A numerical method is said to be stable if errors introduced at any step of the computation do not grow significantly in subsequent steps. In other words, small perturbations in input data lead to small changes in output.

Example: Bisection Method is numerically stable because dividing an interval in half reliably reduces the error by half at each step—errors don't amplify.

Unstable Example: Some ill-conditioned problems (like solving nearly singular systems) can cause round-off errors to grow exponentially.

The first 3 significant digits are $5, 6, 7$.

Form the mantissa: $f = 0.567$

The original number $0.005672$ requires moving the decimal 3 places to get $0.567$, so the exponent is $p = -2$.

Therefore: $0.005672 = 0.567 \times 10^{-2}$

Check: $0.567 \times 0.01 = 0.00567$ ✓

$$|E| = |x_t - x_a| = \left|\frac{1}{3} - 0.33\right|$$

$$= \left|0.333... - 0.33\right| = \left|0.00333...\right|$$

$$\approx 0.00333 \text{ or } \frac{1}{300}$$

Disadvantage 1: Slow Convergence
The Bisection Method has linear convergence—each iteration reduces the error by approximately half. This means you need many iterations for high accuracy. For example, to achieve 10 decimal places of accuracy from an initial interval of width 1, you need about $n \geq \log_2(10^{10}) \approx 33$ iterations!

Disadvantage 2: Cannot Find Roots Where Function Only Touches the Axis
If the function approaches the x-axis but doesn't cross it (like $f(x) = (x-1)^2$ at $x=1$), the signs $f(a)$ and $f(b)$ never change, and Bisection cannot detect the root. The method requires a sign change, not just tangency.

An iterative process is a computational technique that:

1. Starts with an initial guess or approximation $x_0$

2. Applies a fixed formula repeatedly to improve the approximation. Each application is called an iteration.

3. Terminates when a stopping criterion is met, such as:
- The function value becomes sufficiently close to zero: $|f(x_n)| < \epsilon$
- Consecutive approximations differ by less than a tolerance: $|x_{n+1} - x_n| < \epsilon$
- A maximum number of iterations is reached

Example: In Newton-Raphson, we iteratively apply $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ until convergence.

Given $f(x) = x^2 - 5$, the derivative is $f'(x) = 2x$.

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

$$x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n}$$

Simplifying: $$x_{n+1} = \frac{1}{2}\left(x_n + \frac{5}{x_n}\right)$$

This is the famous Babylonian method for computing square roots!

Classification: TRANSCENDENTAL

Reason: The equation contains the exponential function $e^x$, which is not an algebraic function (not a polynomial). Transcendental equations involve functions that 'transcend' algebra, such as:
- Trigonometric: $\sin x, \cos x$
- Exponential: $e^x, a^x$
- Logarithmic: $\log x, \ln x$

Transcendental equations cannot be solved using simple algebraic factoring or the quadratic formula. We must use numerical methods like Bisection or Newton-Raphson.

The first 4 significant digits are: $1, 2, 3, 4$

The 5th significant digit (which will be discarded) is: $5$

Since this digit is exactly 5, we apply the symmetric rounding rule: round UP the 4th digit from 4 to 5.

Result: $12.35$

Note: Some conventions round 5 to the nearest even number (bankers' rounding), but the standard approach here is to round up when the next digit is 5.

Iteration in Bisection stops when ANY of the following criteria is satisfied:

1. Function value condition: $|f(c)| < \epsilon$ (the function value is very close to zero)

2. Interval width condition: $b - a < \epsilon$ (the interval has shrunk to desired accuracy)

3. Relative error condition: $\left|\frac{c_{k+1} - c_k}{c_{k+1}}\right| < \epsilon$ (consecutive approximations agree closely)

4. Maximum iterations: We have reached a predetermined maximum number of iterations (e.g., $n = 100$)

In practice, we usually check condition 2 or 3 along with a maximum iteration limit to ensure the algorithm terminates.

Newton-Raphson is 'Open-End' because:

It requires only a single initial guess $x_0$ (not an interval). The method doesn't need to 'bracket' or confine the root within two points. You simply start somewhere and hope the iterations converge to the root.

Contrast with Bisection:

Bisection is a 'bracketing' method that requires TWO starting values $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs. This 'brackets' the root within the interval $[a, b]$, guaranteeing its presence (by Bolzano's Theorem).

