Specific case of Mean Value Theorem for partial derivatives
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Let $f: Omega subseteq mathbb{R}^{n} longrightarrow mathbb{R}$ be a continuous function in the closed segment $[x,y] subset Omega $, such that the partial derivative with respect to the j-th variable $( frac{partial f(x)}{ partial x_{j}})$ is defined in the segment $ (x,y) $. Prove that $ exists z in (x,y) $ such that:
$$f(y) - f(x) = frac{partial f(z)}{ partial x_{j}} (y_{j} - x_{j})$$
Honestly I'm really surprised that I ended up having to ask this, because at first I thought the prove would just closely follow the same structure from similar theorems. But my main problem is that the conditions I'm given restrict me from using the theorems that I am comfortable with. Most similar questions I've found (like this or this) refer to different MVTs (which I actually already know), but can't seemingly be applied here. The proof for the first one (it's Theorem 36, just using this as a reference) doesn't work here because it only proves the existence of a directional derivative, where the direction is the one from the segment, so for example in my case it could only prove the directional derivative in the direction $ frac{y-x}{||y-x||} $. In fact, the proof does require that the limit $ lim_{t rightarrow t_{0}} g(t)$ is only evaluated for points that are in the segment, because otherwise you can't guarantee that the composition of f and g are continuous, and can't apply the single-variable MVT.
The second one does imply existence of all partial derivatives, but it requires differentiability so that's out of the question.
What I gathered from both proofs is that they are usually revolved around reducing the multivariable functions down to functions in $ mathbb{R} $, where we can use the MVT for the single-variable case. However, I don't know how to do that in this case. I think that my main problem comes from the fact that I don't know what is the direction of the segment $[x,y]$.
Could anyone please give me a hint on how to build a function that lets me reduce this problem to a single-variable case? Or should I take a completely different approach?
real-analysis multivariable-calculus partial-derivative
asked Nov 19 at 7:13
Joaquin C.
154
add a comment |
up vote
1
down vote
favorite
Let $f: Omega subseteq mathbb{R}^{n} longrightarrow mathbb{R}$ be a continuous function in the closed segment $[x,y] subset Omega $, such that the partial derivative with respect to the j-th variable $( frac{partial f(x)}{ partial x_{j}})$ is defined in the segment $ (x,y) $. Prove that $ exists z in (x,y) $ such that:
$$f(y) - f(x) = frac{partial f(z)}{ partial x_{j}} (y_{j} - x_{j})$$
Honestly I'm really surprised that I ended up having to ask this, because at first I thought the prove would just closely follow the same structure from similar theorems. But my main problem is that the conditions I'm given restrict me from using the theorems that I am comfortable with. Most similar questions I've found (like this or this) refer to different MVTs (which I actually already know), but can't seemingly be applied here. The proof for the first one (it's Theorem 36, just using this as a reference) doesn't work here because it only proves the existence of a directional derivative, where the direction is the one from the segment, so for example in my case it could only prove the directional derivative in the direction $ frac{y-x}{||y-x||} $. In fact, the proof does require that the limit $ lim_{t rightarrow t_{0}} g(t)$ is only evaluated for points that are in the segment, because otherwise you can't guarantee that the composition of f and g are continuous, and can't apply the single-variable MVT.
The second one does imply existence of all partial derivatives, but it requires differentiability so that's out of the question.
What I gathered from both proofs is that they are usually revolved around reducing the multivariable functions down to functions in $ mathbb{R} $, where we can use the MVT for the single-variable case. However, I don't know how to do that in this case. I think that my main problem comes from the fact that I don't know what is the direction of the segment $[x,y]$.
Could anyone please give me a hint on how to build a function that lets me reduce this problem to a single-variable case? Or should I take a completely different approach?
real-analysis multivariable-calculus partial-derivative
asked Nov 19 at 7:13
Joaquin C.
154
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f: Omega subseteq mathbb{R}^{n} longrightarrow mathbb{R}$ be a continuous function in the closed segment $[x,y] subset Omega $, such that the partial derivative with respect to the j-th variable $( frac{partial f(x)}{ partial x_{j}})$ is defined in the segment $ (x,y) $. Prove that $ exists z in (x,y) $ such that:
$$f(y) - f(x) = frac{partial f(z)}{ partial x_{j}} (y_{j} - x_{j})$$
Honestly I'm really surprised that I ended up having to ask this, because at first I thought the prove would just closely follow the same structure from similar theorems. But my main problem is that the conditions I'm given restrict me from using the theorems that I am comfortable with. Most similar questions I've found (like this or this) refer to different MVTs (which I actually already know), but can't seemingly be applied here. The proof for the first one (it's Theorem 36, just using this as a reference) doesn't work here because it only proves the existence of a directional derivative, where the direction is the one from the segment, so for example in my case it could only prove the directional derivative in the direction $ frac{y-x}{||y-x||} $. In fact, the proof does require that the limit $ lim_{t rightarrow t_{0}} g(t)$ is only evaluated for points that are in the segment, because otherwise you can't guarantee that the composition of f and g are continuous, and can't apply the single-variable MVT.
