Maclaurin Series Differential Equations Quiz

1) Which of the following is the Maclaurin series representation of cos(x)?

N=0 (-1)ⁿ x^(2n+1)/(2n+1)!

N=0 (-1)ⁿ x²n/(2n)!

N=0 x^(2n+1)/(2n+1)!

N=0 x²n/(2n)!

The Maclaurin series for cos(x) contains only even powers of x with alternating signs. The general term is (-1)ⁿ x²n/(2n)!, where n starts from 0: n=0 gives 1, n=1 gives -x²/2!, n=2 gives x⁴/4!, etc.

Explanation

The Maclaurin series for cos(x) contains only even powers of x with alternating signs. The general term is (-1)ⁿ x²n/(2n)!, where n starts from 0: n=0 gives 1, n=1 gives -x²/2!, n=2 gives x⁴/4!, etc.

2) Find the first five non-zero terms of the Maclaurin series for sin(x²).

X² - x⁶/3! + x¹⁰/5! - x¹⁴/7! + x¹⁸/9!

X² - x⁶/6 + x¹⁰/120 - x¹⁴/5040 + x¹⁸/362880

X² + x⁶/3! + x¹⁰/5! + x¹⁴/7! + x¹⁸/9!

X² + x⁶/6 + x¹⁰/120 + x¹⁴/5040 + x¹⁸/362880

Starting with the Maclaurin series for sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ..., we substitute x → x² to get sin(x²) = x² - (x²)³/3! + (x²)⁵/5! - (x²)⁷/7! + ... = x² - x⁶/6 + x¹⁰/120 - x¹⁴/5040 + x¹⁸/362880.

Explanation

Starting with the Maclaurin series for sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ..., we substitute x → x² to get sin(x²) = x² - (x²)³/3! + (x²)⁵/5! - (x²)⁷/7! + ... = x² - x⁶/6 + x¹⁰/120 - x¹⁴/5040 + x¹⁸/362880.

3) Find the Maclaurin series for (1 + x)^(-½).

Σ [(-1)ⁿ (2n)!/(4ⁿ (n!)²)] xⁿ

Σ [(-1)ⁿ (2n)!/(2ⁿ n!)] xⁿ

Σ [(-1)ⁿ 4ⁿ (n!)²/(2n)!] xⁿ

Σ [(-1)ⁿ n!/(2ⁿ n!)] xⁿ

Using the binomial series (1 + x)^p = Σ [p(p-1)...(p-n+1)/n!]xⁿ from n=0 to ∞, with p = -1/2, we get [(-½)(-3/2)(-5/2)...(-(2n-1)/2)/n!]xⁿ = [(-1)ⁿ (1·3·5·...·(2n-1))/(2ⁿ n!)]xⁿ. Since 1·3·5·...·(2n-1) = (2n)!/(2·4·6·...·2n) = (2n)!/(2ⁿ n!), we get [(-1)ⁿ (2n)!/(2ⁿ n! · 2ⁿ n!)]xⁿ = [(-1)ⁿ (2n)!/(4ⁿ (n!)²)]xⁿ.

Explanation

Using the binomial series (1 + x)^p = Σ [p(p-1)...(p-n+1)/n!]xⁿ from n=0 to ∞, with p = -1/2, we get [(-½)(-3/2)(-5/2)...(-(2n-1)/2)/n!]xⁿ = [(-1)ⁿ (1·3·5·...·(2n-1))/(2ⁿ n!)]xⁿ. Since 1·3·5·...·(2n-1) = (2n)!/(2·4·6·...·2n) = (2n)!/(2ⁿ n!), we get [(-1)ⁿ (2n)!/(2ⁿ n! · 2ⁿ n!)]xⁿ = [(-1)ⁿ (2n)!/(4ⁿ (n!)²)]xⁿ.

4) After substituting the Maclaurin series and simplifying lim[x→0] (eˣ - 1 - x)/x², we obtain the expression (x²/2 + x³/6 + x⁴/24 + ...)/x². What is the coefficient of the x term after dividing each term of the series by x²?

0

1/2

1/6

1/24

We divide each term in the numerator by x². The first term is (x²/2) / x² = 1/2, which is a constant. The second term is (x³/6) / x² = x/6. The third term is (x⁴/24) / x² = x²/24. The resulting series is 1/2 + x/6 + x²/24 + ... . The question asks for the coefficient of the x term, which is the number multiplying x. In the term x/6, the coefficient is 1/6. Option A would be correct if there were no x term at all. Option B is the constant term, not the x term. Option D is the coefficient of the x² term in the final series.

