Legendre (polynomials)and Chebyshev (polynomials of first kind) united

Putting $x=\sin y$ gives,

$\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^4 y +a\sin^3 y+b\sin^2 y+c\sin y+d)^2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^4 y -a\sin^3 y+b\sin^2 y-c\sin y+d)^2$

Thus, $\displaystyle 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^4 y +a\sin^3 y+b\sin^2 y+c\sin y+d)^2 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^4 y -a\sin^3 y+b\sin^2 y-c\sin y+d)^2$

After a little simplification we have,

$\displaystyle I =2J_8 + (2a^2+4b)J_6+(2b^2+4d+4ac)J_4 + (2c^2+4bd)J_2 + \pi d^2$ , where $\displaystyle J_n = \int_{0}^{\frac{\pi}{2}} \sin^n x dx$

Use Beta to evaluate $J_n's$ and hence deduce that,

$\displaystyle I = \frac{\pi}{128}(40a^2+48b^2+64c^2+128d^2+80b+96ac+96d+128bd+35) = \frac{\pi}{128}f_{a,b,c,d}$ which is left to minimize.

By Completing the square , $\displaystyle f_{a,b,c,d} = 8(a^2 + (2a+3c)^2 - c^2 + b^2 + 5(b+1)^2+16d^2+16bd+12d -\frac{5}{8})\ge 8(1+16d^2-16d+12d-\frac{5}{8})=8g_d$ where minimum occurs if $a=0,2a+3c=0,b+1=0$ and putting the value of these we get the last expression in $'d'$

$\displaystyle g_d = 8(16d^2-4d+\frac{3}{8})=8((4d-\frac{1}{2})^2+\frac{1}{8})\ge 1$ and minimum occurs at $\displaystyle d=\frac{1}{8}$

So, $I$ is minimized when $\displaystyle (a,b,c,d)=(0,-1,0,\frac{1}{8})$ and the answer $a+b+c+d=\boxed{-0.875}$

The Chebyshev polynomials (of the first kind) $T_n(x) = \cos(n \cos^{-1}x)$ are such that $\int_{-1}^1 \frac{T_m(x)T_n(x)}{\sqrt{1-x^2}}\,dx \; = \; \left\{ \begin{array}{ll} 0 & m \neq n \\ \tfrac12\pi & m = n > 0 \\ \pi & m = n = 0 \end{array} \right.$ and standard inner-product space theory tells us that $\int_{-1}^1 \big[x^4 - p(x)\big]^2\,dx \hspace{2cm} p(x) \mbox{ a cubic polynomial}$ is minimized when $p(x) \; = \; \alpha T_3(x) + \beta T_2(x) + \gamma T_1(x) + \delta T_0(x)$ where $\begin{array}{rcl} \alpha & = & \displaystyle\tfrac{2}{\pi}\int_{-1}^1 x^4 T_3(x)\,dx \; = \; 0 \\ \beta & = & \displaystyle\tfrac{2}{\pi}\int_{-1}^1 x^4 T_2(x)\,dx \; = \; \tfrac12 \\ \gamma & = & \displaystyle\tfrac{2}{\pi}\int_{-1}^1 x^4 T_1(x)\,dx \; = \; 0 \\ \delta & = & \displaystyle\tfrac{1}{\pi}\int_{-1}^1 x^4 T_0(x)\,dx \; = \; \tfrac38 \end{array}$ so that $p(x) \; = \; \tfrac38 + \tfrac12(2x^2 - 1) \; = \; x^2 - \tfrac18$ and hence $a_1x^3 + b_1x^2 + c_1x + d_1 \; = \; -p(x) \; = \; \tfrac18 - x^2$ so that $a_1+b_1+c_1+d_1=-\tfrac78=\boxed{-0.875}$ .

Since the (suitably normalized) Chebyshev polynomials are what are obtained by applying the Gram-Schmidt orthogonalisation procedure to the sequence $1,x,x^2,x^3,\ldots$ , with respect to the inner product $\langle f,g \rangle \; = \; \int_{-1}^1 \frac{f(x) g(x)}{\sqrt{1-x^2}}\,dx$ we see that $q(x) \;= \; x^4 + a_1x^3 + b_1x^2 + c_1x + d_1 \; = \; x^4 - \frac{\langle x^4,T_0 \rangle}{\Vert T_0\Vert^2}T_0(x) - \frac{\langle x^4,T_1 \rangle}{\Vert T_1\Vert^2}T_1(x) - \frac{\langle x^4,T_2 \rangle}{\Vert T_2\Vert^2}T_2(x) - \frac{\langle x^4,T_3 \rangle}{\Vert T_3\Vert^2}T_3(x)$
and so $\tfrac{q(x)}{\Vert q \Vert}$ is the (suitably normalized) $\sqrt{\tfrac{2}{\pi}}T_4(x)$ , it follows that $q(x) = \tfrac18T_4(x)$ is the fourth Chebyshev polynomial, scaled so as to become monic.

The solution is got with $\frac{1}{8} \cdot T_4(x) = x^4 - x^2 + 1/8$ where $T_4(x)$ is the polynomial of Chebyshev of $4^{th}$ degree of first kind.

