Mixing
Differentiate Legendre’s equation $m$ times using Leibniz' rule for differentiating products
$begingroup$
$$(1-x^2)u''(x)-2xu'(x) + ell(ell+1)u(x)=0tag{1}$$
Assume that $m$ is non-negative, differentiate $(1)$ (Legendre’s equation) $m$ times using Leibniz' theorem for differentiation to show that
$$(1-x^2)v''(x)-2x(m+1)v'(x)+(ell -m)(ell + m + 1)v(x)=0tag{*}$$
where $$v(x)equiv frac{d^mu}{dx^m}$$
From Leibniz' theorem the $m^{text{th}}$ devivative of a product $fg$ is
$$(fg)^m=f^{(m)}g+mf^{(m-1)}g^{(1)}+cdots+frac{m!}{n!(m-n)!}f^{(m-n)}g^{(n)}$$
where $f^{(m)}$ is the $m^{text{th}}$ derivative of $f$, etc.
Hence, differentiating $(1)$ term by term,
$$frac{d^m}{dx^m}left[(1-x^2)u''right]=(1-x^2)u^{(m+2)}-2mxu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}$$
I find the next term to be
$$frac{d^m}{dx^m}left[-2xu'right]=-2xu^{(m+1)}-2mu^{(m)}$$
and the last term is
$$frac{d^m}{dx^m}left[ell(ell+1)uright]=ell(ell+1)u^{(m)}$$
Substituting these into $(1)$ gives
$$(1-x^2)u^{(m+2)}-2xmu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}-2xu^{(m+1)}-2mu^{(m)} + ell(ell+1)u^{(m)}=0$$
which on simplification is
$$(1-x^2)v''(x)-2x(m+1)v'(x)+left[ell(ell+1)-2m^2right]v(x)=0$$
which is not the same as $(*)$.
In the solution the author gives, everything in my working is exactly the same except from the parts marked in red, according to the author's solution the $2$ should not be there (it should be unity). But the derivative of $2x$ is $2$. So I fail to understand why it should not be $2$.
Can someone please explain where I have gone wrong here?
ordinary-differential-equations derivatives legendre-polynomials
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
$endgroup$
add a comment |
$begingroup$
$$(1-x^2)u''(x)-2xu'(x) + ell(ell+1)u(x)=0tag{1}$$
Assume that $m$ is non-negative, differentiate $(1)$ (Legendre’s equation) $m$ times using Leibniz' theorem for differentiation to show that
$$(1-x^2)v''(x)-2x(m+1)v'(x)+(ell -m)(ell + m + 1)v(x)=0tag{*}$$
where $$v(x)equiv frac{d^mu}{dx^m}$$
From Leibniz' theorem the $m^{text{th}}$ devivative of a product $fg$ is
$$(fg)^m=f^{(m)}g+mf^{(m-1)}g^{(1)}+cdots+frac{m!}{n!(m-n)!}f^{(m-n)}g^{(n)}$$
where $f^{(m)}$ is the $m^{text{th}}$ derivative of $f$, etc.
Hence, differentiating $(1)$ term by term,
$$frac{d^m}{dx^m}left[(1-x^2)u''right]=(1-x^2)u^{(m+2)}-2mxu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}$$
I find the next term to be
$$frac{d^m}{dx^m}left[-2xu'right]=-2xu^{(m+1)}-2mu^{(m)}$$
and the last term is
$$frac{d^m}{dx^m}left[ell(ell+1)uright]=ell(ell+1)u^{(m)}$$
Substituting these into $(1)$ gives
$$(1-x^2)u^{(m+2)}-2xmu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}-2xu^{(m+1)}-2mu^{(m)} + ell(ell+1)u^{(m)}=0$$
which on simplification is
$$(1-x^2)v''(x)-2x(m+1)v'(x)+left[ell(ell+1)-2m^2right]v(x)=0$$
which is not the same as $(*)$.
In the solution the author gives, everything in my working is exactly the same except from the parts marked in red, according to the author's solution the $2$ should not be there (it should be unity). But the derivative of $2x$ is $2$. So I fail to understand why it should not be $2$.
Can someone please explain where I have gone wrong here?
ordinary-differential-equations derivatives legendre-polynomials
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
$endgroup$
add a comment |
$begingroup$
$$(1-x^2)u''(x)-2xu'(x) + ell(ell+1)u(x)=0tag{1}$$
Assume that $m$ is non-negative, differentiate $(1)$ (Legendre’s equation) $m$ times using Leibniz' theorem for differentiation to show that
$$(1-x^2)v''(x)-2x(m+1)v'(x)+(ell -m)(ell + m + 1)v(x)=0tag{*}$$
where $$v(x)equiv frac{d^mu}{dx^m}$$
From Leibniz' theorem the $m^{text{th}}$ devivative of a product $fg$ is
$$(fg)^m=f^{(m)}g+mf^{(m-1)}g^{(1)}+cdots+frac{m!}{n!(m-n)!}f^{(m-n)}g^{(n)}$$
where $f^{(m)}$ is the $m^{text{th}}$ derivative of $f$, etc.
