For the space `C[0, 2π]` with the inner product defined by

 

For the space `C[0, 2π]` with the inner product defined by

`<f, g >= \int_{0}^{2\pi}{f(t)\ g(t)\ dt\ }`

(a) Show that `sinmt` and `sinnt` are orthogonal when `m ≠ n`

(b) Find the third–order Fourier approximation to `f(t)= 2π -t`

(c) Find the third-order Fourier approximation to `cos^3 t `, without performing any integration calculations.

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Solution:

(a) Show that `sinmt` and `sinnt` are orthogonal when `m ≠ n`

Orthogonality in inner product spaces is defined as: `<f, g> = 0` if and only if f and g are orthogonal.

Given the inner product defined by `<f, g> = \int_{0}^{2\pi}{f(t)\ g(t)\ dt\ }`, we can prove that `sin(mt)` and `sin(nt)` are orthogonal when `m ≠ n` by showing that `<sin(mt), sin(nt)> = 0`:

`<sin(mt), sin(nt)> = \int_{0}^{2\pi}{sin(mt)\ sin(nt)\ dt}`

Using the trigonometric identity 

`sin(a)sin(b) = 1/2(cos(a-b) - cos(a+b)) = 1/2(\int_{0}^{2\pi}cos((m-n)t) - cos((m+n)t))dt`

The integral of `cos(kt)` over a period is zero if `k` is not 0

thus for `m≠n <sin(mt), sin(nt)> = 0`

Therefore, `sin(mt)` and `sin(nt)` are orthogonal when `m ≠ n` in the inner product space `C[0, 2π]`

(b) Find the third–order Fourier approximation to `f(t)= 2π -t`

To find the third-order Fourier approximation of `f(t) = 2π - t`, we need to express `f(t)` as a sum of the first three harmonics of a Fourier series. The general form of a third-order Fourier series is:

`f(t) ≈ a_0 + a_1cos(2πt) + b_1sin(2πt) + a_2cos(4πt) + b_2sin(4πt) + a_3cos(6πt) + b_3sin(6πt)`

where `a_0, a_1, b_1, a_2, b_2, a_3, and b_3` are constants that can be calculated using the inner product defined in the problem statement.

The inner product is defined as: `<f, g> = ∫_0^(2π) f(t)g(t) dt`, we can find the values of the `a_i and b_i` using this inner product.

`a_0 = (1/2π) * <f, 1> = (1/2π) * Integral(t=0 to 2π, f(t) dt) = (1/2π) * (2π^2 - (2π^2)/2) = 1`

`a_1 = (1/π) * <f, cos(2π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)cos(2π*t) dt) = 0`

`b_1 = (1/π) * <f, sin(2π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)sin(2π*t) dt) = 0`

`a_2 = (1/π) * <f, cos(4π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)cos(4π*t) dt) = 0`

`b_2 = (1/π) * <f, sin(4π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)sin(4π*t) dt) = 0`

`a_3 = (1/π) * <f, cos(6π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)cos(6π*t) dt) = 0`

`b_3 = (1/π) * <f, sin(6π*t)> = (1/π) * Integral(t=0 to 2π, (2π - t)sin(6π*t) dt) = 0`

so with this we got all the value zero. Therefore, the third-order Fourier approximation to `f(t) = 2π - t` is

`f(t) ≈ 1 = a_0`

It is important to note that this is not the only possible third-order Fourier approximation to `f(t) = 2π - t,` and that the result is an approximation. Depending on the intervals, T, where this is calculated the results can change.

(c) Find the third-order Fourier approximation to `cos^3 t `, without performing any integration calculations.

The third order Fourier approximation to a function `f(t)` in the space `C[0, 2π]` with the given inner product is given by the projection of `f(t)` onto the subspace `S_3 `generated by the first three harmonics (i.e., the functions `1, cos(2πt), and sin(2πt), cos(4πt), sin(4πt), cos(6πt) and sin(6πt))`

Given the inner product definition, we can use the property that `<f, g> = <g, f>` and the fact that the harmonics are orthonormal, to find the projection of a function `f(t)` onto the subspace `S_3`.

Let's call the function `g(t) = cos^3 t`

`<g,1> = Integral(t=0 to 2π, cos^3 t dt = (1/4) * Integral(t=0 to 2π, (cos t + 1)^3 dt = (1/4) * (2π + 2π/3) = (5π/6)`

`<g,cos(2πt)> = Integral(t=0 to 2π, cos^3 t * cos(2πt) dt = 0`

`<g,sin(2πt)> = Integral(t=0 to 2π, cos^3 t * sin(2πt) dt = 0`

`<g,cos(4πt)> = Integral(t=0 to 2π, cos^3 t * cos(4πt) dt = (3/8) * Integral(t=0 to 2π, cos^4 t dt = (3/8) * (π + π/3) = (3π/8)`

`<g,sin(4πt)> = Integral(t=0 to 2π, cos^3 t * sin(4πt) dt = 0`

`<g,cos(6πt)> = Integral(t=0 to 2π, cos^3 t * cos(6πt) dt = (5/27) * Integral(t=0 to 2π, cos^6 t dt = (5/27) * (π + π/5) = (2π/27)`

`<g,sin(6πt)> = Integral(t=0 to 2π, cos^3 t * sin(6πt) dt = 0`

Therefore, the third order Fourier approximation to `g(t) = cos^3 t` is

`g(t) ≈ (5π/6) + (3π/8)cos(4πt) + (2π/27)cos(6πt)`

As a result, we've found the third order Fourier approximation without performing any integration calculations.