Prove that the even (odd) degree Legendre polynomials are even > (odd) functions of $t$.
$begingroup$
a.) Prove that the even (odd) degree Legendre polynomials are even
(odd) functions of $t$.
b.) Prove that if $p(t) = p(-t)$ is an even polynomial, then all the
odd order coefficents $c_{2j+1} = 0$ in its Legendre expansion $p(t) =
c_0q_0(t) + ... + c_nq_n(t)$ vanish.
Is "a" similar to proving that Legendre polynomials of even and odd degree are orthogonal and for "b" I do not know how to do.
legendre-polynomials
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
$endgroup$
add a comment |
$begingroup$
a.) Prove that the even (odd) degree Legendre polynomials are even
(odd) functions of $t$.
b.) Prove that if $p(t) = p(-t)$ is an even polynomial, then all the
odd order coefficents $c_{2j+1} = 0$ in its Legendre expansion $p(t) =
c_0q_0(t) + ... + c_nq_n(t)$ vanish.
Is "a" similar to proving that Legendre polynomials of even and odd degree are orthogonal and for "b" I do not know how to do.
legendre-polynomials
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
$endgroup$
add a comment |
$begingroup$
a.) Prove that the even (odd) degree Legendre polynomials are even
(odd) functions of $t$.
b.) Prove that if $p(t) = p(-t)$ is an even polynomial, then all the
odd order coefficents $c_{2j+1} = 0$ in its Legendre expansion $p(t) =
c_0q_0(t) + ... + c_nq_n(t)$ vanish.
Is "a" similar to proving that Legendre polynomials of even and odd degree are orthogonal and for "b" I do not know how to do.
legendre-polynomials
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
$endgroup$
a.) Prove that the even (odd) degree Legendre polynomials are even
(odd) functions of $t$.
b.) Prove that if $p(t) = p(-t)$ is an even polynomial, then all the
odd order coefficents $c_{2j+1} = 0$ in its Legendre expansion $p(t) =
c_0q_0(t) + ... + c_nq_n(t)$ vanish.
Is "a" similar to proving that Legendre polynomials of even and odd degree are orthogonal and for "b" I do not know how to do.
legendre-polynomials
legendre-polynomials
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
edited Dec 7 '18 at 15:36
Nosrati
26.5k62354
26.5k62354
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
asked Nov 2 '12 at 5:08
diimensiondiimension
1,46532340
1,46532340
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
a)
$$
P_n(x)=frac{1}{2^nn!}frac{d^n}{dx^n}left[(x^2-1)^n right].
$$
Here $(x^2-1)^n$ is an even polynomial. Taking the $n$-th derivative the general terms have the form $a_k x^{2k-n}$ which is even iff $n$ is even, and odd iff $n$ is odd.
b) Writing out the Legendre-Fourier coefficients by integral on $[-1,1]$, then the integrand $p(t)P_{2n+1}(t)$ is an odd function so the value of the integral is $0$. (The situation is similar to the classical trigonometric Fourier series case on the interval $[-pi,pi]$.
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
$endgroup$
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
a)
$$
P_n(x)=frac{1}{2^nn!}frac{d^n}{dx^n}left[(x^2-1)^n right].
$$
Here $(x^2-1)^n$ is an even polynomial. Taking the $n$-th derivative the general terms have the form $a_k x^{2k-n}$ which is even iff $n$ is even, and odd iff $n$ is odd.
b) Writing out the Legendre-Fourier coefficients by integral on $[-1,1]$, then the integrand $p(t)P_{2n+1}(t)$ is an odd function so the value of the integral is $0$. (The situation is similar to the classical trigonometric Fourier series case on the interval $[-pi,pi]$.
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
$endgroup$
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
add a comment |
$begingroup$
a)
$$
P_n(x)=frac{1}{2^nn!}frac{d^n}{dx^n}left[(x^2-1)^n right].
$$
Here $(x^2-1)^n$ is an even polynomial. Taking the $n$-th derivative the general terms have the form $a_k x^{2k-n}$ which is even iff $n$ is even, and odd iff $n$ is odd.
b) Writing out the Legendre-Fourier coefficients by integral on $[-1,1]$, then the integrand $p(t)P_{2n+1}(t)$ is an odd function so the value of the integral is $0$. (The situation is similar to the classical trigonometric Fourier series case on the interval $[-pi,pi]$.
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
$endgroup$
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
add a comment |
$begingroup$
a)
$$
P_n(x)=frac{1}{2^nn!}frac{d^n}{dx^n}left[(x^2-1)^n right].
$$
Here $(x^2-1)^n$ is an even polynomial. Taking the $n$-th derivative the general terms have the form $a_k x^{2k-n}$ which is even iff $n$ is even, and odd iff $n$ is odd.
b) Writing out the Legendre-Fourier coefficients by integral on $[-1,1]$, then the integrand $p(t)P_{2n+1}(t)$ is an odd function so the value of the integral is $0$. (The situation is similar to the classical trigonometric Fourier series case on the interval $[-pi,pi]$.
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
$endgroup$
a)
$$
P_n(x)=frac{1}{2^nn!}frac{d^n}{dx^n}left[(x^2-1)^n right].
$$
Here $(x^2-1)^n$ is an even polynomial. Taking the $n$-th derivative the general terms have the form $a_k x^{2k-n}$ which is even iff $n$ is even, and odd iff $n$ is odd.
b) Writing out the Legendre-Fourier coefficients by integral on $[-1,1]$, then the integrand $p(t)P_{2n+1}(t)$ is an odd function so the value of the integral is $0$. (The situation is similar to the classical trigonometric Fourier series case on the interval $[-pi,pi]$.
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
answered Nov 2 '12 at 9:24
vesszabovesszabo
2,9281521
2,9281521
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
add a comment |
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
$begingroup$
Vesszabo, sry to bother you on this question again but can you further elaborate on b.? I understood a but I did not understand b ?
$endgroup$
– diimension
Nov 16 '12 at 23:02
add a comment |
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