Numerical Method

With $h = 1$ the interior grid points are $(1,1), (2,1), (1,2), (2,2)$. Let $f_{11} = a,\ f_{21} = b,\ f_{12} = c,\ f_{22} = d$. All boundary values are $0$.

The five-point formula is

$f_{i+1,j} + f_{i-1,j} + f_{i,j+1} + f_{i,j-1} - 4f_{i,j} = h^2 g(x_i, y_j), \qquad g = 2x^2y^2$

Right-hand sides: $g(1,1) = 2,\ g(2,1) = 8,\ g(1,2) = 8,\ g(2,2) = 32$.

$$\begin{aligned} (1,1):&\quad -4a + b + c = 2\\ (2,1):&\quad a - 4b + d = 8\\ (1,2):&\quad a - 4c + d = 8\\ (2,2):&\quad b + c - 4d = 32 \end{aligned}$$

Equations 2 and 3 give $b = c$. Then $-4a + 2b = 2,\qquad a - 4b + d = 8,\qquad 2b - 4d = 32$

From the first, $a = (2b-2)/4 = (b-1)/2$. In the second, $(b-1)/2 - 4b + d = 8 \Rightarrow -3.5b + d = 8.5$. From the third, $d = (b-16)/2$, so $-3.5b + 0.5b - 8 = 8.5 \Rightarrow -3b = 16.5 \Rightarrow b = -5.5$.

Hence $b = c = -5.5,\quad d = \frac{-5.5 - 16}{2} = -10.75,\quad a = \frac{-5.5-1}{2} = -3.25$

$f(1,1) = -3.25,\quad f(2,1) = f(1,2) = -5.5,\quad f(2,2) = -10.75$

With $h = 1$ the interior grid points are $(1,1), (2,1), (1,2), (2,2)$. Let $f_{11} = a,\ f_{21} = b,\ f_{12} = c,\ f_{22} = d$. All boundary values are $0$.

The five-point formula is

$f_{i+1,j} + f_{i-1,j} + f_{i,j+1} + f_{i,j-1} - 4f_{i,j} = h^2 g(x_i, y_j), \qquad g = 2x^2y^2$

Right-hand sides: $g(1,1) = 2,\ g(2,1) = 8,\ g(1,2) = 8,\ g(2,2) = 32$.

$$\begin{aligned} (1,1):&\quad -4a + b + c = 2\\ (2,1):&\quad a - 4b + d = 8\\ (1,2):&\quad a - 4c + d = 8\\ (2,2):&\quad b + c - 4d = 32 \end{aligned}$$

Equations 2 and 3 give $b = c$. Then $-4a + 2b = 2,\qquad a - 4b + d = 8,\qquad 2b - 4d = 32$

From the first, $a = (2b-2)/4 = (b-1)/2$. In the second, $(b-1)/2 - 4b + d = 8 \Rightarrow -3.5b + d = 8.5$. From the third, $d = (b-16)/2$, so $-3.5b + 0.5b - 8 = 8.5 \Rightarrow -3b = 16.5 \Rightarrow b = -5.5$.

Hence $b = c = -5.5,\quad d = \frac{-5.5 - 16}{2} = -10.75,\quad a = \frac{-5.5-1}{2} = -3.25$

$f(1,1) = -3.25,\quad f(2,1) = f(1,2) = -5.5,\quad f(2,2) = -10.75$

The second equation is read as $4x - y + z = 2$.

Augmented matrix:

$\left[\begin{array}{ccc|c} 3 & -5 & 2 & 15 \\ 4 & -1 & 1 & 2 \\ 1 & -3 & 7 & 22 \end{array}\right]$

Swap $R_1 \leftrightarrow R_3$ to get a pivot of 1:

$\left[\begin{array}{ccc|c} 1 & -3 & 7 & 22 \\ 4 & -1 & 1 & 2 \\ 3 & -5 & 2 & 15 \end{array}\right]$

$R_2 \to R_2 - 4R_1,\ R_3 \to R_3 - 3R_1$:

$\left[\begin{array}{ccc|c} 1 & -3 & 7 & 22 \\ 0 & 11 & -27 & -86 \\ 0 & 4 & -19 & -51 \end{array}\right]$

$R_2 \to R_2/11$:

$\left[\begin{array}{ccc|c} 1 & -3 & 7 & 22 \\ 0 & 1 & -27/11 & -86/11 \\ 0 & 4 & -19 & -51 \end{array}\right]$

