m inline poly polyder fzero

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.