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f(n) = n + 1 + \frac{2}{n + 1}.
Let $ x = n + 1 $, so $ x > 1 $, and $ f(n) = x + \frac{2}{x} $. Define $ g(x) = x + \frac{2}{x} $. Take derivative:
g'(x) = 1 - \frac{2}{x^2}.
Set $ g'(x) = 0 $:
- \frac{2}{x^2} = 0 \Rightarrow x^2 = 2 \Rightarrow x = \sqrt{2}.
Since $ g''(x) = \frac{4}{x^3} > 0 $ for $ x > 0 $, this is a minimum. Thus, the minimum value of $ f(n) $ is $ \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} $.
\boxed{2\sqrt{2}}
Question: A paleobotanist is studying the symmetry of a fossilized flower with radial structure. If the flower has 7 equally spaced petals, and a vector $ \vec{v} $ from the center to a petal tip has components $ ( \cos \theta, \sin \theta ) $, and a second vector to an adjacent petal is $ \vec{w} = (\cos(\theta + \frac{2\pi}{7}), \sin(\theta + \frac{2\pi}{7})) $, find the angle between $ \vec{v} $ and $ \vec{w} $.
Solution: The angle between two unit vectors $ \vec{v} $ and $ \vec{w} $ is given by the cosine of the difference of their angles:
\cos \phi = \vec{v} \cdot \vec{w} = \cos \theta \cos\left(\theta + \frac{2\pi}{7}\right) + \sin \theta \sin\left(\theta + \frac{2\pi}{7}\right).