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:
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
