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Mean value theorem for multivariable functions [duplicate]
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Generalization of Mean Value Theorem to functions, $f: mathbbR^n rightarrow mathbbR^n$
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I`m trying to generalize Lagrange's mean value theorem for $f:A to mathbbR^n$
, differentiable in $A subset mathbbR^m$.
What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..tx_i,..,x_n$), where $x_j$ are any real numbers and $x_i$ is 1. We have $g'(t)=D_f(t)cdot(0,..,1,...0)=frac partial f partial x_i(t)$, And since $g_i(t):mathbbR to mathbbR$ , we can apply the single variable mean value theorem to say that there exists a point $c_x_iin [a,b] $ s.t $frac partial f partial x_i(c_x_i)=g_i'(c_x_i)=frac g_i(a)-g_i(b) a-b$ , for any partial derivative. However, For any partial derivative we get a different $c_x_i$, And I couldn`t find a way to relate $g$ back to $f$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, I know I can break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same,So I got stuck here too.
I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking?
calculus real-analysis multivariable-calculus
asked Aug 2 at 12:06
Sar
3529
marked as duplicate by mfl, 5xum, Community♦ Aug 2 at 12:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
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This question already has an answer here:
Generalization of Mean Value Theorem to functions, $f: mathbbR^n rightarrow mathbbR^n$
2 answers
I`m trying to generalize Lagrange's mean value theorem for $f:A to mathbbR^n$
, differentiable in $A subset mathbbR^m$.
What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..tx_i,..,x_n$), where $x_j$ are any real numbers and $x_i$ is 1. We have $g'(t)=D_f(t)cdot(0,..,1,...0)=frac partial f partial x_i(t)$, And since $g_i(t):mathbbR to mathbbR$ , we can apply the single variable mean value theorem to say that there exists a point $c_x_iin [a,b] $ s.t $frac partial f partial x_i(c_x_i)=g_i'(c_x_i)=frac g_i(a)-g_i(b) a-b$ , for any partial derivative. However, For any partial derivative we get a different $c_x_i$, And I couldn`t find a way to relate $g$ back to $f$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, I know I can break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same,So I got stuck here too.
I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking?
calculus real-analysis multivariable-calculus
asked Aug 2 at 12:06
Sar
3529
marked as duplicate by mfl, 5xum, Community♦ Aug 2 at 12:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31
add a comment |Â
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0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Generalization of Mean Value Theorem to functions, $f: mathbbR^n rightarrow mathbbR^n$
2 answers
I`m trying to generalize Lagrange's mean value theorem for $f:A to mathbbR^n$
, differentiable in $A subset mathbbR^m$.
What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..tx_i,..,x_n$), where $x_j$ are any real numbers and $x_i$ is 1. We have $g'(t)=D_f(t)cdot(0,..,1,...0)=frac partial f partial x_i(t)$, And since $g_i(t):mathbbR to mathbbR$ , we can apply the single variable mean value theorem to say that there exists a point $c_x_iin [a,b] $ s.t $frac partial f partial x_i(c_x_i)=g_i'(c_x_i)=frac g_i(a)-g_i(b) a-b$ , for any partial derivative. However, For any partial derivative we get a different $c_x_i$, And I couldn`t find a way to relate $g$ back to $f$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, I know I can break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same,So I got stuck here too.
I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking?
calculus real-analysis multivariable-calculus
asked Aug 2 at 12:06
Sar
3529
This question already has an answer here:
Generalization of Mean Value Theorem to functions, $f: mathbbR^n rightarrow mathbbR^n$
2 answers
I`m trying to generalize Lagrange's mean value theorem for $f:A to mathbbR^n$
, differentiable in $A subset mathbbR^m$.
