Least-Squares Approximation Property. The Nth partial sum

Chapter 10, Problem 37E

(choose chapter or problem)

Least-Squares Approximation Property. The \(N \text { th }\) partial sum of the Fourier series

                                \(f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left\{a_{n} \cos n x+b_{n} \sin n x\right\}\)

gives the best mean-square approximation of \(f\) by trigonometric polynomial. To prove this, let \(F_{N}(x)\) denote an arbitrary trigonometric polynomial of degree \(N\):

                            \(F_{N}(x)=\frac{\alpha_{0}}{2}+\sum_{n=1}^{N}\left\{\alpha_{n} \cos n x+\beta_{n} \sin n x\right\}\)

and define

                                       \(E:=\int_{-\pi}^{\pi}\left[f(x)-F_{N}(x)\right]^{2} d x\)

which is the total square error. Expanding the integrand, we get

                              \(E=\int_{-\pi}^{\pi} f^{2}(x) d x-2 \int_{-\pi}^{\pi} f(x) F_{N}(x) d x+\int_{-\pi}^{\pi} F_{N}^{2}(x) d x\)

(a) Use the orthogonality of the functions \(\{1, \cos x, \sin x, \cos 2 x, \ldots\}\) to show that

                            \( \int_{-\pi}^{\pi} F_{N}^{2}(x) d x=\pi\left(\frac{a_{0}^{2}}{2}     a_{1}^{2}\right.\)

                                    \(\left.+\cdots+\alpha_{N}^{2}+\beta_{1}^{2}+\cdots+\beta_{N}^{2}\right)\)

and

                \(\int_{-\pi}^{\pi} f(x) F_{N}(x) d x=\pi\left(\frac{\alpha_{0} a_{0}}{2}+\alpha_{1} a_{1}+\cdots+\alpha_{N} a_{N}\right.\)

                                                                \(\left.+\beta_{1} b_{1}+\cdots+\beta_{N} b_{N}\right)\)

(b) Let \(E^{*}\) be the error when we approximate \(f\) by the \(\text { Nth }\) partial sum of its Fourier series, that is, when we choose \(\alpha_{n}=a_{n}\) and \(\beta_{n}=b_{n}\). Show that

                 \(E^{*}=\int_{-\pi}^{\pi}f^{2}(x)dx-\pi\left(\frac{a_{0}^{2}}{2}+a_{1}^{2}+\cdots+a_{N}^{2}+b_{1}^{2}+\cdots+b_{N}^{2}\right)\)

(c) Using the results of parts (a) and (b), show that \(E-E^{*} \geq 0\), that is, \(E \geq E^{*}\), by proving that

           \(E-E^{*}=\pi\left\{\frac{\left(\underline{\alpha}_{0}-\alpha_{0}\right)^{2}}{2}+\left(\alpha_{1}-\alpha_{1}\right)^{2}\right.\)

                       \(+\cdots+\left(\alpha_{N}-\alpha_{N}\right)^{2}+\left(\beta_{1}-b_{1}\right)^{2}\)

                    \(\left.+\cdots+\left(\beta_{N}-b_{N}\right)^{2}\right\}\)

                                                           

Hence, the \(N \text { th }\) partial sum of the Fourier series gives the least total square error, since \(E \geq E^{*}\)

.

Equation Transcription:

th

 ≔

       

                   

                              )

th

{

                 

                 }

th

Text Transcription:

Nth

f(x)sim a_0/2+ sum over n=1 ^infty {a_n cos nx +b_n sin nx}

f

F_N(x)

N

F_N(x)=alpha_0/2+ sum over n=1 ^N {a_n cos nx+beta_n sin nx}

E colon equals integral over -pi ^pi f(x)-F_N(x)]^2 dx

E=integral over -pi ^pi f^2(x)dx-2 integral over -pi ^pi f(x)F_N(x)dx + integral over -pi ^pi F_N^2 (x)dx

{1,cos,x,sin x,cos 2x,...}

Integralover-pi^piF_N^2(x)dx=pi(alpha_0^2/2 a_1 ^2+cdot+alpha_ N^2+beta_1 ^2+cdot+beta_N ^2)

Integralover-pi^pif(x)F_N(x)dx=pi(alpha_0a_0/2+a_1alpha_1+cdot+a_Na_N+beta_1 b_1+cdot+beta_N b_N)

E*

f

Nth

alpha_n=a_n

beta_n=b_n

E*=integral over -pi ^pi f^2(x)dx-pi (a_0 ^2/2+a_1 ^2+...+a_N ^2+b_1 ^2+...+b_N ^2)

E-E* geq 0

E geq E*

E-E*=pi{(a_0-a_0)^2/2+((alpha_1-a_1)^2+cdot(alpha_N-a_N)^2+(beta_1-b_1)^2+cdot(beta_N-b_N)^2}

Nth

E geq E*