How do I integrate something like (x + y) (or an expression with any number of variables) with respect to just one of the variables, let's say x from 0 to 1, and get a function of the other variable (1/2 + y in this case) back?
If you mean symbolically, then use:
syms x y
f=int(x+y,x,0,1)
which gives
f =
y + 1/2
then get f(y=4) from subs(f,4) which gives
ans =
9/2
If you have more than two variables, use:
syms x y z
f=int(x+y+z,x,0,1)
which gives
f =
y + z + 1/2
then for f(y=4,z=5) use subs(f,{y,z},[4,5]) gives
ans =
19/2
Related
I am testing MATLAB capabilities in solving equations for a project that I intend to do, so I gave it a test run with something simple, but the results that it gives me are incorrect. I tried to solve two non-linear equations with two unknowns, one of the solutions is correct the other is not.
syms theta d x y
eq1 = d * cos(theta) == x;
eq2 = d * sin(theta) == y;
sol = solve(eq1, eq2, theta, d)
sol.theta
sol.d
The solutions for d are correct, but for theta I get:
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*atan((x + (x^2 + y^2)^(1/2))/y)
And the correct answer for theta is simply atan(y/x)
Then when I evaluate these solutions with x = 1, y = 0, I get:
eval(sol.d)
eval(sol.theta)
d = 1, -1
theta = NaN, -3.1416
Solutions for d are correct, but theta in that scenario should be 0.
What am I doing wrong?
EDIT: solving it by hand it looks like this: Divide the y equation by the x equation
y/x = (d * sin(theta)) / (d * cos(theta))
y/x = sin(theta)/cos(theta)
y/x = tan(theta)
theta = atan(y/x)
Even if matlab solves it in some other way and gets a different expression, it should still yield the same final result when I use numbers and it PARTIALLY does.
For x = 1 and y = 0, theta should be 0, => this doesnt work, it gives NaN (explanation bellow)
for x = 1 and y = 1, theta should be 45 degrees => this works
for x = 0 and y = 1 theta should be 90 degrees => this works
And I just checked it again with the 45 and 90 degree values for x and y and it works, but for x = 1 and y = 0 it still gives NaN as one of the answers and that is because it gets a 0/0 from the way it is expressing it
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*(1 - (1^2 + 0^2))^(1/2)/0
-2*(1 - 1)^(1/2)/0
0/0
but if its in the form of atan(y/x) the result is
theta = atan(0/1)
theta = atan(0)
theta = 0
Did you mean to solve this:
syms a b theta d real
eq1 = a==d * cos(theta) ;
eq2 = b==d * sin(theta) ;
[sol] = solve([eq1 eq2],[d theta] ,'IgnoreAnalyticConstraints', true,'Real',true,'ReturnConditions',true);
When solving the equations with symbolic x and y, the solver will find a solution with a certain condition, which can be obtained using the argument 'ReturnCondition':
syms x y theta d real
eq1 = d*cos(theta) == x;
eq2 = d*sin(theta) == y;
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
This gives the following result for sol
>> sol.d
(x^2 + y^2)^(1/2)
-(x^2 + y^2)^(1/2)
>> sol.theta
2*pi*k - 2*atan((x - (x^2 + y^2)^(1/2))/y)
2*pi*k - 2*atan((x + (x^2 + y^2)^(1/2))/y)
>> sol.parameters
k
>> sol.conditions
y ~= 0 & in(k, 'integer')
y ~= 0 & in(k, 'integer')
As you can see, y = 0 does not fulfill this general solution given by the solver, resulting in your problem for y = 0. You can find solutions for y = 0 by either making y numeric instead of symbolic, or by adding an assumption:
syms x y theta d real
assume(y==0)
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
I guess its easier to just set y=0 numeric, for this one condition, since there are already 4 possible solutions and conditions for the three lines above.
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);
This question is connected to this one. Suppose again the following code:
syms x
f = 1/(x^2+4*x+9)
Now taylor allows the function f to be expanded about infinity:
ts = taylor(f,x,inf,'Order',100)
But the following code
c = coeffs(ts)
produces errors, because the series does not contain positive powers of x (it contains negative powers of x).
In such a case, what code should be used?
Since the Taylor Expansion around infinity was likely performed with the substitution y = 1/x and expanded around 0, I would explicitly make that substitution to make the power positive for use on coeffs:
syms x y
f = 1/(x^2+4x+9);
ts = taylor(f,x,inf,'Order',100);
[c,ty] = coeffs(subs(ts,x,1/y),y);
tx = subs(ty,y,1/x);
The output from taylor is not a multivariate polynomial, so coeffs won't work in this case. One thing you can try is using collect (you may get the same or similar result from using simplify):
syms x
f = 1/(x^2 + 4*x + 9);
ts = series(f,x,Inf,'Order',5) % 4-th order Puiseux series of f about 0
c = collect(ts)
which returns
ts =
1/x^2 - 4/x^3 + 7/x^4 + 8/x^5 - 95/x^6
c =
(x^4 - 4*x^3 + 7*x^2 + 8*x - 95)/x^6
Then you can use numden to extract the numerator and denominator from either c or ts:
[n,d] = numden(ts)
which returns the following polynomials:
n =
x^4 - 4*x^3 + 7*x^2 + 8*x - 95
d =
x^6
coeffs can then be used on the numerator. You may find other functions listed here helpful as well.
I'm using octave's ezmesh to plot a linear regression defined as follows:
f = #(x,y) 1 * theta(1) + x * theta(2) + y * theta(3) + x * y * theta(4)
For some fixed vector theta:
octave:275> theta
theta =
9.4350e+00
1.7410e-04
3.3702e-02
1.6498e-07
I'm using a domain of [0 120000 0 1400], and can evaluate:
octave:276> f(0, 0)
ans = 9.4350
octave:277> f(120000, 1400)
ans = 105.23
However, if I run:
octave:278> ezmesh(f, [0 120000 0 1400])
The resulting mesh has a z value of around 570 for (0, 0) and just under 640 for (120000, 1400). I'm baffled. What could be causing this?
EDIT: Even if I simplify f to the following, similar behavior occurs:
octave:308> f = #(x, y) (x * y)
Why is ezmesh not handling multiplication as expected (by me), so that the function evaluates as I expect, and the values change when the function is used inside of ezmesh?
ezmesh invokes the function handle on a matrix of values (to benefit from vectorization performance). Use .* for multiplication.
From here, I'm trying to solve a symbolic system of equations like this
syms x y z;
[x, y, z] = solve('z = 4*x', 'x = y', 'z = x^2 + y^2')
x =
0
2
y =
0
2
z =
0
8
except that my equations are generated at different points in the m-file and with random coefficients. My question is how can I accomplish the following...
// Generate the first equation.
n = *random number generated here*;
E1 = (z == n*x + 2*n); // <--- How to save this symbolic equation to use in "solve(...)" later?
// Other work, generate other eqs.
...
// Solve system of eqs.
[x, y, z] = solve( E1 , E2, E3) // What/How to call/use the previously saved symbolic equations.
Thanks for the help!
EDIT
Updated to better explain objective.
Use sprintf if you want to keep the randomly generated value of n for later use with solve:
n = *random number generated here*;
E1 = (z == n*x + 2*n);
Eq1 = sprintf('z == %f*x + 2*%f',n,n);
You can play with the parameters on %f to see how much precision you want to include.