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Obtain the Fourier series of the function๏€จ ๏€ฉf x given as follows:๐‘“(๐‘ฅ) = { ๐‘ฅ, 0 โ‰ค ๐‘ฅ โ‰ค ๐œ‹0, ๐œ‹ โ‰ค ๐‘ฅ โ‰ค 2๐œ‹ }

Question

Obtain the Fourier series of the function๏€จ ๏€ฉf x given as follows:๐‘“(๐‘ฅ) = { ๐‘ฅ, 0 โ‰ค ๐‘ฅ โ‰ค ๐œ‹0, ๐œ‹ โ‰ค ๐‘ฅ โ‰ค 2๐œ‹ }

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Solution

To obtain the Fourier series of the function f(x), we need to find the coefficients of the sine and cosine terms.

Step 1: Determine the period of the function. In this case, the function is periodic with a period of 2ฯ€ since it repeats every 2ฯ€ units.

Step 2: Find the average value of the function over one period. Since f(x) is defined differently for different intervals, we need to calculate the average value separately for each interval.

For the interval 0 โ‰ค x โ‰ค ฯ€, the function f(x) is equal to x. To find the average value, we integrate f(x) over this interval and divide by the length of the interval: average value = (1/ฯ€) โˆซ[0,ฯ€] x dx = (1/ฯ€) [x^2/2] from 0 to ฯ€ = (1/ฯ€) (ฯ€^2/2) = ฯ€/2

For the interval ฯ€ โ‰ค x โ‰ค 2ฯ€, the function f(x) is equal to 0. The average value over this interval is simply 0.

Step 3: Calculate the coefficients of the sine and cosine terms using the formulas:

a0 = average value = ฯ€/2 an = (2/ฯ€) โˆซ[0,ฯ€] x cos(nฯ‰x) dx, for n > 0 bn = (2/ฯ€) โˆซ[0,ฯ€] x sin(nฯ‰x) dx, for n > 0

where ฯ‰ = 2ฯ€/T is the angular frequency and T is the period.

For the cosine terms: a0 = ฯ€/2 an = (2/ฯ€) โˆซ[0,ฯ€] x cos(nฯ€x/ฯ€) dx = (2/ฯ€) โˆซ[0,ฯ€] x cos(nx) dx = (2/ฯ€) [x sin(nx)/n] from 0 to ฯ€ = (2/ฯ€) (ฯ€ sin(nฯ€)/n) = 2 sin(nฯ€)/n

For the sine terms: bn = (2/ฯ€) โˆซ[0,ฯ€] x sin(nฯ€x/ฯ€) dx = (2/ฯ€) โˆซ[0,ฯ€] x sin(nx) dx = -(2/ฯ€) [x cos(nx)/n] from 0 to ฯ€ = -(2/ฯ€) (ฯ€ cos(nฯ€)/n) = -2 cos(nฯ€)/n

Step 4: Write the Fourier series using the calculated coefficients. Since the function is defined differently for different intervals, we need to write the series separately for each interval.

For the interval 0 โ‰ค x โ‰ค ฯ€, the Fourier series is: f(x) = a0/2 + โˆ‘[n=1,โˆž] an cos(nฯ‰x) + bn sin(nฯ‰x) = (ฯ€/4) + โˆ‘[n=1,โˆž] (2 sin(nฯ€)/n) cos(nx) - (2 cos(nฯ€)/n) sin(nx)

For the interval ฯ€ โ‰ค x โ‰ค 2ฯ€, the Fourier series is simply 0.

So, the Fourier series of the function f(x) is given by: f(x) = (ฯ€/4) + โˆ‘[n=1,โˆž] (2 sin(nฯ€)/n) cos(nx) - (2 cos(nฯ€)/n) sin(nx) for 0 โ‰ค x โ‰ค ฯ€ 0 for ฯ€ โ‰ค x โ‰ค 2ฯ€

This problem has been solved

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