What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..tx_i,..,x_n$), where $x_j$ are any real numbers and $x_i$ is 1. We have $g'(t)=D_f(t)cdot(0,..,1,...0)=frac partial f partial x_i(t)$, And since $g_i(t):mathbbR to mathbbR$ , we can apply the single variable mean value theorem to say that there exists a point $c_x_iin [a,b] $ s.t $frac partial f partial x_i(c_x_i)=g_i'(c_x_i)=frac g_i(a)-g_i(b) a-b$ , for any partial derivative. However, For any partial derivative we get a different $c_x_i$, And I couldn`t find a way to relate $g$ back to $f$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, I know I can break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same,So I got stuck here too.
I wonder if this is a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?Can anyone help me complete the idea, or suggest a different way of thinking?
This question already has an answer here:
Generalization of Mean Value Theorem to functions, $f: mathbbR^n rightarrow mathbbR^n$
2 answers
calculus real-analysis multivariable-calculus
asked Aug 2 at 12:06
Sar
3529
asked Aug 2 at 12:06
Sar
3529
asked Aug 2 at 12:06
Sar
3529
asked Aug 2 at 12:06
Sar
3529
3529
marked as duplicate by mfl, 5xum, Community♦ Aug 2 at 12:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by mfl, 5xum, Community♦ Aug 2 at 12:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31
add a comment |Â
1
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31
1
1
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31
add a comment |Â
1 Answer
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You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_x_i$'s for different $i$.
As a counterexample, choose $f: [0,2pi] to mathbb R^2$ with $f(x) = left( cos(x), sin(x)right)$. Then $f(2pi) - f(0) = left(0, 0 right)$, but $f'(x) = left( - sin(x), cos(x)right)$ never assumes this value, as $sin$ and $cos$ have no mutual roots.
answered Aug 2 at 12:13
Stefan
3,0041116
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_x_i$'s for different $i$.
As a counterexample, choose $f: [0,2pi] to mathbb R^2$ with $f(x) = left( cos(x), sin(x)right)$. Then $f(2pi) - f(0) = left(0, 0 right)$, but $f'(x) = left( - sin(x), cos(x)right)$ never assumes this value, as $sin$ and $cos$ have no mutual roots.
answered Aug 2 at 12:13
Stefan
3,0041116
add a comment |Â
up vote
1
down vote
You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_x_i$'s for different $i$.
As a counterexample, choose $f: [0,2pi] to mathbb R^2$ with $f(x) = left( cos(x), sin(x)right)$. Then $f(2pi) - f(0) = left(0, 0 right)$, but $f'(x) = left( - sin(x), cos(x)right)$ never assumes this value, as $sin$ and $cos$ have no mutual roots.
answered Aug 2 at 12:13
Stefan
3,0041116
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_x_i$'s for different $i$.
As a counterexample, choose $f: [0,2pi] to mathbb R^2$ with $f(x) = left( cos(x), sin(x)right)$. Then $f(2pi) - f(0) = left(0, 0 right)$, but $f'(x) = left( - sin(x), cos(x)right)$ never assumes this value, as $sin$ and $cos$ have no mutual roots.
answered Aug 2 at 12:13
Stefan
3,0041116
You are stuck, because there is no solution to this problem! As you already mentioned, you will get different $c_x_i$'s for different $i$.
As a counterexample, choose $f: [0,2pi] to mathbb R^2$ with $f(x) = left( cos(x), sin(x)right)$. Then $f(2pi) - f(0) = left(0, 0 right)$, but $f'(x) = left( - sin(x), cos(x)right)$ never assumes this value, as $sin$ and $cos$ have no mutual roots.
answered Aug 2 at 12:13
Stefan
3,0041116
answered Aug 2 at 12:13
Stefan
3,0041116
answered Aug 2 at 12:13
Stefan
3,0041116
answered Aug 2 at 12:13
Stefan
3,0041116
3,0041116
add a comment |Â
add a comment |Â
1
See the following question, where an explicit solution is given: math.stackexchange.com/questions/2863266/…
– Christian Blatter
Aug 2 at 13:31