Continuation of the theory of my previous exercise

$\large \boxed{ \text{Legendre polynomials} }$

"Monic Legendre polynomials" are: $t_0 (t) = 1 \\ t_1 (t) = t \\ t_2 (t) = t^2 - 1/3 \\ t_3 (t) = t^3 - (3/5)t \\ t_4(t) = t^4 - (6/7)t^2 + (3/35)...$ Does this remember you something?

Proposition 1.-

Let $\langle \cdot , \cdot \rangle$ be an inner real product in $\mathcal{C}([a, b]) = \{f : [a , b] \to \mathbb{R}; \text{ f is a continuous function}\}$ fulfilling $\langle f, g \cdot h \rangle\ = \langle f \cdot g , h \rangle$ ) for all $f, g , h \in \mathcal{C}([a, b])$ .

Then, the sequence of polynomials defined recursively as follows is orthogonal, (and the polynomials are monic): $p_n (x) = (x - a_n) \cdot p_{n - 1} (x) - b_n \cdot p_{n - 2} (x), \space (n \ge 2)$ with $p_0 (x) = 1, \quad p_1 (x) = x - a_1$ and $a_n = \frac{\langle xp_{n - 1}, p_{n -1} \rangle}{\langle p_{n - 1}, p_{n -1} \rangle} , \\ b_n = \frac{\langle xp_{n - 1}, p_{n -2} \rangle}{\langle p_{n - 2}, p_{n -2} \rangle}.$

Proof.-

By looking at the formulas, it is evident that each polynomial is monic of degree n, and therefore non-zero. Furthemore, the denominators in the formulas are non-zero.

Let's prove by induction that $\langle p_n, p_i \rangle = 0$ for $i = 0 , 1, ... n - 1$ .

a) For $n = 0$ there is nothing to prove.

b) For $n = 1$ , $\langle p_1, p_0 \rangle = \langle (x - a_1)p_0, p_0 \rangle = \langle xp_0, p_0 \rangle - a_1 \langle p_0, p_0 \rangle = 0$ due to the definition of $a_1$ .

Now suppose the validity of our statement for an index $n - 1$ , with $n \ge 2$ , then $\langle p_n, p_{n - 1} \rangle = \langle xp_{n - 1}, p_{n - 1} \rangle - a_n \langle p_{n - 1}, p_{n - 1} \rangle - b_n \langle p_{n - 2}, p_{n - 1} \rangle = \langle xp_{n - 1}, p_{n - 1} \rangle - a_n \langle p_{n - 1}, p_{n - 1} \rangle = 0$ $\langle p_n, p_{n - 2} \rangle = \langle xp_{n - 1}, p_{n - 2} \rangle - a_n \langle p_{n - 1}, p_{n - 1} \rangle - b_n \langle p_{n - 2}, p_{n - 2} \rangle = \langle xp_{n - 1}, p_{n - 2} \rangle - b_n \langle p_{n - 2}, p_{n - 2} \rangle = 0$ Now, we are going to use the hypothesis $\langle f, g \cdot h \rangle\ = \langle f \cdot g , h \rangle$ ) for all $f, g , h \in \mathcal{C}([a, b])$ . For $i = 0, 1, 2, ... , n - 3$ , $\langle p_n, p_i \rangle = \langle xp_{n - 1}, p_i \rangle - a_n \langle p_{n - 1}, p_i \rangle - b_n \langle p_{n - 2}, p_i \rangle = \langle p_{n - 1}, xp_i \rangle =$ $\langle p_{n - 1}, p_{i + 1} + a_{i + 1} p_i + b_{i + 1} p_{i - 1} \rangle = 0 .$ For $i = 0$ we should write $xp_0 = p_1 + a_1 p_0. \space \square$ .

Notes before the following key proposition.-

a) If in $\mathcal{C}([-1, 1]) = \{f : [-1 , 1] \to \mathbb{R}; \text{ f is a continuous function}\}$ we use the inner real product $\displaystyle \langle f , g \rangle = \int_{-1}^{1} f(x) \cdot g(x) \, dx$ then the orthogonal monic polynomials are Legendre polynomials .

$\large \boxed{\text{Chebyshev Polynomials}}$

b) If in $\mathcal{C}([-1, 1]) = \{f : [-1 , 1] \to \mathbb{R}; \text{ f is a continuous function}\}$ we use the inner real product $\displaystyle \langle f , g \rangle = \int_{-1}^{1} \frac{f(x) \cdot g(x)}{\sqrt{1 - x^2}} \, dx$ then Chebyshev polynomials are orthogonal:

Proof.-

Let's make the change $x = \cos \theta \Rightarrow$ , $\displaystyle \langle f, g \rangle = \int_{0}^{\pi} f(\cos \theta) g(\cos \theta) \, d\theta$ (Chebyshev polynomials of first kind are $T_n (x) = \cos (n \cdot \cos^{-1} (x))$ and if $n \neq m$ ), $\langle T_n, T_m \rangle = \int_{0}^{\pi} \cos (n\theta) \cos (m\theta) d\theta = \frac{1}{2}\left(\int_{0}^{\pi} \cos ((n + m)\theta) + \cos ((n - m)\theta) \, d\theta \right) =$ $\frac{1}{2}\left(\frac{\sin ((n+m)\theta)}{n +m} + \frac{\sin ((n - m)\theta)}{n - m} \right)_{0}^{\pi} = 0, \square$