Explanation

We divide each term in the numerator by x². The first term is (x²/2) / x² = 1/2, which is a constant. The second term is (x³/6) / x² = x/6. The third term is (x⁴/24) / x² = x²/24. The resulting series is 1/2 + x/6 + x²/24 + ... . The question asks for the coefficient of the x term, which is the number multiplying x. In the term x/6, the coefficient is 1/6. Option A would be correct if there were no x term at all. Option B is the constant term, not the x term. Option D is the coefficient of the x² term in the final series.

5) Evaluate the sum n=0 [(-1)ⁿ/(2n+1)] using Maclaurin series.

π/4

π/2

π/3

π/6

This is the alternating series representation of arctan(1). Using the Maclaurin series for arctan(x) = Σ (-1)ⁿ x^(2n+1)/(2n+1) from n=0 to ∞, we substitute x = 1 to get arctan(1) = Σ (-1)ⁿ (1)^(2n+1)/(2n+1) = Σ (-1)ⁿ/(2n+1) from n=0 to ∞. Since arctan(1) = π/4, this series converges to π/4.

Explanation

This is the alternating series representation of arctan(1). Using the Maclaurin series for arctan(x) = Σ (-1)ⁿ x^(2n+1)/(2n+1) from n=0 to ∞, we substitute x = 1 to get arctan(1) = Σ (-1)ⁿ (1)^(2n+1)/(2n+1) = Σ (-1)ⁿ/(2n+1) from n=0 to ∞. Since arctan(1) = π/4, this series converges to π/4.

6) Find the Maclaurin series for ln(1-x).

N=1 (-1)ⁿ xⁿ/n for |x| < 1

-n=1 xⁿ/n for |x| < 1

N=0 (-1)ⁿ xⁿ/n for |x| < 1

N=0 xⁿ/n for |x| < 1

Since d/dx[ln(1-x)] = -1/(1-x), we integrate both sides:

ln(1-x) = ∫[-1/(1-x)]dx = -∫0∞[Σn=0∞ xⁿ]dx. Integrating term by term:

ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)] + C. To determine C, we substitute x = 0. So

ln(1-0) = ln(1) = 0 = -Σ[0] + C, so C = 0. Therefore, ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)]. Reindexing by letting k = n+1 gives ln(1-x) = -Σk=1∞[xᵏ/k]. This can be written as:

ln(1-x) = -x - x²/2 - x³/3 - x⁴/4 - … This series converges for |x| < 1, and also at x = -1 (by the alternating series test), but diverges at x = 1.

Explanation

Since d/dx[ln(1-x)] = -1/(1-x), we integrate both sides:

ln(1-x) = ∫[-1/(1-x)]dx = -∫0∞[Σn=0∞ xⁿ]dx. Integrating term by term:

ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)] + C. To determine C, we substitute x = 0. So

ln(1-0) = ln(1) = 0 = -Σ[0] + C, so C = 0. Therefore, ln(1-x) = -Σn=0∞[xⁿ⁺¹/(n+1)]. Reindexing by letting k = n+1 gives ln(1-x) = -Σk=1∞[xᵏ/k]. This can be written as:

ln(1-x) = -x - x²/2 - x³/3 - x⁴/4 - … This series converges for |x| < 1, and also at x = -1 (by the alternating series test), but diverges at x = 1.

7) Find the Maclaurin series for arcsin(x).

Σ [(2n)!/(4ⁿ (n!)²)] x^(2n+1)/(2n+1)

Σ [(2n)!/(4ⁿ (n!)²)] x²n/(2n+1)

Σ [(-1)ⁿ (2n)!/(4ⁿ (n!)²)] x^(2n+1)/(2n+1)

Σ [(-1)ⁿ (2n)!/(4ⁿ (n!)²)] x²n/(2n+1)

To find the Maclaurin series for arcsin(x), we integrate the series for 1/√(1-x²) = Σ [(2n)!/(4ⁿ (n!)²)]x²n from n=0 to ∞. Integrating term by term: ∫ Σ [(2n)!/(4ⁿ (n!)²)]x²n dx = Σ [(2n)!/(4ⁿ (n!)²)] x^(2n+1)/(2n+1) + C. At x = 0, arcsin(0) = 0, so C = 0.

Explanation

To find the Maclaurin series for arcsin(x), we integrate the series for 1/√(1-x²) = Σ [(2n)!/(4ⁿ (n!)²)]x²n from n=0 to ∞. Integrating term by term: ∫ Σ [(2n)!/(4ⁿ (n!)²)]x²n dx = Σ [(2n)!/(4ⁿ (n!)²)] x^(2n+1)/(2n+1) + C. At x = 0, arcsin(0) = 0, so C = 0.

