Let's go through the steps and explanations for **Exercise Set 14** (from the images) with **step-by-step solutions and explanations.** --- ## **1. Maclaurin Series for sin(x)** Recall from the Maclaurin series definition (Slide 19): \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] For \( f(x) = \sin(x) \): - \( f(0) = 0 \) - \( f'(x) = \cos(x) \), \( f'(0) = 1 \) - \( f''(x) = -\sin(x) \), \( f''(0) = 0 \) - \( f'''(x) = -\cos(x) \), \( f'''(0) = -1 \) - \( f^{(4)}(x) = \sin(x) \), \( f^{(4)}(0) = 0 \) - and so on... Plugging into the Maclaurin formula: \[ \sin(x) = 0 + 1\cdot x + 0\cdot x^2 + \frac{-1}{3!}x^3 + 0\cdot x^4 + \frac{1}{5!}x^5 + \cdots \] \[ = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \] Or using summation notation (as in the slide): \[ \sin(x) \approx \sum_{k=1}^{n} (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} \] --- ## **2. MATLAB Function for Maclaurin Series (sin_series)** **Function outline in MATLAB:** ```matlab function s = sin_series(x, n) s = 0; for k = 1:n term = ((-1)^(k-1)) * x^(2*k-1) / factorial(2*k-1); s = s + term; end end ``` - Input: `x` (value to evaluate), `n` (number of terms) - Output: sum of first `n` terms in the Maclaurin expansion --- ## **3. Approximations for sin(π/6) and sin(π/3)** ### **a. sin(π/6) by hand (π/6 ≈ 0.5236)** - **1-term:** \( x = 0.5236 \) \[ s_1 = x = 0.5236 \] - **3-term:** \( x - \frac{x^3}{6} + \frac{x^5}{120} \) \[ s_3 = 0.5236 - \frac{0.5236^3}{6} + \frac{0.5236^5}{120} \] \[ = 0.5236 - 0.0239 + 0.0002 = 0.4999 \] The actual value: \( \sin(\pi/6) = 0.5 \) ### **b. sin(π/3) by hand (π/3 ≈ 1.0472)** - **1-term:** \( x = 1.0472 \) - **3-term:** \( x - x^3/6 + x^5/120 \) \[ s_3 = 1.0472 - \frac{1.0472^3}{6} + \frac{1.0472^5}{120} \approx 1.0472 - 0.1914 + 0.0088 = 0.8646 \] The actual value: \( \sin(\pi/3) = 0.8660 \) --- ## **4. Modify Code for Default n=5** MATLAB code with default `n=5`: ```matlab function s = sin_series(x, n) if nargin < 2 n = 5; end s = 0; for k = 1:n term = ((-1)^(k-1)) * x^(2*k-1) / factorial(2*k-1); s = s + term; end end ``` **Test:** - `sin_series(pi/6, 2)` - `sin_series(pi/6)` --- ## **5. Function big_enuf** Checks if all elements' magnitude is at least 5: ```matlab function big_enuf(x) mags = abs(x); if all(mags >= 5) disp(mags) else disp('Sorry this number (or some of these numbers) are too small.') end end ``` **Test cases:** - `big_enuf(-8)` → 8 - `big_enuf(3)` → Sorry... - `big_enuf([0 2 7])` → Sorry... - `big_enuf([5 7 -9])` → [5 7 9] --- ## **6. Function spheres** Returns volume, or volume and area, depending on output arguments: ```matlab function [V, A] = spheres(r) V = (4/3) * pi * r^3; if nargout > 1 A = 4 * pi * r^2; end end ``` - `spheres(3)` → just volume - `[V, A] = spheres(3)` → volume and area --- # **Summary of Answers** - The Maclaurin series for \(\sin(x)\) is \(x - x^3/3! + x^5/5! - ...\) - The MATLAB function for the \(n\)-term sum is provided above. - For \(\sin(\pi/6)\), the 3-term approximation is **0.4999** (very close to 0.5). - For \(\sin(\pi/3)\), the 3-term approximation is **0.8646** (very close to 0.8660). - Functions for big_enuf and spheres are above as well. **Let me know if you want to see the output for any other cases or need MATLAB code explanations!**

Certainly! Let's go through **Question 1** in detail, including each step of the calculation and the final answer. --- ### **Question 1: Derive the Maclaurin Series for \(\sin(x)\)** #### **Recall the Maclaurin series formula:** \[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \] where \(f^{(n)}(x)\) is the \(n^{th}\) derivative of \(f(x)\). --- ### **Step 1: Find the derivatives of \(\sin(x)\) and evaluate at \(x=0\).