergence
dist_c = norm(new_c - c0);
c0 = new_c;
it_c = it_c + 1;
end
When I run this code, for some indexes $i$, it runs fine and fzero can find the solution. But for indexes it just returns NaN and exits without finding the root. This is a somewhat well-behaved problem in Economics and the solution we are looking indeed exists and the algorithm I tried to implement is guaranteed to work. But I don't have much experience with solving this in MATLAB and I guess I have a silly mistake somewhere. Any ideas on how to procede?
This is the typical error message:
Exiting fzero: aborting search for an interval containing a sign change
because complex function value encountered during search.
(Function value at -2.61092 is 0.74278-0.30449i.)
Check function or try again with a different starting value.
Thanks a lot in advance!
The only term that can produce complex numbers is
k'^(alpha - 1) = k'^(-2/3)
You probably want the result according to the real variant of the cube root, which you could get as
sign(k') * abs(k')^(-2/3)
or more generally and avoiding divisions by zero
k' * (1e-16+abs(k'))^(alpha - 2)
I wanted to compute a finite difference with respect to the change of the function in Matlab. In other words
f(x+e_i) - f(x)
is what I want to compute. Note that its very similar to the first order numerical partial differentiation (forward differentiation in this case) :
(f(x+e_i) - f(x)) / (e_i)
Currently I am using for loops to compute it but it seems that Matlab is much slower than I thought. I am doing it as follows:
function [ dU ] = numerical_gradient(W,f,eps)
%compute gradient or finite difference update numerically
[D1, D2] = size(W);
dU = zeros(D1, D2);
for d1=1:D1
for d2=1:D2
e = zeros([D1,D2]);
e(d1,d2) = eps;
f_e1 = f(W+e);
f_e2 = f(W-e);
%numerical_derivative = (f_e1 - f_e2)/(2*eps);
%dU(d1,d2) = numerical_derivative
numerical_difference = f_e1 - f_e2;
dU(d1,d2) = numerical_difference;
end
end
it seems that its really difficult to vectorize the above code because for numerical differences follow the definition of the gradient and partial derivatives which is:
df_dW = [ ..., df_dWi, ...]
where df_dWi assumes the other coordinates are fixed and it only worries about the change of the variable Wi. Thus, I can't just change all the coordinates at once.
Is there a better way to do this? My intuition tells me that the best way to do this is to implement this not in matlab but in some other language, say C and then have matlab call that library. Is that true? Does it mean that the best solution is some Matlab library that does this for me?
I did see:
https://www.mathworks.com/matlabcentral/answers/332414-what-is-the-quickest-way-to-find-a-gradient-or-finite-difference-in-matlab-of-a-real-function-in-hig
but unfortunately, it computes exact derivatives, which isn't what I am looking for. I am explicitly looking for differences or "bad approximation" to the gradient.
Since it seems this code is not easy to vectorize (in fact my intuition tells me its not possible to do so) my only other idea is to implement this finite difference function in C and then have C call the function. Is this a good idea? Anyone know how to do this?
I did try reading the following:
https://www.mathworks.com/help/matlab/matlab_external/standalone-example.html
but it was too difficult to understand for me because I have no idea what a mex file is, if I need to have a arrayProduct.c file as well as a mex.h file, if I also needed a matlab file, etc. If there just existed a way to simply download a working example with all the functions they suggest there and some instructions to compile it, then it would be super helpful. But just reading the hmtl/article like that its impossible for me to infer what they want me to do.
For the sake of completness it seems reddit has some comments in its discussion of this:
https://www.reddit.com/r/matlab/comments/623m7i/how_does_one_compute_a_single_finite_differences/
Here is a more efficient doing so:
function [ vNumericalGrad ] = CalcNumericalGradient( hInputFunc, vInputPoint, epsVal )
numElmnts = size(vInputPoint, 1);
vNumericalGrad = zeros([numElmnts, 1]);
refVal = hInputFunc(vInputPoint);
for ii = 1:numElmnts
% Set the perturbation vector
refInVal = vInputPoint(ii);
vInputPoint(ii) = refInVal + epsVal;
% Compute Numerical Gradient
vNumericalGrad(ii) = (hInputFunc(vInputPoint) - refVal) / epsVal;
% Reset the perturbation vector
vInputPoint(ii) = refInVal;
end
end
This code allocate less memory.
