题目
题目

MTH1030 -1035 - S1 2025 MTH1030/35 Week 9 lesson quiz: Representing functions by infinite series

单项选择题

Which among the following statements is the strongest that is true?A If a function is defined for all x and has a Maclaurin series, then this Maclaurin series converges for all x.B If a function is defined for all x and has a Maclaurin series, then this Maclaurin series is equal to the function for all x.C If a function is defined for all x and has a Maclaurin series, then this Maclaurin series is equal to the function for infinitely many values of \(x\)D If a function is defined for all x and has a Maclaurin series, then this Maclaurin series is equal to the function at \(x=0\).

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思路分析
Let’s parse the question first: we’re comparing statements about a function f defined for all real x that has a Maclaurin series. The Maclaurin series is the power series obtained by expanding f around x = 0, i.e., f(x) = sum_{n=0}^∞ f^{(n)}(0)/n! * x^n in its radius of convergence. Now evaluate each option in turn. Option A: 'If a function is defined for all x and has a Maclaurin series, then this Maclaurin series converges for all x.' This is not necessarily true. A Maclaurin series may have a finite radius of convergen......Login to view full explanation

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Question text 9Marks a) The Maclaurin series of [math: f(x)=2cos⁡(x)]f(x) = 2\cos(x) is [math: f(x)=]Answer 1[input][math: +]Answer 2[input][math: x+]Answer 3[input][math: x2+⋯]x^2+\cdotsb) The Maclaurin series of [math: g(x)=6ln⁡(1−x)]g(x) = 6\ln(1-x) is [math: g(x)=]Answer 4[input][math: +]Answer 5[input][math: x+]Answer 6[input][math: x2+⋯]x^2+\cdotsc) The Maclaurin series of [math: h(x)=2cos⁡(3x)+6ln⁡(1−x2)]h(x) = 2\cos(3x) + 6\ln(1-x^2) is[math: h(x)=]Answer 7[input][math: +]Answer 8[input][math: x+]Answer 9[input][math: x2+⋯]x^2+\cdotsPlease answer all parts of the question.Notes Report question issue Question 7 Notes

Question text 9Marks a) The Maclaurin series of [math: f(x)=2cos⁡(x)]f(x) = 2\cos(x) is [math: f(x)=]Answer 1[input][math: +]Answer 2[input][math: x+]Answer 3[input][math: x2+⋯]x^2+\cdotsb) The Maclaurin series of [math: g(x)=6ln⁡(1−x)]g(x) = 6\ln(1-x) is [math: g(x)=]Answer 4[input][math: +]Answer 5[input][math: x+]Answer 6[input][math: x2+⋯]x^2+\cdotsc) The Maclaurin series of [math: h(x)=2cos⁡(3x)+6ln⁡(1−x2)]h(x) = 2\cos(3x) + 6\ln(1-x^2) is[math: h(x)=]Answer 7[input][math: +]Answer 8[input][math: x+]Answer 9[input][math: x2+⋯]x^2+\cdotsNotes Report question issue Question 7 Notes

Question texta) The Maclaurin series of [math: ex2]e^{x^2} is [math: ex2=]e^{x^2} =Answer 1 Question 9[input] [math: +] Answer 2 Question 9[input][math: x] [math: +] Answer 3 Question 9[input][math: x2]x^2 [math: +] Answer 4 Question 9[input][math: x3]x^3 [math: +…]+\ldotsb) The Maclaurin series of [math: ∫0x6ln⁡(1+t)sin⁡(t)dt]\int_0^x 6\ln(1+t)\sin(t) dt is [math: ∫0x6ln⁡(1+t)sin⁡(t)dt=]\int_0^x 6\ln(1+t)\sin(t) dt =Answer 5 Question 9[input] [math: +] Answer 6 Question 9[input][math: x] [math: +] Answer 7 Question 9[input][math: x2]x^2 [math: +] Answer 8 Question 9[input][math: x3]x^3 [math: +…]+\ldotsPlease answer all parts of the question.

Question texta) The Maclaurin series of [math: ex2]e^{x^2} is [math: ex2=]e^{x^2} =Answer 1 Question 9[input] [math: +] Answer 2 Question 9[input][math: x] [math: +] Answer 3 Question 9[input][math: x2]x^2 [math: +] Answer 4 Question 9[input][math: x3]x^3 [math: +…]+\ldotsb) The Maclaurin series of [math: ∫0x6ln⁡(1+t)sin⁡(t)dt]\int_0^x 6\ln(1+t)\sin(t) dt is [math: ∫0x6ln⁡(1+t)sin⁡(t)dt=]\int_0^x 6\ln(1+t)\sin(t) dt =Answer 5 Question 9[input] [math: +] Answer 6 Question 9[input][math: x] [math: +] Answer 7 Question 9[input][math: x2]x^2 [math: +] Answer 8 Question 9[input][math: x3]x^3 [math: +…]+\ldots

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