$R_1 \to R_1 + 3R_2,\ R_3 \to R_3 - 4R_2$:

$\left[\begin{array}{ccc|c} 1 & 0 & -4/11 & -16/11 \\ 0 & 1 & -27/11 & -86/11 \\ 0 & 0 & -101/11 & -217/11 \end{array}\right]$

$R_3 \to -\tfrac{11}{101}R_3$ gives $z = 217/101$. Then $R_1 \to R_1 + \tfrac{4}{11}R_3,\ R_2 \to R_2 + \tfrac{27}{11}R_3$:

$\left[\begin{array}{ccc|c} 1 & 0 & 0 & -68/101 \\ 0 & 1 & 0 & -257/101 \\ 0 & 0 & 1 & 217/101 \end{array}\right]$

$x = -\frac{68}{101} \approx -0.6733,\quad y = -\frac{257}{101} \approx -2.5446,\quad z = \frac{217}{101} \approx 2.1485$

Check in the first equation: $3(-0.6733) - 5(-2.5446) + 2(2.1485) = -2.020 + 12.723 + 4.297 = 15.0$.

$\left[\begin{array}{ccc|c} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \end{array}\right]$

$R_1 \to R_1/2$:

$\left[\begin{array}{ccc|c} 1 & 1.5 & 2 & 2.5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \end{array}\right]$

$R_2 \to R_2 - 3R_1,\ R_3 \to R_3 - 4R_1$:

$\left[\begin{array}{ccc|c} 1 & 1.5 & 2 & 2.5 \\ 0 & -0.5 & -1 & -1.5 \\ 0 & -1 & -2 & -3 \end{array}\right]$

$R_2 \to -2R_2$:

$\left[\begin{array}{ccc|c} 1 & 1.5 & 2 & 2.5 \\ 0 & 1 & 2 & 3 \\ 0 & -1 & -2 & -3 \end{array}\right]$

$R_1 \to R_1 - 1.5R_2,\ R_3 \to R_3 + R_2$:

$\left[\begin{array}{ccc|c} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]$

The rank of $A$ and of the augmented matrix are both $2 < 3$, so the system is consistent but has infinitely many solutions. Taking $z = t$:

$x = t - 2,\qquad y = 3 - 2t,\qquad z = t$

Check in the first equation: $2(t-2) + 3(3-2t) + 4t = 2t - 4 + 9 - 6t + 4t = 5$. ✓ (For example $t = 0$ gives $(-2, 3, 0)$ and $t = 1$ gives $(-1, 1, 1)$.)

Augmented matrix:

$\left[\begin{array}{ccc|c}1&1&1&6\\2&3&1&10\\1&2&3&13\end{array}\right]$

$R_2\to R_2-2R_1$, $R_3\to R_3-R_1$:

$\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-2\\0&1&2&7\end{array}\right]$

Eliminate $y$ above and below the second pivot: $R_1\to R_1-R_2$, $R_3\to R_3-R_2$:

$\left[\begin{array}{ccc|c}1&0&2&8\\0&1&-1&-2\\0&0&3&9\end{array}\right]$

$R_3\to R_3/3$ gives $z=3$. Then $R_1\to R_1-2R_3$, $R_2\to R_2+R_3$:

$\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&1\\0&0&1&3\end{array}\right]$

Solution: $x=2,\;y=1,\;z=3$. Check: $2+1+3=6$, $4+3+3=10$, $2+2+9=13$.

Bisection method. If $f$ is continuous on $[a, b]$ and $f(a)f(b) < 0$, a root lies in $(a, b)$. Repeatedly halve the interval and keep the half where the sign change persists.

Algorithm

1. Define $f(x)$. Read $a$, $b$, tolerance $\varepsilon$ and maximum iterations $N$.
2. If $f(a)\cdot f(b) > 0$, print "invalid interval" and stop.
3. Repeat for $i = 1$ to $N$:
   1. $c = (a + b)/2$.
   2. If $f(c) = 0$ or $(b - a)/2 < \varepsilon$, print $c$ as the root and stop.
   3. If $f(a)\cdot f(c) < 0$ set $b = c$, else set $a = c$.
4. Print $c$ as the approximate root.

