Exercise 4.5.2

$F(\lambda )={\lambda }^2$, then equation (3) gives the limit of the average eigenvalue of $A^2$.

• 1.

  $A(𝜃)=a_1{e}^{\mathit{i𝜃}}+a_0+a_{-1}{e}^{-\mathit{i𝜃}}$, $A^2(𝜃)=a_1^2{e}^{2\mathit{i𝜃}}+2a_0{a}_1{e}^{\mathit{i𝜃}}+(a_0^2+2a_1{a}_{-1})+2a_0{a}_{-1}{e}^{-\mathit{i𝜃}}+a_{-1}^2{e}^{-2\mathit{i𝜃}}$

On the other hand, for $A^2=\mathit{AA}$, we have the $A_{\mathit{ii}}^2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_k{a}_{\mathit{ik}}{a}_{\mathit{ki}}=a_{i,i-1}{a}_{i-1,i}+a_{i,i}{a}_{i,i}+a_{i,i+1}{a}_{i+1,i}=a_1{a}_{-1}+a_0^2+a_{-1}{a}_1=a_0^2+2a_1{a}_{-1}$

Similarly we have $A_{i,i+1}^2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_k{a}_{\mathit{ik}}{a}_{k,i+1}=a_{i,i-1}{a}_{i-1,i+1}+a_{i,i}{a}_{i,i+1}+a_{i,i+1}{a}_{i+1,i+1}=a_1\times 0+a_0{a}_{-1}+a_{-1}{a}_0=2a_0{a}_{-1}$

$A_{i,i+2}^2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_k{a}_{\mathit{ik}}{a}_{k,i+2}=a_{i,i-1}{a}_{i-1,i+2}+a_{i,i}{a}_{i,i+2}+a_{i,i+1}{a}_{i+1,i+2}=a_1\times 0+a_0\times 0+a_{-1}{a}_{-1}=a_{-1}^2$

So we see that $A^2$ symbol is ${(A(𝜃))}^2$.

• 1.

  ${\int }_0^{2\pi }{A}^2(𝜃)\mathit{d𝜃}={\int \; <!--\; nolimits\; -->}_0^{2\pi }\left(a_1^2{e}^{2\mathit{i𝜃}}+2a_0{a}_1{e}^{\mathit{i𝜃}}+(a_0^2+2a_1{a}_{-1})+2a_0{a}_{-1}{e}^{-\mathit{i𝜃}}+a_{-1}^2{e}^{-2\mathit{i𝜃}}\right)\mathit{d𝜃}={\int \; <!--\; nolimits\; -->}_0^{2\pi }\left(e^{2\mathit{i𝜃}}-4e^{\mathit{i𝜃}}+6-4e^{-\mathit{i𝜃}}+e^{-2\mathit{i𝜃}}\right)\mathit{d𝜃}={\int \; <!--\; nolimits\; -->}_0^{2\pi }(6+2\cos <!--\; FUNCTION\; APPLICATION\; -->(2𝜃)+8\cos <!--\; FUNCTION\; APPLICATION\; -->(𝜃)\mathit{d𝜃}=12\pi$