Trade-off:
- Newton-Raphson: Fast (quadratic convergence) but risky (might diverge)
- Bisection: Slow (linear convergence) but safe (always converges)

Verification that a root exists:

$f(2) = 2^3 - 4(2) - 1 = 8 - 8 - 1 = -1$ (negative)
$f(3) = 3^3 - 4(3) - 1 = 27 - 12 - 1 = 14$ (positive)

Since $f(2) < 0$ and $f(3) > 0$, a root exists in $[2, 3]$ by Bolzano's Theorem. ✓

Iteration Table:

Result after 3 iterations:
The root is confined to the interval $[2, 2.125]$. The best approximation is the most recent midpoint $c_3 = 2.125$, though we could also use the interval midpoint $\frac{2 + 2.125}{2} = 2.0625$.

Setup:

We want to solve $f(x) = x^2 - 5 = 0$, which gives $\sqrt{5}$.
Derivative: $f'(x) = 2x$

Formula: $$x_{n+1} = \frac{1}{2}\left(x_n + \frac{5}{x_n}\right)$$

Iteration Table:

Convergence Analysis:

Notice how the error decreases extremely rapidly:
- After iteration 0: Error $\approx 0.236$
- After iteration 1: Error $\approx 0.014$ (reduced by factor $\approx 17$)
- After iteration 2: Error $\approx 0.000043$ (reduced by factor $\approx 300$)

This is quadratic convergence—the number of correct digits roughly doubles with each iteration! By iteration 3, we have $\sqrt{5} \approx 2.236068$ (accurate to 6 decimal places).

Given:
- Initial interval: $[a, b] = [1, 3]$
- Interval width: $b - a = 3 - 1 = 2$
- Desired accuracy: $\epsilon = 10^{-4} = 0.0001$

Formula:
$$n \geq \frac{\log(b-a) - \log(\epsilon)}{\log(2)}$$

Step 1: Calculate $\log(b-a)$
$$\log(2) = \log(10^{0.301}) \approx 0.3010$$

Step 2: Calculate $\log(\epsilon)$
$$\log(10^{-4}) = -4$$

Step 3: Apply the formula
$$n \geq \frac{0.3010 - (-4)}{0.3010} = \frac{0.3010 + 4}{0.3010} = \frac{4.3010}{0.3010}$$

Step 4: Calculate the division
$$n \geq \frac{4.3010}{0.3010} \approx 14.29$$

Step 5: Round up to the next integer
$$n = 15$$

Verification:
After 15 iterations, the interval width becomes:
$$\frac{2}{2^{15}} = \frac{2}{32768} \approx 0.00006103 < 0.0001$$ ✓

Conclusion: At least 15 iterations are required to achieve the desired accuracy.

Geometric Interpretation:

The Newton-Raphson Method uses the tangent line at the current point to find the next approximation. Here's how it works geometrically:

1. Start at $(x_n, f(x_n))$ on the curve

2. Draw the tangent line to the curve at this point. The slope of the tangent is $f'(x_n)$.

3. Find where the tangent crosses the x-axis. This is the point $(x_{n+1}, 0)$.

4. Move up from $x_{n+1}$ to the curve to get the new point $(x_{n+1}, f(x_{n+1}))$.

5. Repeat the process from this new point.

Description of a Diagram:

```
y
| curve: y = f(x)
| / • | / (x_n, f(x_n))
| /
\ | tangent line
\|
/|\
/ | \
/ | x_{n+1}
----+----+---- x
x_n
```

Why It Converges Quickly (Quadratic Convergence):

The tangent line is a very good local approximation of the curve near the point $(x_n, f(x_n))$. This means:

1. The tangent-x-intercept $x_{n+1}$ is much closer to the true root than the previous guess.

2. Each iteration roughly squares the number of correct digits. This is called quadratic convergence.

3. For example, if $x_n$ is correct to 2 decimal places, then $x_{n+1}$ might be correct to 4 decimal places, and $x_{n+2}$ to 8 decimal places!

Comparison:
- Bisection: Reduces error by factor of 2 per iteration (linear, slow)
- Newton-Raphson: Reduces error by squaring it per iteration (quadratic, fast)

This dramatic speed difference is why Newton-Raphson is preferred in most modern applications, provided a good initial guess is available.