Program (C)

#include <stdio.h>
#include <math.h>

double f(double x) {
    return x * x * x - 4 * x + 1;      /* change as required */
}

int main() {
    double a, b, c, eps;
    int i, maxit;

    printf("Enter a, b, tolerance and max iterations: ");
    scanf("%lf %lf %lf %d", &a, &b, &eps, &maxit);

    if (f(a) * f(b) > 0) {
        printf("Root is not bracketed in [a, b]\n");
        return 1;
    }

    for (i = 1; i <= maxit; i++) {
        c = (a + b) / 2;
        printf("Iter %d: c = %.6f, f(c) = %.6f\n", i, c, f(c));
        if (f(c) == 0 || (b - a) / 2 < eps)
            break;
        if (f(a) * f(c) < 0)
            b = c;
        else
            a = c;
    }
    printf("Approximate root = %.6f\n", c);
    return 0;
}

The number of iterations needed for error below $\varepsilon$ is $n \ge \log_2\!\big((b-a)/\varepsilon\big)$.

Why errors matter. Almost every numerical computation gives an approximate answer. A computational scientist must know where the error comes from, how large it is and how it grows, because:

• the reliability of a result depends on its error bound; a result without an error estimate is of little use;
• errors can accumulate over thousands of operations and make a stable-looking program give nonsense;
• it guides the choice of algorithm, step size, number of iterations and precision needed;
• it tells us when to stop an iterative method (tolerance) and how to avoid ill-conditioned or unstable formulations.

Inherent errors are errors already present in the input data before the computation begins. They arise from measurement limits, imperfect instruments, or from the mathematical model/data being approximated (for example taking $\pi = 3.14$ or a measured length of $5.6$ cm). Their size is fixed by the data and cannot be reduced by a better algorithm, only by better data.

Numerical errors are introduced by the computation itself. They consist of:

• Round-off error: caused by representing numbers with a finite number of digits, e.g. $1/3 = 0.3333$.
• Truncation error: caused by replacing an infinite process by a finite one, e.g. cutting a Taylor series after a few terms or replacing a derivative by a finite difference.

Let $f(x) = x^2 + 4x - 10$.

At the given end points $f(5) = 35 > 0$ and $f(6) = 62 > 0$. There is no sign change, so no root lies in $[5, 6]$ (the actual roots are $-2 \pm \sqrt{14} \approx 1.7417,\ -5.7417$). The bisection method needs a sign change, so the interval is corrected to $[1, 2]$, where $f(1) = -5 < 0$ and $f(2) = 2 > 0$.

Midpoint $c = (a+b)/2$; keep the half in which the sign changes.

The root is approximately $x \approx 1.74$ (exact value $-2+\sqrt{14} = 1.7417$).

The table is $f(x) = \sqrt{x}$, so we want $\sqrt{3.5}$. For second order (three points) Lagrange interpolation choose the points nearest to $x = 3.5$: $x_0 = 2,\ x_1 = 3,\ x_2 = 4$ with $f_0 = 1.4142,\ f_1 = 1.7321,\ f_2 = 2$.

$P_2(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f_0 + \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f_1 + \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f_2$

At $x = 3.5$:

$L_0 = \frac{(0.5)(-0.5)}{(-1)(-2)} = -0.125,\quad L_1 = \frac{(1.5)(-0.5)}{(1)(-1)} = 0.75,\quad L_2 = \frac{(1.5)(0.5)}{(2)(1)} = 0.375$

(check: $L_0+L_1+L_2 = 1$)

$P_2(3.5) = -0.125(1.4142) + 0.75(1.7321) + 0.375(2) = -0.17678 + 1.29908 + 0.75 = 1.8723$

So $\sqrt{3.5} \approx 1.8723$ (true value $1.8708$).

Trapezoidal rule: divide $[a, b]$ into $n$ equal parts of width $h = (b-a)/n$:

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n)\right]$

Program (C) for $\int_a^b f(x)\,dx$ with $f(x) = x^2 + 1$ as an example function:

#include <stdio.h>
#include <math.h>

double f(double x) {
    return x * x + 1;          /* change to the required function */
}

int main() {
    double a, b, h, sum, result;
    int n, i;

    printf("Enter lower limit a, upper limit b: ");
    scanf("%lf %lf", &a, &b);
    printf("Enter number of sub-intervals n: ");
    scanf("%d", &n);

    h = (b - a) / n;
    sum = f(a) + f(b);
    for (i = 1; i < n; i++)
        sum += 2 * f(a + i * h);

    result = (h / 2) * sum;
    printf("Integral = %.6f\n", result);
    return 0;
}

For $a=0,\ b=1,\ n=10$ the program prints $1.335000$ (exact $4/3$).

