Construct the Fourier series of the following periodic function.
f(x)= cosx, -π≤x<0
1, 0≤x<π
To construct the Fourier series of a periodic function, you need to compute the coefficients of the sine and cosine terms in the series. The Fourier series representation of a function f(x) with period 2π is given by the equation:
f(x) = a₀/2 + Σ[aₙcos(nx) + bₙsin(nx)]
Where a₀ is the constant term, aₙ and bₙ are the coefficients for the cosine and sine terms, and n is the harmonic number.
To find the coefficients for our given function f(x)=cosx on the interval -π≤x<0 and f(x)=1 on the interval 0≤x<π, we need to consider the odd and even behavior of the function in each interval.
For -π≤x<0:
Since f(x) = cosx, which is an even function, the coefficient of the sine term bₙ will be zero for all n.
To find the coefficient of the cosine term aₙ, we can use the formula:
aₙ = (2/π) * ∫[f(x)cos(nx)dx] over the interval -π≤x<0
Evaluating this integral, we have:
aₙ = (2/π) * ∫[cosx*cos(nx)dx] from -π to 0
Integrating cosx*cos(nx) gives:
aₙ = (2/π) * [(sinx*sin(nx))/(n+1)] from -π to 0
Substituting the limits, we get:
aₙ = (2/π) * [(0-0)/(n+1)] = 0 for all n
So, for -π≤x<0, there are no cosine terms in the Fourier series.
For 0≤x<π:
Since f(x) = 1, which is an even function, similar to the previous interval, the coefficient of the sine term bₙ will be zero for all n.
To find the coefficient of the cosine term aₙ, we can use the formula:
aₙ = (2/π) * ∫[f(x)cos(nx)dx] over the interval 0≤x<π
Evaluating this integral, we have:
aₙ = (2/π) * ∫[cosx*cos(nx)dx] from 0 to π
Integrating cosx*cos(nx) gives:
aₙ = (2/π) * [(sinx*sin(nx))/(n+1)] from 0 to π
Substituting the limits, we get:
aₙ = (2/π) * [((sin(π)*sin(nπ))/(n+1))-((sin(0)*sin(n0))/(n+1))]
Since sin(π) and sin(0) are both zero, this simplifies to:
aₙ = (2/π) * (0-0) = 0 for all n
So, for 0≤x<π, there are also no cosine terms in the Fourier series.
Therefore, the Fourier series of the given function f(x)=cosx, -π≤x<0 and f(x)=1, 0≤x<π is:
f(x) = b₁sinx + b₃sin(3x) + b₅sin(5x) + ...
You can see that the Fourier series only has sine terms, and the coefficients bₙ can be found using a similar integration process as before but for the sine terms.