Proving that $D^k_textbf{h}f(textbf{x}) =sum_{|alpha|=k}frac{k!}{alpha!}D^alpha f(textbf{x})h^alpha$
I'm trying to prove this identity involving the k-th order directional derivative along a given vector $textbf{h}$:
$$D^k_textbf{h},f(textbf{x}) =sum_{|alpha|=k}frac{k!}{alpha!}D^alpha f(textbf{x})h^alpha,$$
using multi-index notation. (Here,
$D^k_textbf{h},f(textbf{x})equivsum^n_{i_1,...,i_k=1}D_{i_1,...,i_k}, f(textbf{x}),h_{i_1}...h_{i_k}$)
To begin with, I'd like to prove that
$$D^k_textbf{h},f(textbf{x})=(h_1D_1,f(textbf{x})+,...+h_nD_nf(textbf{x}))^k,$$
then I know from my lecture notes I can use the multinomial theorem to obtain the desired result. I'll proceed by induction. I know this equality holds for $k=1$, so I'll suppose it holds for $k$ and see if it is true for $k+1$. Using the induction hypothesis, I get
$$D^{k+1}_textbf{h},f(textbf{x})=D_textbf{h}(D^k_textbf{h},f(textbf{x}))=sum_{j=1}^nD_j(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^kh_j=(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^{k+1}.$$
I'm not sure of the last step though. I've used that, in general, $sum^n_{j=1}x_j(h_1x_1+,...+h_nx^n)^kh_j=(h_1x_1+,...+h_nx_n)^{k+1}$, but given that in this case the partial derivative is an operator, I'm having doubts about the applicability of this. Is the proof correct?
analysis multivariable-calculus
asked 2 days ago
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945
add a comment |
I'm trying to prove this identity involving the k-th order directional derivative along a given vector $textbf{h}$:
$$D^k_textbf{h},f(textbf{x}) =sum_{|alpha|=k}frac{k!}{alpha!}D^alpha f(textbf{x})h^alpha,$$
using multi-index notation. (Here,
$D^k_textbf{h},f(textbf{x})equivsum^n_{i_1,...,i_k=1}D_{i_1,...,i_k}, f(textbf{x}),h_{i_1}...h_{i_k}$)
To begin with, I'd like to prove that
$$D^k_textbf{h},f(textbf{x})=(h_1D_1,f(textbf{x})+,...+h_nD_nf(textbf{x}))^k,$$
then I know from my lecture notes I can use the multinomial theorem to obtain the desired result. I'll proceed by induction. I know this equality holds for $k=1$, so I'll suppose it holds for $k$ and see if it is true for $k+1$. Using the induction hypothesis, I get
$$D^{k+1}_textbf{h},f(textbf{x})=D_textbf{h}(D^k_textbf{h},f(textbf{x}))=sum_{j=1}^nD_j(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^kh_j=(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^{k+1}.$$
I'm not sure of the last step though. I've used that, in general, $sum^n_{j=1}x_j(h_1x_1+,...+h_nx^n)^kh_j=(h_1x_1+,...+h_nx_n)^{k+1}$, but given that in this case the partial derivative is an operator, I'm having doubts about the applicability of this. Is the proof correct?
analysis multivariable-calculus
asked 2 days ago
guest_user
945
add a comment |
I'm trying to prove this identity involving the k-th order directional derivative along a given vector $textbf{h}$:
$$D^k_textbf{h},f(textbf{x}) =sum_{|alpha|=k}frac{k!}{alpha!}D^alpha f(textbf{x})h^alpha,$$
using multi-index notation. (Here,
$D^k_textbf{h},f(textbf{x})equivsum^n_{i_1,...,i_k=1}D_{i_1,...,i_k}, f(textbf{x}),h_{i_1}...h_{i_k}$)
To begin with, I'd like to prove that
$$D^k_textbf{h},f(textbf{x})=(h_1D_1,f(textbf{x})+,...+h_nD_nf(textbf{x}))^k,$$
then I know from my lecture notes I can use the multinomial theorem to obtain the desired result. I'll proceed by induction. I know this equality holds for $k=1$, so I'll suppose it holds for $k$ and see if it is true for $k+1$. Using the induction hypothesis, I get
$$D^{k+1}_textbf{h},f(textbf{x})=D_textbf{h}(D^k_textbf{h},f(textbf{x}))=sum_{j=1}^nD_j(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^kh_j=(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^{k+1}.$$
I'm not sure of the last step though. I've used that, in general, $sum^n_{j=1}x_j(h_1x_1+,...+h_nx^n)^kh_j=(h_1x_1+,...+h_nx_n)^{k+1}$, but given that in this case the partial derivative is an operator, I'm having doubts about the applicability of this. Is the proof correct?
analysis multivariable-calculus
asked 2 days ago
guest_user
945
I'm trying to prove this identity involving the k-th order directional derivative along a given vector $textbf{h}$:
$$D^k_textbf{h},f(textbf{x}) =sum_{|alpha|=k}frac{k!}{alpha!}D^alpha f(textbf{x})h^alpha,$$
using multi-index notation. (Here,
$D^k_textbf{h},f(textbf{x})equivsum^n_{i_1,...,i_k=1}D_{i_1,...,i_k}, f(textbf{x}),h_{i_1}...h_{i_k}$)
To begin with, I'd like to prove that
$$D^k_textbf{h},f(textbf{x})=(h_1D_1,f(textbf{x})+,...+h_nD_nf(textbf{x}))^k,$$
then I know from my lecture notes I can use the multinomial theorem to obtain the desired result. I'll proceed by induction. I know this equality holds for $k=1$, so I'll suppose it holds for $k$ and see if it is true for $k+1$. Using the induction hypothesis, I get
$$D^{k+1}_textbf{h},f(textbf{x})=D_textbf{h}(D^k_textbf{h},f(textbf{x}))=sum_{j=1}^nD_j(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^kh_j=(h_1D_1,f(textbf{x})+,...+h_nD_n,f(textbf{x}))^{k+1}.$$
I'm not sure of the last step though. I've used that, in general, $sum^n_{j=1}x_j(h_1x_1+,...+h_nx^n)^kh_j=(h_1x_1+,...+h_nx_n)^{k+1}$, but given that in this case the partial derivative is an operator, I'm having doubts about the applicability of this. Is the proof correct?
analysis multivariable-calculus
analysis multivariable-calculus
asked 2 days ago
guest_user
945
asked 2 days ago
guest_user
945
asked 2 days ago
guest_user
945
asked 2 days ago
guest_user
945
asked 2 days ago
guest_user
945
945
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