for any $e$ prove that the difference between $f(x)$ and $f(x+e)$ is the same as the directional derivative of $f(x+te)$ in the direction of $e$
Clash Royale CLAN TAG#URR8PPP
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Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.
So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.
I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.
And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$
real-analysis differential-equations derivatives proof-verification
edited Sep 11 at 16:15
Harry49
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
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Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.
So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.
I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.
And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$
real-analysis differential-equations derivatives proof-verification
edited Sep 11 at 16:15
Harry49
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21
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up vote
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down vote
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Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.
So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.
I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.
And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$
real-analysis differential-equations derivatives proof-verification
edited Sep 11 at 16:15
Harry49
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.
So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.
I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.
And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$
real-analysis differential-equations derivatives proof-verification
real-analysis differential-equations derivatives proof-verification
edited Sep 11 at 16:15
Harry49
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
edited Sep 11 at 16:15
Harry49
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
edited Sep 11 at 16:15
Harry49
5,5502929
edited Sep 11 at 16:15
Harry49
5,5502929
edited Sep 11 at 16:15
Harry49
5,5502929
5,5502929
asked Sep 11 at 4:19
Big_Tubbz
234
asked Sep 11 at 4:19
Big_Tubbz
234
asked Sep 11 at 4:19
Big_Tubbz
234
234
Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21
add a comment |
Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21
Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21
Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21
add a comment |
1 Answer
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Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".
answered Sep 11 at 18:49
Christian Blatter
169k7111321
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".
answered Sep 11 at 18:49
Christian Blatter
169k7111321
add a comment |
up vote
1
down vote
Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".
answered Sep 11 at 18:49
Christian Blatter
169k7111321
add a comment |
up vote
1
down vote
up vote
1
down vote
Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".
answered Sep 11 at 18:49
Christian Blatter
169k7111321
Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".
answered Sep 11 at 18:49
Christian Blatter
169k7111321
answered Sep 11 at 18:49
Christian Blatter
169k7111321
answered Sep 11 at 18:49
Christian Blatter
169k7111321
answered Sep 11 at 18:49
Christian Blatter
169k7111321
169k7111321
add a comment |
add a comment |
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Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21