The above code performance will be totally controlled by the speed of the hInputFunction.
The small tricks compared to original code are:
No memory reallocation of e each iteration.
Instead of addition of vectors W + e there are 2 assignments to the array.
Decreasing the calls to hInputFunction() by half by defining the reference value outside the loop (This only works for Forward / Backward difference).
Probably this will be very close to C code unless you can code in C more efficiently the function which computes the value (hInputFunction).
A full implementation can be found in StackOverflow Q44984132 Repository (It was Posted in StackOverflow Q44984132).
See CalcFunGrad( vX, hObjFun, difMode, epsVal ).
A way better approach (numerically more stable, no issue of choosing the perturbation hyperparameter, accurate up to machine precision) is to use algorithmic/automatic differentiation. For this you need the Matlab Deep Learning Toolbox. Then you can use dlgradient to compute the gradient. Below you find the source code attached corresponding to your example.
Most importantly, you can examine the error and observe that the deviation of the automatic approach from the analytical solution is indeed machine precision, while for the finite difference approach (I choose second order central differences) the error is orders of magnitude higher. For 100 points and a range of $[-10, 10]$ this errors are somewhat tolerable, but if you play a bit with Rand_Max and n_points you observe that the errors become larger and larger.
Error of algorithmic / automatic diff. is: 1.4755528111219851e-14
Error of finite difference diff. is: 1.9999999999348703e-01 for perturbation 1.0000000000000001e-01
Error of finite difference diff. is: 1.9999999632850161e-03 for perturbation 1.0000000000000000e-02
Error of finite difference diff. is: 1.9999905867860374e-05 for perturbation 1.0000000000000000e-03
Error of finite difference diff. is: 1.9664569947425062e-07 for perturbation 1.0000000000000000e-04
Error of finite difference diff. is: 1.0537897883625319e-07 for perturbation 1.0000000000000001e-05
Error of finite difference diff. is: 1.5469326944467290e-06 for perturbation 9.9999999999999995e-07
Error of finite difference diff. is: 1.3322061696937969e-05 for perturbation 9.9999999999999995e-08
Error of finite difference diff. is: 1.7059535957436630e-04 for perturbation 1.0000000000000000e-08
Error of finite difference diff. is: 4.9702408787320664e-04 for perturbation 1.0000000000000001e-09
Source Code:
f2.m
function y = f2(x)
x1 = x(:, 1);
x2 = x(:, 2);
x3 = x(:, 3);
y = x1.^2 + 2*x2.^2 + 2*x3.^3 + 2*x1.*x2 + 2*x2.*x3;
f2_grad_analytic.m:
function grad = f2_grad_analytic(x)
x1 = x(:, 1);
x2 = x(:, 2);
x3 = x(:, 3);
grad(:, 1) = 2*x1 + 2*x2;
grad(:, 2) = 4*x2 + 2*x1 + 2 * x3;
grad(:, 3) = 6*x3.^2 + 2*x2;
f2_grad_AD.m:
function grad = f2_grad_AD(x)
x1 = x(:, 1);
x2 = x(:, 2);
x3 = x(:, 3);
y = x1.^2 + 2*x2.^2 + 2*x3.^3 + 2*x1.*x2 + 2*x2.*x3;
grad = dlgradient(y, x);
CalcNumericalGradient.m:
function NumericalGrad = CalcNumericalGradient(InputPoints, eps)
% (Central, second order accurate FD)
NumericalGrad = zeros(size(InputPoints) );
for i = 1:size(InputPoints, 2)
perturb = zeros(size(InputPoints));
perturb(:, i) = eps;
NumericalGrad(:, i) = (f2(InputPoints + perturb) - f2(InputPoints - perturb)) / (2 * eps);
end
main.m:
clear;
close all;
clc;
n_points = 100;