Proposition 2.- ( key proposition )

If $\langle \cdot, \cdot \rangle$ is an inner product in $\mathcal{C}([a,b])$ and $p_0, p_1, ..., p_n, ...$ is a sequence of monic orthogonal polynomials with degree of $p_i = i, \space \forall i = 0, 1, 2, ... , n, ...$ . Then , if $q$ is a monic polynomial of degree $n$ , then $|| q ||_{\langle \cdot, \cdot \rangle} \ge ||p_n||_{\langle \cdot, \cdot \rangle}$ . Furthemore, if other monic polynomial $p$ of degree $n$ , fulfills $|| p ||_{\langle \cdot, \cdot \rangle} \leq ||p_n||_{\langle \cdot, \cdot \rangle}$ , then $p = p_n$

Proof.-

It's clear that $\text{ Span } \{p_0, p_1, ... , p_n\} \subseteq \text {Span } \{1, x , x^2, ... x^n\}$ . We are going to see that $\text{ dim ( Span } \{p_0, p_1, ... , p_n\}) = n + 1$ and this will imply that $\text{ Span } \{p_0, p_1, ... , p_n\} = \text {Span } \{1, x , x^2, ... x^n\}$ . For proving this, we are going to see that these monic polynomials are linearly independent. Let's suppose that $\lambda_0 p_0 + \lambda_1 p_1 + \lambda_2 p_2 + ... + \lambda_n p_n = 0 \Rightarrow$ with $\lambda_i$ constant $\forall i = 0, 1, 2, ... , n$ , and for $i = 0, 1, 2, ... , n$ fixed but arbitrary $\langle \lambda_0 p_0 + \lambda_1 p_1 + \lambda_2 p_2 + ... + \lambda_n p_n, p_i \rangle = 0 = \lambda_i \langle p_i, p_i \rangle \Rightarrow \lambda_i = 0$ . Therefore, $\text{ Span } \{p_0, p_1, ... , p_n\} = \text {Span } \{1, x , x^2, ... x^n\}$ . Hence, $q = \lambda_0 p_0 + \lambda_1 p_1 + \lambda_2 p_2 + ... + \lambda_n p_n, \Rightarrow \langle q , p_i \rangle =$ $= \lambda_i \langle p_i, p_i \rangle \Rightarrow \lambda_i = \frac{\langle q, p_i \rangle}{\langle p_i, p_i \rangle}$ . Thus, $\displaystyle q = \sum_{i = 0}^n \frac{\langle q, p_i \rangle}{\langle p_i, p_i \rangle} \cdot p_i$ and because of $q$ is a monic polynomial $\displaystyle q = p_n + \sum_{i = 0}^{n - 1} \frac{\langle q, p_i \rangle}{\langle p_i, p_i \rangle} \cdot p_i \Rightarrow$ $\displaystyle || q ||_{\langle \cdot, \cdot \rangle}^2 = || p_n ||_{\langle \cdot, \cdot \rangle}^2 + \sum_{i = 0}^{n - 1} \left|\frac{\langle q, p_i \rangle}{\langle p_i, p_i \rangle}\right|^2 \ge || p_n ||_{\langle \cdot, \cdot \rangle}^2 \Rightarrow$ $|| q ||_{\langle \cdot, \cdot \rangle} \ge ||p_n||_{\langle \cdot, \cdot \rangle}.$ Furthemore, if other monic $p$ polynomial of degree $n$ , fulfills $|| p ||_{\langle \cdot, \cdot \rangle} \leq ||p_n||_{\langle \cdot, \cdot \rangle}$ , we have just seen that $|| p_n ||_{\langle \cdot, \cdot \rangle} \leq || p ||_{\langle \cdot, \cdot \rangle}$ and this implies that $|| p_n ||_{\langle \cdot, \cdot \rangle} = || p ||_{\langle \cdot, \cdot \rangle}$ but then due to the previous linear combination of $q$ , and using it for $p$ we conclude $p = p_n \square$ .

Solution of my previous exercise and this exercise

Using proposition 1 and 2 here, Legendre monic polynomial of degree 3, minimize the integral of my previous exercise.

Using proposition 1 and 2 here,again, Chebyshev polynomial of degree 4 divided by 8 ,minimize the integral of this exercise

Proposition 3.- (Not necssary for this exercise and the previous exercise)

If $p$ is a monic polynomial of degree $n$ in $[-1, 1]$ , then $|| p ||_{\infty} \ge \frac{1}{2^{n - 1}} = || \frac{1}{2^{n - 1}} \cdot T_n ||_{\infty}$ in [-1, 1], whith $T_n$ being Chebyshev polynomial of degree n and $|| f ||_{\infty} = \text{ supremum } \{|f(x)|; x \in [-1, 1]\}$ .

To be continued