Problem 8 - Calculus for Biology and Medicine

Question

Show that \(\cos x=x\) has a solution in \((0.5,1)\). Use the bisection method to find a solution that is accurate to two decimal places.

Step-by-Step Solution

Answer

There is a solution at approximately 0.74 in the interval (0.5, 1).

Step 1: Verify the Intermediate Value Theorem Conditions

To use the Intermediate Value Theorem (IVT) to show that there is a solution to \( \cos x = x \) in \((0.5, 1)\), we define the function \( f(x) = \cos x - x \). We need to check if the function changes sign in the interval:- Compute \( f(0.5) = \cos(0.5) - 0.5 \). Since \( \cos(0.5) \approx 0.8776 \), we get \( f(0.5) = 0.8776 - 0.5 = 0.3776 \).- Compute \( f(1) = \cos(1) - 1 \). Since \( \cos(1) \approx 0.5403 \), we get \( f(1) = 0.5403 - 1 = -0.4597 \).Since \( f(0.5) > 0 \) and \( f(1) < 0 \), \( f(x) \) changes sign and IVT guarantees at least one root between \( 0.5 \) and \( 1 \).

Step 2: Initial Bisection Method Setup

Based on Step 1, we choose the initial interval \([a, b] = [0.5, 1]\). The midpoint, \( c \), is calculated as \( c = \frac{a+b}{2} = \frac{0.5 + 1}{2} = 0.75 \).Evaluate \( f(0.75) = \cos(0.75) - 0.75 \). Since \( \cos(0.75) \approx 0.7317 \), we have \( f(0.75) = 0.7317 - 0.75 = -0.0183 \).Since \( f(0.75) < 0 \), it means the root is between \( 0.5 \) and \( 0.75 \). We update the interval to \([0.5, 0.75]\).

Step 3: First Bisection

Compute the new midpoint: \( c = \frac{0.5 + 0.75}{2} = 0.625 \).Evaluate \( f(0.625) = \cos(0.625) - 0.625 \). With \( \cos(0.625) \approx 0.8101 \), we get \( f(0.625) = 0.8101 - 0.625 = 0.1851 \).Now, \( f(0.625) > 0 \), indicating the root lies between \( 0.625 \) and \( 0.75 \). Update the interval to \([0.625, 0.75]\).

Step 4: Second Bisection

Compute the new midpoint: \( c = \frac{0.625 + 0.75}{2} = 0.6875 \).Evaluate \( f(0.6875) = \cos(0.6875) - 0.6875 \). Since \( \cos(0.6875) \approx 0.7721 \), \( f(0.6875) = 0.7721 - 0.6875 = 0.0846 \).Again, \( f(0.6875) > 0 \), so the root is now between \( 0.6875 \) and \( 0.75 \). Update the interval to \([0.6875, 0.75]\).

Step 5: Third Bisection

Compute the new midpoint: \( c = \frac{0.6875 + 0.75}{2} = 0.71875 \).Evaluate \( f(0.71875) = \cos(0.71875) - 0.71875 \). Since \( \cos(0.71875) \approx 0.7540 \), we find \( f(0.71875) = 0.7540 - 0.71875 = 0.03525 \).Because \( f(0.71875) > 0 \), the root is between \( 0.71875 \) and \( 0.75 \). Update the interval to \([0.71875, 0.75]\).

Step 6: Fourth Bisection

Compute the new midpoint: \( c = \frac{0.71875 + 0.75}{2} = 0.734375 \).Evaluate \( f(0.734375) = \cos(0.734375) - 0.734375 \). With \( \cos(0.734375) \approx 0.7427 \), we have \( f(0.734375) = 0.7427 - 0.734375 = 0.008325 \).Since \( f(0.734375) > 0 \), the root is between \( 0.734375 \) and \( 0.75 \). Update the interval to \([0.734375, 0.75]\).

