Question

In the space \(C\left( { - 1,1} \right)\),we define the inner product \(\left\langle {{\bf{f,g}}} \right\rangle {\bf{ = }}\int\limits_{{\bf{ - 1}}}^{\bf{1}} {\frac{{\bf{2}}}{{\bf{\pi }}}} \sqrt {{\bf{1 - }}{{\bf{t}}^{\bf{2}}}} {\bf{f}}\left( {\bf{t}} \right){\bf{g}}\left( {\bf{t}} \right){\bf{dt = }}\frac{{\bf{2}}}{{\bf{\pi }}}\int\limits_{{\bf{ - 1}}}^{\bf{1}} {\sqrt {{\bf{1 - }}{{\bf{t}}^{\bf{2}}}} } {\bf{f}}\left( {\bf{t}} \right){\bf{g}}\left( {\bf{t}} \right){\bf{dt}}\). See Exercise 33; here we let \(w\left( t \right) = \frac{2}{\pi }\sqrt {1 - {t^2}} \). (This function \(w\left( t \right)\)is called a Wigner semicircle distribution, after the Hungarian physicist and mathematician E.P. Wigner \(\left( {{\bf{1902 - 1995}}} \right)\), who won the \(1963\)Nobel Prize in Physics.) Since this is not a course in calculus, here are some inner products that will turn out to be useful:\(\left\langle {1,{t^2}} \right\rangle = \frac{1}{4},\left\langle {t,{t^3}} \right\rangle = \frac{1}{8}, and \left\langle {{t^3},{t^3}} \right\rangle = \frac{5}{{64}}\).

(a) Find \(\int\limits_{{\bf{ - 1}}}^{\bf{1}} {{\bf{w}}\left( {\bf{t}} \right)} {\bf{dt}}\). Sketch a rough graph of the weight function \(w\left( t \right)\).

(b) Find the norm of the constant function \(f\left( t \right) = 1\).

(c) Find \(\left\langle {{{\bf{t}}^{\bf{2}}}{\bf{,}}{{\bf{t}}^{\bf{3}}}} \right\rangle \); explain More generally, find \(\left\langle {{t^n},{t^m}} \right\rangle \)for positive integers n and m whose sum is odd.

(d) Find \(\left\langle {t,t} \right\rangle \)and \(\left\langle {{t^2},{t^2}} \right\rangle \).Also, find the norm of the functions \(t\)and \({t^2}\).

(e) Applying the Gram-Schmidt process to the standard basis \({\bf{1,t,}}{{\bf{t}}^{\bf{2}}}{\bf{,}}{{\bf{t}}^{\bf{3}}}\)of \({{\bf{P}}_{\bf{3}}}\),construct an orthonormal basis \({{\bf{g}}_{\bf{0}}}\left( {\bf{t}} \right){\bf{,}}...{\bf{,}}{{\bf{g}}_{\bf{3}}}\left( {\bf{t}} \right)\)of \({{\bf{P}}_{\bf{3}}}\)for the given inner product. (The polynomials \({{\bf{g}}_{\bf{0}}}\left( {\bf{t}} \right){\bf{,}}...{\bf{,}}{{\bf{g}}_{\bf{3}}}\left( {\bf{t}} \right)\)are the first few Chebyshev polynomials of the second kind, named after the Russian mathematician Pafnuty Chebyshev (1821 - 1894). They have a wide range of applications in math, physics, and engineering.)

(f)Find the polynomial \({\bf{g}}\left( {\bf{t}} \right)\)in \({{\bf{P}}_{\bf{3}}}\)that best approximates the function \({\bf{f}}\left( {\bf{t}} \right){\bf{ = }}{{\bf{t}}^{\bf{4}}}\)on the interval \(\left( {{\bf{ - 1,1}}} \right)\),for the inner product introduced in this exercise.

Short answer

(a) The value of \(\int\limits_{ - 1}^1 {w\left( t \right)dt} \) is 1, and the graph of the weight function \(w\left( t \right)\)is

(b) The norm of the constant function \(f\left( t \right) = 1\)is 1

(c) The value of \(\left\langle {{t^2},{t^3}} \right\rangle \)is 0 and the value of \(\left\langle {{t^n},{t^m}} \right\rangle \)is 0 for positive integers n and m whose sum is odd.

(d) The value of \(\left\langle {t,t} \right\rangle is \frac{1}{4}\)and the value of \(\left\langle {{t^2},{t^2}} \right\rangle is \frac{1}{8}\)and the norm of the functions \(t and {t^2}\)is \(\frac{1}{2} and \sqrt {\frac{1}{8}} \).

