Let's work through each part of the problem step by step: --- ### **Given Function** \[ f(h) = \frac{\cosh(h^{3/2}) - 1}{h^3} \] --- ### **(a) MATLAB Code and Numerical Exploration** #### **Step 1: MATLAB Code** Here is a simple MATLAB script to compute \( f(h) \) and plot it for a range of \( h \): ```matlab % Range of h values (logarithmically spaced for good resolution near 0) h = logspace(-10, -1, 100); f = (cosh(h.^(3/2)) - 1) ./ (h.^3); % Plot f(h) vs h figure; loglog(h, f, 'bo-'); xlabel('h'); ylabel('f(h)'); title('Plot of f(h) vs h'); grid on; ``` #### **Step 2: Estimating the Limit Numerically** - As \( h \to 0 \), observe the value of \( f(h) \) for the smallest \( h \). - The function should approach the limit \( L \). #### **Step 3: Robustness and Round-off Error** - For very small \( h \), round-off errors in floating-point arithmetic affect the computation, especially when subtracting nearly equal numbers (`cosh(h^{3/2})` approaches 1). - **Underflow**: Extremely small values of \( h \) may lead to values indistinguishable from zero due to floating-point underflow. #### **Step 4: Identifying Regions** - **Stable region**: \( h \) is not too small, so round-off errors are negligible. - **Error-prone region**: \( h \) is small; round-off errors dominate. - **Underflow region**: \( h \) is so small that the numerator or denominator underflows to zero. By inspecting the plot, you should be able to see a plateau close to the limit, then deviations as \( h \) gets extremely small. --- ### **(b) Analytical Calculation of the Limit** #### **Step 1: Taylor Expansion** Recall: \[ \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \] Let \( x = h^{3/2} \): \[ \cosh(h^{3/2}) = 1 + \frac{h^3}{2} + \frac{h^6}{24} + \cdots \] So, \[ \cosh(h^{3/2}) - 1 = \frac{h^3}{2} + \frac{h^6}{24} + \cdots \] Now, plug into \( f(h) \): \[ f(h) = \frac{\frac{h^3}{2} + \frac{h^6}{24} + \cdots}{h^3} = \frac{1}{2} + \frac{h^3}{24} + \cdots \] So, \[ \lim_{h \to 0} f(h) = \frac{1}{2} \] That is, \( L = 0.5 \). --- #### **Step 2: Rate of Convergence** \[ |f(h) - L| \approx \frac{h^3}{24} \] So \( p = 3 \) (since the error is \( O(h^3) \)). --- #### **Step 3: Plotting the Rate of Convergence** Add to your MATLAB code: ```matlab L = 0.5; % The analytical limit error = abs(f - L); figure; loglog(h, error, 'ro-'); xlabel('h'); ylabel('|f(h) - L|'); title('Convergence Rate of f(h) to L'); grid on; % Fit a straight line to log-log data to estimate p coeffs = polyfit(log(h), log(error), 1); p = coeffs(1); % Should be close to 3 fprintf('Estimated convergence rate p: %f\n', p); ``` - The slope \( p \) of the line in the log-log plot of \(|f(h)-L|\) vs \( h \) should be close to 3 **in the stable region** (not too small \( h \)). - You should mention the range of \( h \) values where the slope matches the theoretical value. --- ### **Summary of Answers:** #### **(a)** - Use MATLAB to compute and plot \( f(h) \) for a range of \( h \). - Estimate the limit as \( h \to 0 \) numerically (\( \approx 0.5 \)). - The algorithm is not robust for very small \( h \) due to round-off error and underflow. The stable region is for "moderately small" \( h \). Underflow occurs for very small \( h \) (e.g., \( h < 10^{-8} \) in double precision). - Plot should show: stable region (plateau), error-prone region (deviation), and underflow region (collapse to zero). #### **(b)** - Analytically, \( \lim_{h \to 0} f(h) = 0.5 \). - The error \( |f(h) - 0.5| \sim O(h^3) \). - Plot \( |f(h) - 0.5| \) vs \( h \) on log-log scale, estimate slope \( p \approx 3 \) in the stable region. --- **Let me know if you want the complete MATLAB code or further explanation of any step!**

