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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+\cdotsNotes Report question issue Question 7 Notes

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We are given three Maclaurin expansion tasks and a set of fills to complete. I will restate the setup for each part, then inspect what the correct coefficients are, and explain why each possible fill is correct or incorrect. a) The Maclaurin series of f(x) = 2 cos(x) is f(x) = [blank] + [blank] x + [blank] x^2 + … - To expand cos(x) around 0: cos(x) = 1 − x^2/2 + x^4/24 − … - Multiply by 2: f(x) = 2(1 − x^2/2 + x^4/24 − …) = 2 − x^2 + (2/24)x^4 − … - Therefore the constant term is 2, the coefficient of x is 0, and the ......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 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

What are the constants in the first four terms of the Maclaurin series of \(\cos(x^2)\)? Input in the form \((a_0,a_1,a_2,a_3)\).

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