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The coefficient of x4 in the power series expansion of esin x about x = 0 is _______ (correct up to three decimal places).
Correct answer is '-0.125'. Can you explain this answer?
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The coefficient of x4 in the power series expansion of esin x about x ...
The power series expansion of a function f(x) about x = 0 is given by:
f(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + ...
To find the coefficient of x⁴ in the power series expansion of esin x about x = 0, we need to find the value of a₄.
The power series expansion of esin x can be written as:
esin x = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + ...
We know that the power series expansion of esin x is given by:
esin x = 1 + sin x + (1/2!)(sin x)² + (1/3!)(sin x)³ + (1/4!)(sin x)⁴ + ...
Comparing the terms on both sides of the equation, we can see that a₀ = 1, a₁ = 1, a₂ = 1/2!, a₃ = 1/3!, and so on.
Therefore, the coefficient of x⁴ is given by:
a₄ = (1/4!)(sin x)⁴ = (1/4!)(sin x)⁴ = (1/24)(sin x)⁴ = (1/24)(sin⁴ x)
Since we are looking for the coefficient of x⁴ in the power series expansion of esin x about x = 0, we need to evaluate this expression at x = 0.
a₄ = (1/24)(sin⁴ 0) = (1/24)(0)⁴ = (1/24)(0) = 0
Therefore, the coefficient of x⁴ in the power series expansion of esin x about x = 0 is 0.
However, the given correct answer is -0.125, which means there might be a mistake in the question or in the answer. It is not possible to obtain a coefficient of x⁴ as -0.125 in the power series expansion of esin x about x = 0.
f(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + ...
To find the coefficient of x⁴ in the power series expansion of esin x about x = 0, we need to find the value of a₄.
The power series expansion of esin x can be written as:
esin x = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + ...
We know that the power series expansion of esin x is given by:
esin x = 1 + sin x + (1/2!)(sin x)² + (1/3!)(sin x)³ + (1/4!)(sin x)⁴ + ...
Comparing the terms on both sides of the equation, we can see that a₀ = 1, a₁ = 1, a₂ = 1/2!, a₃ = 1/3!, and so on.
Therefore, the coefficient of x⁴ is given by:
a₄ = (1/4!)(sin x)⁴ = (1/4!)(sin x)⁴ = (1/24)(sin x)⁴ = (1/24)(sin⁴ x)
Since we are looking for the coefficient of x⁴ in the power series expansion of esin x about x = 0, we need to evaluate this expression at x = 0.
a₄ = (1/24)(sin⁴ 0) = (1/24)(0)⁴ = (1/24)(0) = 0
Therefore, the coefficient of x⁴ in the power series expansion of esin x about x = 0 is 0.
However, the given correct answer is -0.125, which means there might be a mistake in the question or in the answer. It is not possible to obtain a coefficient of x⁴ as -0.125 in the power series expansion of esin x about x = 0.
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The coefficient of x4 in the power series expansion of esin x about x ...
-0.305 or -0.120
are correct answer
are correct answer
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The coefficient of x4 in the power series expansion of esin x about x = 0 is _______ (correct up to three decimal places).Correct answer is '-0.125'. Can you explain this answer?
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The coefficient of x4 in the power series expansion of esin x about x = 0 is _______ (correct up to three decimal places).Correct answer is '-0.125'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The coefficient of x4 in the power series expansion of esin x about x = 0 is _______ (correct up to three decimal places).Correct answer is '-0.125'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The coefficient of x4 in the power series expansion of esin x about x = 0 is _______ (correct up to three decimal places).Correct answer is '-0.125'. Can you explain this answer?.
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