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Taylor approximation of function in two variables
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Let $f(x,y)$ be a real-valued function in two variables.
Let $x_n to x_0$ and $y_n to y_0$ as $n to infty$.
We assume $f(x,y)$ is differentiable in $x$ at $(x_0, y_0)$ with derivative $f'_x(x_0,y_0)$. Further suppose $f'_x(x,y)$ is continuous at $(x_0,y_0)$.
Question: Consider the following Taylor expansion about $(x_0,y_n)$:
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
Is it neccesary true that $R_n(x_n,y_n) = o(|x_n - x_0|)$?
Clearly if all instances of $y_n$ are replaced by $y_0$, the Taylor expansion is correct based on the definition of differentiability. And if $f(x,y)$ is differentiable in both $x$ and $y$ at $(x_0,y_0)$, we have
$$f(x_n,y_n) = f(x_0,y_0) + f'_x(x_0,y_0)(x_n - x_0) + f'_y(x_0,y_0)(y_n - y_0) + R_n^*(x_n,y_n)$$
with $R^*_n(x_n,y_n) = o(|x_n - x_0| + |y_n - y_0|)$.
However, if we don't have differentiability in the two variables, I wonder if the Taylor expansion in the question is correct.
derivatives continuity taylor-expansion partial-derivative
asked Jul 15 at 4:22
Guillaume F.
351211
add a comment |Â
up vote
1
down vote
favorite
Let $f(x,y)$ be a real-valued function in two variables.
Let $x_n to x_0$ and $y_n to y_0$ as $n to infty$.
We assume $f(x,y)$ is differentiable in $x$ at $(x_0, y_0)$ with derivative $f'_x(x_0,y_0)$. Further suppose $f'_x(x,y)$ is continuous at $(x_0,y_0)$.
Question: Consider the following Taylor expansion about $(x_0,y_n)$:
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
Is it neccesary true that $R_n(x_n,y_n) = o(|x_n - x_0|)$?
Clearly if all instances of $y_n$ are replaced by $y_0$, the Taylor expansion is correct based on the definition of differentiability. And if $f(x,y)$ is differentiable in both $x$ and $y$ at $(x_0,y_0)$, we have
$$f(x_n,y_n) = f(x_0,y_0) + f'_x(x_0,y_0)(x_n - x_0) + f'_y(x_0,y_0)(y_n - y_0) + R_n^*(x_n,y_n)$$
with $R^*_n(x_n,y_n) = o(|x_n - x_0| + |y_n - y_0|)$.
However, if we don't have differentiability in the two variables, I wonder if the Taylor expansion in the question is correct.
derivatives continuity taylor-expansion partial-derivative
asked Jul 15 at 4:22
Guillaume F.
351211
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f(x,y)$ be a real-valued function in two variables.
Let $x_n to x_0$ and $y_n to y_0$ as $n to infty$.
We assume $f(x,y)$ is differentiable in $x$ at $(x_0, y_0)$ with derivative $f'_x(x_0,y_0)$. Further suppose $f'_x(x,y)$ is continuous at $(x_0,y_0)$.
Question: Consider the following Taylor expansion about $(x_0,y_n)$:
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
Is it neccesary true that $R_n(x_n,y_n) = o(|x_n - x_0|)$?
Clearly if all instances of $y_n$ are replaced by $y_0$, the Taylor expansion is correct based on the definition of differentiability. And if $f(x,y)$ is differentiable in both $x$ and $y$ at $(x_0,y_0)$, we have
$$f(x_n,y_n) = f(x_0,y_0) + f'_x(x_0,y_0)(x_n - x_0) + f'_y(x_0,y_0)(y_n - y_0) + R_n^*(x_n,y_n)$$
with $R^*_n(x_n,y_n) = o(|x_n - x_0| + |y_n - y_0|)$.
However, if we don't have differentiability in the two variables, I wonder if the Taylor expansion in the question is correct.
derivatives continuity taylor-expansion partial-derivative
asked Jul 15 at 4:22
Guillaume F.
351211
Let $f(x,y)$ be a real-valued function in two variables.
Let $x_n to x_0$ and $y_n to y_0$ as $n to infty$.
We assume $f(x,y)$ is differentiable in $x$ at $(x_0, y_0)$ with derivative $f'_x(x_0,y_0)$. Further suppose $f'_x(x,y)$ is continuous at $(x_0,y_0)$.
Question: Consider the following Taylor expansion about $(x_0,y_n)$:
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
Is it neccesary true that $R_n(x_n,y_n) = o(|x_n - x_0|)$?