Here $f(x, y) = x^2 + y^2$, $x_0 = 0$, $y_0 = 0$, $h = 0.2$.

$k_1 = f(x_0, y_0) = 0$ $k_2 = f\!\left(x_0 + \tfrac h2,\ y_0 + \tfrac h2 k_1\right) = f(0.1, 0) = 0.01$ $k_3 = f\!\left(x_0 + \tfrac h2,\ y_0 + \tfrac h2 k_2\right) = f(0.1, 0.001) = 0.01 + 0.000001 = 0.010001$ $k_4 = f(x_0 + h,\ y_0 + h k_3) = f(0.2, 0.0020002) = 0.04 + 0.000004 = 0.040004$

$y(0.2) = y_0 + \frac h6\left(k_1 + 2k_2 + 2k_3 + k_4\right) = \frac{0.2}{6}\left(0 + 0.02 + 0.020002 + 0.040004\right)$

$y(0.2) = \frac{0.2}{6}(0.080006) \approx 0.002667$

(a) Cholesky's method factorizes a symmetric positive definite matrix as $A = LL^T$. First check the pivots.

$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 22 \\ 3 & 22 & 82 \end{bmatrix}$

$l_{11} = \sqrt{1} = 1,\quad l_{21} = 2/1 = 2,\quad l_{31} = 3/1 = 3$

$l_{22}^2 = a_{22} - l_{21}^2 = 3 - 4 = -1 < 0$

The second leading minor is $1\cdot 3 - 2\cdot 2 = -1 < 0$, so $A$ is not positive definite and a real Cholesky factor $LL^T$ does not exist. It can be completed only with complex entries, $l_{22} = i$:

$l_{32} = \frac{22 - 3\cdot 2}{i} = -16i,\quad l_{33}^2 = 82 - 9 - (-16i)^2 = 73 + 256 = 329,\ l_{33} = \sqrt{329} \approx 18.14$

(Sign of $l_{32}$: $l_{32}\,l_{22} = 16 \Rightarrow l_{32} = 16/i = -16i$.)

Equivalently, in real arithmetic, $A = LDL^T$ with $L = \begin{bmatrix}1&0&0\\2&1&0\\3&-16&1\end{bmatrix},\quad D = \mathrm{diag}(1,\,-1,\,329)$ The negative entry in $D$ confirms $A$ is indefinite.

(b) $x^2 + x - 2 = 0$ (roots $1$ and $-2$). Rewrite in the form $x = g(x)$:

$x = g(x) = \sqrt{2 - x}$

Convergence needs $|g'(x)| < 1$ near the root: $g'(x) = -\dfrac{1}{2\sqrt{2-x}}$, and $|g'(1)| = 0.5 < 1$, so the iteration converges to $x = 1$. (The form $x = 2 - x^2$ has $|g'(1)| = 2$ and diverges.)

Take $x_0 = 0.5$ and iterate $x_{n+1} = \sqrt{2 - x_n}$:

The values oscillate about and approach $x \approx 1.00$, so the root is $x = 1$.

Data: $(x_0, f_0) = (2, 4),\ (x_1, f_1) = (3, 9),\ (x_2, f_2) = (4, 16)$, with $h = 1$. Use the natural cubic spline ($M_0 = M_2 = 0$), where $M_i = S''(x_i)$.

For the interior node:

$h M_0 + 2(h + h)M_1 + h M_2 = 6\left[\frac{f_2 - f_1}{h} - \frac{f_1 - f_0}{h}\right]$

$4M_1 = 6\,[\,7 - 5\,] = 12 \Rightarrow M_1 = 3$

On $[x_0, x_1] = [2, 3]$:

$S(x) = \frac{M_0 (x_1 - x)^3 + M_1 (x - x_0)^3}{6h} + \left(\frac{f_0}{h} - \frac{M_0 h}{6}\right)(x_1 - x) + \left(\frac{f_1}{h} - \frac{M_1 h}{6}\right)(x - x_0)$

$S(x) = \frac{3(x-2)^3}{6} + 4(3 - x) + 8.5\,(x - 2)$

At $x = 2.5$:

$S(2.5) = \frac{3(0.5)^3}{6} + 4(0.5) + 8.5(0.5) = 0.0625 + 2 + 4.25 = 6.3125$

So $f(2.5) \approx 6.3125$ (the true value of $x^2$ is $6.25$).

