I am trying to compute the taylor series of ln(x) for any value of x.
What I have so far is:
clear
clc
n = input('Enter number of iiterations (n): ' );
x = input('enter value of x (x): ');
y = zeros(1,n);
for i = 0:n
y(i+1)=sum + (-1)^(n+1)*(x-1)^n/n;
end
But this code seems to be broken and I can't figure out why. Any suggestions on how to improve?
This is a one liner in addition to the for-loop answer provided by #farbiondriven
For 0<x<1 :
sumLn = #(x, n)(sum(((-1).^(0:n-1)).*((x-1).^(1:n))./(1:n)));
sumLn(0.5,10)
ans =
-0.6931
>> log(0.5)
ans =
-0.6931
For x>0.5 :
sumLn = #(x, n)(sum( ((x-1)/x).^(1:n) ./ (1:n) ));
sumLn(2,10)
ans =
0.6931
log(2) =
0.6931
Note: The variable x in this formula is bounded as mentioned in this link.
Try this:
clear
clc
n = input('Enter number of iterations (n): ' );
x = input('enter value of x with abs value < 1 (x): ');
y = zeros(1,n+1);
y(1)=0;
for i = 1:n
y(i+1)= y(i) + ((-1)^(i+1)*(x-1)^i/i);
end
txt = sprintf('The output is: %f', y(n+1))
I suggest using built-in function and hopefully there is one. taylor(f,var) approximates f with the Taylor series expansion of f up to the fifth order at the point var = 0.
Specify Expansion Point :
Find the Taylor series expansions at x = 1 for these functions. The default expansion point is 0. To specify a different expansion point, use 'ExpansionPoint':
syms x
taylor(log(x), x, 'ExpansionPoint', 1)
ans =
x - (x - 1)^2/2 + (x - 1)^3/3 - (x - 1)^4/4 + (x - 1)^5/5 - 1
Specify Truncation Order :
The default truncation order is 6.
syms x
f = log(x);
t6 = taylor(f, x);
Use 'Order' to control the truncation order. For example, approximate the same expression up to the orders 8.
syms x
taylor(log(x), x, 'ExpansionPoint', 1, 'Order', 8);
Related
I am trying to use the fixed point iteration method with initial approximation x(1)=0 to obtain an approximation to the root of the equation f(x)=3x+sin(x)e^x=0.
The stopping criterion is
|x(k+1)-x(k)|<0.0001
x(1) = 0;
n = 100;
for k = 1:n
f(k) = 3*x(k) +sin(x(k))-exp(x(k));
if (abs(f(k))<0.0001)
break;
end
syms x
diff(f(k));
x(k+1) = x(1)- (f(k))/(diff(f(k)));
end
[x' f']
This is the error I am getting: Error using / Matrix dimensions must
agree. Error in prac2Q2 (line 15)
x(k+1) = x(1)- (f(k))/(diff(f(k)));
I would suggest to calculate the derivative by hand and use that term as denominator or to save the derivative in another variable and use this as the denominator.
Derivative as Variable
f(k) = ...;
df(k) = diff(f(k));
x(k+1) = x(k) - f(k) / df(k);
PS: I cannot test this, because I do not have access to the Symbolic Toolbox right now.
If you're looking for the root of 3*x +sin(x)-exp(x) you want to resolve this equation:
3*x + sin(x) - exp(x) = 0
The easiest way will be to isolate x in one side of the equation:
x = (exp(x) - sin(x))/3 % now iterate until x = (exp(x) - sin(x))/3
Now I would recommand to use an easier fixed point method: x(k+1) = (x(k)+f(x(k)))/2
x = 1 % x0
while 1
y = (exp(x)-sin(x))/3; % we are looking for the root not for a fixed point !!! y = f(x)
x = (x+y)/2 % after a few iterations x == y, so x = (x+y)/2 or x = 2x/2
if abs(x-y) < 1e-10
break
end
end
And you obtain the correct result:
x = 0.36042
No need of symbolic math.
I have a vector x = [x_1 x_2 ... x_n], a vector y = [y_1 y_2 y_3] and a matrix X = [x_11 x_12 ... x_1n; x_21 x_22 ... x_2n; x_31 x_32 ... x_3n].
For i = 1, 2, ..., n, I want to compute the following sum in MATLAB:
sum((x(i) - y*X(:,i))^2)
What I have tried to write is the following MATLAB code:
vv = (x(1) - y*X(:,1))^2; % as an initialization for i=1
for i = 2 : n
vv = vv + (x(i) - y * X(:,i))^2
end
But I am wondering if I can compute that without for loop in order to potentially reduce the computational time especially if n is very high... So are there any other much more optimal possibilities to do that in MATLAB?
Any help will be very appreciated!