Hence, differentiating $(1)$ term by term,
$$frac{d^m}{dx^m}left[(1-x^2)u''right]=(1-x^2)u^{(m+2)}-2mxu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}$$
I find the next term to be
$$frac{d^m}{dx^m}left[-2xu'right]=-2xu^{(m+1)}-2mu^{(m)}$$
and the last term is
$$frac{d^m}{dx^m}left[ell(ell+1)uright]=ell(ell+1)u^{(m)}$$
Substituting these into $(1)$ gives
$$(1-x^2)u^{(m+2)}-2xmu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}-2xu^{(m+1)}-2mu^{(m)} + ell(ell+1)u^{(m)}=0$$
which on simplification is
$$(1-x^2)v''(x)-2x(m+1)v'(x)+left[ell(ell+1)-2m^2right]v(x)=0$$
which is not the same as $(*)$.
In the solution the author gives, everything in my working is exactly the same except from the parts marked in red, according to the author's solution the $2$ should not be there (it should be unity). But the derivative of $2x$ is $2$. So I fail to understand why it should not be $2$.
Can someone please explain where I have gone wrong here?
ordinary-differential-equations derivatives legendre-polynomials
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
$endgroup$
$$(1-x^2)u''(x)-2xu'(x) + ell(ell+1)u(x)=0tag{1}$$
Assume that $m$ is non-negative, differentiate $(1)$ (Legendre’s equation) $m$ times using Leibniz' theorem for differentiation to show that
$$(1-x^2)v''(x)-2x(m+1)v'(x)+(ell -m)(ell + m + 1)v(x)=0tag{*}$$
where $$v(x)equiv frac{d^mu}{dx^m}$$
From Leibniz' theorem the $m^{text{th}}$ devivative of a product $fg$ is
$$(fg)^m=f^{(m)}g+mf^{(m-1)}g^{(1)}+cdots+frac{m!}{n!(m-n)!}f^{(m-n)}g^{(n)}$$
where $f^{(m)}$ is the $m^{text{th}}$ derivative of $f$, etc.
Hence, differentiating $(1)$ term by term,
$$frac{d^m}{dx^m}left[(1-x^2)u''right]=(1-x^2)u^{(m+2)}-2mxu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}$$
I find the next term to be
$$frac{d^m}{dx^m}left[-2xu'right]=-2xu^{(m+1)}-2mu^{(m)}$$
and the last term is
$$frac{d^m}{dx^m}left[ell(ell+1)uright]=ell(ell+1)u^{(m)}$$
Substituting these into $(1)$ gives
$$(1-x^2)u^{(m+2)}-2xmu^{(m+1)}-color{red}{2}m(m-1)u^{(m)}-2xu^{(m+1)}-2mu^{(m)} + ell(ell+1)u^{(m)}=0$$
which on simplification is
$$(1-x^2)v''(x)-2x(m+1)v'(x)+left[ell(ell+1)-2m^2right]v(x)=0$$
which is not the same as $(*)$.
In the solution the author gives, everything in my working is exactly the same except from the parts marked in red, according to the author's solution the $2$ should not be there (it should be unity). But the derivative of $2x$ is $2$. So I fail to understand why it should not be $2$.
Can someone please explain where I have gone wrong here?
ordinary-differential-equations derivatives legendre-polynomials
ordinary-differential-equations derivatives legendre-polynomials
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
asked Jan 6 at 23:17
BLAZEBLAZE
6,076112756
6,076112756
add a comment |
add a comment |
1 Answer
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$begingroup$
$$frac{m!}{n!(m-n)!}=frac{m(m-1)}{2}$$ and this cancels the $2$ in red.
answered Jan 6 at 23:38
Ben WBen W
2,234615
$endgroup$
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
add a comment |
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1 Answer
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$begingroup$
$$frac{m!}{n!(m-n)!}=frac{m(m-1)}{2}$$ and this cancels the $2$ in red.
answered Jan 6 at 23:38
Ben WBen W
2,234615
$endgroup$
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
add a comment |
$begingroup$
$$frac{m!}{n!(m-n)!}=frac{m(m-1)}{2}$$ and this cancels the $2$ in red.
answered Jan 6 at 23:38
Ben WBen W
2,234615
$endgroup$
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
add a comment |
$begingroup$
$$frac{m!}{n!(m-n)!}=frac{m(m-1)}{2}$$ and this cancels the $2$ in red.
answered Jan 6 at 23:38
Ben WBen W
2,234615
$endgroup$
$$frac{m!}{n!(m-n)!}=frac{m(m-1)}{2}$$ and this cancels the $2$ in red.
answered Jan 6 at 23:38
Ben WBen W
2,234615
answered Jan 6 at 23:38
Ben WBen W
2,234615
answered Jan 6 at 23:38
Ben WBen W
2,234615
answered Jan 6 at 23:38
Ben WBen W
2,234615
2,234615
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
add a comment |
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
$begingroup$
Perfect answer, thank you very much
$endgroup$
– BLAZE
Jan 6 at 23:43
1
1
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
$begingroup$
Sure thing .. reminds me of my own "oops" moments, which are plentiful ; )
$endgroup$
– Ben W
Jan 6 at 23:44
add a comment |
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