Error in numerical methods:
The difference between the exact (true) value and the approximate value obtained by a numerical method is called error. If $x$ is the true value and $\tilde{x}$ is the approximate value, then error is usually taken as
$E = x - \tilde{x} \quad \text{or} \quad E = \tilde{x} - x.$

Different kinds of error:

• Absolute error: $E_a = |x - \tilde{x}|$
• Relative error: $E_r = \dfrac{|x - \tilde{x}|}{|x|}, \quad x \neq 0$
• Percentage error: $E_p = \dfrac{|x - \tilde{x}|}{|x|} \times 100\%$

By source, errors are also classified as:

• Inherent error: due to inexact input data.
• Truncation error: due to truncating an infinite process or series.
• Round-off error: due to finite-digit arithmetic.
• Discretization error: due to approximating a continuous problem by a discrete one.

Numerical part:
True value, $x = 625.4835$.
Approximating to 4 significant digits: the first four significant digits are $6,2,5,4$; the next digit is $8 \ge 5$, so we round up.
$\tilde{x} = 625.5$

Absolute error: $E_a = |625.5 - 625.4835| = 0.0165$

Relative error: $E_r = \frac{0.0165}{625.4835} \approx 2.63796 \times 10^{-5}$

Percentage error: $E_p = 2.63796 \times 10^{-5} \times 100 \approx 0.002638\%$

$\boxed{\text{Percentage error} \approx 0.002638\%}$

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OR

MATLAB program for RK-2 method to solve
$y' = f(x,y), \quad y(x_0)=y_0 \quad \text{at } x=x_p$

% RK-2 (Heun's method) for IVP y' = f(x,y), y(x0)=y0 at x=xp
clc; clear;

% Define the function f(x,y)
f = @(x,y) x + y;      % Change this to the given f(x,y)

% Inputs
x0 = 0;                % initial x
y0 = 1;                % initial y
xp = 1;                % target x
h  = 0.1;              % step size

x = x0;
y = y0;

while x < xp
    if x + h > xp
        h = xp - x;    % adjust last step
    end
    
    k1 = h * f(x, y);
    k2 = h * f(x + h, y + k1);
    
    y = y + (k1 + k2)/2;
    x = x + h;
end

fprintf('At x = %f, y = %f\n', x, y);

Note: For the midpoint RK-2 method, replace the two stages by

k1 = h * f(x, y);
k2 = h * f(x + h/2, y + k1/2);
y = y + k2;

Write the system as (Ax=b):

[ A=\begin{bmatrix} 1&1&1\ 2&-3&4\ 3&4&5 \end{bmatrix},\qquad b=\begin{bmatrix} 9\13\40 \end{bmatrix} ]

Using Doolittle’s LU-decomposition, take (A=LU), where (L) is unit lower triangular and (U) is upper triangular.

[ L=\begin{bmatrix} 1&0&0\ 2&1&0\ 3&-\frac15&1 \end{bmatrix},\qquad U=\begin{bmatrix} 1&1&1\ 0&-5&2\ 0&0&\frac{12}{5} \end{bmatrix} ]

Check:

[ LU= \begin{bmatrix} 1&1&1\ 2&-3&4\ 3&4&5 \end{bmatrix}=A ]

Now solve (Ly=b):

[ \begin{bmatrix} 1&0&0\ 2&1&0\ 3&-\frac15&1 \end{bmatrix} \begin{bmatrix} y_1\y_2\y_3 \end{bmatrix}

\begin{bmatrix} 9\13\40 \end{bmatrix} ]

From the first row:

[ y_1=9 ]

From the second row:

[ 2y_1+y_2=13 ]

[ 18+y_2=13 \Rightarrow y_2=-5 ]

From the third row:

[ 3y_1-\frac15y_2+y_3=40 ]

[ 27-\frac15(-5)+y_3=40 ]

[ 27+1+y_3=40 \Rightarrow y_3=12 ]

So,

[ y=\begin{bmatrix} 9\-5\12 \end{bmatrix} ]

Now solve (Ux=y):

[ \begin{bmatrix} 1&1&1\ 0&-5&2\ 0&0&\frac{12}{5} \end{bmatrix} \begin{bmatrix} x\y\z \end{bmatrix}

\begin{bmatrix} 9\-5\12 \end{bmatrix} ]

