Infinite Sequences & Series unit test
A test on this unit alone, marked as a percentage and a letter grade — for the test your class is actually sitting, rather than for May. Answer everything, then submit once: seeing the answer to question 3 before attempting question 4 makes the final percentage meaningless.
Alternating error bound
Taylor polynomial vs Taylor series
Which error bound to use
Power series centered at a
Taylor series centered elsewhere
Root test
Endpoints can disagree
Series for ln(1 + x)
Alternating series error bound
Radius survives term-by-term operations
Term-by-term differentiation and integration
Taylor series
Short answer 1. Define or explain: Radius then endpoints
3 ptsShort answer 2. Define or explain: Lagrange error bound
3 ptsShort answer 3. Define or explain: Common series to recognize instantly
3 ptsShort answer 4. Define or explain: A Taylor polynomial is finite
3 ptsFree response
9 ptsNO CALCULATOR. The Taylor series for a function f about x = 4 is given by Σ (from n = 1 to ∞) (x − 4)ⁿ⁺¹ / ((n + 1)·3ⁿ) = (x − 4)²/(2·3) + (x − 4)³/(3·3²) + (x − 4)⁴/(4·3³) + ⋯ + (x − 4)ⁿ⁺¹/((n + 1)·3ⁿ) + ⋯ and converges to f(x) on its interval of convergence.
A. Using the ratio test, find the interval of convergence of the Taylor series for f about x = 4. Justify your answer.
B. Find the first three nonzero terms and the general term of the Taylor series for f′, the derivative of f, about x = 4.
C. The Taylor series for f′ described in part B is a geometric series. For all x in the interval of convergence of the Taylor series for f′, show that f′(x) = (x − 4)/(7 − x).
D. It is known that the radius of convergence of the Taylor series for f about x = 4 is the same as the radius of convergence of the Taylor series for f′ about x = 4. Does the Taylor series for f′ described in part B converge to f′(x) = (x − 4)/(7 − x) at x = 8? Give a reason for your answer.