Rand_Max = 20;
x_test_FD = rand(n_points, 3) * Rand_Max - Rand_Max/2;
% Calculate analytical solution
grad_analytic = f2_grad_analytic(x_test_FD);
grad_AD = zeros(n_points, 3);
for i = 1:n_points
x_test_dl = dlarray(x_test_FD(i,:) );
grad_AD(i,:) = dlfeval(#f2_grad_AD, x_test_dl);
end
Err_AD = norm(grad_AD - grad_analytic);
fprintf("Error of algorithmic / automatic diff. is: %.16e\n", Err_AD);
eps_range = [1e-1, 1e-2, 1e-3, 1e-4, 1e-5, 1e-6, 1e-7, 1e-8, 1e-9];
for i = 1:length(eps_range)
eps = eps_range(i);
grad_FD = CalcNumericalGradient(x_test_FD, eps);
Err_FD = norm(grad_FD - grad_analytic);
fprintf("Error of finite difference diff. is: %.16e for perturbation %.16e\n", Err_FD, eps);
end
I would like to solve the following equation: tan(x) = 1/x
What I did:
syms x
eq = tan(x) == 1/x;
sol = solve(eq,x)
But this gives me only one numerical approximation of the solution. After that I read about the following:
[sol, params, conds] = solve(eq, x, 'ReturnConditions', true)
But this tells me that it can't find an explicit solution.
How can I find numerical solutions to this equation within some given range?
I've never liked using solvers "blindly", that is, without some sort of decent initial value selection scheme. In my experience, the values you will find when doing things blindly, will be without context as well. Meaning, you'll often miss solutions, think something is a solution while in reality the solver exploded, etc.
For this particular case, it is important to realize that fzero uses numerical derivatives to find increasingly better approximations. But, derivatives for f(x) = x · tan(x) - 1 get increasingly difficult to accurately compute for increasing x:
As you can see, the larger x becomes, the better f(x) approximates a vertical line; fzero will simply explode! Therefore it is imperative to get an estimate as closely to the solution as possible before even entering fzero.
So, here's a way to get good initial values.
Consider the function
f(x) = x · tan(x) - 1
Knowing that tan(x) has Taylor expansion:
tan(x) ≈ x + (1/3)·x³ + (2/15)·x⁵ + (7/315)·x⁷ + ...
we can use that to approximate the function f(x). Truncating after the second term, we can write:
f(x) ≈ x · (x + (1/3)·x³) - 1
Now, key to realize is that tan(x) repeats with period π. Therefore, it is most useful to consider the family of functions:
fₙ(x) ≈ x · ( (x - n·π) + (1/3)·(x - n·π)³) - 1
Evaluating this for a couple of multiples and collecting terms gives the following generalization:
f₀(x) = x⁴/3 - 0π·x³ + ( 0π² + 1)x² - (0π + (0π³)/3)·x - 1
f₁(x) = x⁴/3 - 1π·x³ + ( 1π² + 1)x² - (1π + (1π³)/3)·x - 1
f₂(x) = x⁴/3 - 2π·x³ + ( 4π² + 1)x² - (2π + (8π³)/3)·x - 1
f₃(x) = x⁴/3 - 3π·x³ + ( 9π² + 1)x² - (3π + (27π³)/3)·x - 1
f₄(x) = x⁴/3 - 4π·x³ + (16π² + 1)x² - (4π + (64π³)/3)·x - 1
⋮
fₙ(x) = x⁴/3 - nπ·x³ + (n²π² + 1)x² - (nπ + (n³π³)/3)·x - 1
Implementing all this in a simple MATLAB test:
% Replace this with the whole number of pi's you want to
% use as offset
n = 5;
% The coefficients of the approximating polynomial for this offset
C = #(npi) [1/3
-npi
npi^2 + 1
-npi - npi^3/3
-1];
% Find the real, positive polynomial roots
R = roots(C(n*pi));
R = R(imag(R)==0);
R = R(R > 0);
% And use these as initial values for fzero()
x_npi = fzero(#(x) x.*tan(x) - 1, R)
In a loop, this can produce the following table:
% Estimate (polynomial) Solution (fzero)
0.889543617524132 0.860333589019380 0·π