Step 7: Fifth Bisection

Compute the new midpoint: \( c = \frac{0.734375 + 0.75}{2} = 0.7421875 \).Evaluate \( f(0.7421875) = \cos(0.7421875) - 0.7421875 \). Since \( \cos(0.7421875) \approx 0.7375 \), we obtain \( f(0.7421875) = 0.7375 - 0.7421875 = -0.0046875 \).Now, \( f(0.7421875) < 0 \), meaning the root is between \( 0.734375 \) and \( 0.7421875 \). Update the interval to \([0.734375, 0.7421875]\).

Step 8: Sixth Bisection

Compute the new midpoint: \( c = \frac{0.734375 + 0.7421875}{2} = 0.73828125 \).Evaluate \( f(0.73828125) = \cos(0.73828125) - 0.73828125 \). Since \( \cos(0.73828125) \approx 0.7390 \), it evaluates to \( f(0.73828125) = 0.7390 - 0.73828125 \approx 0.00071875 \).Since \( f(0.73828125) > 0 \), the root is between \( 0.73828125 \) and \( 0.7421875 \). Update the interval to \([0.73828125, 0.7421875]\).

Step 9: Solution with Required Precision

Keep performing the bisection process until the interval equals or is less than 0.01 (precision requirement for two decimal places). Continuously narrow down to centers until the final approximation:Final approximation: \( x \approx 0.74 \) is the root with two decimal places. The midpoint between the last interval \([0.73828125, 0.7421875]\) is \( 0.7392578125 \), which rounds off to two decimal places as 0.74.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that helps determine whether a continuous function has a root in a specific interval. It asserts that if a function is continuous on an interval \([a, b]\) and if it takes values of opposite signs at the endpoints, then it must cross the x-axis somewhere between \(a\) and \(b\). This means there is at least one root in that interval.

To apply the IVT, consider a function like \( f(x) = \cos x - x \). You check the function's values at the endpoints of your interval to see if they have opposite signs. If so, the theorem tells us there's a root between them.

• For instance, with \( f(0.5) = 0.3776 \) and \( f(1) = -0.4597 \), the change in sign confirms a root exists between \(0.5\) and \(1\).

Understanding IVT is crucial for confirming the presence of roots before embarking on a numerical solution like bisection.

Cosine Function

The cosine function, represented as \( \cos(x) \), is a fundamental trigonometric function that describes the x-coordinate of a point on the unit circle as the angle \(x\) varies. This periodic function repeats every \(2\pi\) radians and ranges from -1 to 1.

The cosine function is smooth and continuous, important attributes that allow us to apply the IVT. In the context of the equation \( \cos x = x \), we're essentially looking for the angle \( x \) where the cosine of the angle equals the angle itself. This is not geometrically intuitive, thus requiring numerical methods to pinpoint its exact value.

• The function is decreasing as \(x\) moves from 0 to \(\pi\).
• This property of being greater than or equal to its angle in the initial domain is pivotal in our analysis for finding roots.

Knowing these properties lets us better grasp why and how specific intervals are selected for calculations.

Root Finding

Root finding is a common task in mathematics and engineering, aiming to determine where a function intersects the x-axis (i.e., where it equals zero).

For the equation \( \cos x = x \), the problem reduces to finding zeros of the function \( f(x) = \cos x - x \). Various methods are available, from exact algebraic manipulations to numerical techniques.

• The bisection method, a simple and robust numerical approach, narrows down intervals to pinpoint the root as opposed to explicit mathematical solutions.
• It systematically halves the suspected interval, then checks which subinterval contains the root based on sign changes.

This technique is considered reliable because it compels us to confirm sign changes exist, precisely sussing out where the function crosses the axis.

Numerical Analysis

Numerical analysis involves the development and use of methods for solving complex mathematical problems numerically. Root finding, such as locating the solution for \( \cos x = x \), is a classic task.

The bisection method serves as a simple yet effective tool in numerical analysis. This method repeatedly narrows the interval where a function changes sign until a satisfactory precision is achieved.

• It's particularly useful when an exact solution is difficult to determine or doesn't exist.
• The method hinges on assessing interval midpoints and altering boundaries based on whether the sign changes, ensuring the root lies within the refined space.

Numerical analysis proves vital for handling problems that defy simple algebraic solutions, empowering practitioners to tackle practical problems efficiently.