You do not need the loop at all,
for i = 2:n
y*X(:,i)
end
is the same as just y*X, so x(i) - yX(:,i) is simply x - yX so basically, its:
vv = sum((x - y * X).^2);
Thanks for #beaker for pointing the mistake.
Assume we have three equations:
eq1 = x1 + (x1 - x2) * t - X == 0;
eq2 = z1 + (z1 - z2) * t - Z == 0;
eq3 = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1 == 0;
while six of known variables are:
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
So we are looking for three unknown variables that are:
X , Z and t
I wrote two method to solve it. But, since I need to run these code for 5.7 million data, it become really slow.
Method one (using "solve"):
tic
S = solve( eq1 , eq2 , eq3 , X , Z , t ,...
'ReturnConditions', true, 'Real', true);
toc
X = double(S.X(1))
Z = double(S.Z(1))
t = double(S.t(1))
results of method one:
X = 316190;
Z = 234060;
t = -2.9280;
Elapsed time is 0.770429 seconds.
Method two (using "fsolve"):
coeffs = [a,b,x1,x2,z1,z2]; % Known parameters
x0 = [ x2 ; z2 ; 1 ].'; % Initial values for iterations
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
options = optimoptions('fsolve','Display','none');
tic
M = fsolve(f_d,x0,options);
toc
results of method two:
X = 316190; % X = M(1)
Z = 234060; % Z = M(2)
t = -2.9280; % t = M(3)
Elapsed time is 0.014 seconds.
Although, the second method is faster, but it still needs to be improved. Please let me know if you have a better solution for that. Thanks
* extra information:
if you are interested to know what those 3 equations are, the first two are equations of a line in 2D and the third equation is an ellipse equation. I need to find the intersection of the line with the ellipse. Obviously, we have two points as result. But, let's forget about the second answer for simplicity.
My suggestion it's to use the second approce,which it's the recommended by matlab for nonlinear equation system.
Declare a M-function
function Y=mysistem(X)
%X(1) = X
%X(2) = t
%X(3) = Z
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
Y(1,1) = x1 + (x1 - x2) * X(2) - X(1);
Y(2,1) = z1 + (z1 - z2) * X(2) - X(3);
Y(3,1) = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1;
end
Then for solving use
x0 = [ x2 , z2 , 1 ];
M = fsolve(#mysistem,x0,options);
If you may want to reduce the default precision by changing StepTolerance (default 1e-6).
Also for more increare you may want to use the jacobian matrix for greater efficencies.
For more reference take a look in official documentation:
fsolve Nonlinear Equations with Analytic Jacobian
Basically giving the solver the Jacobian matrix of the system(and special options) you can increase method efficency.
I'm basicaly trying to find the product of an expression that goes like this:
(x-(N-1)/2).....(x+(N-1)/2) for even value of N
x is a value that I will set at the beginning that changes too but that is a different problem...
let's say for the sake of argument that for now x is a constant (ex x=1)
example for N=6
(x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)*(x+5/2)
the idea was to create a row vector every element of which is each individual term (P(1)=x-5/2) (P(2)=x-3/2)...etc and then calculate its product
N=6;
x=1;
P=ones(1,N);
for k=(-N-1)/2:(N-1)/2
for n=1:N
P(n)=(x-k);
end
end
y=prod(P);
instead this creates a vector that takes only the first value of the epxression and then
repeats the same value at each cell.
there is obviously a fundamental problem with my loop but I just can't see it.
So if anyone can help with that OR suggest a better way to calculate the product I would be grateful.
Use vectorized commands
Why use a loop when you can use vectorized commands like prod?
y = prod(2 * x + [-N + 1 : 2 : N - 1]) / 2;
For convenience, you may want to define an anonymous function for it:
f = #(N,x) reshape(prod(bsxfun(#plus, 2 * x(:), -N + 1 : 2 : N - 1) / 2, 2), size(x));
Note that the function is compatible with a (row or column) vector input x.
Tests in MATLAB's Command Window
>> f(6, [2,2]')
ans =
-14.7656
4.9219
-3.5156
4.9219
-14.7656
>> f(6, [2,2])
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Benchmark
Here is a comparison of rayreng's approach versus mine. The former emerges as the clear winner... :'( ...at least as N increases.
