Problem Analysis #20

To solve the problem, we are given the function:

$$f(\theta) = \sin\theta - \theta + \frac{\theta^3}{6}$$

on the interval $[-1, 1]$, and we are to compute $M + m$, where $M$ is the maximum value and $m$ is the minimum value of $f(\theta)$ on this interval.

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Step 1: Analyze the Function's Behavior

We begin by examining the derivative of $f(\theta)$:

$$f'(\theta) = \cos\theta - 1 + \frac{\theta^2}{2}$$

This derivative is continuous and differentiable everywhere. To find critical points within the interval $(-1, 1)$, we set $f'(\theta) = 0$:

$$\cos\theta - 1 + \frac{\theta^2}{2} = 0$$

Let’s analyze whether there are any solutions other than $\theta = 0$. We observe that:

• At $\theta = 0$:

  $$f'(0) = \cos(0) - 1 + \frac{0^2}{2} = 1 - 1 + 0 = 0$$

• For $\theta \neq 0$, consider the Taylor expansion of $\cos\theta$ around 0:

  $$\cos\theta = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \cdots$$

  So,

  $$f'(\theta) = \left(1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \cdots \right) - 1 + \frac{\theta^2}{2} = \frac{\theta^4}{24} - \frac{\theta^6}{720} + \cdots$$

  The leading term is $\frac{\theta^4}{24}$, which is positive for all $\theta \neq 0$. Therefore, $f'(\theta) > 0$ for all $\theta \in [-1, 1] \setminus \{0\}$, and $f'(\theta) = 0$ only at $\theta = 0$.

Thus, the function $f(\theta)$ is strictly increasing on $[-1, 1]$, as its derivative is non-negative (and strictly positive except at one point).

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Step 2: Use Monotonicity to Locate Extrema

Since $f(\theta)$ is strictly increasing on $[-1, 1]$, the maximum occurs at the right endpoint $\theta = 1$, and the minimum occurs at the left endpoint $\theta = -1$.

Therefore:

$$M = f(1), \quad m = f(-1)$$

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Step 3: Evaluate $f(1)$ and $f(-1)$

We compute:

$$f(1) = \sin(1) - 1 + \frac{1^3}{6} = \sin(1) - 1 + \frac{1}{6} = \sin(1) - \frac{5}{6}$$ $$f(-1) = \sin(-1) - (-1) + \frac{(-1)^3}{6} = -\sin(1) + 1 - \frac{1}{6} = -\sin(1) + \frac{5}{6}$$

Now, compute $M + m$:

$$M + m = f(1) + f(-1) = \left(\sin(1) - \frac{5}{6}\right) + \left(-\sin(1) + \frac{5}{6}\right) = 0$$

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Step 4: Confirm via Symmetry

Notice that the function $f(\theta)$ is odd, since:

$$f(-\theta) = \sin(-\theta) - (-\theta) + \frac{(-\theta)^3}{6} = -\sin\theta + \theta - \frac{\theta^3}{6} = -f(\theta)$$

So, $f(-\theta) = -f(\theta)$. Thus, on the symmetric interval $[-1, 1]$, it follows that:

$$f(1) = -f(-1) \Rightarrow M + m = f(1) + f(-1) = f(1) - f(1) = 0$$

This confirms our earlier calculation.

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Final Answer

$$\boxed{0}$$