tyring to prove product rule for higher-order partial derivatives
$begingroup$
Prove that
$$ partial^{alpha} (fg) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
Here we are using multi index notation $alpha = (alpha_1,...,alpha_n)$ and $partial^{alpha}(fg) = dfrac{ partial (fg)^{alpha_1+...+alpha_n} }{partial x_1^{alpha_1} ... partial x_n^{alpha_n} } $
I think we can use induction and we start with $n=1$ that is when we have a function of a single variable. So, we have $alpha = alpha_1$ and we have the usual leibniz rule:
$$ partial^{alpha_1} (fg) = frac{ partial (fg)^{alpha_1} }{partial x_1 ^{alpha_1} } = sum_{i=1}^{alpha_1} frac{alpha_1!}{i! (alpha_1-i)!} (partial^if)(partial^{alpha_1-i}g) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
where $beta = i$ and $gamma = alpha_1-i$ and $alpha = alpha_1$. So it holds for this case. Is this a correct approach to solve this problem?
calculus
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
add a comment |
$begingroup$
Prove that
$$ partial^{alpha} (fg) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
Here we are using multi index notation $alpha = (alpha_1,...,alpha_n)$ and $partial^{alpha}(fg) = dfrac{ partial (fg)^{alpha_1+...+alpha_n} }{partial x_1^{alpha_1} ... partial x_n^{alpha_n} } $
I think we can use induction and we start with $n=1$ that is when we have a function of a single variable. So, we have $alpha = alpha_1$ and we have the usual leibniz rule:
$$ partial^{alpha_1} (fg) = frac{ partial (fg)^{alpha_1} }{partial x_1 ^{alpha_1} } = sum_{i=1}^{alpha_1} frac{alpha_1!}{i! (alpha_1-i)!} (partial^if)(partial^{alpha_1-i}g) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
where $beta = i$ and $gamma = alpha_1-i$ and $alpha = alpha_1$. So it holds for this case. Is this a correct approach to solve this problem?
calculus
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15
add a comment |
$begingroup$
Prove that
$$ partial^{alpha} (fg) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
Here we are using multi index notation $alpha = (alpha_1,...,alpha_n)$ and $partial^{alpha}(fg) = dfrac{ partial (fg)^{alpha_1+...+alpha_n} }{partial x_1^{alpha_1} ... partial x_n^{alpha_n} } $
I think we can use induction and we start with $n=1$ that is when we have a function of a single variable. So, we have $alpha = alpha_1$ and we have the usual leibniz rule:
$$ partial^{alpha_1} (fg) = frac{ partial (fg)^{alpha_1} }{partial x_1 ^{alpha_1} } = sum_{i=1}^{alpha_1} frac{alpha_1!}{i! (alpha_1-i)!} (partial^if)(partial^{alpha_1-i}g) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
where $beta = i$ and $gamma = alpha_1-i$ and $alpha = alpha_1$. So it holds for this case. Is this a correct approach to solve this problem?
calculus
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
Prove that
$$ partial^{alpha} (fg) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
Here we are using multi index notation $alpha = (alpha_1,...,alpha_n)$ and $partial^{alpha}(fg) = dfrac{ partial (fg)^{alpha_1+...+alpha_n} }{partial x_1^{alpha_1} ... partial x_n^{alpha_n} } $
I think we can use induction and we start with $n=1$ that is when we have a function of a single variable. So, we have $alpha = alpha_1$ and we have the usual leibniz rule:
$$ partial^{alpha_1} (fg) = frac{ partial (fg)^{alpha_1} }{partial x_1 ^{alpha_1} } = sum_{i=1}^{alpha_1} frac{alpha_1!}{i! (alpha_1-i)!} (partial^if)(partial^{alpha_1-i}g) = sum_{beta + gamma = alpha} frac{
alpha!}{beta! gamma!} (partial^{beta}f)(partial^{gamma} g) $$
where $beta = i$ and $gamma = alpha_1-i$ and $alpha = alpha_1$. So it holds for this case. Is this a correct approach to solve this problem?
calculus
calculus
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
asked Jan 20 at 8:00
Jimmy SabaterJimmy Sabater
2,666323
2,666323
$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15
add a comment |
$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15
$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15
$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15
add a comment |
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$begingroup$
I'd prove this by induction. But not induction on $n$; it looks to me like induction on $|alpha|=alpha_1`+dots+alpha_n$.
$endgroup$
– David C. Ullrich
Jan 20 at 18:15