Question: A computer graphics specialist models light intensity $ I(x) $ along a surface as $ I(x) = \frac{1}{x^2 + 4x + 5} $. If the cumulative visual effect over points $ x = 1 $ to $ x = n $ is modeled by $ C(n) = \sum_{x=1}^{n} I(x) $, find $ C(4) $.

["Optimizing Visual Illumination: Calculating Cumulative Light Intensity from $ x = 1 $ to $ x = 4 $", "Understanding how light behaves across physical surfaces is essential in computer graphics, particularly in rendering realistic shading and material responses. A key function in this analysis is the light intensity $ I(x) $, modeled as a rational function:", "$$\nI(x) = \frac{1}{x^2 + 4x + 5}\n$$", "This function describes how light intensity diminishes along a surface coordinate $ x $, with quadratic behavior in the denominator introducing natural decay patterns. In visual simulations, computing the cumulative light impact from $ x = 1 $ to $ x = n $—represented as $ C(n) = \sum_{x=1}^{n} I(x) $—enables accurate assessment of perceived brightness and lighting continuity.", "We are tasked with finding $ C(4) $, the sum of light intensities at discrete positions $ x = 1 $ through $ x = 4 $.", "---", "### Step 1: Simplify the Denominator", "First, complete the square for the quadratic expression in the denominator:", "$$\nx^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x+2)^2 + 1\n$$", "Thus, the intensity becomes:", "$$\nI(x) = \frac{1}{(x+2)^2 + 1}\n$$", "This form is familiar from calculus and signal processing—it resembles the standard arctangent derivative, though here we compute a discrete sum rather than an integral.", "---", "### Step 2: Evaluate $ I(x) $ for $ x = 1 $ to $ 4 $", "We now calculate each term individually.", "- For $ x = 1 $:\n $$\n I(1) = \frac{1}{(1+2)^2 + 1} = \frac{1}{9 + 1} = \frac{1}{10}\n $$", "- For $ x = 2 $:\n $$\n I(2) = \frac{1}{(2+2)^2 + 1} = \frac{1}{16 + 1} = \frac{1}{17}\n $$", "- For $ x = 3 $:\n $$\n I(3) = \frac{1}{(3+2)^2 + 1} = \frac{1}{25 + 1} = \frac{1}{26}\n $$", "- For $ x = 4 $:\n $$\n I(4) = \frac{1}{(4+2)^2 + 1} = \frac{1}{36 + 1} = \frac{1}{37}\n $$", "---", "### Step 3: Compute the Cumulative Effect $ C(4) $", "Now sum all contributions:", "$$\nC(4) = \frac{1}{10} + \frac{1}{17} + \frac{1}{26} + \frac{1}{37}\n$$", "To combine these fractions, find the least common denominator (LCD). Since 10 = 2·5, 17 is prime, 26 = 2·13, and 37 is prime, the LCD is:", "$$\n\ ext{LCD} = 2 \cdot 5 \cdot 13 \cdot 17 \cdot 37 = 43790\n$$", "Now convert each term:", "- $ \frac{1}{10} = \frac{4379}{43790} $\n- $ \frac{1}{17} = \frac{2575}{43790} $\n- $ \frac{1}{26} = \frac{1685}{43790} $\n- $ \frac{1}{37} = \frac{1180}{43790} $", "Add:", "$$\nC(4) = \frac{4379 + 2575 + 1685 + 1180}{43790} = \frac{9929}{43790}\n$$", "This fraction is already in simplest form (check GCD: 9929 and 43790 share no common factors >1).", "---", "### Final Answer", "The cumulative visual effect of light intensity from $ x = 1 $ to $ x = 4 $ is:", "$$\n\boxed{C(4) = \frac{9929}{43790}}\n$$", "This precise computational approach reflects how computer graphics specialists model and optimize visual realism by quantifying light distribution across surfaces—turning mathematical functions into immersive digital experiences."]