8) Develop the Maclaurin series for a function y(x) that satisfies y'' - y = 0 with y(0) = 1 and y'(0) = 0.

Σ x²n/(2n)!

Σ x^(2n+1)/(2n+1)!

Σ (-1)ⁿ x²n/(2n)!

Σ (-1)ⁿ x^(2n+1)/(2n+1)!

The differential equation y'' - y = 0 with y(0) = 1 and y'(0) = 0 describes the function y(x) = cosh(x). Assuming a Maclaurin series y = Σ aₙ xⁿ, we find y' = Σ n aₙ xⁿ⁻¹ and y'' = Σ n(n-1)aₙ x^(n-2). Substituting into y'' - y = 0 gives Σ n(n-1)aₙ x^(n-2) - Σ aₙ xⁿ = 0. Reindexing and equating coefficients shows that a_0 = 1, a₁ = 0, and the pattern a_(2n) = 1/(2n)!, a_(2n+1) = 0, giving cosh(x) = Σ x²n/(2n)!.

Explanation

The differential equation y'' - y = 0 with y(0) = 1 and y'(0) = 0 describes the function y(x) = cosh(x). Assuming a Maclaurin series y = Σ aₙ xⁿ, we find y' = Σ n aₙ xⁿ⁻¹ and y'' = Σ n(n-1)aₙ x^(n-2). Substituting into y'' - y = 0 gives Σ n(n-1)aₙ x^(n-2) - Σ aₙ xⁿ = 0. Reindexing and equating coefficients shows that a_0 = 1, a₁ = 0, and the pattern a_(2n) = 1/(2n)!, a_(2n+1) = 0, giving cosh(x) = Σ x²n/(2n)!.

9) Find the Maclaurin series for sinh(x).

Σ x^(2n+1)/(2n+1)!

Σ (-1)ⁿ x^(2n+1)/(2n+1)!

Σ x²n/(2n)!

Σ (-1)ⁿ x²n/(2n)!

The hyperbolic sine function is defined as sinh(x) = (eˣ - e⁻ˣ)/2. Using the Maclaurin series for eˣ = Σ xⁿ/n! and e⁻ˣ = Σ (-1)ⁿ xⁿ/n!, we get sinh(x) = (½)[Σ xⁿ/n! - Σ (-1)ⁿ xⁿ/n!] = (½)Σ [1 - (-1)ⁿ]xⁿ/n!. This gives non-zero terms only when n is odd, so sinh(x) = Σ x^(2n+1)/(2n+1)! from n=0 to ∞.

Explanation

The hyperbolic sine function is defined as sinh(x) = (eˣ - e⁻ˣ)/2. Using the Maclaurin series for eˣ = Σ xⁿ/n! and e⁻ˣ = Σ (-1)ⁿ xⁿ/n!, we get sinh(x) = (½)[Σ xⁿ/n! - Σ (-1)ⁿ xⁿ/n!] = (½)Σ [1 - (-1)ⁿ]xⁿ/n!. This gives non-zero terms only when n is odd, so sinh(x) = Σ x^(2n+1)/(2n+1)! from n=0 to ∞.

10) After substituting the Maclaurin series for cos(x) into limx→0 (1 - cos(x))/x² and simplifying, what is the lowest power of x remaining in the numerator?

X⁰

X¹

X²

X⁴

1 − cos(x) = x²/2 − x⁴/24 + ... lowest power is x².

Explanation

1 − cos(x) = x²/2 − x⁴/24 + ... lowest power is x².

11) Find the Maclaurin series for f(x) = 1/(1-x)² by differentiating the geometric series.

N=0 n * xⁿ⁻¹

N=0 xⁿ

N=0 (-1)ⁿ * n * xⁿ⁻¹

N=0 xⁿ⁺¹ / (n+1)

We know that 1/(1-x) = Sum of xⁿ. Differentiating both sides with respect to x, the derivative of 1/(1-x) is 1/(1-x)². The derivative of the series Sum of xⁿ is Sum of n*xⁿ⁻¹. Note that the index starts at n=1 because the constant term (n=0) becomes 0 upon differentiation. This can also be written as Sum from n=0 to infinity of (n+1)xⁿ.

Explanation

We know that 1/(1-x) = Sum of xⁿ. Differentiating both sides with respect to x, the derivative of 1/(1-x) is 1/(1-x)². The derivative of the series Sum of xⁿ is Sum of n*xⁿ⁻¹. Note that the index starts at n=1 because the constant term (n=0) becomes 0 upon differentiation. This can also be written as Sum from n=0 to infinity of (n+1)xⁿ.