Step 1: Verify that a root exists

$f(0) = 0 - \cos(0) = 0 - 1 = -1$ (negative)
$f(1) = 1 - \cos(1) \approx 1 - 0.5403 = 0.4597$ (positive)

Since the signs are opposite, a root exists in $[0, 1]$. ✓

Iteration 1:

Midpoint: $c_1 = \frac{0 + 1}{2} = 0.5$

Function value: $f(0.5) = 0.5 - \cos(0.5) \approx 0.5 - 0.8776 = -0.3776$ (negative)

Sign analysis:
- $f(0) \times f(0.5) = (-1) \times (-0.3776) = 0.3776 > 0$ → same signs, root NOT in $[0, 0.5]$
- $f(0.5) \times f(1) = (-0.3776) \times 0.4597 < 0$ → opposite signs, root IS in $[0.5, 1]$ ✓

New interval: $[0.5, 1]$

Iteration 2:

Midpoint: $c_2 = \frac{0.5 + 1}{2} = 0.75$

Function value: $f(0.75) = 0.75 - \cos(0.75) \approx 0.75 - 0.7317 = 0.0183$ (positive)

Sign analysis:
- $f(0.5) \times f(0.75) = (-0.3776) \times 0.0183 < 0$ → opposite signs, root IS in $[0.5, 0.75]$ ✓

New interval: $[0.5, 0.75]$

Results Table:

Final Result:

After 2 iterations, the root is trapped in $[0.5, 0.75]$. Our best single estimate is $c_2 = 0.75$ (or the midpoint of the final interval: $\frac{0.5 + 0.75}{2} = 0.625$). The actual root is approximately $0.7391$.

Setup:

Function: $f(x) = x^3 - x - 1$
Derivative: $f'(x) = 3x^2 - 1$

Formula: $$x_{n+1} = x_n - \frac{x_n^3 - x_n - 1}{3x_n^2 - 1}$$

Iteration 1:

Starting value: $x_0 = 1.3$

Calculate $f(1.3)$:
$$f(1.3) = (1.3)^3 - 1.3 - 1 = 2.197 - 1.3 - 1 = -0.103$$

Calculate $f'(1.3)$:
$$f'(1.3) = 3(1.3)^2 - 1 = 3(1.69) - 1 = 5.07 - 1 = 4.07$$

Update:
$$x_1 = 1.3 - \frac{-0.103}{4.07} = 1.3 + 0.0253 = 1.3253$$

Iteration 2:

Starting value: $x_1 = 1.3253$:

Calculate $f(1.3253)$:
$$f(1.3253) = (1.3253)^3 - 1.3253 - 1 \approx 2.3251 - 1.3253 - 1 = -0.0002$$

Calculate $f'(1.3253)$:
$$f'(1.3253) = 3(1.3253)^2 - 1 \approx 3(1.7564) - 1 = 5.2692 - 1 = 4.2692$$

Update:
$$x_2 = 1.3253 - \frac{-0.0002}{4.2692} = 1.3253 + 0.00005 = 1.32535$$

Convergence Analysis:

Calculate relative error between $x_1$ and $x_2$:
$$\left|\frac{x_2 - x_1}{x_2}\right| = \left|\frac{1.32535 - 1.3253}{1.32535}\right| = \left|\frac{0.00005}{1.32535}\right| \approx 3.77 \times 10^{-5}$$

Results Table:

Conclusion:

The method converges very rapidly! By iteration 2, $|f(x_2)| \approx 0$ (essentially at the root), and the relative error is tiny ($\approx 0.004\%$). The root is approximately $x \approx 1.32535$. This demonstrates the quadratic convergence of Newton-Raphson.

Round-off Error:

Round-off error occurs because we cannot represent all real numbers exactly in decimal form with a limited number of digits. When we round or chop a number, we introduce error.

Example in $e^x$ calculation:

The exact value of $e = 2.718281828459045...$

If we round to 5 decimal places: $e \approx 2.71828$

The round-off error is: $|2.718281828... - 2.71828| \approx 0.0000018$

This error arises purely from the limitation of representing the irrational number $e$ with finitely many digits.

---

Truncation Error:

Truncation error occurs when we approximate an infinite process with a finite one. For example, when we approximate an infinite series by using only a finite number of terms.

Example in $e^x$ calculation:

The Taylor series for $e^x$ is:
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \quad (\text{infinite})$$

If we approximate by truncating after 4 terms:
$$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$

For $x = 1$:
$$e \approx 1 + 1 + \frac{1}{2} + \frac{1}{6} = 2.666...$$

The truncation error is: $|2.718281... - 2.666...| \approx 0.052$

This error arises because we dropped the infinite tail of the series: $\frac{1}{24} + \frac{1}{120} + \cdots$

---

Key Difference:

Detailed Comparison Table:

Example Scenario:

Problem: Find the root of $x - \cos(x) = 0$

- Bisection: Since we can find $f(0) < 0$ and $f(1) > 0$, we have a bracketing interval. Use Bisection—guaranteed to converge in about 10 iterations for $\epsilon = 10^{-3}$.

