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Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0?
- a)sin (x3)
- b)sin (x2 )
- c)cos (x3 )
- d)cos (x2 )
Correct answer is option 'A'. Can you explain this answer?
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Which of the following functions would have only odd powers of x in it...
Explanation:
The Taylor series expansion of a function about the point x = 0 is given by:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
In order for the Taylor series expansion to contain only odd powers of x, all the even terms in the expansion need to be zero. This means that the derivative of the function with respect to x evaluated at x = 0 should be zero for all even powers of x.
a) sin (x^3):
Taking the derivatives of sin (x^3) with respect to x, we get:
f(x) = sin (x^3)
f'(x) = 3x^2 * cos (x^3)
f''(x) = 6x * cos (x^3) - 9x^4 * sin (x^3)
f'''(x) = 6 * cos (x^3) - 54x^3 * sin (x^3) - 36x^5 * cos (x^3)
Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 3(0)^2 * cos (0) = 0
f''(0) = 6(0) * cos (0) - 9(0)^4 * sin (0) = 0
f'''(0) = 6 * cos (0) - 54(0)^3 * sin (0) - 36(0)^5 * cos (0) = 6
Since all the even derivatives are zero, the Taylor series expansion of sin (x^3) about x = 0 contains only odd powers of x.
b) sin (x^2):
Taking the derivatives of sin (x^2) with respect to x, we get:
f(x) = sin (x^2)
f'(x) = 2x * cos (x^2)
f''(x) = 2cos (x^2) - 4x^2 * sin (x^2)
f'''(x) = -8x * sin (x^2) - 8x^3 * cos (x^2)
Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 2(0) * cos (0) = 0
f''(0) = 2cos (0) - 4(0)^2 * sin (0) = 2
f'''(0) = -8(0) * sin (0) - 8(0)^3 * cos (0) = 0
The third derivative is zero, so the Taylor series expansion of sin (x^2) about x = 0 contains even powers of x and not only odd powers of x.
c) cos (x^3):
Taking the derivatives of cos (x^3) with respect to x, we get:
f(x) = cos (x^3)
f'(x) = -3x^2 *
The Taylor series expansion of a function about the point x = 0 is given by:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
In order for the Taylor series expansion to contain only odd powers of x, all the even terms in the expansion need to be zero. This means that the derivative of the function with respect to x evaluated at x = 0 should be zero for all even powers of x.
a) sin (x^3):
Taking the derivatives of sin (x^3) with respect to x, we get:
f(x) = sin (x^3)
f'(x) = 3x^2 * cos (x^3)
f''(x) = 6x * cos (x^3) - 9x^4 * sin (x^3)
f'''(x) = 6 * cos (x^3) - 54x^3 * sin (x^3) - 36x^5 * cos (x^3)
Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 3(0)^2 * cos (0) = 0
f''(0) = 6(0) * cos (0) - 9(0)^4 * sin (0) = 0
f'''(0) = 6 * cos (0) - 54(0)^3 * sin (0) - 36(0)^5 * cos (0) = 6
Since all the even derivatives are zero, the Taylor series expansion of sin (x^3) about x = 0 contains only odd powers of x.
b) sin (x^2):
Taking the derivatives of sin (x^2) with respect to x, we get:
f(x) = sin (x^2)
f'(x) = 2x * cos (x^2)
f''(x) = 2cos (x^2) - 4x^2 * sin (x^2)
f'''(x) = -8x * sin (x^2) - 8x^3 * cos (x^2)
Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 2(0) * cos (0) = 0
f''(0) = 2cos (0) - 4(0)^2 * sin (0) = 2
f'''(0) = -8(0) * sin (0) - 8(0)^3 * cos (0) = 0
The third derivative is zero, so the Taylor series expansion of sin (x^2) about x = 0 contains even powers of x and not only odd powers of x.
c) cos (x^3):
Taking the derivatives of cos (x^3) with respect to x, we get:
f(x) = cos (x^3)
f'(x) = -3x^2 *
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