3.425836967935954 3.425618459481728 1·π
6.437309348195653 6.437298179171947 2·π
9.529336042900365 9.529334405361963 3·π
12.645287627956868 12.645287223856643
15.771285009691695 15.771284874815882
18.902410011613000 18.902409956860023
22.036496753426441 22.036496727938566 ⋮
25.172446339768143 25.172446326646664
28.309642861751708 28.309642854452012
31.447714641852869 31.447714637546234
34.586424217960058 34.586424215288922 11·π
As you can see, the approximant is basically equal to the solution. Corresponding plot:
To find a numerical solution to a function within some range, you can use fzero like this:
fun = #(x)x*tan(x)-1; % Multiplied by x so fzero has no issue evaluating it at x=0.
range = [0 pi/2];
sol = fzero(fun,range);
The above would return just one solution (0.8603). If you want additional solutions, you will have to call fzero more times. This can be done, for example, in a loop:
fun = #(x)tan(x)-1/x;
RANGE_START = 0;
RANGE_END = 3*pi;
RANGE_STEP = pi/2;
intervals = repelem(RANGE_START:RANGE_STEP:RANGE_END,2);
intervals = reshape(intervals(2:end-1),2,[]).';
sol = NaN(size(intervals,1),1);
for ind1 = 1:numel(sol)
sol(ind1) = fzero(fun, mean(intervals(ind1,:)));
end
sol = sol(~isnan(sol)); % In case you specified more intervals than solutions.
Which gives:
[0.86033358901938;
1.57079632679490; % Wrong
3.42561845948173;
4.71238898038469; % Wrong
6.43729817917195;
7.85398163397449] % Wrong
Note that:
The function is symmetric, and so are its roots. This means you can solve on just the positive interval (for example) and get the negative roots "for free".
Every other entry in sol is wrong because this is where we have asymptotic discontinuities (tan transitions from +Inf to -Inf), which is mistakenly recognized by MATLAB as a solution. So you can just ignore them (i.e. sol = sol(1:2:end);.
Multiply the equation by x and cos(x) to avoid any denominators that can have the value 0,
f(x)=x*sin(x)-cos(x)==0
Consider the normalized function
h(x)=(x*sin(x)-cos(x)) / (abs(x)+1)
For large x this will be increasingly close to sin(x) (or -sin(x) for large negative x). Indeed, plotting this this is already visually true, up to an amplitude factor, for x>pi.
For the first root in [0,pi/2] use the Taylor approximation at x=0 of second degree x^2-(1-0.5x^2)==0 to get x[0]=sqrt(2.0/3) as root approximation, for the higher ones take the sine roots x[n]=n*pi, n=1,2,3,... as initial approximations in the Newton iteration xnext = x - f(x)/f'(x) to get
n initial 1. Newton limit of Newton
0 0.816496580927726 0.863034004302817 0.860333589019380
1 3.141592653589793 3.336084918413964 3.425618459480901
2 6.283185307179586 6.403911810682199 6.437298179171945
3 9.424777960769379 9.512307014150883 9.529334405361963
4 12.566370614359172 12.635021895208379 12.645287223856643
5 15.707963267948966 15.764435036320542 15.771284874815882
6 18.849555921538759 18.897518573777646 18.902409956860023
7 21.991148575128552 22.032830614521892 22.036496727938566
8 25.132741228718345 25.169597069842926 25.172446326646664
9 28.274333882308138 28.307365162331923 28.309642854452012
10 31.415926535897931 31.445852385744583 31.447714637546234
11 34.557519189487721 34.584873343220551 34.586424215288922
I am currently trying to implement a machine learning algorithm that involves the logistic loss function in MATLAB. Unfortunately, I am having some trouble due to numerical overflow.