$f(x) = x^2 - 5x + 6,\quad f'(x) = 2x - 5$. Newton-Raphson formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Start with $x_0 = 5$: $f(5) = 6,\ f'(5) = 5$.

The iteration converges to the root $x = 3$ (the other root of $(x-2)(x-3) = 0$ is $2$).

Algorithm (Simpson's 1/3 rule)

1. Read $a$, $b$ and an even number of sub-intervals $n$.
2. Compute $h = (b - a)/n$.
3. Set $sum = f(a) + f(b)$.
4. For $i = 1, 2, \dots, n-1$: compute $x = a + ih$; if $i$ is odd add $4f(x)$ to $sum$, otherwise add $2f(x)$.
5. Result $I = \dfrac{h}{3}\,sum$.
6. Print $I$ and stop.

$\int_a^b f(x)\,dx \approx \frac h3\left[f_0 + 4(f_1 + f_3 + \dots) + 2(f_2 + f_4 + \dots) + f_n\right]$

Evaluation: $f(x) = 3x^2 + 2x - 5$, $a = 0$, $b = 3$, take $n = 4$, $h = 0.75$.

$I = \frac{0.75}{3}\left[-5 + 28 + 4(-1.8125 + 14.6875) + 2(4.75)\right] = 0.25\,[23 + 51.5 + 9.5] = 0.25 \times 84 = 21$

Exact value: $[x^3 + x^2 - 5x]_0^3 = 27 + 9 - 15 = 21$. Simpson's rule is exact for polynomials up to degree 3, so the result is exact.

Ill-conditioned system. A linear system $Ax = b$ is ill-conditioned if a very small change (or rounding error) in the coefficients or the right-hand side produces a large change in the solution. The coefficient matrix is then nearly singular: its determinant is close to zero relative to its entries and its condition number $\kappa(A) = \|A\|\,\|A^{-1}\|$ is very large. Such systems give unreliable answers even with a stable algorithm.

Cholesky decomposition $A = LL^T$ (valid since $A$ is symmetric positive definite: leading minors $4,\ 19,\ 21 > 0$):

$l_{11} = \sqrt{4} = 2,\quad l_{21} = \frac{1}{2} = 0.5,\quad l_{31} = \frac12 = 0.5$

$l_{22} = \sqrt{5 - 0.5^2} = \sqrt{4.75} = 2.1794$

$l_{32} = \frac{2 - (0.5)(0.5)}{2.1794} = \frac{1.75}{2.1794} = 0.8030$

$l_{33} = \sqrt{3 - 0.5^2 - 0.8030^2} = \sqrt{3 - 0.25 - 0.6448} = \sqrt{2.1052} = 1.4510$

$L = \begin{bmatrix} 2 & 0 & 0 \\ 0.5 & 2.1794 & 0 \\ 0.5 & 0.8030 & 1.4510 \end{bmatrix},\qquad A = LL^T$

$f(x, y) = x^2 + y^2$, $y(0) = 0$, $h = 0.2$. Two steps are needed.

Step 1 ($x_0 = 0,\ y_0 = 0$):

$k_1 = f(0, 0) = 0$

$k_2 = f(0.1,\ 0) = 0.01$

$k_3 = f(0.1,\ 0.001) = 0.010001$

$k_4 = f(0.2,\ 0.0020002) = 0.040004$

$y_1 = y(0.2) = 0 + \frac{0.2}{6}\,(0 + 0.02 + 0.020002 + 0.040004) = 0.002667$

Step 2 ($x_1 = 0.2,\ y_1 = 0.002667$):

$k_1 = f(0.2,\ 0.002667) = 0.04 + 0.0000071 = 0.040007$

$k_2 = f(0.3,\ 0.002667 + 0.1(0.040007)) = f(0.3,\ 0.006668) = 0.09 + 0.0000445 = 0.090044$

$k_3 = f(0.3,\ 0.002667 + 0.1(0.090044)) = f(0.3,\ 0.011671) = 0.09 + 0.000136 = 0.090136$

$k_4 = f(0.4,\ 0.002667 + 0.2(0.090136)) = f(0.4,\ 0.020694) = 0.16 + 0.000428 = 0.160428$

$y_2 = y(0.4) = 0.002667 + \frac{0.2}{6}\,(0.040007 + 0.180088 + 0.180272 + 0.160428) = 0.002667 + \frac{0.2}{6}(0.560795)$

$y(0.4) \approx 0.02136$

(a) Types of Errors

Errors in numerical computation are of the following kinds.

