Just started acquaintance with Octave. Trying to calculate the sum of an infinite series. Math is bad, can anyone help? In the code for x I took one.
row formula
sum((-1)^([1:inf]+2)/factorial([1:inf])*1^[1:inf]);
Mathematically, the formula is
exp(-1/x)-1
which can derived from Taylor expansion.
Symbolic verification
pkg load symbolic
syms x n
simplify(symsum((-1/x).^n./(factorial(n)),n,1,inf))
which gives
>> simplify(symsum((-1/x).^n./(factorial(n)),n,1,inf))
ans = (sym)
-1
---
x
-1 + e
Numeric verification
n = 20;
x = 2;
such that
>> sum((-1/x).^(1:n)./(factorial(1:n)))
ans = -0.39347
>> exp(-1/x)-1
ans = -0.39347
Related
x=[-4 -2; 3 1] [x y] + [-t -2t-1]
x(0)=[3 -5]
As I know firstly, when the matrix is denoted by A, we must compute exp(At) by diagonalizing A: if A=PDP^(-1) for a diagonal D then exp(At) = P exp(Dt) P^(-1) where exp(Dt) is a diagonal matrix with (exp(Dt))_{ii} = exp(D_(ii) t)...
How can I write The Maple code? ( Sorry, I write in here since I didn't find maple.stackexchange)
restart: with(LinearAlgebra):
A := Matrix(2,2,[-4,-2,3,1]);
....
Hi have made this code to plot a function.
I need to mark with an red X all the crossings between x = 0 and the blue wave line in the graph.
I have made some tries but with '-xr' in the plot function but it places X marks out of the crossings.
Anyone knows how to do it. Many thanks.
Code:
% entrada
a = input('Introduza o valor de a: ');
% ficheiro fonte para a função
raizes;
% chamada à função
x = 0:.1:50;
or = x;
or(:) = 0;
h = #(x) cos(x);
g = #(x) exp(a*x)-1;
f = #(x) h(x) - g(x);
zeros = fzero(f,0);
plot(x,f(x));
hold on
plot(zeros,f(zeros),'-xr')
hold off
Graph (it only marks one zero, i need all the zero crossings):
As mentioned in the comments above, you need to look for the zeros of your function before you can plot them. You can do this mathematically (in this case set f(x) = g(x) and solve for x) or you can do this analytically with something like fsolve.
If you read the documentation for fsolve, you will see that it searches for the zero closest to the provided x0 if passed a scalar or the first zero if passed an interval. What we can do for a quick attempt at a solution is to pass our x values into fsolve as initial guesses and filter out the unique values.
% Set up sample data
a = .05;
x = 0:.1:50;
% Set up equations
h = #(x) cos(x);
g = #(x) exp(a*x)-1;
f = #(x) h(x) - g(x);
% Find zeros of f(x)
crossingpoints = zeros(length(x), 1); % Initialize array
for ii = 1:length(x) % Use x data points as guesses for fzero
try
crossingpoints(ii) = fzero(f, x(ii)); % Find zero closest to guess
end
end
crossingpoints(crossingpoints < 0) = []; % Throw out zeros where x < 0
% Find unique zeros
tol = 10^-8;
crossingpoints = sort(crossingpoints(:)); % Sort data for easier diff
temp = false(size(crossingpoints)); % Initialize testing array
% Find where the difference between 'zeros' is less than or equal to the
% tolerance and throw them out
temp(1:end-1) = abs(diff(crossingpoints)) <= tol;
crossingpoints(temp) = [];
% Sometimes catches beginning of the data set, filter it out if this happens
if abs(f(crossingpoints(1))) >= (0 + tol)
crossingpoints(1) = [];
end
% Plot data
plot(x, f(x))
hold on
plot(crossingpoints, f(crossingpoints), 'rx')
hold off
grid on
axis([0 20 -2 2]);
Which gives us the following:
Note that due to errors arising from floating point arithmetic we have to utilize a tolerance to filter our zeros rather than utilizing a function like unique.
I have a linear program where I can put in numbers for n and it gives me the ouput of the LP for this specific n. I want to do this now for various n=10...1000. Is there a technique where I dont have to do it manually fpr each n and instead does this automatically and outputs the solution of the LP for each n in a file? I like to plot the graph later on.
