Examples C

Section 4.6 Examples C

Legendre’s Function.

Example 4.6.1.

Express \(f(x)=4x^{3}+6x^{2}+7x+2 \) in terms of Legendre’s Polynomials.

Solution.

Let

\begin{equation} 4x^{3}+6x^{2}+7x+2\equiv aP_{3}(x)+bP_{2}(x)+cP_{1}(x)+dP_{0}(x) \tag{4.6.1} \end{equation}

\begin{equation*} =a\left[\frac{5}{2}x^{3}-\frac{3x}{2}\right]+b\left[\frac{3}{2}x^{2}-\frac{1}{2}\right]+c(x)+d(1) \end{equation*}

\begin{equation*} =\frac{5ax^{3}}{2}-\frac{3ax}{2}+\frac{3bx^{2}}{2}-\frac{b}{2}+cx+d \end{equation*}

\begin{equation*} =\frac{5ax^{3}}{2}+\frac{3bx^{2}}{2}+\left(-\frac{3a}{2}+c\right)x-\frac{b}{2}+d \end{equation*}

Equating the coefficients of like powers of \(x\text{,}\) we have

\begin{equation*} \frac{5a}{2}=4 \quad \Rightarrow a =\frac{8}{5}; \end{equation*}

\begin{equation*} \frac{3b}{2}=6 \quad \Rightarrow b =4 \end{equation*}

\begin{equation*} \frac{-3a}{2}+c=7 \quad \Rightarrow -\frac{3}{2}\cdot\frac{8}{5}+c =7 \quad \Rightarrow c=\frac{47}{5} \end{equation*}

\begin{equation*} \frac{-b}{2}+d=2 \quad \Rightarrow -\frac{4}{2}+d=2 \quad \Rightarrow d=4. \end{equation*}

Putting these values in eqn. (4.6.1), we get -

\begin{equation*} \frac{8}{5}P_{3}(x)+4P_{2}(x)+\frac{47}{5}P_{1}(x)+4P_{0}(x) \end{equation*}

Example 4.6.2.

Show that \(P_{n}(-x)=(-1)^{n}P_{n}(x)\) and hence prove that

\begin{equation*} P_{2m}(-x)=P_{2m}(x) \end{equation*}
\begin{equation*} P_{2m+1}(-x)=-P_{2m+1}(x) \end{equation*}

Solution.

We know that

\begin{equation*} P_{n}(x)=\sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}x^{n-2r}; \end{equation*}

where,

\begin{equation*} N= \begin{cases} \frac{n}{2},\quad \text{for n even}\\ =\frac{n-1}{2},\quad \text{for n odd.} \end{cases} \end{equation*}

Thus

\begin{equation*} P_{n}(-x)=\sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}(-1)^{n-2r}x^{n-2r} \end{equation*}

\begin{equation*} =(-1)^{n}\sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}x^{n-2r} =(-1)^{n} P_{n}(x) \end{equation*}

\begin{equation*} \therefore \quad P_{n}(-x)=(-1)^{n}P_{n}(x) \end{equation*}

Now,

if \(n=2m\text{,}\) we have
\begin{equation*} P_{2m}(-x)=(-1)^{2m}P_{2m}(x) \end{equation*}
or,
\begin{equation*} P_{2m}(-x)= P_{2m}(x) \end{equation*}
if \(n=2m+1\text{,}\) we have
\begin{equation*} P_{2m+1}(-x)=(-1)^{2m+1}P_{2m+1}(x) \end{equation*}
or,
\begin{equation*} P_{2m+1}(-x)= - P_{2m+1}(x) \end{equation*}

Example 4.6.3.

Assuming that a polynomial \(f(x) \) of degree \(n\) can be written as

\begin{equation*} f(x)=\sum\limits_{n=0}^{\infty}a_{n}P_{n}(x), \end{equation*}

show that

\begin{equation*} a_{n}\frac{2}{2n+1}= \int\limits_{-1}^{+1}f(x)P_{n}(x)\,dx. \end{equation*}

Solution.

\begin{equation*} f(x) =\sum\limits_{n=0}^{\infty}a_{n}P_{n}(x)=a_{0}P_{0}(x)+a_{1}P_{1}(x)+a_{2}P_{2}(x)+\cdots+a_{n}P_{n}(x)+\cdots \end{equation*}