In general, for a given an input s, the value of the logistic function is:
log(1 + exp(s))
and the slope of the logistic loss function is:
exp(s)./(1 + exp(s)) = 1./(1 + exp(-s))
In my algorithm, the value of s = X*beta. Here X is a matrix with N data points and P features per data point (i.e. size(X)=[N,P]) and beta is a vector of P coefficients for each feature such that size(beta)=[P 1].
I am specifically interested in calculating the average value and gradient of the Logistic function for given value of beta.
The average value of the Logistic function w.r.t to a value of beta is:
L = 1/N * sum(log(1+exp(X*beta)),1)
The average value of the slope of the Logistic function w.r.t. to a value of b is:
dL = 1/N * sum((exp(X*beta)./(1+exp(X*beta))' X, 1)'
Note that size(dL) = [P 1].
My issue is that these expressions keep producing numerical overflows. The problem effectively comes from the fact that exp(s)=Inf when s>1000 and exp(s)=0 when s<-1000.
I am looking for a solution such that s can take on any value in floating point arithmetic. Ideally, I would also really appreciate a solution that allows me to evaluate the value and gradient in a vectorized / efficient way.
How about the following approximations:
– For computing L, if s is large, then exp(s) will be much larger than 1:
1 + exp(s) ≅ exp(s)
and consequently
log(1 + exp(s)) ≅ log(exp(s)) = s.
If s is small, then using the Taylor series of exp()
exp(s) ≅ 1 + s
and using the Taylor series of log()
log(1 + exp(s)) ≅ log(2 + s) ≅ log(2) + s / 2.
– For computing dL, for large s
exp(s) ./ (1 + exp(s)) ≅ 1
and for small s
exp(s) ./ (1 + exp(s)) ≅ 1/2 + s / 4.
– The code to compute L could look for example like this:
s = X*beta;
l = log(1+exp(s));
ind = isinf(l);
l(ind) = s(ind);
ind = (l == 0);
l(ind) = log(2) + s(ind) / 2;
L = 1/N * sum(l,1)
I found a good article about this problem.
Cutting through a lot of words, we can simplify the argument to stating that the original expression
log(1 + exp(s))
can be rewritten as
log(exp(s)*(exp(-s) + 1))
= log(exp(s)) + log(exp(-s) + 1)
= s + log(exp(-s) + 1)
This stops overflow from occurring - it doesn't prevent underflow, but by the time that occurs, you have your answer (namely, s). You can't just use this instead of the original, since it will still give you problems. However, we now have the basis for a function that can be written that will be accurate and won't produce over/underflow:
function LL = logistic(s)
if s<0
LL = log(1 + exp(s));
else
LL = s + logistic(-s);
I think this maintains reasonably good accuracy.
EDIT now to the meat of your question - making this vectorized, and allowing the calculation of the slope as well. Let's take these one at a time:
function LL = logisticVec(s)
LL = zeros(size(s));
LL(s<0) = log(1 + exp(s(s<0)));
LL(s>=0) = s(s>=0) + log(1 + exp(-s(s>=0)));
To obtain the average you wanted:
L = logisticVec(X*beta) / N;
The slope is a little bit trickier; note I believe you may have a typo in your expression (missing a multiplication sign).
dL/dbeta = sum(X * exp(X*beta) ./ (1 + exp(X*beta))) / N;
If we divide top and bottom by exp(X*beta) we get
dL = sum(X ./ (exp(-X*beta) + 1)) / N;
Once again, the overflow has gone away and we are left with underflow - but since the underflowed value has 1 added to it, the error this creates is insignificant.