Exercises
Explore how Taylor and Maclaurin polynomials approximate functions near a chosen center. This quiz assesses series construction, coefficient formulas, standard expansions, intervals of convergence, Lagrange error bounds, approximation accuracy, limits, and integration using truncated series. Questions range from direct calculations to graphical interpretation and practical estimation.
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A Maclaurin series is the special case of a Taylor series with center a = 0. Its coefficients are determined by derivatives evaluated at zero.
The coefficient of each power is determined by the corresponding derivative at the center: f^(k)(a)/k!. The sum begins with k = 0.
Every derivative of e^x equals e^x and has value 1 at zero. Therefore, its Maclaurin coefficients are 1/k!.
The Maclaurin series for sine alternates signs and contains only odd powers: sin(x) = x - x³/3! + x⁵/5! - ...
Cosine has only even powers with alternating signs. Through degree 4, the expansion is 1 - x²/2! + x⁴/4!.
The expansion is ln(1+x) = x - x²/2 + x³/3 - x⁴/4 + .... Retaining terms through degree 3 gives the stated polynomial.
This is a geometric series with first term 1 and common ratio x. It converges to 1/(1-x) when |x| is less than 1.
The radius is 1. At x = 1, the resulting alternating harmonic series converges. At x = -1, the resulting negative harmonic series diverges. Thus the interval is (-1, 1].
For a degree-2 approximation, the remainder uses the third derivative: |R_2(0.1)| ≤ M|0.1|³/3!. Since the third derivative is e^x, one may take M = e^0.1.
The Lagrange remainder depends on the (n+1)st derivative at some intermediate point c and includes the factor (x-a)^(n+1)/(n+1)!.
A degree-6 Taylor polynomial matches the function's derivatives through order 6 at the center. This added local information often gives a useful approximation over a wider interval.
A Taylor polynomial is constructed to match the function's value and its first n derivatives at the center a. The equality is not generally true away from that center.
The coefficient of (x-a)^k is f^(k)(a)/k!. For k = 1 and a = 2, this is f'(2)/1! = f'(2).
For f(x) = sqrt(x), f(1) = 1 and f'(1) = 1/2. Thus L(x) = 1 + (x-1)/2, so L(1.04) = 1.02.
Using the generalized binomial coefficients gives 1 + (-1/2)x + [(-1/2)(-3/2)/2]x². The quadratic coefficient is 3/8.
The radius of convergence is 1. At both x = 1 and x = -1, the series becomes an alternating series whose term magnitudes decrease to zero, so both endpoints are included.
Since e^x = 1 + x + x²/2 + O(x³), the numerator is x²/2 + O(x³). Dividing by x² and taking the limit gives 1/2.
The next nonzero term in the sine expansion is x⁵/120. Therefore, after retaining terms through x³, the approximation error is O(x⁵).
Multiply e^x = 1 + x + x²/2 + x³/6 + ... by cos(x) = 1 - x²/2 + .... The x³ coefficient is 1/6 - 1/2 = -1/3.
Integrating term by term gives [x - x³/3 + x⁵/10] from 0 to 0.2. This equals approximately 0.197365.
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