(e) An orthonormal basis \({g_0}\left( t \right),...,{g_3}\left( t \right)\)of \({P_3}\)for the given inner product is \(1,2t,\sqrt 8 {t^2},\frac{8}{{\sqrt 5 }}{t^3}\).

(f) The polynomial \(g\left( t \right)\)in \({P_3}\)that best approximates the function \(f\left( t \right) = {t^4}\)on the interval \(\left( { - 1,1} \right)\)for the inner product introduced in this exercise is \(\frac{2}{\pi }\left( {\frac{3}{4} + \frac{{15}}{{2\sqrt 3 }}\pi {t^3}} \right)\).

Step-by-step solution

Find the value of \(\int\limits_{{\bf{ - 1}}}^{\bf{1}} {{\bf{w}}\left( {\bf{t}} \right){\bf{dt}}} \)and the graph of the weight function \(w\left( t \right)\)

Consider

\(\int\limits_{ - 1}^1 {w\left( t \right)dt} \)

Substitute the value of \(w\left( t \right)\)as follows:

\(\int\limits_{ - 1}^1 {\frac{2}{\pi }\sqrt {1 - {t^2}} } dt\)

Simplify it as follows:

\(\begin{aligned}{}\int\limits_{ - 1}^1 {\frac{2}{\pi }\sqrt {1 - {t^2}} dt} &= \frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } dt\\ &= \frac{2}{\pi } \times 2\int\limits_0^1 {\sqrt {1 - {t^2}} } dt\\ &= \frac{4}{\pi } \times I - - - - \left( 1 \right), where I is \int\limits_0^1 {\sqrt {1 - {t^2}} } dt\end{aligned}\)

Consider,

\(\begin{aligned}{}I &= \int\limits_0^1 {\sqrt {1 - {t^2}} } dt\\Let t &= sin\theta \\dt &= cos\theta d\theta \\I &= \int\limits_0^1 {\sqrt {1 - si{n^2}\theta } } cos\theta d\theta \end{aligned}\)

Further simplify as follows:

\(\begin{aligned}{}I &= \int\limits_0^1 {\sqrt {co{s^2}\theta } } cos\theta d\theta \\ &= \int\limits_0^1 {cos\theta cos\theta d\theta } \\ &= \int\limits_0^1 {co{s^2}\theta d\theta } \\ = \int\limits_0^1 {\left( {\frac{{1 + cos2\theta }}{2}} \right)d\theta } \end{aligned}\)

Further simplify as follows:

\(\begin{aligned}{}I &= \frac{1}{2}\int\limits_0^1 {\left( {1 + cos2\theta } \right)d\theta } \\ &= \frac{1}{2}\int\limits_0^1 {d\theta + \frac{1}{2}\int\limits_0^1 {cos2\theta d\theta } } \\ &= \frac{1}{2}\left( \theta \right)_0^1 + \frac{1}{2}\left( {\frac{{sin2\theta }}{2}} \right)_0^1\\ &= \frac{1}{2}\left( \theta \right)_0^1 + \frac{1}{4}\left( {2sin\theta cos\theta } \right)_0^1\end{aligned}\)

Substitute the value of \(\theta \)we get,

\(I = \frac{1}{2}\left( {si{n^{ - 1}}t} \right) + \frac{1}{4}\left( {2t\sqrt {1 - {t^2}} } \right)_0^1\)

Simplify further as follows:

\(\begin{aligned}{}I &= \frac{1}{2}\left( {si{n^{ - 1}}1 - si{n^1}0} \right) + \frac{1}{4}\left( {2 \times 1\sqrt {1 - {1^2}} - 2 \times 0\sqrt {1 - {0^2}} } \right)\\ &= \frac{1}{2}\left( {\frac{\pi }{2} - 0} \right) + \frac{1}{4}\left( {0 - 0} \right)\\ &= \frac{1}{2} \times \frac{\pi }{2}\\ &= \frac{\pi }{4}\end{aligned}\)

Now, substitute the value of \(I\)in equation \(\left( 1 \right)\) we get,

\(\int\limits_{ - 1}^1 {\frac{2}{\pi }\sqrt {1 - {t^2}} dt = \frac{4}{\pi } \times \frac{\pi }{4}} = 1\)

Thus, the value of \(\int\limits_{ - 1}^1 {w\left( t \right)dt is 1} \).

Now, the graph of the weight function \(w\left( t \right)\)is

Thus, value of \(\int\limits_{ - 1}^1 {w\left( t \right)} \)is 1.