** | Derivative order | Function \(f^{(n)}(x)\) | At \(x=0\) | Value \(f^{(n)}(0)\) | |------------------|---------------------------|------------|---------------------| | 0 | \(\sin(x)\) | 0 | 0 | | 1 | \(\cos(x)\) | 1 | 1 | | 2 | \(-\sin(x)\) | 0 | 0 | | 3 | \(-\cos(x)\) | -1 | -1 | | 4 | \(\sin(x)\) | 0 | 0 | | 5 | \(\cos(x)\) | 1 | 1 | | 6 | \(-\sin(x)\) | 0 | 0 | | 7 | \(-\cos(x)\) | -1 | -1 | The derivatives repeat every 4 steps, with values at 0 cycling through 0, 1, 0, -1, 0, 1, 0, -1, ... --- ### **Step 2: Write out the Maclaurin series terms** Using the formula: \[ \sin(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \] and the derivatives evaluated: - For \(n=0\): \(f^{(0)}(0) = 0\) - For \(n=1\): \(f^{(1)}(0) = 1\) - For \(n=2\): \(f^{(2)}(0) = 0\) - For \(n=3\): \(f^{(3)}(0) = -1\) - For \(n=4\): \(f^{(4)}(0) = 0\) - For \(n=5\): \(f^{(5)}(0) = 1\) - For \(n=6\): \(f^{(6)}(0) = 0\) - For \(n=7\): \(f^{(7)}(0) = -1\) So, the non-zero derivatives occur at odd \(n\): \[ \begin{aligned} \sin(x) &= \frac{f^{(1)}(0)}{1!} x + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(5)}(0)}{5!} x^5 + \frac{f^{(7)}(0)}{7!} x^7 + \cdots \\ &= \frac{1}{1!} x + \frac{-1}{3!} x^3 + \frac{1}{5!} x^5 + \frac{-1}{7!} x^7 + \cdots \end{aligned} \] This can be written more compactly as: \[ \boxed{ \sin(x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} } \] --- ### **Final Answer:** \[ \boxed{ \boxed{ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots } } \] or explicitly: \[ \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \frac{x^9}{362880} - \cdots \] --- ### **Summary:** - The Maclaurin series for \(\sin(x)\) contains only odd powers of \(x\). - The signs alternate, starting with a positive \(x\). - The coefficients are reciprocals of odd factorials. --- Let me know if you'd like further explanation or specific terms!

Certainly! Let's derive the Maclaurin series for \(\sin(x)\) step-by-step, focusing on detailed calculations without using a table. --- ### **Step 1: Recall the Maclaurin Series Formula** The Maclaurin series expansion of a function \(f(x)\) is: \[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \] where \(f^{(n)}(0)\) is the \(n^{th}\) derivative of \(f(x)\) evaluated at \(x=0\). --- ### **Step 2: Compute derivatives of \(\sin(x)\) and evaluate at \(x=0\)** - \(f(x) = \sin(x)\) Calculate derivatives: 1. \(f^{(0)}(x) = \sin(x)\) 2. \(f^{(1)}(x) = \cos(x)\) 3. \(f^{(2)}(x) = -\sin(x)\) 4. \(f^{(3)}(x) = -\cos(x)\) 5. \(f^{(4)}(x) = \sin(x)\) 6. \(f^{(5)}(x) = \cos(x)\) 7. \(f^{(6)}(x) = -\sin(x)\) 8. \(f^{(7)}(x) = -\cos(x)\) And so on, with derivatives repeating every 4 steps. Now, evaluate each at \(x=0\): - \(\sin(0) = 0\) - \(\cos(0) = 1\) - \(-\sin(0) = 0\) - \(-\cos(0) = -1\) - \(\sin(0) = 0\) - \(\cos(0) = 1\) - \(-\sin(0) = 0\) - \(-\cos(0) = -1\) ### **Step 3: Identify non-zero derivatives at \(x=0\)** The derivatives at 0 are: - \(f^{(0)}(0) = 0\) - \(f^{(1)}(0) = 1\) - \(f^{(2)}(0) = 0\) - \(f^{(3)}(0) = -1\) - \(f^{(4)}(0) = 0\) - \(f^{(5)}(0) = 1\) - \(f^{(6)}(0) = 0\) - \(f^{(7)}(0) = -1\) - And so on... The pattern repeats every 4 derivatives, with non-zero derivatives only at odd orders. ### **Step 4: Write the series using only non-zero terms** Based on the derivatives, the series includes only odd powers: \[ \sin(x) = \frac{f^{(1)}(0)}{1!} x + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(5)}(0)}{5!} x^5 + \frac{f^{(7)}(0)}{7!} x^7 + \cdots \] Substituting the derivative values: \[ \sin(x) = \frac{1}{1!} x + \frac{-1}{3!} x^3 + \frac{1}{5!} x^5 + \frac{-1}{7!} x^7 + \cdots \] ### **Step 5: Express the pattern with summation notation** Notice the signs alternate and the powers increase by 2 each time: \[ \boxed{ \sin(x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} } \] ### **Final Answer:** \[ \boxed{ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots } \] or explicitly: \[ \boxed{ \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \frac{x^9}{362880} - \cdots } \] --- ### **Summary:** - The Maclaurin series for \(\sin(x)\) involves only odd powers of \(x\). - The coefficients are reciprocals of odd factorials. - The signs alternate starting with positive. --- Let me know if you'd like more details or specific terms!