Clearly if all instances of $y_n$ are replaced by $y_0$, the Taylor expansion is correct based on the definition of differentiability. And if $f(x,y)$ is differentiable in both $x$ and $y$ at $(x_0,y_0)$, we have
$$f(x_n,y_n) = f(x_0,y_0) + f'_x(x_0,y_0)(x_n - x_0) + f'_y(x_0,y_0)(y_n - y_0) + R_n^*(x_n,y_n)$$
with $R^*_n(x_n,y_n) = o(|x_n - x_0| + |y_n - y_0|)$.
However, if we don't have differentiability in the two variables, I wonder if the Taylor expansion in the question is correct.
derivatives continuity taylor-expansion partial-derivative
asked Jul 15 at 4:22
Guillaume F.
351211
edited Jul 15 at 21:01
asked Jul 15 at 4:22
Guillaume F.
351211
asked Jul 15 at 4:22
Guillaume F.
351211
asked Jul 15 at 4:22
Guillaume F.
351211
351211
add a comment |Â
add a comment |Â
2 Answers
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Counterexample: Define $f$ to be $0$ on the set $xcup (x,y):yle 0,$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$
answered Jul 15 at 7:11
zhw.
66k42870
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up vote
0
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accepted
Yes. continuity of $f_x'(x,y)$ at $(x_0,y_0)$ is enough.
Proof:
Because $f_x'(x,y)$ is continuous at $(x_0,y_0)$, there exists a closed ball $mathcalB$ centered at $(x_0,y_0)$ such as $f'_x(x,y)$ exists. For $n$ high enough, both $x_n$ and $y_n$ are in the ball, and we can thus write
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
with
$$fracR_n(x_n,y_n)x_n - x_0 = fracf(x_n,y_n) - f(x_0,y_n)x_n - x_0 - f'_x(x_0,y_n) $$
We need to find a sufficient condition for
$$ fracR_n(x_n,y_n)x_n - x_0 = o(1)$$
For $n$ high enough so that $x_n$ and $y_n$ are in the ball $mathcalB$, we have, from the Mean Value Theorem,
$$frac f(x_n,y_n) - f(x_0,y_n)x_n - x_0 = f'_x(tildex_n,y_n)$$
where $tildex_n$ is between $x_n$ and $x_0$.
Hence, together with the triangle inequality,
$$beginalign
left|fracR_n(x_n,y_n)x_n - x_0right|
&= left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| \
&le left| f'_x(tildex_n,y_n) - f'_x(x_0,y_0) right| + left| f'_x(x_0,y_n) - f'_x(x_0,y_0) right|\
&= o(1)
endalign $$
with the last true because of the continuity of the derivative at $(x_0,y_0)$.
This proves the result.
Note 1: Essentially the same proof can be used when $x$ and $y$ are vectors. However, if $x$ is a vector, we need that $f$ be differentiable in $x$ in a neighborhood of $(x_0,y_0)$ to justify the use of the mean value theorem.
Note 2: Instead of the continuity of $f_x'(x,y)$ at $(x_0,y_0)$, we can assume that at $x_0$ and $y$ on a closed ball $mathcalB$ centered at $y_0$ , we have $f_x'(x,y)$ continuous in $x$ uniformly in $y$, so that, for any $x_n to 0$,
$$sup_y in mathcalBleft| f'_x(x_n,y) - f'_x(x_0,y) right| = o(1)$$
In which case, for $n$ large enough so that $y_n$ is in the ball,
$$left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| le sup_y in mathcalBleft| f'_x(tildex_n,y) - f'_x(x_0,y) right| = o(1)$$
answered Jul 15 at 4:37
Guillaume F.
351211
add a comment |Â
2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Counterexample: Define $f$ to be $0$ on the set $xcup (x,y):yle 0,$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$
answered Jul 15 at 7:11
zhw.
66k42870
add a comment |Â
up vote
1
down vote
Counterexample: Define $f$ to be $0$ on the set $xcup (x,y):yle 0,$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$
answered Jul 15 at 7:11
zhw.
66k42870
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Counterexample: Define $f$ to be $0$ on the set $xcup (x,y):yle 0,$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$
answered Jul 15 at 7:11
zhw.
66k42870
Counterexample: Define $f$ to be $0$ on the set $xcup (x,y):yle 0,$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$
answered Jul 15 at 7:11
zhw.
66k42870
answered Jul 15 at 7:11
zhw.
66k42870
answered Jul 15 at 7:11
zhw.
66k42870
answered Jul 15 at 7:11
zhw.
66k42870
66k42870
add a comment |Â
add a comment |Â
up vote
0
down vote
accepted
Yes. continuity of $f_x'(x,y)$ at $(x_0,y_0)$ is enough.