       Step2: Find the value of the norm of the constant function \({\bf{f}}\left( {\bf{t}} \right){\bf{ = 1}}\)

Consider,

\(\begin{aligned}{}\left\| f \right\| &= \sqrt {\left\langle {f\left( t \right),f\left( t \right)} \right\rangle } \\ &= \sqrt {\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } f\left( t \right)f\left( t \right)dt} \end{aligned}\)

Substitute \(f\left( t \right) = 1\)we get,

\(\begin{aligned}{}\left\| f \right\| &= \sqrt {\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } 1 \times 1dt} \\ &= \sqrt {\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } dt} \end{aligned}\)

Now, substitute the calculated value of \(\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } dt\)in part (a) we get,

\(\left\| f \right\| = \sqrt 1 = 1\)

Find the value of \(\left\langle {{t^2},{t^3}} \right\rangle \)and the value of \(\left\langle {{t^n},{t^m}} \right\rangle \)

Consider,

\(\left\langle {{t^2},{t^3}} \right\rangle \)

Use the given value of inner product we get,

\(\begin{aligned}{}\left\langle {{t^2},{t^3}} \right\rangle &= \frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } \times {t^2} \times {t^3}dt\\ &= \frac{2}{\pi }\int\limits_{ - 1}^1 {{t^5}} \sqrt {1 - {t^2}} dt\end{aligned}\)

As the above function is odd

Therefore, its value is zeros

Thus,

\(\begin{aligned}{}\left\langle {{t^2},{t^3}} \right\rangle &= \frac{2}{\pi } \times 0\\ &= 0\end{aligned}\)

Thus, the value of \(\left\langle {{t^2},{t^3}} \right\rangle is 0\).

Now,

Consider,

\(\left\langle {{t^n},{t^m}} \right\rangle \)

Use the given value of inner product we get,

\(\begin{aligned}{}\left\langle {{t^n},{t^m}} \right\rangle &= \frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } \times {t^n} \times {t^m}dt\\ &= \frac{2}{\pi }\int\limits_{ - 1}^1 {{t^{n + m}}\sqrt {1 - {t^2}} } dt\end{aligned}\)

As given,

The sum of \(n{\rm{ and }}m\)is odd so, the above integral is odd

Thus,

\(\begin{aligned}{}\left\langle {{t^n},{t^m}} \right\rangle &= \frac{2}{\pi } \times 0\\ &= 0\end{aligned}\)

Thus, the value of \(\left\langle {{t^n},{t^m}} \right\rangle is 0\).

Find the value of \(\left\langle {t,t} \right\rangle \)and \(\left\langle {{t^2},{t^2}} \right\rangle \)

Consider,

\(\left\langle {t,t} \right\rangle \)

Use the given value of inner product we get,

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } \times t \times tdt\\ &= \frac{2}{\pi }\int\limits_{ - 1}^1 {{t^2}} \sqrt {1 - {t^2}} dt\\ &= \frac{2}{\pi } \times 2\int\limits_0^1 {{t^2}} \sqrt {1 - {t^2}} dt\\ &= \frac{4}{\pi }\int\limits_0^1 {{t^2}\sqrt {1 - {t^2}} } dt\end{aligned}\)

Further, simplify as follows:

\(\begin{aligned}{}put t &= sin\theta , when t \to 0,\theta \to 0 and t \to 1,\theta \to \frac{\pi }{2}\\dt &= cos\theta d\theta \\\left\langle {t,t} \right\rangle &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^2}\theta \sqrt {1 - si{n^2}\theta } } cos\theta d\theta \\ &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^2}\theta \sqrt {co{s^2}\theta } } cos\theta d\theta \end{aligned}\)

Further, simplify as follows:

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^2}\theta co{s^2}\theta d\theta } \\ &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{1 - cos2\theta }}{2}} \right)\left( {\frac{{1 + cos2\theta }}{2}} \right)d\theta } \\ &= \frac{4}{\pi } \times \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\left( {{1^2} - co{s^2}2\theta } \right)d\theta } \\ &= \frac{1}{\pi }\int\limits_0^{\frac{\pi }{2}} {d\theta - \frac{1}{\pi }\int\limits_0^{\frac{\pi }{2}} {co{s^2}2\theta d\theta } } \end{aligned}\)

Again, simplify as follows:

