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Random Talk on Random Thoughts

Found an Upper Bound for the Modulus of Legendre Polynomials

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Suppose that $w: [a,b] \to \C$ is piecewise continuous, then we have

\begin{equation} \abslr{\int_{a}^{b} w(t) \ud t} \le \int_{a}^{b} \abs{w(t)} \ud t. \label{lemma} \end{equation}

The trick is to treat the integral on the L.H.S. of \eqref{lemma} as a number.

\begin{align} \int_{a}^{b} w(t) \ud t &= r_0 e^{i\theta_0} \text{ for some } r \in \R \text{ and } \theta_0 \in [-\pi,\pi) \label{trick1}\\ r_0 &= \int_{a}^{b} e^{-i\theta_0} w(t) \ud t \label{trick2} \end{align}

Then from \eqref{lemma} and \eqref{trick1}, it suffices to show that

\begin{equation} r_0 = \abslr{r_0 e^{i\theta_0}} \le \int_{a}^{b} \abs{w(t)} \ud t. \label{new_goal1} \end{equation}

Since \eqref{trick2} consists of $e^{-i\theta_0}$, which is absent on the R.H.S. of \eqref{lemma} and \eqref{new_goal1}, we add it back. Thus, we need to show

\begin{equation} r_0 \le \int_{a}^{b} \abslr{e^{-i\theta_0} w(t)} \ud t. \label{new_goal2} \end{equation}

Looking at \eqref{trick2} again, we’re almost there! Using elementary properties of complex numbers and definite integrals for real-valued functions and the fact/definition that

\begin{equation} \Re\left(\int_{a}^{b} w(t) \ud t\right) = \int_{a}^{b} \Re[w(t)] \ud t, \label{Re-int} \end{equation}

we can write

\begin{equation} \begin{aligned} r_0 &= \Re\left(\int_{a}^{b} e^{-i\theta_0} w(t) \ud t\right) \\ &= \int_{a}^{b} \Re[e^{-i\theta_0} w(t)] \ud t\\ &\le \int_{a}^{b} \abslr{e^{-i\theta_0} w(t)} \ud t. \end{aligned} \label{fin} \end{equation}

Therefore, \eqref{new_goal2} is satisfied, so as \eqref{lemma}.

An upper bound for Legendre polynomials

I have never evaluated

\begin{equation} P_n(x) = \frac{1}{\pi} \int_0^\pi (x + i \sqrt{1 - x^2} \cos{\theta})^n \ud\theta, \label{pnx} \end{equation}

where $-1 \le x \le 1$ and $n = 0,1,2,\dots$

Applying \eqref{lemma}, one gets

\begin{equation} \begin{aligned} P_n(x) &= \frac{1}{\pi} \int_0^\pi (x + i \sqrt{1 - x^2} \cos{\theta})^n \ud\theta\\ &\le \frac{1}{\pi} \int_0^\pi \ud\theta\\ &= 1 \end{aligned} \label{crux} \end{equation}


\begin{equation} \begin{aligned} &\quad\; \abs{x + i \sqrt{1 - x^2} \cos{\theta}}\\ &= \sqrt{x^2 + (1 - x^2) \cos^2{\theta}}\\ &\le \sqrt{x^2 + (1 - x^2)}\\ &= 1. \end{aligned} \label{norm-ub} \end{equation}