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The Fourier series for the function f(x)=sin2x is
- a)sinx+sin2x
- b)1-cos2x
- c)sin2x+cos2x
- d)0.5-0.5cos2x
Correct answer is option 'D'. Can you explain this answer?
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The Fourier series for the function f(x)=sin2x is a)sinx+sin2xb)1-cos2...
Here f(x ) = sin2 x is even function, hence f( x ) has no sine term.
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The Fourier series for the function f(x)=sin2x is a)sinx+sin2xb)1-cos2...
Fourier Series of Function f(x)=sin2x
Fourier series is a way of representing a periodic function as a sum of sine and cosine functions.
Formula for Fourier Series Coefficients
The coefficients of the Fourier series can be calculated using the following formula:
an = (2/T) ∫f(x) cos(nωx) dx
bn = (2/T) ∫f(x) sin(nωx) dx
where T is the period of the function, ω = 2π/T is the angular frequency, and n is the harmonic number.
Fourier Series of Function f(x)=sin2x
The function f(x) = sin2x is a periodic function with a period of π. Therefore, we can use the formula for Fourier series coefficients to find its Fourier series representation.
an = (2/π) ∫sin2x cos(nx) dx
bn = (2/π) ∫sin2x sin(nx) dx
Solving these integrals, we get
an = 0
bn = (4/(nπ)) [(-1)n-1/(n+2) + (-1)n+1/(n-2)]
The Fourier series of f(x) is given by
f(x) = (4/π) [sin2x/2 - sin(4x)/8 + sin(6x)/18 - sin(8x)/32 + ...]
Since the Fourier series is a sum of sine functions, we can simplify it using the identity sin2x = (1/2) - (1/2)cos2x.
f(x) = (2/π) [1/2 - (1/2)cos2x - (1/8)sin(4x) + (1/18)sin(6x) - (1/32)sin(8x) + ...]
f(x) = 0.5 - 0.5cos2x + (2/π) [(1/18)sin(6x) - (1/8)sin(4x) + (1/32)sin(8x) - ...]
Therefore, the correct answer is option D, 0.5 - 0.5cos2x.
Fourier series is a way of representing a periodic function as a sum of sine and cosine functions.
Formula for Fourier Series Coefficients
The coefficients of the Fourier series can be calculated using the following formula:
an = (2/T) ∫f(x) cos(nωx) dx
bn = (2/T) ∫f(x) sin(nωx) dx
where T is the period of the function, ω = 2π/T is the angular frequency, and n is the harmonic number.
Fourier Series of Function f(x)=sin2x
The function f(x) = sin2x is a periodic function with a period of π. Therefore, we can use the formula for Fourier series coefficients to find its Fourier series representation.
an = (2/π) ∫sin2x cos(nx) dx
bn = (2/π) ∫sin2x sin(nx) dx
Solving these integrals, we get
an = 0
bn = (4/(nπ)) [(-1)n-1/(n+2) + (-1)n+1/(n-2)]
The Fourier series of f(x) is given by
f(x) = (4/π) [sin2x/2 - sin(4x)/8 + sin(6x)/18 - sin(8x)/32 + ...]
Since the Fourier series is a sum of sine functions, we can simplify it using the identity sin2x = (1/2) - (1/2)cos2x.
f(x) = (2/π) [1/2 - (1/2)cos2x - (1/8)sin(4x) + (1/18)sin(6x) - (1/32)sin(8x) + ...]
f(x) = 0.5 - 0.5cos2x + (2/π) [(1/18)sin(6x) - (1/8)sin(4x) + (1/32)sin(8x) - ...]
Therefore, the correct answer is option D, 0.5 - 0.5cos2x.
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The Fourier series for the function f(x)=sin2x is a)sinx+sin2xb)1-cos2...
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The Fourier series for the function f(x)=sin2x is a)sinx+sin2xb)1-cos2xc)sin2x+cos2xd)0.5-0.5cos2xCorrect answer is option 'D'. Can you explain this answer?
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