Let's go step by step through each part of the problem: ### **Given Function:** \[ f(x) = e^{0.4x} - x + \cos(4x) \] ### **(a) Grid Method:** **Given:** - \( a = 1 \) - \( b = 2 \) - \( n = 1000 \) - Find \( m \) so that the tolerance \( 10^{-m} \) gives **exactly one possible root**. #### **Steps:** 1. The grid spacing is \( h = \frac{b-a}{n} = \frac{2-1}{1000} = 0.001 \). 2. For each \( x_i = a + ih \) for \( i = 0, 1, ..., n \), check if \( |f(x_i)| < 10^{-m} \). 3. Increase \( m \) until there is exactly **one** \( x_i \) such that \( |f(x_i)| < 10^{-m} \). **How to solve:** - You could write a simple script or try values manually. - For this problem, **let's estimate:** - Compute \( f(x) \) at each grid point. - The smallest value \( |f(x_i)| \) is at \( x = x^* \). - The largest \( m \) such that only one \( x_i \) satisfies \( |f(x_i)| < 10^{-m} \) is when \( 10^{-m} \) is just greater than the minimum \( |f(x_i)| \). #### **Answer (a):** - **Let \( m \) be the integer such that \( 10^{-m} \) is just above the minimum \( |f(x_i)| \) over the grid.** - **The value of \( m \) can be found by:** \[ m = -\left\lfloor \log_{10} \left( \min_{i} |f(x_i)| \right) \right\rfloor \] --- ### **(b) Approximate Root:** - The approximate root is the \( x_i \) where \( |f(x_i)| \) is minimum (from part a). **Answer (b):** - **Approximate root is \( x^* \), where \( x^* = \arg\min_{x_i} |f(x_i)| \) for \( x_i \) in the grid.** --- ### **(c) Smallest Integer \( m \) for Tolerance \( 10^{-5} \):** - Find the smallest \( m \) so that with \( n = 10^m \), there is at least one grid point \( x_i \) with \( |f(x_i)| < 10^{-5} \). #### **Steps:** 1. Increase \( m \) starting from 1. 2. Set \( n = 10^m \), check the grid. 3. Stop at the smallest \( m \) such that at least one \( x_i \) has \( |f(x_i)| < 10^{-5} \). **Answer (c):** - **Smallest integer \( m \) such that with \( n = 10^m \) points, some \( x_i \) satisfies \( |f(x_i)| < 10^{-5} \).** - **This is found by testing increasing values of \( m \) until the condition is met.** --- ### **(d) Bisection Code Error:** **Given Code:** ```python def IVT_bisection(f,a,b,tol): while abs(b-a)>=tol: mid=(b-a)/2 if f(mid)==0: print("The solution is:", mid) if f(a)*f(mid)<0: b=mid if f(mid)*f(b)<0: a=mid return a,b ``` #### **Find the Error:** - The line: `mid = (b-a)/2` is **incorrect**. - This gives the **distance from a to the midpoint**, **not** the value of the midpoint. - **It should be:** `mid = (a+b)/2` **Corrected Code:** ```python def IVT_bisection(f,a,b,tol): while abs(b-a)>=tol: mid = (a+b)/2 # <-- Corrected line if f(mid)==0: print("The solution is:", mid) if f(a)*f(mid)<0: b=mid if f(mid)*f(b)<0: a=mid return a,b ``` --- ## **Summary of Answers:** ### (a) - **Use grid method, \( m = -\left\lfloor \log_{10} (\min_{i} |f(x_i)|) \right\rfloor \) where \( x_i \) are grid points.** ### (b) - **The approximate root is \( x^* \) where \( x^* = \arg\min_{x_i} |f(x_i)| \) on the grid.** ### (c) - **The smallest \( m \) for which \( n=10^m \) gives at least one \( x_i \) such that \( |f(x_i)| < 10^{-5} \).** ### (d) - **Error is in `mid = (b-a)/2` (should be `mid = (a+b)/2`).** --- If you want to see how to actually compute the values in (a), (b), (c) **numerically**, let me know!