Approximating Functions with Maclaurin Series: eˣ, sin x, cos x, ln(1+x), arctan x

The function 1/(1-x) is a geometric series with first term 1 and common ratio x. The sum of an infinite geometric series a/(1-r) is Σ  from n=0 to infinity of a*rⁿ. Here a=1 and r=x, giving sum of xⁿ. This series converges when |x| < 1. Option B is for 1/(1+x). Option C is for eˣ. Option D is for cos x.

The Maclaurin series for sin x contains only odd powers of x with alternating signs. The pattern is x - x³/3! + x⁵/5! - x⁷/7! + ... The general term uses (-1)ⁿ to create the alternating signs and (2n+1) to generate odd numbers. For n=0, we get x/1! = x. For n=1, we get -x³/3!. For n=2, we get x⁵/5!. This matches option A. Option B represents cos x (even powers). Option C represents eˣ (all powers). Option D has the right powers but lacks the alternating signs.

We can obtain 1/(1+x) by substituting -x for x in the series for 1/(1-x). This gives Σ  from n=0 to infinity of (-x)ⁿ. Since (-x)ⁿ = (-1)ⁿ xⁿ, this is equivalent to Σ  from n=0 to infinity of (-1)ⁿ xⁿ, which expands to 1 - x + x² - x³ + x⁴ - ... Option B is for 1/(1-x). Both A and C are mathematically equivalent representations, so D is correct.

Substitute x² for x in the series for 1/(1-x). This gives Σ  from n=0 to infinity of (x²)ⁿ. Since (x²)ⁿ = x^(2n), we get Σ  from n=0 to infinity of x^(2n). Writing out the terms: when n=0, x⁰= 1. When n=1, x². When n=2, x⁴. When n=3, x⁶. So the series is 1 + x² + x⁴ + x⁶ + ... Therefore both A and C are correct representations. Option B has alternating signs, which would correspond to 1/(1+x²).

The Maclaurin series for ln(1+x) is x - x²/2 + x³/3 - x⁴/4 + x⁵/5 - ... This can be written as sum from n=1 to infinity of (-1)ⁿ⁺¹ xⁿ / n. Note the series starts at n=1 because the n=0 term would be undefined. Option B is the geometric series for 1/(1+x). Option C is for eˣ. Option D is for arctan x.

Multiply (1 + x + x²/2 + ...) by (x - x³/6 + ...). Distributing terms: 1(x) = x. x(x) = x². 1(-x³/6) + (x²/2)(x) = -x³/6 + x³/2 = 2x³/6 = x³/3. Summing these gives x + x² + x³/3.

The Maclaurin series for arctan x is x - x³/3 + x⁵/5 - x⁷/7 + ... This contains only odd powers with alternating signs and denominators equal to the power. The general form is Σ  from n=0 to infinity of (-1)ⁿ x^(2n+1) / (2n+1). Option B is for ln(1+x). Option C has the right form but missing alternating signs. Option D is for cos x.

The first two nonzero terms of the Maclaurin series of sin(x²) is x² - (x²)³/3! = x² - x^6/6. Integrating term-wise gives ∫(x² - x^6/6) dx = x³/3 - x^7/42. Evaluating from 0 to 1, we get (1/3 - 1/42) - 0.

The Maclaurin series for eˣ is derived by evaluating all derivatives of eˣ at x=0. Since every derivative of eˣ is eˣ, and e⁰= 1, we have f⁽ⁿ⁾(0) = 1 for all n. Plugging these into the general Maclaurin series formula Σ  from n=0 to infinity of f⁽ⁿ⁾(0) * xⁿ / n! gives us exactly 1 + x + x²/2! + x³/3! + x⁴/4! + ... This is the complete, correct series. Option B incorrectly omits the constant term 1. Option C shows a geometric series, not the exponential series. Option D has the wrong coefficients and factorial placement.

The series for 1/(1-x) is a geometric series sum of xⁿ. A geometric series converges when the absolute value of the common ratio is less than 1. Here the ratio is x, so we require |x| < 1. This can also be verified using the ratio test: limit as n→∞ of |xⁿ⁺¹/xⁿ| = |x| < 1. Option B is true for eˣ, sin x, and cos x, but not for this geometric series. Option C is too restrictive. Option D is the opposite of the correct condition.

We know from calculus that d/dx[ln(1+x)] = 1/(1+x). Therefore, to get the series for ln(1+x), we integrate the series for 1/(1+x) term by term. The series for 1/(1+x) is 1 - x + x² - x³ + x⁴ - ... Integrating gives x - x²/2 + x³/3 - x⁴/4 + x⁵/5 - ..., which is exactly the series for ln(1+x). Option B describes the reverse relationship. Options C and D are false.

Substituting x² for x in the series for 1/(1+x) gives Σ  from n=0 to infinity of (-x²)ⁿ = Σ  from n=0 to infinity of (-1)ⁿ x^(2n). This expands to 1 - x² + x⁴ - x⁶ + x⁸- ... The student's series is actually correct. Option B would give 1 + x² + x⁴ + x⁶ + ..., which is for 1/(1-x²). Option C would be odd powers, which doesn't match. Option D is incorrect because geometric series don't have factorials.

The expansion for (1+x)^(½) is 1 + x/2 - x²/8 + ... . To find √(1.2), we set x = 0.2. The first term is 1. The second term is 0.2/2 = 0.1. The third term is -(0.2)²/8 = -0.04/8 = -0.005. Summing these gives 1 + 0.1 - 0.005 = 1.095.

The eˣ series (and also sin x, cos x series) converges for all real numbers. The 1/(1-x) series converges only for |x| < 1. The ln(1+x) and arctan x series both converge for |x| < 1 and at x = 1, but diverge for |x| > 1. Therefore, the eˣ series has the largest interval of convergence (all real numbers). Option B has the smallest interval. Options C and D have intermediate intervals.

The binomial series for (1+x)^p uses generalized binomial coefficients. The coefficient of xⁿ is (p choose n) = p(p-1)(p-2)...(p-n+1)/n!. This gives the series 1 + p x + p(p-1) x² / 2! + p(p-1)(p-2) x³ / 3! + ... Option B would be for e^(p x) if expanded differently. Option C is for eˣ. Option D is for 1/(1+x).