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \frac{1}{\pi }\int\limits_0^{\frac{\pi }{2}} {d\theta } - \frac{1}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{1 + cos4\theta }}{2}} \right)d\theta } \\ &= \frac{1}{\pi }\left( \theta \right)_0^{\frac{\pi }{2}} - \frac{1}{{2\pi }}\left( {\theta + \frac{{sin4\theta }}{4}} \right)_0^{\frac{\pi }{2}}\\ &= \frac{1}{\pi }\left( {\frac{\pi }{2} - 0} \right) - \frac{1}{{2\pi }}\left( {\frac{\pi }{2} + \frac{{sin4 \times \frac{\pi }{2}}}{4} - 0 - \frac{{sin4 \times 0}}{4}} \right)\\ &= \frac{1}{2} - \frac{1}{{2\pi }}\left( {\frac{\pi }{2} + 0 - 0 - 0} \right)\end{aligned}\)

On further simplification, we get

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \frac{1}{2} - \frac{1}{{2\pi }} \times \frac{\pi }{2}\\ &= \frac{1}{2} - \frac{1}{4}\\ &= \frac{{2 - 1}}{4}\\ &= \frac{1}{4}\end{aligned}\)

Thus, the value of \(\left\langle {t,t} \right\rangle \)is \(\frac{1}{4}\).

Now, consider

\(\left\langle {{t^2},{t^2}} \right\rangle \)

Use the given value of inner product we get,

\(\begin{aligned}{}\left\langle {{t^2},{t^2}} \right\rangle &= \frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } {t^2}{t^2}dt\\ &= \frac{2}{\pi } \times 2\int\limits_0^1 {{t^4}} \sqrt {1 - {t^2}} dt\\put t &= sin\theta , when t \to 0,\theta \to 0 and t \to 1,\theta \to \frac{\pi }{2}\\dt &= cos\theta d\theta \end{aligned}\)

\(\begin{aligned}{}\left\langle {{t^2},{t^2}} \right\rangle &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^4}\theta \sqrt {1 - si{n^2}\theta } } cos\theta d\theta \\ &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^4}\theta \sqrt {co{s^2}\theta } } cos\theta d\theta \\ &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^4}\theta co{s^2}\theta d\theta } - - - - - - \left( 1 \right)\end{aligned}\)

Apply the identity,\(\int\limits_0^a {F\left( x \right)dx = \int\limits_0^a {F\left( {a - x} \right)} dx} \)we get,

\(\begin{aligned}{}\left\langle {{t^2},{t^2}} \right\rangle &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^4}\left( {\frac{\pi }{2} - \theta } \right)co{s^2}\left( {\frac{\pi }{2} - \theta } \right)d\theta } \\ &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {co{s^4}\theta si{n^2}\theta d\theta - - - - - - - - - \left( 2 \right)} \end{aligned}\)

Now, add equation (1) and equation (2) we get,

\(\begin{aligned}{}2\left\langle {{t^2},{t^2}} \right\rangle &= \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {si{n^4}\theta co{s^2}\theta + co{s^4}\theta si{n^2}\theta } \right)d\theta } \\\left\langle {{t^2},{t^2}} \right\rangle &= \frac{4}{{2\pi }}\int\limits_0^{\frac{\pi }{2}} {si{n^2}\theta co{s^2}\theta \left( {si{n^2}\theta + co{s^2}\theta } \right)d\theta } \\ &= \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {si{n^2}\theta co{s^2}\theta d\theta } \\ &= \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {{{\left( {sin\theta cos\theta } \right)}^2}d\theta } \end{aligned}\)

Simplify further as follows:

\(\begin{aligned}{}\left\langle {{t^2},{t^2}} \right\rangle &= \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {{{\left( {\frac{{2sin\theta cos\theta }}{2}} \right)}^2}d\theta } \\ &= \frac{2}{\pi } \times \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {{{\left( {sin2\theta } \right)}^2}d\theta } \\ &= \frac{1}{{2\pi }}\int\limits_0^{\frac{\pi }{2}} {si{n^2}2\theta d\theta } \\ &= \frac{1}{{2\pi }}\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{1 - cos4\theta }}{2}} \right)d\theta } \end{aligned}\)

Again, simplify as follows:

\(\begin{aligned}{}\left\langle {{t^2},{t^2}} \right\rangle &= \frac{1}{{2\pi }} \times \frac{1}{2}\left( {\theta - \frac{{sin4\theta }}{4}} \right)_0^{\frac{\pi }{2}}\\ &= \frac{1}{{4\pi }}\left( {\frac{\pi }{2} - \frac{{sin4 \times \frac{\pi }{2}}}{4} - 0 - sin0} \right)\\ &= \frac{1}{{4\pi }}\left( {\frac{\pi }{2} - 0 - 0 - 0} \right)\\ = \frac{1}{8}\end{aligned}\)

Thus, the value of \(\left\langle {t,t} \right\rangle \)is \(\frac{1}{4}\)and the value of \(\left\langle {{t^2},{t^2}} \right\rangle \)is \(\frac{1}{8}\).

