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The trigonometric Fourier series of an even function does not have the
Select one:
  • a)
    Cosine terms
  • b)
    Sine terms
  • c)
    Odd harmonic terms
  • d)
    dc term
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The trigonometric Fourier series of an even function does not have the...
The trigonometric Fourier series of an even function has cosine terms which are even functions.
It has dc term if its average value is finite and no dc term if average value is zero.
So it does not have sine terms.
The correct answer is: Sine terms
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Most Upvoted Answer
The trigonometric Fourier series of an even function does not have the...
Explanation:

The trigonometric Fourier series represents a periodic function as a sum of sine and cosine functions. It is given by the formula:

\[f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)\]

where \(a_0\), \(a_n\), and \(b_n\) are the Fourier coefficients.

An even function is a function that satisfies the condition \(f(x) = f(-x)\) for all \(x\) in its domain. This means that the function is symmetrical about the y-axis. Examples of even functions include \(f(x) = \cos(x)\) and \(f(x) = x^2\).

Even function and its Fourier series:

When a function is even, its Fourier series simplifies because the odd harmonic terms vanish. This is because the sine function is an odd function, satisfying the condition \(\sin(-x) = -\sin(x)\). Therefore, for an even function, the sine terms in the Fourier series will always be zero.

Reason for absence of sine terms:

The absence of sine terms in the trigonometric Fourier series of an even function can be understood as follows:

1. For an even function \(f(x)\), we have \(f(x) = f(-x)\) for all \(x\) in the domain.
2. When we substitute \(-x\) into the Fourier series formula, we get:

\[f(-x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(-nx) + b_n \sin(-nx) \right)\]

3. Since the cosine function is an even function, it satisfies the condition \(\cos(-x) = \cos(x)\). Therefore, the cosine terms in the Fourier series remain the same when substituting \(-x\).
4. However, the sine function is an odd function, satisfying the condition \(\sin(-x) = -\sin(x)\). Therefore, the sine terms in the Fourier series change sign when substituting \(-x\).
5. Since \(f(-x) = f(x)\), the sine terms in the Fourier series must be zero to satisfy this condition. Otherwise, the function would not be symmetrical about the y-axis.

Conclusion:

In conclusion, the trigonometric Fourier series of an even function does not have sine terms. This is because even functions are symmetrical about the y-axis, and the sine terms change sign when substituting \(-x\), leading to their cancellation in the Fourier series.
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The trigonometric Fourier series of an even function does not have theSelect one:a)Cosine termsb)Sine termsc)Odd harmonic termsd)dc termCorrect answer is option 'B'. Can you explain this answer?
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