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The series expansion of sin⁡x/x near origin is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The series expansion ofsinx/x near origin isa)b)c)d)Correct answer is ...
Concept:
Taylor series:
The Taylor series of a real or complex-valued function f (x) that is infinitely differentiable at a real or complex number ‘a’ is the power series.
Expression of Taylor series is:

Calculation:
Given:
We have to find the series expansion of sin⁡x/x near origin, or a = 0.
Let f(x) = sin x
f(0) = sin (0) = 0,
f'(0) = cos (0) = 1,
f''(0) = -sin(0) = 0,
f'''(0) = -1 .... so on
Putting all the values in Taylor series expansion, we get:
Series expansion of sin x  will be:
sinx = x − x3/3!+…
Therefore the series expansion of sin⁡ x/x near origin will be:
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Most Upvoted Answer
The series expansion ofsinx/x near origin isa)b)c)d)Correct answer is ...
Concept:
Taylor series:
The Taylor series of a real or complex-valued function f (x) that is infinitely differentiable at a real or complex number ‘a’ is the power series.
Expression of Taylor series is:

Calculation:
Given:
We have to find the series expansion of sin⁡x/x near origin, or a = 0.
Let f(x) = sin x
f(0) = sin (0) = 0,
f'(0) = cos (0) = 1,
f''(0) = -sin(0) = 0,
f'''(0) = -1 .... so on
Putting all the values in Taylor series expansion, we get:
Series expansion of sin x  will be:
sinx = x − x3/3!+…
Therefore the series expansion of sin⁡ x/x near origin will be:
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The series expansion ofsinx/x near origin isa)b)c)d)Correct answer is option 'C'. Can you explain this answer?
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