To find the value of the norm of the functions \(t and {t^2}\).

Consider,

\(\begin{aligned}{}\left\| t \right\| &= \sqrt {\left\langle {t,t} \right\rangle } \\ &= \sqrt {\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} } \times t \times tdt} \\ &= \sqrt {\frac{1}{4}} \\ = \frac{1}{2}\end{aligned}\)

And

\(\begin{aligned}{}\left\| {{t^2}} \right\| &= \sqrt {\left\langle {{t^2},{t^2}} \right\rangle } \\ &= \sqrt {\frac{2}{\pi }\int\limits_{ - 1}^1 {\sqrt {1 - {t^2}} {t^2}{t^2}dt} } \\ &= \sqrt {\frac{1}{8}} \end{aligned}\)

Thus, the value of \(\left\| t \right\|\)is \(\frac{1}{2}\)and \(\left\| {{t^2}} \right\|\)is \(\sqrt {\frac{1}{8}} \).

To find the orthonormal basis \({{\bf{g}}_{\bf{0}}}\left( {\bf{t}} \right){\bf{,}}...{\bf{,}}{{\bf{g}}_{\bf{3}}}\left( {\bf{t}} \right)\)of \({{\bf{P}}_{\bf{3}}}\).

As given, the standard basis of \({P_3}\)are \({g_0}\left( t \right),...,{g_3}\left( t \right)\). These basis are orthogonal so to find the orthonormal basis we need to divide every basis element by its norm.

Therefore,

\(\begin{aligned}{}{g_0}\left( t \right) &= \frac{1}{{\left\| 1 \right\|}}\\ &= \frac{1}{1}\\ &= 1\end{aligned}\)

\(\begin{aligned}{}{g_1}\left( t \right) &= \frac{t}{{\left\| t \right\|}}\\ &= \frac{t}{{\frac{1}{2}}}\\ &= 2t\end{aligned}\)

\(\begin{aligned}{}{g_2}\left( t \right) &= \frac{{{t^2}}}{{\left\| {{t^2}} \right\|}}\\ &= \frac{{{t^2}}}{{\sqrt {\frac{1}{8}} }}\\ &= \sqrt 8 {t^2}\end{aligned}\)

\(\begin{aligned}{}{g_3}\left( t \right) &= \frac{{{t^3}}}{{\left\| {{t^3}} \right\|}}\\ &= \frac{{{t^3}}}{{\frac{{\sqrt 5 }}{8}}}\\ &= \frac{8}{{\sqrt 5 }}{t^3}\end{aligned}\)

Thus, the orthogonal basis is \(1,2t,\sqrt 8 {t^2},\frac{8}{{\sqrt 5 }}{t^3}\).

To find the polynomial \({\bf{g}}\left( {\bf{t}} \right)\)in \({{\bf{P}}_{\bf{3}}}\).

The polynomial \(g\left( t \right)\)in \({P_3}\)that best approximates the function \(f\left( t \right) = {t^4}\)on the interval \(\left( { - 1,1} \right)\)is

\(g\left( t \right) = 1\left\langle {1,f} \right\rangle + 2t\left\langle {t,f} \right\rangle + \sqrt 8 {t^2}\left\langle {{t^2},f} \right\rangle + \frac{8}{{\sqrt 5 }}{t^3}\left\langle {{t^3},f} \right\rangle \)

Use the given formula of inner product we get,

\(\begin{aligned}{}g\left( t \right) &= \frac{2}{\pi }\left( {\int\limits_{ - 1}^1 {{t^4}\sqrt {1 - {t^2}} dt + 2t\int\limits_{ - 1}^1 {{t^5}\sqrt {1 - {t^2}} dt + \sqrt 8 {t^2}\int\limits_{ - 1}^1 {{t^6}\sqrt {1 - {t^2}} dt + \frac{8}{{\sqrt 5 }}{t^3}\int\limits_{ - 1}^1 {{t^7}\sqrt {1 - {t^2}} dt} } } } } \right)\\ &= \frac{2}{\pi }\left( {\frac{3}{4} + \frac{{15}}{{2\sqrt 3 }}\pi {t^3}} \right)\end{aligned}\)

Thus, the value of the polynomial \(g\left( t \right)\)is \(\frac{2}{\pi }\left( {\frac{3}{4} + \frac{{15}}{{2\sqrt 3 }}\pi {t^3}} \right)\).