The second one does imply existence of all partial derivatives, but it requires differentiability so that's out of the question.
What I gathered from both proofs is that they are usually revolved around reducing the multivariable functions down to functions in $ mathbb{R} $, where we can use the MVT for the single-variable case. However, I don't know how to do that in this case. I think that my main problem comes from the fact that I don't know what is the direction of the segment $[x,y]$.
Could anyone please give me a hint on how to build a function that lets me reduce this problem to a single-variable case? Or should I take a completely different approach?
real-analysis multivariable-calculus partial-derivative
asked Nov 19 at 7:13
Joaquin C.
154
Let $f: Omega subseteq mathbb{R}^{n} longrightarrow mathbb{R}$ be a continuous function in the closed segment $[x,y] subset Omega $, such that the partial derivative with respect to the j-th variable $( frac{partial f(x)}{ partial x_{j}})$ is defined in the segment $ (x,y) $. Prove that $ exists z in (x,y) $ such that:
$$f(y) - f(x) = frac{partial f(z)}{ partial x_{j}} (y_{j} - x_{j})$$
Honestly I'm really surprised that I ended up having to ask this, because at first I thought the prove would just closely follow the same structure from similar theorems. But my main problem is that the conditions I'm given restrict me from using the theorems that I am comfortable with. Most similar questions I've found (like this or this) refer to different MVTs (which I actually already know), but can't seemingly be applied here. The proof for the first one (it's Theorem 36, just using this as a reference) doesn't work here because it only proves the existence of a directional derivative, where the direction is the one from the segment, so for example in my case it could only prove the directional derivative in the direction $ frac{y-x}{||y-x||} $. In fact, the proof does require that the limit $ lim_{t rightarrow t_{0}} g(t)$ is only evaluated for points that are in the segment, because otherwise you can't guarantee that the composition of f and g are continuous, and can't apply the single-variable MVT.
The second one does imply existence of all partial derivatives, but it requires differentiability so that's out of the question.
What I gathered from both proofs is that they are usually revolved around reducing the multivariable functions down to functions in $ mathbb{R} $, where we can use the MVT for the single-variable case. However, I don't know how to do that in this case. I think that my main problem comes from the fact that I don't know what is the direction of the segment $[x,y]$.
Could anyone please give me a hint on how to build a function that lets me reduce this problem to a single-variable case? Or should I take a completely different approach?
real-analysis multivariable-calculus partial-derivative
real-analysis multivariable-calculus partial-derivative
asked Nov 19 at 7:13
Joaquin C.
154
asked Nov 19 at 7:13
Joaquin C.
154
asked Nov 19 at 7:13
Joaquin C.
154
asked Nov 19 at 7:13
Joaquin C.
154
asked Nov 19 at 7:13
Joaquin C.
154
154
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
We may have $x_j=y_j$ but $f(x) neq f(y)$. So the assertion is false.
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We may have $x_j=y_j$ but $f(x) neq f(y)$. So the assertion is false.
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
add a comment |
up vote
1
down vote
accepted
We may have $x_j=y_j$ but $f(x) neq f(y)$. So the assertion is false.
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We may have $x_j=y_j$ but $f(x) neq f(y)$. So the assertion is false.
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
We may have $x_j=y_j$ but $f(x) neq f(y)$. So the assertion is false.
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
answered Nov 19 at 7:19
Kavi Rama Murthy
45.9k31854
45.9k31854
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
add a comment |
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
Thanks for making it explicit, now it makes sense. But since that's the case, then what's a minimal condition that could be added to make it true? Continuity in the whole domain? Or a restriction to the segment?
– Joaquin C.
Nov 19 at 7:28
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
@JoaquinC. You can expect such a result to be true only when the line segment is parallel to $x_j-$ axis and in this case it reduces to MVT in $mathbb R$.
– Kavi Rama Murthy
Nov 19 at 7:31
add a comment |
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