Proof:
Because $f_x'(x,y)$ is continuous at $(x_0,y_0)$, there exists a closed ball $mathcalB$ centered at $(x_0,y_0)$ such as $f'_x(x,y)$ exists. For $n$ high enough, both $x_n$ and $y_n$ are in the ball, and we can thus write
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
with
$$fracR_n(x_n,y_n)x_n - x_0 = fracf(x_n,y_n) - f(x_0,y_n)x_n - x_0 - f'_x(x_0,y_n) $$
We need to find a sufficient condition for
$$ fracR_n(x_n,y_n)x_n - x_0 = o(1)$$
For $n$ high enough so that $x_n$ and $y_n$ are in the ball $mathcalB$, we have, from the Mean Value Theorem,
$$frac f(x_n,y_n) - f(x_0,y_n)x_n - x_0 = f'_x(tildex_n,y_n)$$
where $tildex_n$ is between $x_n$ and $x_0$.
Hence, together with the triangle inequality,
$$beginalign
left|fracR_n(x_n,y_n)x_n - x_0right|
&= left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| \
&le left| f'_x(tildex_n,y_n) - f'_x(x_0,y_0) right| + left| f'_x(x_0,y_n) - f'_x(x_0,y_0) right|\
&= o(1)
endalign $$
with the last true because of the continuity of the derivative at $(x_0,y_0)$.
This proves the result.
Note 1: Essentially the same proof can be used when $x$ and $y$ are vectors. However, if $x$ is a vector, we need that $f$ be differentiable in $x$ in a neighborhood of $(x_0,y_0)$ to justify the use of the mean value theorem.
Note 2: Instead of the continuity of $f_x'(x,y)$ at $(x_0,y_0)$, we can assume that at $x_0$ and $y$ on a closed ball $mathcalB$ centered at $y_0$ , we have $f_x'(x,y)$ continuous in $x$ uniformly in $y$, so that, for any $x_n to 0$,
$$sup_y in mathcalBleft| f'_x(x_n,y) - f'_x(x_0,y) right| = o(1)$$
In which case, for $n$ large enough so that $y_n$ is in the ball,
$$left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| le sup_y in mathcalBleft| f'_x(tildex_n,y) - f'_x(x_0,y) right| = o(1)$$
answered Jul 15 at 4:37
Guillaume F.
351211
add a comment |Â
up vote
0
down vote
accepted
Yes. continuity of $f_x'(x,y)$ at $(x_0,y_0)$ is enough.
Proof:
Because $f_x'(x,y)$ is continuous at $(x_0,y_0)$, there exists a closed ball $mathcalB$ centered at $(x_0,y_0)$ such as $f'_x(x,y)$ exists. For $n$ high enough, both $x_n$ and $y_n$ are in the ball, and we can thus write
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
with
$$fracR_n(x_n,y_n)x_n - x_0 = fracf(x_n,y_n) - f(x_0,y_n)x_n - x_0 - f'_x(x_0,y_n) $$
We need to find a sufficient condition for
$$ fracR_n(x_n,y_n)x_n - x_0 = o(1)$$
For $n$ high enough so that $x_n$ and $y_n$ are in the ball $mathcalB$, we have, from the Mean Value Theorem,
$$frac f(x_n,y_n) - f(x_0,y_n)x_n - x_0 = f'_x(tildex_n,y_n)$$
where $tildex_n$ is between $x_n$ and $x_0$.
Hence, together with the triangle inequality,
$$beginalign
left|fracR_n(x_n,y_n)x_n - x_0right|
&= left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| \
&le left| f'_x(tildex_n,y_n) - f'_x(x_0,y_0) right| + left| f'_x(x_0,y_n) - f'_x(x_0,y_0) right|\
&= o(1)
endalign $$
with the last true because of the continuity of the derivative at $(x_0,y_0)$.
This proves the result.
Note 1: Essentially the same proof can be used when $x$ and $y$ are vectors. However, if $x$ is a vector, we need that $f$ be differentiable in $x$ in a neighborhood of $(x_0,y_0)$ to justify the use of the mean value theorem.
Note 2: Instead of the continuity of $f_x'(x,y)$ at $(x_0,y_0)$, we can assume that at $x_0$ and $y$ on a closed ball $mathcalB$ centered at $y_0$ , we have $f_x'(x,y)$ continuous in $x$ uniformly in $y$, so that, for any $x_n to 0$,
$$sup_y in mathcalBleft| f'_x(x_n,y) - f'_x(x_0,y) right| = o(1)$$
In which case, for $n$ large enough so that $y_n$ is in the ball,
$$left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| le sup_y in mathcalBleft| f'_x(tildex_n,y) - f'_x(x_0,y) right| = o(1)$$
answered Jul 15 at 4:37
Guillaume F.
351211
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes. continuity of $f_x'(x,y)$ at $(x_0,y_0)$ is enough.
