Proving the Product Rule with Multiindex Notation
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Let $f, g : mathbb{R}^n rightarrow mathbb{R}$ be smooth functions (that is, all partial derivatives exist to arbitrary orders). Show that for all multiindices $alpha in mathbb{N}^{n}_{0}$ we have
$$large partial^alpha(f cdot g)(x)= sum_{beta in mathbb{N}^{n}_{0} :beta leqslant alpha} binom{alpha}{beta} partial^{beta}f(x)partial^{alpha-beta}g(x)$$
for all $xinmathbb{R}^n$, where $Large binom{alpha}{beta}=frac{alpha!}{beta!(alpha-beta)!}=frac{alpha_{1}!cdotcdotcdotalpha_{n}!}{beta_{1}!cdotcdotcdotbeta_{n}!(alpha_{1}-beta_{1}!)cdotcdotcdot(alpha_{n}-beta_{n})!}.$
I am able to decipher the base case where $|alpha|=1$. Since for either $|beta|=0$ and $|beta|=1$, we have that $binom{alpha}{beta}=1$ and $partial^{(0,...,0)}f(x)=f(x)$. Therefore, we have that
$$large partial^{alpha}(fcdot g)(x)=sum^{n}_{i=1}partial_i f(x)g(x) + partial_i g(x)f(x).$$
However, I have no idea about the inductive step. Anyone have any solution or hint? I would greatly appreciate it. Thank you in advance.
real-analysis multivariable-calculus derivatives
asked Nov 19 at 8:50
Gaby Alfonso
660315
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up vote
1
down vote
favorite
Let $f, g : mathbb{R}^n rightarrow mathbb{R}$ be smooth functions (that is, all partial derivatives exist to arbitrary orders). Show that for all multiindices $alpha in mathbb{N}^{n}_{0}$ we have
$$large partial^alpha(f cdot g)(x)= sum_{beta in mathbb{N}^{n}_{0} :beta leqslant alpha} binom{alpha}{beta} partial^{beta}f(x)partial^{alpha-beta}g(x)$$
for all $xinmathbb{R}^n$, where $Large binom{alpha}{beta}=frac{alpha!}{beta!(alpha-beta)!}=frac{alpha_{1}!cdotcdotcdotalpha_{n}!}{beta_{1}!cdotcdotcdotbeta_{n}!(alpha_{1}-beta_{1}!)cdotcdotcdot(alpha_{n}-beta_{n})!}.$
I am able to decipher the base case where $|alpha|=1$. Since for either $|beta|=0$ and $|beta|=1$, we have that $binom{alpha}{beta}=1$ and $partial^{(0,...,0)}f(x)=f(x)$. Therefore, we have that
$$large partial^{alpha}(fcdot g)(x)=sum^{n}_{i=1}partial_i f(x)g(x) + partial_i g(x)f(x).$$
However, I have no idea about the inductive step. Anyone have any solution or hint? I would greatly appreciate it. Thank you in advance.
real-analysis multivariable-calculus derivatives
asked Nov 19 at 8:50
Gaby Alfonso
660315
1
For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f, g : mathbb{R}^n rightarrow mathbb{R}$ be smooth functions (that is, all partial derivatives exist to arbitrary orders). Show that for all multiindices $alpha in mathbb{N}^{n}_{0}$ we have
$$large partial^alpha(f cdot g)(x)= sum_{beta in mathbb{N}^{n}_{0} :beta leqslant alpha} binom{alpha}{beta} partial^{beta}f(x)partial^{alpha-beta}g(x)$$
for all $xinmathbb{R}^n$, where $Large binom{alpha}{beta}=frac{alpha!}{beta!(alpha-beta)!}=frac{alpha_{1}!cdotcdotcdotalpha_{n}!}{beta_{1}!cdotcdotcdotbeta_{n}!(alpha_{1}-beta_{1}!)cdotcdotcdot(alpha_{n}-beta_{n})!}.$
I am able to decipher the base case where $|alpha|=1$. Since for either $|beta|=0$ and $|beta|=1$, we have that $binom{alpha}{beta}=1$ and $partial^{(0,...,0)}f(x)=f(x)$. Therefore, we have that
$$large partial^{alpha}(fcdot g)(x)=sum^{n}_{i=1}partial_i f(x)g(x) + partial_i g(x)f(x).$$
However, I have no idea about the inductive step. Anyone have any solution or hint? I would greatly appreciate it. Thank you in advance.
real-analysis multivariable-calculus derivatives
asked Nov 19 at 8:50
Gaby Alfonso
660315
Let $f, g : mathbb{R}^n rightarrow mathbb{R}$ be smooth functions (that is, all partial derivatives exist to arbitrary orders). Show that for all multiindices $alpha in mathbb{N}^{n}_{0}$ we have
$$large partial^alpha(f cdot g)(x)= sum_{beta in mathbb{N}^{n}_{0} :beta leqslant alpha} binom{alpha}{beta} partial^{beta}f(x)partial^{alpha-beta}g(x)$$
for all $xinmathbb{R}^n$, where $Large binom{alpha}{beta}=frac{alpha!}{beta!(alpha-beta)!}=frac{alpha_{1}!cdotcdotcdotalpha_{n}!}{beta_{1}!cdotcdotcdotbeta_{n}!(alpha_{1}-beta_{1}!)cdotcdotcdot(alpha_{n}-beta_{n})!}.$
I am able to decipher the base case where $|alpha|=1$. Since for either $|beta|=0$ and $|beta|=1$, we have that $binom{alpha}{beta}=1$ and $partial^{(0,...,0)}f(x)=f(x)$. Therefore, we have that
$$large partial^{alpha}(fcdot g)(x)=sum^{n}_{i=1}partial_i f(x)g(x) + partial_i g(x)f(x).$$
However, I have no idea about the inductive step. Anyone have any solution or hint? I would greatly appreciate it. Thank you in advance.
real-analysis multivariable-calculus derivatives
real-analysis multivariable-calculus derivatives
asked Nov 19 at 8:50
Gaby Alfonso
660315
asked Nov 19 at 8:50
Gaby Alfonso
660315
edited Nov 19 at 9:06
asked Nov 19 at 8:50
Gaby Alfonso
660315
asked Nov 19 at 8:50
Gaby Alfonso
660315
asked Nov 19 at 8:50
Gaby Alfonso
660315
660315
1
For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17
add a comment |
1
For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17
1
1
For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17
add a comment |
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For every multiindex $alpha$ with $|alpha|=k+1$ there exists a multiindex $gamma$ and an index $1leq ileq n$ such that $|gamma|=k$ and $partial^alpha=partial_ipartial^gamma$. Using this you can do induction by $|alpha|$.
– maxmilgram
Nov 19 at 9:14
Also, note that the sum in the "base case" is not correctly given. You do not sum over all partial derivatives, only over those specified by $alpha$.
– maxmilgram
Nov 19 at 9:17