Certainly! Let's carefully develop **Part 2** of the exercise, which involves writing a MATLAB function called `sine_series` that computes the Maclaurin series approximation of \(\sin(x)\) using a specified number of terms. --- ## **Part 2: Write a MATLAB function `sine_series`** ### **Objective:** - The function should accept two inputs: `x` (the point at which to evaluate \(\sin(x)\)) and `n` (the number of terms in the series). - The output should be the sum of the first `n` terms of the Maclaurin series for \(\sin(x)\). ### **Step-by-step explanation:** ### **Step 1: Recall the series formula** From **Part 1**, the Maclaurin series for \(\sin(x)\) is: \[ \sin(x) \approx \sum_{k=1}^{n} (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} \] ### **Step 2: Set up the function** In MATLAB, the function will: - Take inputs `x` and `n`. - Initialize a variable `sum_series` to zero. - Loop from `k=1` to `n`. - Calculate each term: \[ \text{term}_k = (-1)^{k-1} \times \frac{x^{2k-1}}{(2k-1)!} \] - Add each term to `sum_series`. ### **Step 3: MATLAB code** ```matlab function s = sine_series(x, n) % Check if n is provided; if not, default to 5 if nargin < 2 n = 5; end % Initialize sum s = 0; % Loop to compute sum of series terms for k = 1:n numerator = (-1)^(k-1) * x^(2*k - 1); denominator = factorial(2*k - 1); term = numerator / denominator; s = s + term; end end ``` ### **Step 4: Explanation of code** - `nargin < 2`: Checks if the user provided `n`. If not, defaults to 5. - Loop from `k=1` to `n`. - For each `k`, computes the term using the formula. - Adds the term to the sum `s`. - Returns the sum after the loop. --- ## **Part 2: Final answer and example usage** ### **Example:** Calculate the 3-term Maclaurin series approximation for \(\sin(\pi/6)\): ```matlab approx_value = sine_series(pi/6, 3); disp(approx_value); ``` **Expected calculation:** \[ \sin(x) \approx x - \frac{x^3}{6} + \frac{x^5}{120} \] Substitute \(x = \pi/6 \approx 0.5236\): - First term: \(0.5236\) - Second term: \(- \frac{(0.5236)^3}{6} \approx - \frac{0.143}{6} \approx -0.0238\) - Third term: \(+ \frac{(0.5236)^5}{120} \approx + \frac{0.038}{120} \approx +0.0003\) Sum: \[ 0.5236 - 0.0238 + 0.0003 \approx 0.4999 \] which is very close to the actual \(\sin(\pi/6) = 0.5\). --- ## **Summary:** - The MATLAB function `sine_series` computes the Maclaurin series for \(\sin(x)\) up to `n` terms. - It uses a `for` loop to sum the series terms. - It defaults to `n=5` if no second argument is provided. - Example usage shows high accuracy even with a small number of terms. --- **Let me know if you'd like me to provide more example tests or further explanations!**

Certainly! Let's carefully go through **Question 3**, which involves evaluating the Maclaurin series approximations for \(\sin(x/6)\) and verifying the accuracy both by hand and using MATLAB. --- ## **Question 3: Check your code for problem 2 by finding the 1-term, 3-term, and 5-term approximations for \(\sin(\pi/6)\) and \(\sin(\pi/3)\). Also verify the 3-term approximation for \(\sin(\pi/6)\) by hand.** --- ## **Part A: Approximate \(\sin(\pi/6)\) using the Maclaurin series** Recall the series: \[ \sin(x) \approx \sum_{k=1}^{n} (-1)^{k-1} \frac{x^{2k-1}}{(2k-1)!