Solution to Form #1

[1,2,3]*[1;2;3]

ans =

   14

[1,2,3]*[1;2;3]=?

The first source of error is roundoff error

Just storing a number in a computer requires rounding

fprintf('realmax = %1.20e\n',realmax)
fprintf('realmin = %1.20e\n',realmin)
fprintf('maximum relative error = %1.20e\n',eps)

realmax = 1.79769313486231570815e+308
realmin = 2.22507385850720138309e-308
maximum relative error = 2.22044604925031308085e-16

s=1;
for i=1:1000
    s=s+eps/10;
end
s==1

ans =  1

Freefall Model (revisited)

Octave solution (will run same on Matlab)

Set default values in Octave for linewidth and text size

%plot --format svg

set (0, "defaultaxesfontname", "Helvetica")
set (0, "defaultaxesfontsize", 18)
set (0, "defaulttextfontname", "Helvetica")
set (0, "defaulttextfontsize", 18) 
set (0, "defaultlinelinewidth", 4)

Define time from 0 to 12 seconds with N timesteps function defined as freefall

function [v_analytical,v_terminal,t]=freefall(N)
    t=linspace(0,12,N)';
    c=0.25; m=60; g=9.81; v_terminal=sqrt(m*g/c);

    v_analytical = v_terminal*tanh(g*t/v_terminal);
    v_numerical=zeros(length(t),1);
    delta_time =diff(t);
    for i=1:length(t)-1
        v_numerical(i+1)=v_numerical(i)+(g-c/m*v_numerical(i)^2)*delta_time(i);
    end
    % Print values near 0,2,4,6,8,10,12 seconds
    indices = round(linspace(1,length(t),7));
    fprintf('time (s)|vel analytical (m/s)|vel numerical (m/s)\n')
    fprintf('-----------------------------------------------\n')
    M=[t(indices),v_analytical(indices),v_numerical(indices)];
    fprintf('%7.1f | %18.2f | %15.2f\n',M(:,1:3)');
    plot(t,v_analytical,'-',t,v_numerical,'o-')
end

[v_analytical,v_terminal,t]=freefall(120);

time (s)|vel analytical (m/s)|vel numerical (m/s)
-----------------------------------------------
    0.0 |               0.00 |            0.00
    2.0 |              18.76 |           18.82
    4.0 |              32.64 |           32.80
    6.1 |              40.79 |           40.97
    8.0 |              44.80 |           44.94
   10.0 |              46.84 |           46.93
   12.0 |              47.77 |           47.82

Types of error

Freefall is example of "truncation error"

Truncation error results from approximating exact mathematical procedure

We approximated the derivative as $\delta v/\delta t\approx\Delta v/\Delta t$

Can reduce error by decreasing step size -> $\Delta t$=delta_time

Another example of truncation error is a Taylor series (or Maclaurin if centered at a=0)

Taylor series: $f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{3!}(x-a)^{3}+...$

We can approximate the next value in a function by adding Taylor series terms:

Approximation	formula
$0^{th}$-order	$f(x_{i+1})=f(x_{i})+R_{1}$
$1^{st}$-order	$f(x_{i+1})=f(x_{i})+f'(x_{i})h+R_{2}$
$2^{nd}$-order	$f(x_{i+1})=f(x_{i})+f'(x_{i})h+\frac{f''(x_{i})}{2!}h^{2}+R_{3}$
$n^{th}$-order	$f(x_{i+1})=f(x_{i})+f'(x_{i})h+\frac{f''(x_{i})}{2!}h^{2}+...\frac{f^{(n)}}{n!}h^{n}+R_{n}$

Where $R_{n}=\frac{f^{(n+1)}(\xi)}{(n+1)!}h^{n+1}$ is the error associated with truncating the approximation at order $n$.

The $n^{th}$-order approximation estimates that the unknown function, $f(x)$, is equal to an $n^{th}$-order polynomial.

In the Freefall example, we estimated the function with a $1^{st}$-order approximation, so

$v(t_{i+1})=v(t_{i})+v'(t_{i})(t_{i+1}-t_{i})+R_{1}$

$v'(t_{i})=\frac{v(t_{i+1})-v(t_{i})}{t_{i+1}-t_{i}}-\frac{R_{1}}{t_{i+1}-t_{i}}$

$\frac{R_{1}}{t_{i+1}-t_{i}}=\frac{v''(\xi)}{2!}(t_{i+1}-t_{i})$

or the truncation error for a first-order Taylor series approximation is

$\frac{R_{1}}{t_{i+1}-t_{i}}=O(\Delta t)$

digital representation of a number is rarely exact
arithmetic (+,-,/,*) causes roundoff error

fprintf('%1.20f\n',double(pi))
fprintf('%1.20f\n',single(pi))

3.14159265358979311600
3.14159274101257324219

ME3255S2017/lecture_02/README.md