Proof:
Because $f_x'(x,y)$ is continuous at $(x_0,y_0)$, there exists a closed ball $mathcalB$ centered at $(x_0,y_0)$ such as $f'_x(x,y)$ exists. For $n$ high enough, both $x_n$ and $y_n$ are in the ball, and we can thus write
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
with
$$fracR_n(x_n,y_n)x_n - x_0 = fracf(x_n,y_n) - f(x_0,y_n)x_n - x_0 - f'_x(x_0,y_n) $$
We need to find a sufficient condition for
$$ fracR_n(x_n,y_n)x_n - x_0 = o(1)$$
For $n$ high enough so that $x_n$ and $y_n$ are in the ball $mathcalB$, we have, from the Mean Value Theorem,
$$frac f(x_n,y_n) - f(x_0,y_n)x_n - x_0 = f'_x(tildex_n,y_n)$$
where $tildex_n$ is between $x_n$ and $x_0$.
Hence, together with the triangle inequality,
$$beginalign
left|fracR_n(x_n,y_n)x_n - x_0right|
&= left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| \
&le left| f'_x(tildex_n,y_n) - f'_x(x_0,y_0) right| + left| f'_x(x_0,y_n) - f'_x(x_0,y_0) right|\
&= o(1)
endalign $$
with the last true because of the continuity of the derivative at $(x_0,y_0)$.
This proves the result.
Note 1: Essentially the same proof can be used when $x$ and $y$ are vectors. However, if $x$ is a vector, we need that $f$ be differentiable in $x$ in a neighborhood of $(x_0,y_0)$ to justify the use of the mean value theorem.
Note 2: Instead of the continuity of $f_x'(x,y)$ at $(x_0,y_0)$, we can assume that at $x_0$ and $y$ on a closed ball $mathcalB$ centered at $y_0$ , we have $f_x'(x,y)$ continuous in $x$ uniformly in $y$, so that, for any $x_n to 0$,
$$sup_y in mathcalBleft| f'_x(x_n,y) - f'_x(x_0,y) right| = o(1)$$
In which case, for $n$ large enough so that $y_n$ is in the ball,
$$left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| le sup_y in mathcalBleft| f'_x(tildex_n,y) - f'_x(x_0,y) right| = o(1)$$
answered Jul 15 at 4:37
Guillaume F.
351211
Yes. continuity of $f_x'(x,y)$ at $(x_0,y_0)$ is enough.
Proof:
Because $f_x'(x,y)$ is continuous at $(x_0,y_0)$, there exists a closed ball $mathcalB$ centered at $(x_0,y_0)$ such as $f'_x(x,y)$ exists. For $n$ high enough, both $x_n$ and $y_n$ are in the ball, and we can thus write
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n - x_0) + R_n(x_n,y_n)$$
with
$$fracR_n(x_n,y_n)x_n - x_0 = fracf(x_n,y_n) - f(x_0,y_n)x_n - x_0 - f'_x(x_0,y_n) $$
We need to find a sufficient condition for
$$ fracR_n(x_n,y_n)x_n - x_0 = o(1)$$
For $n$ high enough so that $x_n$ and $y_n$ are in the ball $mathcalB$, we have, from the Mean Value Theorem,
$$frac f(x_n,y_n) - f(x_0,y_n)x_n - x_0 = f'_x(tildex_n,y_n)$$
where $tildex_n$ is between $x_n$ and $x_0$.
Hence, together with the triangle inequality,
$$beginalign
left|fracR_n(x_n,y_n)x_n - x_0right|
&= left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| \
&le left| f'_x(tildex_n,y_n) - f'_x(x_0,y_0) right| + left| f'_x(x_0,y_n) - f'_x(x_0,y_0) right|\
&= o(1)
endalign $$
with the last true because of the continuity of the derivative at $(x_0,y_0)$.
This proves the result.
Note 1: Essentially the same proof can be used when $x$ and $y$ are vectors. However, if $x$ is a vector, we need that $f$ be differentiable in $x$ in a neighborhood of $(x_0,y_0)$ to justify the use of the mean value theorem.
Note 2: Instead of the continuity of $f_x'(x,y)$ at $(x_0,y_0)$, we can assume that at $x_0$ and $y$ on a closed ball $mathcalB$ centered at $y_0$ , we have $f_x'(x,y)$ continuous in $x$ uniformly in $y$, so that, for any $x_n to 0$,
$$sup_y in mathcalBleft| f'_x(x_n,y) - f'_x(x_0,y) right| = o(1)$$
In which case, for $n$ large enough so that $y_n$ is in the ball,
$$left| f'_x(tildex_n,y_n) - f'_x(x_0,y_n) right| le sup_y in mathcalBleft| f'_x(tildex_n,y) - f'_x(x_0,y) right| = o(1)$$
answered Jul 15 at 4:37
Guillaume F.
351211
edited Jul 17 at 20:18
answered Jul 15 at 4:37
Guillaume F.
351211
answered Jul 15 at 4:37
Guillaume F.
351211
answered Jul 15 at 4:37
Guillaume F.
351211
351211
add a comment |Â
add a comment |Â
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