} \] ### **Step 1: Compute actual values** - \(\pi/6 \approx 0.5236\) - \(\pi/3 \approx 1.0472\) --- ## **Part B: Approximate \(\sin(\pi/6)\)** ### **1-term approximation:** \[ \sin(\pi/6) \approx x = 0.5236 \] **Error:** Actual \(\sin(\pi/6) = 0.5\), so the approximation is **0.5236**, which is an overestimate by **0.0236**. --- ### **3-term approximation:** \[ \sin(x) \approx x - \frac{x^3}{6} + \frac{x^5}{120} \] Compute each term: - \(x = 0.5236\) - \(x^3 = (0.5236)^3 \approx 0.143\) - \(x^5 = (0.5236)^5 \approx 0.038\) Now, substitute: \[ \text{Approx} = 0.5236 - \frac{0.143}{6} + \frac{0.038}{120} \] Calculate each: - \(\frac{0.143}{6} \approx 0.0238\) - \(\frac{0.038}{120} \approx 0.0003\) Sum: \[ 0.5236 - 0.0238 + 0.0003 = 0.4999 \] **Result:** Approximate \(\sin(\pi/6) \approx 0.4999\), which is extremely close to the exact value **0.5**. --- ### **5-term approximation:** Add the next term: \[ + \frac{x^7}{5040} \] Calculate \(x^7\): - \(x^7 = x^5 \times x^2 \approx 0.038 \times (0.5236)^2 \approx 0.038 \times 0.274 = 0.0104\) Then: \[ \frac{0.0104}{5040} \approx 2.06 \times 10^{-6} \] Add this: \[ 0.4999 + 2.06 \times 10^{-6} \approx 0.4999 \] Negligible difference at this precision. ### **Final approximation for \(\sin(\pi/6)\):** \[ \boxed{ \approx 0.4999 } \] which is very close to the true value \(0.5\). --- ## **Part C: Approximate \(\sin(\pi/3)\)** - \(x = 1.0472\) ### **1-term:** \[ \sin(\pi/3) \approx 1.0472 \] **Error:** Actual \(\sin(\pi/3) = \sqrt{3}/2 \approx 0.8660\), so overestimate by about 0.1812. ### **3-term:** Calculate powers: - \(x^3 \approx (1.0472)^3 \approx 1.147\) - \(x^5 \approx (1.0472)^5 \approx 1.268\) Plug into the series: \[ \text{Approx} = 1.0472 - \frac{1.147}{6} + \frac{1.268}{120} \] Calculate: - \(\frac{1.147}{6} \approx 0.1912\) - \(\frac{1.268}{120} \approx 0.0106\) Sum: \[ 1.0472 - 0.1912 + 0.0106 = 0.8666 \] **Result:** \(\approx 0.8666\), very close to the actual \(0.8660\). ### **5-term approximation:** Next term involves \(x^7\), which is roughly: - \(x^7 \approx x^5 \times x^2 \approx 1.268 \times 1.097 \approx 1.392\) and \[ \frac{1.392}{5040} \approx 2.76 \times 10^{-4} \] Adding this: \[ 0.8666 + 2.76 \times 10^{-4} \approx 0.8670 \] which improves accuracy. --- ## **Part D: Summary of hand calculations** | Approximation | \(\sin(\pi/6)\) | Error | \(\sin(\pi/3)\) | Error | |----------------|------------------|--------|----------------|--------| | 1-term | 0.5236 | 0.0236 | 1.0472 | 0.1812 | | 3-term | 0.4999 | 0.0001 | 0.8666 | 0.0004 | | 5-term | ~0.4999 | negligible | ~0.8670 | negligible | --- ## **Part E: MATLAB verification** You can verify these approximations using the MATLAB function `sine_series`: ```matlab % Approximate sin(pi/6) approx1 = sine_series(pi/6, 1); approx3 = sine_series(pi/6, 3); approx5 = sine_series(pi/6, 5); % Approximate sin(pi/3) approx1b = sine_series(pi/3, 1); approx3b = sine_series(pi/3, 3); approx5b = sine_series(pi/3, 5); disp([approx1, approx3, approx5]); disp([approx1b, approx3b, approx5b]); ``` Expected outputs: - For \(\pi/6\): approximately 0.5236, 0.4999, very close to 0.5 - For \(\pi/3\): approximately 1.0472, 0.8666, 0.8660 --- # **Final Summary:** | Functionality | 1-term approximation | 3-term approximation | 5-term approximation | |----------------|------------------------|------------------------|----------------------| | \(\sin(\pi/6)\) | 0.5236 (overestimate) | 0.4999 (very close) | ~0.4999 (negligible error) | | \(\sin(\pi/3)\) | 1.0472 (overestimate) | 0.8666 (very close) | ~0.8660 (excellent accuracy) | These calculations demonstrate how adding more terms improves the accuracy of the Maclaurin series approximation. --- Let me know if you'd like further clarifications!