12) To evaluate lim x→0 (sin(x) - x + x³/6)/x⁵ using Maclaurin series, which terms are essential?

Terms up to x³

Terms up to x⁴

Terms up to x⁵

Terms up to x⁶

The Maclaurin series for sin(x) is x - x³/3! + x^5/5! - x^7/7! + ... When we substitute, sin(x) - x + x³/6 = (x - x³/6 + x^5/120 - ...) - x + x³/6 = x^5/120 - ... Dividing by x^5 gives 1/120 - ..., so we need terms up to x^5 in the sin(x) series to get a non-zero constant term after division.

Explanation

The Maclaurin series for sin(x) is x - x³/3! + x^5/5! - x^7/7! + ... When we substitute, sin(x) - x + x³/6 = (x - x³/6 + x^5/120 - ...) - x + x³/6 = x^5/120 - ... Dividing by x^5 gives 1/120 - ..., so we need terms up to x^5 in the sin(x) series to get a non-zero constant term after division.

13) Find the Maclaurin series for e^(-x²) by substitution.

N=0 (-1)ⁿ x^(2n)/n!

N=0 x^(2n)/n!

N=0 (-1)ⁿ xⁿ/n!

N=0 (-x²)ⁿ/n!

Starting with the Maclaurin series for eˣ = Σ xⁿ/n! from n=0 to ∞, we substitute x → -x² to get e^(-x²) = Σ (-x²)ⁿ/n! = Σ (-1)ⁿ x^(2n)/n! from n=0 to ∞.

Explanation

Starting with the Maclaurin series for eˣ = Σ xⁿ/n! from n=0 to ∞, we substitute x → -x² to get e^(-x²) = Σ (-x²)ⁿ/n! = Σ (-1)ⁿ x^(2n)/n! from n=0 to ∞.

14) If evaluating limx→0 (1 - cos(3x))/x² using Maclaurin series, what coefficient appears in the x² term of the expanded numerator after substitution?

1/2

3/2

9/2

3

For cos(3x), substitute 3x into the series: cos(3x) = 1 - (3x)²/2! + (3x)⁴/4! - ... = 1 - 9x²/2 + 81x⁴/24 - ... . The numerator becomes 1 - (1 - 9x²/2 + ...) = 9x²/2 - ... . The coefficient of x² is 9/2.

Explanation

For cos(3x), substitute 3x into the series: cos(3x) = 1 - (3x)²/2! + (3x)⁴/4! - ... = 1 - 9x²/2 + 81x⁴/24 - ... . The numerator becomes 1 - (1 - 9x²/2 + ...) = 9x²/2 - ... . The coefficient of x² is 9/2.

15) To evaluate limx→0 (e^(2x) - 2e^x + 1)/x² correctly using the Maclaurin series, which statement is true?

Only first two terms needed

At least three terms required

Series approach not applicable

Only constant term matters

The Maclaurin series for e^x = 1 + x + x²/2 + x³/6 + ... So e^(2x) = 1 + 2x + (2x)²/2 + ... = 1 + 2x + 2x² + ... and 2e^x = 2(1 + x + x²/2 + ...) = 2 + 2x + x² + ... Therefore e^(2x) - 2e^x + 1 = (1 + 2x + 2x²) - (2 + 2x + x²) + 1 + higher terms = (1 - 2 + 1) + (2x - 2x) + (2x² - x²) + ... = x² + ... If we only used two terms (up to x), we would get e^(2x) - 2e^x + 1 = (1 + 2x) - 2(1 + x) + 1 = 1 + 2x - 2 - 2x + 1 = 0, which is incorrect. We need at least three terms to get the x² term.

Explanation

The Maclaurin series for e^x = 1 + x + x²/2 + x³/6 + ... So e^(2x) = 1 + 2x + (2x)²/2 + ... = 1 + 2x + 2x² + ... and 2e^x = 2(1 + x + x²/2 + ...) = 2 + 2x + x² + ... Therefore e^(2x) - 2e^x + 1 = (1 + 2x + 2x²) - (2 + 2x + x²) + 1 + higher terms = (1 - 2 + 1) + (2x - 2x) + (2x² - x²) + ... = x² + ... If we only used two terms (up to x), we would get e^(2x) - 2e^x + 1 = (1 + 2x) - 2(1 + x) + 1 = 1 + 2x - 2 - 2x + 1 = 0, which is incorrect. We need at least three terms to get the x² term.

Maclaurin Series from Differential Equations: Modeling with sin, cos, sinh, cosh & Beyond