Varying N, fixed x
Fixed N (= 10), vector x of varying length
Fixed N (= 100), vector x of varying length
Benchmark code
function benchmark
% varying N, fixed x
clear all
n = logspace(2,4,20)';
x = rand(1000,1);
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(n)
% rayreng's approach (poly/polyval)
fr = #() rayreng(n(k), x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n(k), x);
tj(k) = timeit(fj);
end
figure
hold on
plot(n, tr, 'bo')
plot(n, tj, 'ro')
hold off
xlabel('N')
ylabel('time (s)')
legend('rayreng', 'jubobs')
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
function benchmark2
% fixed N, varying x
clear all
n = 100;
nx = round(logspace(2,4,20));
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(nx)
disp(k)
x = rand(nx(k), 1);
% rayreng's approach (poly/polyval)
fr = #() rayreng(n, x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n, x);
tj(k) = timeit(fj);
end
figure
hold on
plot(nx, tr, 'bo')
plot(nx, tj, 'ro')
hold off
xlabel('number of elements in vector x')
ylabel('time (s)')
legend('rayreng', 'jubobs')
title(['n = ' num2str(n)])
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
An alternative
Alternatively, because the terms in your product form an arithmetic progression (each term is greater than the previous one by 1/2), you can use the formula for the product of an arithmetic progression.
I agree with #Jubobs in that you should avoid using for loops for this kind of computation. There are cases where for loops perform fast, but for something as simple as this, avoid using loops if possible.
An alternative approach to what Jubobs has suggested is that you can consider that polynomial equation to be in factored form where each factor denotes a root located at that particular location. You can use poly to convert these factors into a polynomial equation, then use polyval to evaluate the expression at the point you want. First, generate your roots by linspace where the points vary from -(N-1)/2 to (N-1)/2 and there are N of them, then plug this into poly. Finally, for any values of x, put this into polyval with the output of poly. The advantage of this approach is that you can evaluate multiple points of x in a single sweep.
Going with what you have, you would simply do this:
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
out = polyval(p, x);
With your example, supposing that N = 6, this would be the output of the first line:
p =
1.0000 0 -8.7500 0 16.1875 0 -3.5156
As such, this is saying that when we expand out (x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)(x+5/2), we get:
x^6 - 8.75x^4 + 16.1875x^2 - 3.5156
If we take a look at the roots of this equation, this is what we get:
r = roots(p)
r =
-2.5000
2.5000
-1.5000
1.5000
-0.5000
0.5000
As you can see, each term corresponds to one factor in your polynomial equation, so we do have the right mindset here. Now, all you have to do is use p with your values of x into polyval to obtain your results. For example, if I wanted to evaluate that polynomial from -2 <= x <= 2 where x is an integer, this is the result I get:
polyval(p, -2:2)
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Therefore, when x = -2, the result is -14.7656 and so on.
Though I would recommend the solution by #Jubobs, it is also good to check what the issue is with your loop.
The first indication that something is wrong, is that you have a nested loop over 2 variables, and only index with one of them to store the result. Probably you just need a single loop.
Here is a loop that you may be interested in that should do roughly what you need:
N=6;
x=1;
k=(-N-1)/2:(N-1)/2
P = ones(size(k));
for n=1:numel(k)
P(n)=(x-k(n));
end
y=prod(P);
I tried to keep the code close to the original, so hopefully it is easy to understand.
Is there anyway to use a loop index to call symbolic variables in Matlab? For example, consider the following code whose goal is to store the symbolic expression "x1+x2+x3" in "y".
syms x1 x2 x3
y = 0;
for i = 1:3
y = y + xi;
end
The code does not work because on each iteration Matlab reads "y = y + xi" and returns the error "xi is undefined", instead of reading "y = y + x1", "y = y + x2" and "y = y + x3", is there anyway around this?
Thanks.
I'd suggest this, provided that you can create your numbered symbolic variable slightly differently:
x = sym('x',[1 3]); % or: syms x1 x2 x3; x = [x1 x2 x3];
y = x(1);
for i = 2:numel(x)
y = y+x(i);
end
Of course in this simple example, the entire for loop and everything else can be replaced with:
y = sum(sym('x',[1 3]));
See the documentation on sym for more details.
EDIT: Note that, as #pm89 points out, by allocating the 1-by-3 symbolic vector x, you of course won't have direct access to the symbolic variables x1,x2, and x3 in your workspace, but will have to index them as shown. This is similar to working with arrays or cells and has many of the same benefits as my second vectorized example illustrates.
If your Matlab does not support the matrix declaration of symbolics directly (as sym('x',[3 1])) you can write your own function for that:
function out = Matrix_Sym(name, size) %#ok<STOUT>
rows = size(1);
cols = size(2);
S = '';
for k1 = 1:rows
for k2 = 1:cols
if rows == 1
S = [S name int2str(k2) ' '];
elseif cols == 1
S = [S name int2str(k1) ' '];
else
S = [S name int2str(k1) int2str(k2) ' '];
end
end
end
eval(['syms ' S]);
eval (['out = reshape([' S '], [rows, cols]);']);
Then you could get the same result with:
x = Matrix_Sym('x', [3 1])
...