Multiplying both sides by \(P_{n}(x)\text{,}\) we get -

\begin{equation*} f(x)P_{n}(x)= a_{0}P_{0}(x)P_{n}(x)+a_{1}P_{1}(x)P_{n}(x) \end{equation*}

\begin{equation*} +a_{2}P_{2}(x)P_{n}(x)+\cdots+a_{n}P^{2}_{n}(x)+\cdots \end{equation*}

or,

\begin{equation*} \int\limits_{-1}^{+1}f(x)P_{n}(x)\,dx \end{equation*}

\begin{equation*} =\int\limits_{-1}^{+1}\left[a_{0}P_{0}(x)P_{n}(x)+a_{1}P_{1}(x)P_{n}(x)\right. \end{equation*}

\begin{equation*} \left. +a_{2}P_{2}(x)P_{n}(x)+\cdots+a_{n}P^{2}_{n}(x)+\cdots\right]\,dx \end{equation*}

\begin{equation*} =[0+0+\cdots+a_{n}\frac{2}{2n+1}+\cdots] \end{equation*}

\begin{equation*} \therefore \quad a_{n}\frac{2}{2n+1} = \int\limits_{-1}^{+1}f(x)P_{n}(x)\,dx \end{equation*}

Example 4.6.4.

From the values of Legendre’s polynomials, prove that

\begin{equation*} x^{3}=\frac{1}{5}\left[2P_{3}(x) +3P_{1}(x)\right]. \end{equation*}

Solution.

Let

\begin{equation} x^{3} = aP_{3}(x)+bP_{2}(x)+cP_{1}(x)+dP_{0}(x)\tag{4.6.2} \end{equation}

\begin{equation*} =a\frac{1}{2}(5x^{3}-3x)+b\frac{1}{2}(3x^{2}-1)+c(x)+d(1) \end{equation*}

\begin{equation*} =\frac{5ax^{3}}{2}-\frac{3ax}{2}+\frac{3bx^{2}}{2}-\frac{b}{2}+d \end{equation*}

\begin{equation*} =\frac{5ax^{3}}{2}+\frac{3bx^{2}}{2}+\left(-\frac{3a}{2}+c\right)x-\frac{b}{2}+d \end{equation*}

equating the coefficients of like powers of \(x\text{,}\) we get -

\begin{equation*} \frac{5a}{2}=1 \quad \Rightarrow a = \frac{2}{5}; \end{equation*}

\begin{equation*} \frac{3b}{2}=0 \quad \Rightarrow b = 0; \end{equation*}

\begin{equation*} -\frac{3a}{2}+c=0 \quad \Rightarrow c = \frac{3}{2}\cdot\frac{2}{5}=\frac{3}{5}; \end{equation*}

and

\begin{equation*} -\frac{b}{2}+d=0 \quad \Rightarrow d =0. \end{equation*}

Putting these values in eqn. (4.6.2), we get -

\begin{equation*} x^{3} = \frac{2}{5}P_{3}(x)+\frac{3}{5}P_{1}(x) =\frac{1}{5}\left[2P_{3}(x)+3P_{1}(x)\right]. \end{equation*}

Example 4.6.5.

Using Rodrigue’s formula, prove that \(\int\limits_{-1}^{+1}x^{m}P_{n}(x)\,dx =0\) if \(m \lt n\text{.}\)

Solution.

\begin{equation*} \int\limits_{-1}^{+1}x^{m}P_{n}(x)\,dx =\int\limits_{-1}^{+1}x^{m}\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}\,dx \end{equation*}

\begin{equation*} = \frac{1}{2^{n}n!}\int\limits_{-1}^{+1}x^{m}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}\,dx \end{equation*}

On integrating by parts we get -

\begin{equation*} =\frac{1}{2^{n}n!}\left[\left\{x^{m}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}\right\}_{-1}^{+1}-\int\limits_{-1}^{+1}mx^{m-1}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}\,dx\right] \end{equation*}

\begin{equation*} =0-\frac{m}{2^{n}n!}\int\limits_{-1}^{+1}x^{m-1}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}\,dx \end{equation*}

Integrating we have

\begin{equation*} =(-1)^{2}\frac{m(m-1)}{2^{n}n!}\int\limits_{-1}^{+1}x^{m-2}\frac{d^{n-2}}{dx^{n-2}}(x^{2}-1)^{n}\,dx \end{equation*}

Integrating \(m-2 \) times, we get -

\begin{equation*} =(-1)^{m}\frac{m(m-1)\cdots 2\cdot 1}{2^{n}n!}\int\limits_{-1}^{+1}\frac{d^{n-m}}{dx^{n-m}}(x^{2}-1)^{n}\,dx \end{equation*}

\begin{equation*} =\frac{(-1)^{m}m!}{2^{n}n!}\left[\frac{d^{n-m-1}}{dx^{n-m-1}}(x^{2}-1)^{n}\right]_{-1}^{+1}=0. \end{equation*}

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