inline Link to heading

  • Purpose: Define a function in the same .m file.
  • Note: Filename should not be inline.m, which could triggered error message.
  • Example is func_inline.m as follow.
    f = inline('x^2 + 2.34');
fprintf("x\tf(x)\n")
for x = 0:4
 fprintf('%.0f\t%.2f\n', x, f(x)) 
end
  • Result
    >> func_inline
x	f(x)
0	2.34
1	3.34
2	6.34
3	11.34
4	18.34

fzero Link to heading

  • Purpose: Find root of a function with initial guess.
  • Example is as follow.
    f = inline('x^2 + 6.25');
x = fzero(f, 4);
fprintf('Equation \n');
disp(f)
fprintf('root = %0.2f', x)
  • Result
    Equation 

     Inline function:
     f(x) = x^2 - 6.25

root = 2.50

polynomial Link to heading

  • A polynomial of $n$ degree is defined as $$\tag{1} p_n(x) = \sum_{i = 0}^n a_i x^i. $$
  • Its coefficients can be stored in a vector or array as follow $$\tag{2} a = [a_n \ \ a_{n-1} \ \ \dots \ \ a_i \ \ \dots \ \ a_1 \ \ a_0 ]. $$
  • Its symbolic representaion can be obtained from its coefficients using poly2sym function as follow.
    % test_poly2sym

% define coefficients of a polynomial
a = [1 -1 2 -2 3];

% obtain its symbolic representation
ps = poly2sym(a);

% display it
display(ps);
    whose result is
    >> test_poly2sym

ps =

x^4 - x^3 + 2*x^2 - 2*x + 3
    or $x^4 - x^3 + 2x^2 - 2x + 3$ that is related to $c = [1 \ \ -1 \ \ 2 \ \ -2 \ \ 3]$.

derivative of $p_n(x)$ Link to heading

  • From Eqn (1) it can be obtained that its derivative with respect to $x$ is $$\tag{3} p_n '(x) = \sum_{i = 1}^n i \ a_i \ x^{i-1}. $$

  • If $q_m(x) = p '_n(x)$ then

    $$\tag{4} q_ m(x) = \sum _{i = 0}^m b _i x^i, $$

    with

    $$\tag{5} b = [b_m \ \ b_{m-1} \ \ \dots \ \ b_i \ \ \dots \ \ b_1 \ \ b_0 ]. $$

  • And the relation of elements between $b$ and $a$ is

    $$\tag{6} b_i = (i+1) a_{i+1}, \ \ \ \ i = 1, \ 2, \ \dots, \ m, $$

    with $m = n - 1$.

polyder Link to heading

  • It derives the coefficients of a polynomial.
  • Suppose there is a polynomial $$ p_4(x) = x^4 + 2 x^3 - 4 x^2 + x - 5 $$ or $a = [1 \ \ 2 \ \ -4 \ \ 1 \ \ -5]$.
  • Then its derivative with respect to $x$ is $$ q_3(x) = 4 x^3 + 6 x^2 - 8 x + 1 $$ or $b = [4 \ \ 6 \ \ -8 \ \ 1]$.
  • Notice the following lines of code.
    % test_polyder

% define coefficients of a polynomial
a = [1 2 -4 1 -5];

% derive it
b = polyder(a);

% display result
fprintf('p(x) = [');
fprintf('%g, ', a(1:end-1));
fprintf('%g]\n', a(end));

fprintf('q(x) = [');
fprintf('%g, ', b(1:end-1));
fprintf('%g]\n', b(end));
    and the result
    >> test_polyder
p(x) = [1, 2, -4, 1, -5]
q(x) = [4, 6, -8, 1]
    which are previous
    • $p_4(x) = x^4 + 2 x^3 - 4 x^2 + x - 5$, and
    • $q_3(x) = 4 x^3 + 6 x^2 - 8 x + 1$.

wrap it all up Link to heading

  • Result
    p(x) = x^4 + 2*x^3 - 4*x^2 + x - 5

q(x) = 4*x^3 + 6*x^2 - 8*x + 1
  • Code
    % display symbolic
fprintf('p(x) = ')
disp(poly2sym(a));
fprintf('q(x) = ')
disp(poly2sym(b));
    where a and b are as previously defined.