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MAT136H5 S 2025 - All Sections 6.2 preparation check
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Read Theorem 6.2 in the textbook Links to an external site. . Suppose that ∞ ∑ n=0cnxn is a power series with interval of convergence (−3,3) and that ∞ ∑ n=0cnxn converges to f(x)= 1 x2 . a) What is the interval of convergence of ∞ ∑ n=05x⋅cnxn? (-3,3) b) On the interval of convergence, what does the series ∞ ∑ n=05x⋅cnxn converge to? 5 / x
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The problem asks us to analyze a power series ∑_{n=0}^∞ c_n x^n with interval of convergence (−3, 3) and whose sum on that interval is f(x) = 1/x^2. Then we consider ∑_{n=0}^∞ 5x · c_n x^n, which can be rewritten as ∑_{n=0}^∞ 5 c_n x^{n+1}.
Option a) (-3, 3) as the interval of convergence for ∑ 5x · c_n x^n.
- The original power series ∑ c_n x^n has radius of convergence 3, so its domain of conv......Login to view full explanationLog in for full answers
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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 correct?
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.
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