First Three Legendre Polynomials
\[P_0(x)=1.\]
\[P_1(x)=x.\]
\[P_2(x)=\frac{1}{2}(3x^2-1).\]
Legendre Polynomials
Expansion Using the First Three Legendre Polynomials
Legendre polynomials are developed through three IOE past-paper questions. Use the standard forms of P0, P1 and P2, replace the powers of x and collect like terms.
UnitSeries Solutions and Special Functions
TopicLegendre Polynomial Expansion
Question Type2 Mark Problems
Theory Required for These Problems
Expressing \(x^2\)
From the formula for \(P_2(x)\),
\[2P_2(x)=3x^2-P_0(x).\]
Therefore,
\[x^2=\frac{1}{3}P_0(x)+\frac{2}{3}P_2(x).\]
Comparison Method
Write the polynomial as
\[f(x)=A_0P_0(x)+A_1P_1(x)+A_2P_2(x).\]
Replace \(1\), \(x\) and \(x^2\), then collect the coefficients of \(P_0\), \(P_1\) and \(P_2\).
Quick Check
Substitute the standard forms of \(P_0\), \(P_1\) and \(P_2\) into the final expansion. It must simplify back to the given polynomial.
Note: A quadratic polynomial requires only \(P_0\), \(P_1\) and \(P_2\).
Method used: Replace \(1\), \(x\) and \(x^2\) by their Legendre polynomial forms, collect like terms, and verify the expansion.
Continue through the IOE Engineering Mathematics II Digital Workbook. This legendre polynomials page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
60
2 Marks
Expansion of (1+x-x^2)
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 6 • 2 Marks
Express
\[f(x)=1+x-x^2\]
in terms of Legendre polynomials.
Solution
Standard forms of \(P_0,P_1,P_2\)We know that
\[P_0(x)=1,\qquad P_1(x)=x,\qquad x^2=\frac13P_0(x)+\frac23P_2(x).\]
\[P_0(x)=1,\qquad P_1(x)=x.\]\[x^2=\frac13P_0(x)+\frac23P_2(x).\]
Now,
\[1+x-x^2=P_0(x)+P_1(x)-\left[\frac13P_0(x)+\frac23P_2(x)\right].\]
\[1+x-x^2=P_0(x)+P_1(x)\]\[-\left[\frac13P_0(x)+\frac23P_2(x)\right].\]
Collecting like terms,
\[1+x-x^2=\frac23P_0(x)+P_1(x)-\frac23P_2(x).\]\[1+x-x^2=\frac23P_0(x)+P_1(x)-\frac23P_2(x).\]
\[1+x-x^2=\frac23P_0(x)+P_1(x)\]\[{}-\frac23P_2(x).\]
61
2 Marks
Expansion of (4+2x+x^2)
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 6 • 2 Marks
Express
\[f(x)=4+2x+x^2\]
in terms of Legendre polynomials.
Solution
Standard forms of \(P_0,P_1,P_2\)Using
\[1=P_0(x),\qquad x=P_1(x),\qquad x^2=\frac13P_0(x)+\frac23P_2(x).\]
\[1=P_0(x),\qquad x=P_1(x).\]\[x^2=\frac13P_0(x)+\frac23P_2(x).\]
Now,
\[4+2x+x^2=4P_0(x)+2P_1(x)+\frac13P_0(x)+\frac23P_2(x).\]
\[4+2x+x^2=4P_0(x)+2P_1(x)\]\[+\frac13P_0(x)+\frac23P_2(x).\]
Therefore,
\[4+2x+x^2=\frac{13}{3}P_0(x)+2P_1(x)+\frac23P_2(x).\]\[4+2x+x^2=\frac{13}{3}P_0(x)+2P_1(x)+\frac23P_2(x).\]
\[4+2x+x^2=\frac{13}{3}P_0(x)\]\[{}+2P_1(x)+\frac23P_2(x).\]
62
2 Marks
Expansion of (2x^2-4x+2)
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 6 • 2 Marks
Express
\[2x^2-4x+2\]
in terms of Legendre polynomials.
Solution
Standard forms of \(P_0,P_1,P_2\)Using
\[x^2=\frac13P_0(x)+\frac23P_2(x),\qquad x=P_1(x),\qquad 1=P_0(x).\]
\[x^2=\frac13P_0(x)+\frac23P_2(x).\]\[x=P_1(x),\qquad1=P_0(x).\]
Now,
\[2x^2-4x+2=2\left[\frac13P_0(x)+\frac23P_2(x)\right]-4P_1(x)+2P_0(x).\]
\[2x^2-4x+2=2\left[\frac13P_0(x)+\frac23P_2(x)\right]\]\[-4P_1(x)+2P_0(x).\]
Therefore,
\[2x^2-4x+2=\frac83P_0(x)-4P_1(x)+\frac43P_2(x).\]\[2x^2-4x+2=\frac83P_0(x)-4P_1(x)+\frac43P_2(x).\]
\[2x^2-4x+2=\frac83P_0(x)\]\[{}-4P_1(x)+\frac43P_2(x).\]