Math: Taylor and Maclaurin Series
Approximating functions with polynomials
Math: Taylor and Maclaurin Series
Approximating functions with polynomials
Math - Grade 9-12
- 1
Write the first four nonzero terms of the Maclaurin series for e^x.
For e^x, every derivative at x = 0 equals 1.
The Maclaurin series for e^x is 1 + x + x^2/2! + x^3/3! + ... . The first four nonzero terms are 1, x, x^2/2, and x^3/6. - 2
Write the first four nonzero terms of the Maclaurin series for sin x.
The Maclaurin series for sin x is x - x^3/3! + x^5/5! - x^7/7! + ... . The first four nonzero terms are x, -x^3/6, x^5/120, and -x^7/5040. - 3
Write the first four nonzero terms of the Maclaurin series for cos x.
Cosine has only even powers in its Maclaurin series.
The Maclaurin series for cos x is 1 - x^2/2! + x^4/4! - x^6/6! + ... . The first four nonzero terms are 1, -x^2/2, x^4/24, and -x^6/720. - 4
Use the Maclaurin polynomial 1 + x + x^2/2 to approximate e^0.2. Round your answer to four decimal places.
Substituting x = 0.2 gives 1 + 0.2 + (0.2)^2/2 = 1.2 + 0.02 = 1.2200. The approximation is e^0.2 is about 1.2200. - 5
Use the Maclaurin polynomial x - x^3/6 to approximate sin(0.3). Round your answer to four decimal places.
Use radians, not degrees.
Substituting x = 0.3 gives 0.3 - (0.3)^3/6 = 0.3 - 0.027/6 = 0.3 - 0.0045 = 0.2955. The approximation is sin(0.3) is about 0.2955. - 6
Find the third-degree Taylor polynomial for f(x) = ln x centered at x = 1.
Use the formula f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3!.
For f(x) = ln x, the derivatives at x = 1 are f(1) = 0, f'(1) = 1, f''(1) = -1, and f'''(1) = 2. The third-degree Taylor polynomial is (x - 1) - (x - 1)^2/2 + (x - 1)^3/3. - 7
Find the second-degree Taylor polynomial for f(x) = sqrt(x) centered at x = 4.
For f(x) = sqrt(x), f(4) = 2, f'(x) = 1/(2sqrt(x)), so f'(4) = 1/4, and f''(x) = -1/(4x^(3/2)), so f''(4) = -1/32. The second-degree Taylor polynomial is 2 + (x - 4)/4 - (x - 4)^2/64. - 8
The Maclaurin series for 1/(1 - x) is 1 + x + x^2 + x^3 + ... . Use the first four terms to approximate 1/(1 - 0.1).
Substitute x = 0.1 into the polynomial 1 + x + x^2 + x^3.
Using the first four terms gives 1 + 0.1 + 0.1^2 + 0.1^3 = 1 + 0.1 + 0.01 + 0.001 = 1.111. The approximation is 1/(1 - 0.1) is about 1.111. - 9
For the geometric series 1 + x + x^2 + x^3 + ... , state the interval of convergence.
The geometric series 1 + x + x^2 + x^3 + ... converges when |x| < 1. Its interval of convergence is -1 < x < 1. - 10
Use the first three nonzero terms of the Maclaurin series for cos x to approximate cos(0.5). Round your answer to four decimal places.
The first three nonzero terms of cos x include powers x^0, x^2, and x^4.
The first three nonzero terms are 1 - x^2/2! + x^4/4!. Substituting x = 0.5 gives 1 - 0.25/2 + 0.0625/24 = 1 - 0.125 + 0.002604... = 0.877604... . The approximation is cos(0.5) is about 0.8776. - 11
Find the coefficient of x^5 in the Maclaurin series for sin x.
The Maclaurin series for sin x is x - x^3/3! + x^5/5! - x^7/7! + ... . The coefficient of x^5 is 1/5!, which equals 1/120. - 12
Explain why a Taylor polynomial usually gives its best approximation near its center.
Think about what it means for a graph to match slope and curvature at one point.
A Taylor polynomial is built so that its value and several derivatives match the original function at the center. Because of this matching, the polynomial has the same local behavior near the center, but the error often grows as x moves farther away.