Power Series Applications: Taylor Series, Special Functions & DE Models

We replace sin(x) with its Maclaurin series: x - x³/6 + x⁵/120 - ... . The numerator becomes (x - x³/6 + x⁵/120 - ...) - x. The x terms cancel out, leaving -x³/6 + x⁵/120 - ... . Now we divide this by the denominator x³. The expression becomes (-x³/6)/x³ + (x⁵/120)/x³ - ... = -1/6 + x²/120 - ... . As x approaches 0, all terms containing x vanish, leaving only the constant term -1/6.

This is a geometric series Σ (x/2)ⁿ. For x=1, the series is Σ (½)ⁿ. The first term a (when n=0) is (½)⁰ = 1. The common ratio r is 1/2. Since |r| < 1, the series converges. The sum formula is S = a / (1 - r). Plugging in the values: S = 1 / (1 - 1/2) = 1 / (½) = 2.

This is a definition-level recognition problem. The Taylor series for eˣ centered at 0 (Maclaurin series) is derived from the fact that all derivatives of eˣ are eˣ, and at x=0, they equal 1. The coefficients are 1/n!. Thus, eˣ = 1 + x + x²/2! + x³/3! + ... which matches the summation notation Σ xⁿ / n!.

We already know the series for ln(1+x) is x - x²/2 + x³/3 - ... . The function is simply the polynomial (1+x) multiplied by this series. We can distribute the multiplication: 1 * (series) + x * (series). This is much faster than computing derivatives from scratch using the product rule (method A) or trying complex substitutions. We simply shift the powers of the known series and add them to the original series.

We are given the formula sum of n*x^(n-1) = 1/(1-x)². We simply need to substitute x = 1/3 into the right-hand side. 1 / (1 - 1/3)² = 1 / (⅔)² = 1 / (4/9) = 9/4.

We use the Ratio Test. We evaluate lim (n→∞) | a_{n+1} / aₙ |. Here aₙ = (2x)ⁿ / n². The ratio is | [ (2x)ⁿ⁺¹ / (n+1)² ] * [ n² / (2x)ⁿ ] |. Grouping the x terms gives |2x|. Grouping the n terms gives n² / (n+1)², which approaches 1 as n approaches infinity. Combining these, the limit L is |2x| * 1 = |2x|. For the series to converge, the Ratio Test requires that the limit L be strictly less than 1. Thus we need |2x| < 1. Solving for x, we get |x| < 1/2. Thus, the radius of convergence is 1/2.

The Ratio Test on |xⁿ/n| gives a radius of 1. We check endpoints x = -1 and x = 1. At x = 1, the series is Σ (-1)ⁿ / n, which is the alternating harmonic series. It converges by the Alternating Series Test. At x = -1, the series becomes Σ 1/n, the harmonic series, which diverges. So interval is (-1,1].

We know that arctan(x) is the integral of 1/(1+x²) dx. The series for 1/(1+x²) is Σ (-1)ⁿ x²ⁿ. To find the series for arctan(x), we integrate Σ (-1)ⁿ x²ⁿ term-by-term. The integral of x²ⁿ is x²ⁿ⁺¹ / (2n+1). The coefficient (-1)ⁿ remains. So the series is Σ (-1)ⁿ x²ⁿ⁺¹ / (2n+1). The constant of integration is 0 because arctan(0) = 0.

We multiply the series for eˣ (1 + x + x²/2 + x³/6 + x⁴/24 + ...) and cos(x) (1 - x²/2 + x⁴/24 - ...). We look for combinations that produce x⁴:

1 * (x⁴/24) = x⁴/24.

x * (no x³ term in cos x) = 0.

(x²/2) * (-x²/2) = -x⁴/4.

(x³/6) * (no x term in cos x) = 0.

(x⁴/24) * 1 = x⁴/24.

Summing coefficients: 1/24 - 1/4 + 1/24 = 2/24 - 6/24 = -4/24 = -1/6.

If y' = 0, then y must be a constant. In a power series sum aₙ xⁿ, the only constant term is a₀. All terms with x (where n ≥ 1) must have coefficients of 0. Mathematically, y' = sum n*aₙ*x^(n-1). For this to be 0 for all x, every coefficient n*aₙ must be 0. Since n is not 0 for n≥1, aₙ must be 0.

The relationship between the Taylor series coefficient and the derivative is given by aₙ = f⁽ⁿ⁾(0) / n!. For n=4, we have a₄ = f⁽⁴⁾(0) / 4!. We are given a₄ = 2. Therefore, 2 = f⁽⁴⁾(0) / 24 (since 4! = 24). Multiplying both sides by 24, we find f⁽⁴⁾(0) = 2 * 24 = 48.

The series is in the form Σ cₙ (x - a)ⁿ, so the center is a = 3. To find the radius, we can write the term as [5(x-3)]ⁿ. For geometric convergence, the absolute value of the base must be less than 1: |5(x-3)| < 1. Dividing by 5, we get |x-3| < 1/5. This implies the distance from the center 3 must be less than 1/5. Thus, the radius is 1/5.

For a given center point c, if a function f(x) can be represented as a power series in some interval around c, then that power series must be the Taylor series of f centered at c, with coefficients an = f(n)(c)/n!. This uniqueness is a fundamental property of power series representations, meaning there cannot be two different power series centered at c that both converge to f(x) on some interval around c.

The standard series is for sin(u). The function has sin(2x²) as a component. To utilize the standard series, we must substitute the entire argument of the sine function, which is 2x², into the variable u of the standard series. After this substitution, we would multiply the entire result by the outer x term. Therefore, the correct substitution for the argument is u = 2x².