Derivative of $x^{beta}$
Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$
What is the result of $$partial_x^{alpha}x^{beta}$$?
Thanks
real-analysis partial-derivative
edited Dec 4 '18 at 11:10
Harry49
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
add a comment |
Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$
What is the result of $$partial_x^{alpha}x^{beta}$$?
Thanks
real-analysis partial-derivative
edited Dec 4 '18 at 11:10
Harry49
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21
add a comment |
Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$
What is the result of $$partial_x^{alpha}x^{beta}$$?
Thanks
real-analysis partial-derivative
edited Dec 4 '18 at 11:10
Harry49
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$
What is the result of $$partial_x^{alpha}x^{beta}$$?
Thanks
real-analysis partial-derivative
real-analysis partial-derivative
edited Dec 4 '18 at 11:10
Harry49
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
edited Dec 4 '18 at 11:10
Harry49
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
edited Dec 4 '18 at 11:10
Harry49
6,00121031
edited Dec 4 '18 at 11:10
Harry49
6,00121031
edited Dec 4 '18 at 11:10
Harry49
6,00121031
6,00121031
asked Dec 3 '18 at 20:06
aymenaymen
304
asked Dec 3 '18 at 20:06
aymenaymen
304
asked Dec 3 '18 at 20:06
aymenaymen
304
304
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21
add a comment |
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21
add a comment |
1 Answer
1
active
oldest
votes
Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 4 '18 at 15:40
Julián AguirreJulián Aguirre
67.7k24094
67.7k24094
add a comment |
add a comment |
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So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
– Chase Ryan Taylor
Dec 3 '18 at 20:21
yes that is it exactly
– aymen
Dec 3 '18 at 20:21