- Newton-Raphson: Since $f'(x) = 1 + \sin(x)$ is simple and we can guess $x_0 = 0.7$ is close to the root, use Newton-Raphson—converges in 2-3 iterations.

Limitation 1: Division by Zero When $f'(x_n) = 0$

If at any iteration, $f'(x_n) = 0$, the formula $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ becomes undefined (division by zero).

Example: $f(x) = (x - 1)^2$ (root at $x = 1$)
- $f'(x) = 2(x - 1)$
- At $x_n = 1$, we have $f'(1) = 0$ → cannot proceed
- Even slightly off: at $x_n = 1.01$, $f'(1.01) = 0.02$ (very small denominator) → huge step, erratic behavior

---

Limitation 2: Divergence from Poor Initial Guess

If $x_0$ is far from the true root, Newton-Raphson can diverge instead of converging.

Example: $f(x) = \arctan(x)$ (root at $x = 0$)
- If we start at $x_0 = 3$ (far from root)
- The tangent line at $(3, \arctan(3))$ has a small slope, so the x-intercept $x_1$ might be even farther from 0
- Iterations spiral away from the root instead of converging to it

---

Limitation 3: Infinite Loops (Oscillation)

Sometimes Newton-Raphson oscillates between two or more values without converging.

Example: $f(x) = x^3 - 2x + 4$ starting at $x_0 = 0.2$
- Iteration 1: $x_1 \approx 1.2$
- Iteration 2: $x_2 \approx 0.2$ (back near the start)
- The method cycles without getting closer to any root

---

Limitation 4: Convergence to Wrong Root

Newton-Raphson might converge to a different root than expected, depending on the initial guess.

Example: $f(x) = (x - 1)(x - 2)(x - 3) = 0$ has roots at $x = 1, 2, 3$
- If $x_0 = 1.5$, Newton-Raphson might converge to $x = 1$
- If $x_0 = 2.5$, it might converge to $x = 3$
- The choice of initial guess determines which root is found

---

Limitation 5: Slow Convergence for Some Functions

Although Newton-Raphson typically has quadratic convergence, some functions (especially those with small derivatives near the root) can show slower convergence.

Example: $f(x) = \sqrt{x}$ near $x = 0$ has $f'(x) = \frac{1}{2\sqrt{x}}$, which becomes very large as $x \to 0^+$. Newton-Raphson behaves poorly near the root.

---

Mitigation Strategies:

1. Choose $x_0$ wisely: Use graphical analysis or Bisection first to get close to the root
2. Check for $f'(x_n) \approx 0$: If the derivative becomes very small, switch to Bisection
3. Use a hybrid approach: Start with Bisection to bracket the root, then switch to Newton-Raphson for final refinement
4. Monitor divergence: If $|x_{n+1} - x_n|$ is increasing, stop and reconsider your approach

Setup:

Function: $f(x) = 3x - \cos(x) - 1$
Derivative: $f'(x) = 3 + \sin(x)$ (calculator must be in RADIAN mode)

Iterative formula: $$x_{n+1} = x_n - \frac{3x_n - \cos(x_n) - 1}{3 + \sin(x_n)}$$

Iteration 1:

Starting with $x_0 = 0.6$:
- $f(0.6) = 3(0.6) - \cos(0.6) - 1 = 1.8 - 0.825336 - 1 = -0.025336$
- $f'(0.6) = 3 + \sin(0.6) = 3 + 0.564642 = 3.564642$

Update:
$$x_1 = 0.6 - \frac{-0.025336}{3.564642} = 0.6 + 0.007108 = 0.607108$$

Iteration 2:

Starting with $x_1 = 0.607108$:
- $f(0.607108) = 3(0.607108) - \cos(0.607108) - 1 = 1.821324 - 0.821319 - 1 = 0.000005 \approx 0$
- $f'(0.607108) = 3 + \sin(0.607108) = 3 + 0.570601 = 3.570601$

Update:
$$x_2 = 0.607108 - \frac{0.000005}{3.570601} \approx 0.607108 - 0.000001 = 0.607107$$

Iteration Table:

Conclusion:

The method converged extremely fast because our initial guess was highly accurate. After just 2 iterations, the function value is practically zero ($0.000005$), confirming $x \approx 0.607107$ is the root of the equation.