Building New Maclaurin Series: Substitution, Products, Integration & Differentiation

Using the series Σ  from n=0 to infinity of xⁿ / n!, we calculate each term explicitly. For n=0: x⁰/ 0! = 1. For n=1: x^1 / 1! = x. For n=2: x² / 2! = x²/2. For n=3: x³ / 3! = x³/6. Therefore the first four terms are 1 + x + x²/2 + x³/6. Option A shows these numerical values. Option C shows the same values using factorial notation (2! = 2, 3! = 6). Since both represent the same series, D is the correct answer. Option B is incorrect because it omits the factorial denominators.

Using the binomial series with p=1/2: (1+x)^(½) = 1 + (½)x + (½)(-½)x²/2! + ... = 1 + x/2 + (-1/4)x²/2 + ... = 1 + x/2 - x²/8 + ... The second term coefficient is 1/2. The third term coefficient is (½)(-½)/2 = -1/8. Option A matches this. Option B is incorrect. Option C has the wrong sign for x². Option D has the wrong sign for x.

For p=3, using the binomial series: (1+x)³ = 1 + 3x + 3(2)x²/2! + 3(2)(1)x³/3! + ... The coefficients follow the pattern 1, 3, 3, 1, 0, 0, ... since this is a finite polynomial. The x² coefficient is 3*2/2 = 3. So the first three terms are 1 + 3x + 3x². Option B incorrectly calculates the coefficient as 6. Option C divides by 2 unnecessarily. Option D incorrectly squares the coefficient.

Multiply the series for eˣ (sum of xⁿ / n!) by x. This gives x * sum of xⁿ / n! = sum of xⁿ⁺¹ / n!. Expanding: when n=0, x^1/0! = x. When n=1, x²/1! = x². When n=2, x³/2! = x³/2. When n=3, x⁴/3! = x⁴/6. So the series is x + x² + x³/2! + x⁴/3! + ... Both A and C represent this correctly. Option B is invalid due to an undefined factorial when n=0.

Multiply the series for sin x (sum of (-1)ⁿ x^(2n+1) / (2n+1)!) by x². This gives sum of (-1)ⁿ x^(2n+3) / (2n+1)!. Expanding: when n=0, (-1)⁰x³ / 1! = x³. When n=1, (-1)^1 x⁵ / 3! = -x⁵/6. When n=2, (-1)² x⁷ / 5! = x⁷/120. So the series is x³ - x⁵/3! + x⁷/5! - x^9/7! + ... Both A and C represent this correctly. Option B is the series for sin x without the x² multiplication.

The series for eˣ is sum of xⁿ / n!. Differentiating term by term: d/dx[xⁿ / n!] = n*xⁿ⁻¹ / n! = xⁿ⁻¹ / (n-1)!. Reindexing with m = n-1 gives sum of x^m / m!, which is the same series. This makes sense because the derivative of eˣ is eˣ. Option B is the integral of 1/(1+x). Option C is unrelated. Option D is also unrelated.

The series for sin x is x - x³/3! + x⁵/5! - x⁷/7! + ... Differentiating term by term: 1 - 3x²/3! + 5x⁴/5! - 7x⁶/7! + ... = 1 - x²/2! + x⁴/4! - x⁶/6! + ..., which is the Maclaurin series for cos x. This matches the calculus fact that d/dx(sin x) = cos x. Option B would be the result of a sign error. Options C and D are incorrect.

Integrating term by term: ∫(xⁿ / n!) dx = xⁿ⁺¹ / [(n+1)n!] + C = xⁿ⁺¹ / (n+1)! + C. Reindexing gives sum of xⁿ / n! + C, which is the same series plus a constant. This makes sense because ∫eˣ dx = eˣ + C. Option B would be the integral of 1/(1+x). Option C is also the integral of 1/(1-x). Option D is unrelated.

First, cos(x²) has series sum of (-1)ⁿ (x²)^(2n) / (2n)! = sum of (-1)ⁿ x^(4n) / (2n)!. Multiplying by x gives sum of (-1)ⁿ x^(4n+1) / (2n)!. Expanding: when n=0: x^1/0! = x. When n=1: -x⁵/2! = -x⁵/2. When n=2: x^9/4! = x^9/24. So the series is x - x⁵/2! + x^9/4! - x^13/6! + ... Both A and C represent this correctly.

In the simplified series 1 + x/2! + x²/3! + x³/4! + ..., we examine what happens to each term as x approaches 0. The first term is the constant 1, which remains 1 regardless of x. Every other term contains a positive power of x multiplied by a constant (1/2!, 1/3!, etc.). As x approaches 0, any term containing x, x², x³, or any higher power of x will approach 0. Therefore, the entire series approaches 1 + 0 + 0 + 0 + ... = 1. This confirms that the limit equals 1. Option B would ignore the constant term. Option C incorrectly suggests divergence. Option D is false because we can determine the value.

The Maclaurin series for 1/(1+x²) is sum from n=0 to infinity of (-1)ⁿ * x^(2n). This is a geometric series with ratio r = -x², which converges when |r| < 1. Thus, |-x²| < 1 implies |x²| < 1, or |x| < 1. Therefore, the radius of convergence is 1.

We start with the Maclaurin series for eˣ, which is 1 + x + x²/2! + x³/3! + x⁴/4! + ... . Substituting this into the numerator gives (1 + x + x²/2 + x³/6 + x⁴/24 + ... - 1 - x). Next, we simplify by canceling the 1 and the x terms. This leaves us with x²/2 + x³/6 + x⁴/24 + ... . The coefficient of the x³ term in this simplified expression is 1/6. Option A is the coefficient of x². Option C is the coefficient of x⁴. Option D is incorrect because there is a non-zero x³ term.

Substituting sin x = x - x³/6 + x⁵/120 - ... gives numerator (x - x³/6 + ... - x) = -x³/6 + ... . After dividing by x³, the x³ term becomes the constant term -1/6, while higher powers vanish as x→0. The limit is determined by the coefficient of the smallest non-zero power that cancels appropriately.

To find the series for x cos(x), we multiply the series for x by the series for cos(x): x · Σ (-1)ⁿ x²n/(2n)! from n=0 to ∞ = Σ (-1)ⁿ x^(2n+1)/(2n)! from n=0 to ∞. This represents the Maclaurin series for x cos(x).

The sin(x) series is alternating, so the error when truncating after a term is less than the absolute value of the next term. We can also use the Lagrange form of the remainder involving the fifth derivative. Both methods give valid error estimates.