• Inherent (data) errors: present in the input data because of measurement or modelling limits.
• Round-off errors: arise because computers store numbers with a finite number of digits, e.g. $\pi \approx 3.1416$.
• Truncation errors: arise when an infinite process is replaced by a finite one, e.g. stopping a Taylor series after $n$ terms.
• Blunders/gross errors: human mistakes such as wrong data entry or wrong formula.

They are measured as absolute error $E_a = |x_{true} - x_{approx}|$, relative error $E_r = E_a/|x_{true}|$ and percentage error $E_r \times 100\%$.

(b) Convergent Methods

An iterative method $x_{n+1} = g(x_n)$ is convergent if the sequence $x_0, x_1, x_2, \dots$ approaches the true root $\alpha$ as $n \to \infty$, i.e. $|x_n - \alpha| \to 0$. For fixed-point iteration, convergence is guaranteed if $|g'(x)| < 1$ near the root. The speed is described by the order of convergence $p$: $|e_{n+1}| \approx C|e_n|^p$. Bisection converges linearly ($p = 1$) but always, Secant has $p \approx 1.618$ and Newton-Raphson has $p = 2$ (quadratic). Methods that do not satisfy these conditions diverge or oscillate.

(c) PDE

A partial differential equation contains the partial derivatives of an unknown function of two or more independent variables. The common second order linear types are:

• Elliptic (Laplace $\nabla^2 u = 0$, Poisson $\nabla^2 u = g(x, y)$),
• Parabolic (heat equation $u_t = c\,u_{xx}$),
• Hyperbolic (wave equation $u_{tt} = c^2 u_{xx}$).

Numerically they are solved by replacing derivatives by finite differences on a grid, e.g. $u_{xx} \approx \dfrac{u_{i+1,j} - 2u_{i,j} + u_{i-1,j}}{h^2}$, and solving the resulting algebraic equations.

(a) Gauss-Seidel method

The system is not diagonally dominant in the given order, so rearrange the equations (second equation first for $x$, third for $y$, first for $z$):

$x = \frac{14 - y - 3z}{5},\qquad y = \frac{7 - x - 9z}{10},\qquad z = \frac{17 + 2x - 7y}{10}$

Start with $(x, y, z) = (0, 0, 0)$ and always use the latest values.

The iterates converge (slowly, because the dominance is weak) to the exact solution

$x = 1,\quad y = -3,\quad z = 4$

Check: $2(1) - 7(-3) - 10(4) = -17$, $5 + (-3) + 12 = 14$, $1 - 30 + 36 = 7$.

(b) Taylor's theorem

$y(x) = y(x_0) + (x - x_0)y'(x_0) + \frac{(x - x_0)^2}{2!}y''(x_0) + \frac{(x-x_0)^3}{3!}y'''(x_0) + \dots + \frac{(x-x_0)^n}{n!}y^{(n)}(x_0) + R_n$

For $y' = x^2 + y^2$, $y(0) = 1$:

$y' = x^2 + y^2 \Rightarrow y'(0) = 1$ $y'' = 2x + 2yy' \Rightarrow y''(0) = 2$ $y''' = 2 + 2y'^2 + 2yy'' \Rightarrow y'''(0) = 2 + 2 + 4 = 8$ $y^{(4)} = 6y'y'' + 2yy''' \Rightarrow y^{(4)}(0) = 12 + 16 = 28$ $y^{(5)} = 6y''^2 + 8y'y''' + 2yy^{(4)} \Rightarrow y^{(5)}(0) = 24 + 64 + 56 = 144$

$y(x) = 1 + x + x^2 + \frac{4}{3}x^3 + \frac{7}{6}x^4 + \frac{6}{5}x^5 + \dots$

At $x = 0.5$:

$y(0.5) = 1 + 0.5 + 0.25 + 0.16667 + 0.07292 + 0.0375 + \dots \approx 2.027$

Including further terms ($x^6$ term $\approx 0.0193$, and so on) the sum rises slowly and settles near $2.07$, so $y(0.5) \approx 2.07$.