From the third row:

[ \frac{12}{5}z=12 ]

[ z=5 ]

From the second row:

[ -5y+2z=-5 ]

[ -5y+10=-5 ]

[ -5y=-15 \Rightarrow y=3 ]

From the first row:

[ x+y+z=9 ]

[ x+3+5=9 ]

[ x=1 ]

Hence the required solution is

[ \boxed{x=1,\quad y=3,\quad z=5} ]

Verification:

[ 1+3+5=9 ]

[ 2(1)-3(3)+4(5)=2-9+20=13 ]

[ 3(1)+4(3)+5(5)=3+12+25=40 ]

So the solution is correct.

Using Lagrange’s interpolation formula,

[ y(x)=\sum_{i=0}^{3} y_i \frac{(x-x_0)(x-x_1)\cdots (x-x_{i-1})(x-x_{i+1})\cdots (x-x_3)} {(x_i-x_0)(x_i-x_1)\cdots (x_i-x_{i-1})(x_i-x_{i+1})\cdots (x_i-x_3)} ]

Given points:

[ (x_0,y_0)=(1,10),\quad (x_1,y_1)=(4,25),\quad (x_2,y_2)=(9,40),\quad (x_3,y_3)=(11,60) ]

For (x=3),

[ y(3)=10\frac{(3-4)(3-9)(3-11)}{(1-4)(1-9)(1-11)} +25\frac{(3-1)(3-9)(3-11)}{(4-1)(4-9)(4-11)} ]

[ +40\frac{(3-1)(3-4)(3-11)}{(9-1)(9-4)(9-11)} +60\frac{(3-1)(3-4)(3-9)}{(11-1)(11-4)(11-9)} ]

Now,

[ 10\frac{(-1)(-6)(-8)}{(-3)(-8)(-10)} =10\cdot \frac{1}{5}=2 ]

[ 25\frac{(2)(-6)(-8)}{(3)(-5)(-7)} =25\cdot \frac{32}{35}=\frac{160}{7} ]

[ 40\frac{(2)(-1)(-8)}{(8)(5)(-2)} =40\cdot \left(-\frac15\right)=-8 ]

[ 60\frac{(2)(-1)(-6)}{(10)(7)(2)} =60\cdot \frac{3}{35}=\frac{36}{7} ]

Therefore,

[ y(3)=2+\frac{160}{7}-8+\frac{36}{7} ]

[ =2-8+\frac{196}{7} =-6+28=22 ]

[ \boxed{y(3)=22} ]

Let $f(x)=3x-\cos x-1$, so we solve $f(x)=0$.

Locating the root. $f(0.6)=1.8-0.82534-1=-0.02534<0$ and $f(0.7)=2.1-0.76484-1=0.33516>0$, so a root lies between $0.6$ and $0.7$.

Secant formula. $x_{n+1}=x_n-\frac{f(x_n)(x_n-x_{n-1})}{f(x_n)-f(x_{n-1})}$

Take $x_0=0.6$, $x_1=0.7$.

The last two iterates agree to four decimal places ($0.6071$).

Root: $x\approx 0.6071$.

Take $n=4$ sub-intervals (an even number, as Simpson's $\tfrac13$ rule requires), so $h=\dfrac{1-0}{4}=0.25$ and $f(x)=e^{-x^2}$.

Simpson's $\tfrac13$ rule: $I\approx\frac{h}{3}\left[(y_0+y_4)+4(y_1+y_3)+2y_2\right]$

$I\approx\frac{0.25}{3}\left[1.3678794+4(1.5091959)+2(0.7788008)\right]$

$=\frac{0.25}{3}\left[1.3678794+6.0367836+1.5576016\right]=\frac{0.25}{3}(8.9622646)=0.746855$

Result: $\displaystyle\int_0^1e^{-x^2}dx\approx 0.7469$.

Here $f(x,y)=x+\sin y$, $x_0=0$, $y_0=1$, $h=0.1$. Euler's method: $y_{n+1}=y_n+h\,(x_n+\sin y_n)$

(Use $y$ in radians.)

Result: $y(1)\approx 2.375$.