This is my linear program:
#Specify the number of n for the linear program.
param n := 5000;
#This is the set of probabilities of
set N := {1 .. n};
#We specify the variables for the probabilities p_1,...p_n.
var p[<i> in N] real >= 0;
#These are the values of the vector c. It specifies a constant for each p_i.
param c[<i> in N] := i/n ;
#We define the entries a_{ij} of the Matrix A.
defnumb a(i,j) :=
if i < j then 0
else if i == j then i
else 1 end end;
#The objective function.
maximize prob: sum <i> in N : c[i] * p[i];
#The condition which needs to be fulfilled.
subto condition:
forall <i> in N:
sum <j> in N: a(i,j) * p[j] <= 1;
You can give arguments via the console with:
-D n=[number you want] -o [output file]
Then you could iterate several n by just using a shell script, e.g.,
for i in {1..100}
do
zimpl -D n=$i -o 'output_'$i yourfile.zpl
done
I want to create a polynomial with given coefficients. This seems very simple but what I have found till now did not appear to be the thing I desired.
For example in such an environment;
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
Given a list/vector v of length n (I will set this n and v at the begining), I want to get the polynomial v(x) as v[i]*x^i.
(Actually after that I am going to build the quotient ring GF(4,'a')[x] /< x^n-v(x) > after getting this v(x) from above) then I will say;
S = R.quotient(x^n-v(x), 'y')
y = S.gen()
But I couldn't write it.
This is a frequently asked question in many places so it is better to leave it here as an answer although the answer I have is so simple:
I just wrote R(v) and it gave me the polynomial:
sage
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
R(v)
x^10 + a*x^8 + a*x^7 + x^6 + x^5 + x^4 + a*x + 1
Basically (that is, ignoring the specifics of your polynomial ring) you have a list/vector v of length n and you require a polynomial which is the sum of all v[i]*x^i. Note that this sum equals the matrix product V.X where V is a one row matrix (essentially equal to the vector v) and X is a column matrix consisting of powers of x. In Maxima you could write
v: [1,a,0,0,1,1,1,a,a,0,1]$
n: length(v)$
V: matrix(v)$
X: genmatrix(lambda([i,j], x^(i-1)), n, 1)$
V.X;
The output is
x^10+ax^8+ax^7+x^6+x^5+x^4+a*x+1
This may be quite a basic question for someone who knows linear programming.
In most of the problems that I saw on LP has somewhat similar to following format
max 3x+4y
subject to 4x-5y = -34
3x-5y = 10 (and similar other constraints)
So in other words, we have same number of unknown in objective and constraint functions.
My problem is that I have one unknown variable in objective function and 3 unknowns in constraint functions.
The problem is like this
Objective function: min w1
subject to:
w1 + 0.1676x + 0.1692y >= 0.1666
w1 - 0.1676x - 0.1692y >= -0.1666
w1 + 0.3039x + 0.3058y >= 0.3
w1 - 0.3039x - 0.3058y >= -0.3
x + y = 1
x >= 0
y >= 0
As can be seen, the objective function has only one unknown i.e. w1 and constraint functions have 3 (or lets say 2) unknown i.e w1, x and y.
Can somebody please guide me how to solve this problem, especially using R or MATLAB linear programming toolbox.
Your objective only involves w1 but you can still view it as a function of w1,x,y, where the coefficient of w1 is 1, and the coeffs of x,y are zero:
min w1*1 + x*0 + y*0
Once you see this you can formulate it in the usual way as a "standard" LP.
Prasad is correct. The number of unknowns in the objective function does not matter. You can view unknowns that are not present as having a zero coefficient.
This LP is easily solved using Matlab's linprog function. For more
details on linprog see the documentation here.
% We lay out the variables as X = [w1; x; y]
c = [1; 0; 0]; % The objective is w1 = c'*X
% Construct the constraint matrix
% Inequality constraints will be written as Ain*X <= bin
% w1 x y
Ain = [ -1 -0.1676 -0.1692;
-1 0.1676 0.1692;
-1 -0.3039 -0.3058;
-1 0.3039 0.3058;
];
bin = [ -0.166; 0.166; -0.3; 0.3];
% Construct equality constraints Aeq*X == beq
Aeq = [ 0 1 1];
beq = 1;
%Construct lower and upper bounds l <= X <= u
l = [ -inf; 0; 0];
u = inf(3,1);
% Solve the LP using linprog
[X, optval] = linprog(c,Ain,bin,Aeq,beq,l,u);
% Extract the solution
w1 = X(1);
x = X(2);
y = X(3);