题目
题目

MTH1030 -1035 - S1 2025 MTH1030/35 Week 10 lesson quiz: Power series defining functions

单项选择题

Which of the following statements is not correct?

选项
A.a. If two functions have the same Maclaurin series (term by term), the two functions are the same (for all values of [math: x]).
B.b. If a function is equal to two power series (for all values of [math: x]), the two power series are the same term by term.
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思路分析
The question asks which statement is not correct about Maclaurin/taylor series and their relation to the underlying functions. Option a: 'If two functions have the same Maclaurin series (term by term), the two functions are the same (for all values of x).' Here the key issue is the scope of validity. If two analytic functions sha......Login to view full explanation

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Question text 5Marks a) Find the radii of convergence of the following series:[math: ∑n=1∞n2+2n3n2+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3n^2+2}x^n radius = Answer 1[input][math: ∑n=1∞n2+2n3n+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3^n+2}x^n radius = Answer 2[input]b) Suppose a power series [math: ∑n=1∞anxn]\displaystyle \sum_{n=1}^{\infty} a_n x^n is convergent for [math: x=−3]x= -3 and divergent for [math: x=5]x = 5.For [math: x=−1]x = -1, the power series Answer 3[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=−6]x = -6, the power series Answer 4[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=3]x = 3, the power series Answer 5[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]Notes Report question issue Question 8 Notes

Which of the following statements is NOT true?

In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 (x−4)n 3n2    Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| (x−4)n+1 3(n+1)2 (x−4)n 3n2 |      |x-4| b) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|<1  ?    [ Select ] Nothing (because the ratio test inconclusive) The power series converges The power series diverges c) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|>1  ?    [ Select ] The power series converges Nothing (because the ratio test inconclusive) The power series diverges d) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|=1 ?    [ Select ] The power series diverges The power series converges Nothing (because the ratio test inconclusive) e) For what values of x is |x−4|<1 ? (Solve the inequality)  [ Select ] x < 5 -4 < x < 5 -4 < x < 4 3 < x < 5 2 < x < 6 f) When x=5  the power series in the question becomes ∞ ∑ n=0 1 3n2  . Does this series converge or diverge?    [ Select ] Diverges Converges g) When x=3   the power series in the question becomes ∞ ∑ n=0 (−1)n 3n2   . Does this series converge or diverge?    [ Select ] Diverges Converges h) Summarizing the results we have found, for what values of x does the power series in the question converge? [ Select ] Converges for x in the interval (-4,4) Converges for x in the interval [3,5] Converges for x in the interval (-4,5] Converges for x in the interval [3,5) Converges for x in the interval (3,5] Converges for x in the interval [-4,4] Note: This is the interval of convergence.

In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 8n n! xn  Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| 8n+1xn+1 (n+1)! 8nxn n! |      [ Select ] 1 / 4 0 1 infinity 8 1 / 8 b) What does the Ratio Test and your answer in (a) tell you about the series?    [ Select ] The series converges only when -1 < x < 1 The series converges only when -8 < x < 8 The series converges for all values of x The series only converges when x = 0 The series does not converge for any values of x c) What is the radius of convergence for the series?  R = infinity d) What is the interval of convergence for the series?    [ Select ] (-infinity, infinity) [-1, 1] (-8, 8) [-8, 8] x = 0 (-1, 1) Note: The interval of convergence is the set of values of x for which the series converges.  e) Consider your answer to (d). Does the series converge when x=3 ?    [ Select ] Yes There is not enough information No

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