Certainly! Let's proceed step-by-step, providing detailed calculations, explanations, and final answers for each part. --- ## **Given Function** \[ f(h) = \frac{\cosh(h^{3/2}) - 1}{h^3} \] where \( h > 0 \). --- ## **Part (a): Numerical estimation and plotting** ### **Step 1: Behavior of \( f(h) \) as \( h \to 0 \)** Since directly computing \( f(0) \) is impossible (division by zero and \( \cosh(0) = 1 \)), we estimate the limit: \[ L = \lim_{h \to 0} f(h) \] by evaluating \( f(h) \) for small \( h \). ### **Step 2: MATLAB code to compute \( f(h) \)** Here's the MATLAB code snippet: ```matlab % Define a range of h values logarithmically spaced h = logspace(-10, -1, 100); % from 1e-10 to 1e-1 f = (cosh(h.^(3/2)) - 1) ./ (h.^3); % Plot figure; loglog(h, f, 'bo-'); xlabel('h'); ylabel('f(h)'); title('Behavior of f(h) as h approaches 0'); grid on; ``` ### **Step 3: Estimating the limit \( L \)** - For the smallest \( h \), evaluate \( f(h) \). - For example, at \( h = 10^{-10} \): \[ f(10^{-10}) = \frac{\cosh((10^{-10})^{3/2}) - 1}{(10^{-10})^3} \] Calculate numerator: \[ (10^{-10})^{3/2} = 10^{-15} \] Since \( \cosh(x) \approx 1 + \frac{x^2}{2} \) for small \( x \): \[ \cosh(10^{-15}) \approx 1 + \frac{(10^{-15})^2}{2} = 1 + \frac{10^{-30}}{2} \approx 1 \] Thus: \[ \cosh(10^{-15}) - 1 \approx 10^{-30}/2 = 5 \times 10^{-31} \] And denominator: \[ (10^{-10})^3 = 10^{-30} \] So, \[ f(10^{-10}) \approx \frac{5 \times 10^{-31}}{10^{-30}} = 0.5 \] This suggests that as \( h \to 0 \), \( f(h) \to 0.5 \). **Final estimate:** \[ \boxed{ L \approx 0.5 } \] ### **Step 4: Is the algorithm robust?** - **Round-off errors** become significant when subtracting nearly equal large numbers or dealing with extremely small values. - For very small \( h \), \( \cosh(h^{3/2}) \) approaches 1, so \( \cosh(h^{3/2}) - 1 \) becomes very tiny—close to machine epsilon. - **Underflow** occurs when \( \cosh(h^{3/2}) - 1 \) becomes too small to be distinguished from zero, or when \( h^3 \) becomes extremely tiny, causing the division to be unstable. **Estimated underflow point:** - For double precision (~\( 2 \times 10^{-16} \)), the numerator is approximately \( \sim 10^{-30} \). Since \( \cosh(h^{3/2}) - 1 \) is on the order of \( h^3/2 \), as derived analytically, very small \( h \) (like \( 10^{-11} \) or less) will lead to numerical inaccuracies. - In practical terms, **underflow occurs around \( h \sim 10^{-8} \)**. --- ## **Part (b): Analytical derivation and convergence rate** ### **Step 1: Analytical limit of \( f(h) \)** Recall the Taylor expansion of \( \cosh(x) \): \[ \cosh(x) = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \cdots \] Substitute \( x = h^{3/2} \): \[ \cosh(h^{3/2}) = 1 + \frac{(h^{3/2})^2}{2} + \frac{(h^{3/2})^4}{24} + \cdots \] Calculate: \[ (h^{3/2})^2 = h^{3} \] \[ (h^{3/2})^4 = h^{6} \] Thus: \[ \cosh(h^{3/2}) = 1 + \frac{h^3}{2} + \frac{h^6}{24} + \cdots \] Subtract 1: \[ \cosh(h^{3/2}) - 1 = \frac{h^3}{2} + \frac{h^6}{24} + \cdots \] Now, compute \( f(h) \): \[ f(h) = \frac{\cosh(h^{3/2}) - 1}{h^3} = \frac{\frac{h^3}{2} + \frac{h^6}{24} + \cdots}{h^3} \] Divide numerator and denominator: \[ f(h) = \frac{h^3/2}{h^3} + \frac{h^6/24}{h^3} + \cdots = \frac{1}{2} + \frac{h^3}{24} + \cdots \] ### **Step 2: Limit as \( h \to 0 \)** \[ \boxed{ \lim_{h \to 0} f(h) = \frac{1}{2} } \] ### **Step 3: Rate of convergence** The dominant correction term is \( \frac{h^3}{24} \), which indicates: \[ |f(h) - L| \sim \frac{h^3}{24} \] This is of order \( O(h^3) \), meaning: \[ |f(h) - 0.5| \leq C \cdot h^3 \] for some constant \( C \). **Therefore, the rate of convergence exponent \( p \) is 3.** --- ### **Step 4: Plotting \( |f(h) - L| \) and estimating \( p \)** **MATLAB snippet:** ```matlab L = 0.5; error = abs(f - L); % Plot error on log-log scale figure; loglog(h, error, 'ro-'); xlabel('h'); ylabel('|f(h) - L|'); title('Convergence of f(h) to L'); grid on; % Linear fit in log-log to estimate p coeffs = polyfit(log(h), log(error), 1); p_estimated = coeffs(1); % slope disp(['Estimated rate p: ', num2str(p_estimated)]); ``` **Expected result:** - The slope \( p \) should be approximately \(-3\), confirming that: \[ |f(h) - 0.5| \sim h^3 \] --- ## **Final Summary:** | **Part** | **Details** | **Final Result** | |------------|-------------------------|------------------| | **(a)** | Numerical plotting of \( f(h) \) over a range of \( h \), observing the limit as \( h \to 0 \). | \(\boxed{L \approx 0.5}\) | | **Robustness** | The algorithm is sensitive to round-off errors for very small \( h \), especially near \( 10^{-8} \). Underflow occurs roughly around \( h \sim 10^{-8} \). | — | | **(b)** | Analytical limit: \( \lim_{h \to 0} f(h) = 1/2 \). | \(\boxed{L = 0.5}\) | | **Rate of convergence** | Error behaves as \( \sim h^3 \), so \( p = 3 \). | \( \boxed{p = 3} \) | --- If you'd like, I can provide the complete MATLAB code with comments for implementing this analysis practically.