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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...
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Which of the following functions would have only odd powers of x in it...
Solution:
The Taylor series expansion of a function f(x) about a point x = 0 is given by:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
If a function has only odd powers of x in its Taylor series expansion about x = 0, it means that all the even derivatives of the function at x = 0 are equal to zero.
Let's evaluate the even derivatives of the given functions at x = 0 to determine which one has only odd powers of x in its Taylor series expansion:
a) sin(x^3)
f(0) = sin(0) = 0
f'(0) = 3x^2 cos(x^3) = 0 (when x = 0)
f''(0) = 6x(-3x^6) sin(x^3) = 0 (when x = 0)
f'''(0) = 6(-3x^6) cos(x^3) + 9x^4(-3x^6) sin(x^3) = -54x^12 (when x = 0)
f''''(0) = -54(6x^6) sin(x^3) + 54(12x^10) cos(x^3) + 27x^8(-3x^6) sin(x^3) = 0 (when x = 0)
Therefore, sin(x^3) has only odd powers of x in its Taylor series expansion about x = 0.
b) sin(x^2)
f(0) = sin(0) = 0
f'(0) = 2x cos(x^2) = 0 (when x = 0)
f''(0) = 2cos(x^2) - 4x^2sin(x^2) = 2 (when x = 0)
f'''(0) = -4xsin(x^2) - 8x^3cos(x^2) = 0 (when x = 0)
Therefore, sin(x^2) does not have only odd powers of x in its Taylor series expansion about x = 0.
c) cos(x^3)
f(0) = cos(0) = 1
f'(0) = -3x^2 sin(x^3) = 0 (when x = 0)
f''(0) = -6x(-3x^6) cos(x^3) - 9x^4 sin(x^3) = 0 (when x = 0)
f'''(0) = -6(9x^8) sin(x^3) - 54x^6 cos(x^3) - 27x^4(-3x^6) sin(x^3) = -54x^12 (when x = 0)
f''''(0) = -54(54x^8) cos(x^3) + 54(72x^12) sin(x^3) - 27x^8(-3x^6) cos(x^3) = 0 (when x = 0)
Therefore, cos(x^3) does not
The Taylor series expansion of a function f(x) about a point x = 0 is given by:
f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...
If a function has only odd powers of x in its Taylor series expansion about x = 0, it means that all the even derivatives of the function at x = 0 are equal to zero.
Let's evaluate the even derivatives of the given functions at x = 0 to determine which one has only odd powers of x in its Taylor series expansion:
a) sin(x^3)
f(0) = sin(0) = 0
f'(0) = 3x^2 cos(x^3) = 0 (when x = 0)
f''(0) = 6x(-3x^6) sin(x^3) = 0 (when x = 0)
f'''(0) = 6(-3x^6) cos(x^3) + 9x^4(-3x^6) sin(x^3) = -54x^12 (when x = 0)
f''''(0) = -54(6x^6) sin(x^3) + 54(12x^10) cos(x^3) + 27x^8(-3x^6) sin(x^3) = 0 (when x = 0)
Therefore, sin(x^3) has only odd powers of x in its Taylor series expansion about x = 0.
b) sin(x^2)
f(0) = sin(0) = 0
f'(0) = 2x cos(x^2) = 0 (when x = 0)
f''(0) = 2cos(x^2) - 4x^2sin(x^2) = 2 (when x = 0)
f'''(0) = -4xsin(x^2) - 8x^3cos(x^2) = 0 (when x = 0)
Therefore, sin(x^2) does not have only odd powers of x in its Taylor series expansion about x = 0.
c) cos(x^3)
f(0) = cos(0) = 1
f'(0) = -3x^2 sin(x^3) = 0 (when x = 0)
f''(0) = -6x(-3x^6) cos(x^3) - 9x^4 sin(x^3) = 0 (when x = 0)
f'''(0) = -6(9x^8) sin(x^3) - 54x^6 cos(x^3) - 27x^4(-3x^6) sin(x^3) = -54x^12 (when x = 0)
f''''(0) = -54(54x^8) cos(x^3) + 54(72x^12) sin(x^3) - 27x^8(-3x^6) cos(x^3) = 0 (when x = 0)
Therefore, cos(x^3) does not
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