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Prove that second partial derivatives does not depend on the order of differentiation [duplicate]
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This question already has an answer here:
Symmetry of second derivative - Sufficiency of twice-differentiability
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I'm trying to prove that if
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous in an open set containing $a in mathbbR^2$, then they are equal using the definition of derivative:
First, I say
$$ dfracpartial^2fpartial x partial y = dfracpartialpartial x left( dfracpartial fpartial y right) = dfracpartialpartial x left( lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = lim_k to 0 dfrac1k left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)h - lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = \ lim_k to 0 left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right) $$
Then I do the same thing for $ dfracpartial^2 fpartial y partial x $, and I get
$$ lim_h to 0 left( lim_k to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right)$$
My question is: am I allow to say that those limits are equal because both
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous on $a$, or do I need something extra?
Please let me know if the question is clear enough, or if I made some silly mistakes :)
Thanks in advance!
calculus multivariable-calculus derivatives
edited Jul 17 at 16:24
Tiwa Aina
2,576319
asked Jul 17 at 15:33
SantiMontouliu
255
marked as duplicate by Holo, Community♦ Jul 17 at 17:57
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
4
down vote
favorite
This question already has an answer here:
Symmetry of second derivative - Sufficiency of twice-differentiability
1 answer
I'm trying to prove that if
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous in an open set containing $a in mathbbR^2$, then they are equal using the definition of derivative:
First, I say
$$ dfracpartial^2fpartial x partial y = dfracpartialpartial x left( dfracpartial fpartial y right) = dfracpartialpartial x left( lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = lim_k to 0 dfrac1k left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)h - lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = \ lim_k to 0 left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right) $$
Then I do the same thing for $ dfracpartial^2 fpartial y partial x $, and I get
$$ lim_h to 0 left( lim_k to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right)$$
My question is: am I allow to say that those limits are equal because both
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous on $a$, or do I need something extra?
Please let me know if the question is clear enough, or if I made some silly mistakes :)
Thanks in advance!
calculus multivariable-calculus derivatives
edited Jul 17 at 16:24
Tiwa Aina
2,576319
asked Jul 17 at 15:33
SantiMontouliu
255
marked as duplicate by Holo, Community♦ Jul 17 at 17:57
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.
2
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
1
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
Symmetry of second derivative - Sufficiency of twice-differentiability
1 answer
I'm trying to prove that if
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous in an open set containing $a in mathbbR^2$, then they are equal using the definition of derivative:
First, I say
$$ dfracpartial^2fpartial x partial y = dfracpartialpartial x left( dfracpartial fpartial y right) = dfracpartialpartial x left( lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = lim_k to 0 dfrac1k left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)h - lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = \ lim_k to 0 left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right) $$
Then I do the same thing for $ dfracpartial^2 fpartial y partial x $, and I get
$$ lim_h to 0 left( lim_k to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right)$$
My question is: am I allow to say that those limits are equal because both
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous on $a$, or do I need something extra?
Please let me know if the question is clear enough, or if I made some silly mistakes :)
Thanks in advance!
calculus multivariable-calculus derivatives
edited Jul 17 at 16:24
Tiwa Aina
2,576319
asked Jul 17 at 15:33
SantiMontouliu
255
This question already has an answer here:
Symmetry of second derivative - Sufficiency of twice-differentiability
1 answer
I'm trying to prove that if
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous in an open set containing $a in mathbbR^2$, then they are equal using the definition of derivative:
First, I say
$$ dfracpartial^2fpartial x partial y = dfracpartialpartial x left( dfracpartial fpartial y right) = dfracpartialpartial x left( lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = lim_k to 0 dfrac1k left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)h - lim_h to 0 dfracf(x,y+h)-f(x,y)h right) = \ lim_k to 0 left( lim_h to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right) $$
Then I do the same thing for $ dfracpartial^2 fpartial y partial x $, and I get
$$ lim_h to 0 left( lim_k to 0 dfracf(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)kh right)$$
My question is: am I allow to say that those limits are equal because both
$$dfracpartial^2 fpartial x partial y quad textand quad dfracpartial^2 fpartial y partial x$$
are continuous on $a$, or do I need something extra?
Please let me know if the question is clear enough, or if I made some silly mistakes :)
Thanks in advance!
This question already has an answer here:
Symmetry of second derivative - Sufficiency of twice-differentiability
1 answer
calculus multivariable-calculus derivatives
edited Jul 17 at 16:24
Tiwa Aina
2,576319
asked Jul 17 at 15:33
SantiMontouliu
255
edited Jul 17 at 16:24
Tiwa Aina
2,576319
edited Jul 17 at 16:24
Tiwa Aina
2,576319
edited Jul 17 at 16:24
Tiwa Aina
2,576319
2,576319
asked Jul 17 at 15:33
SantiMontouliu
255
asked Jul 17 at 15:33
SantiMontouliu
255
asked Jul 17 at 15:33
SantiMontouliu
255
255
marked as duplicate by Holo, Community♦ Jul 17 at 17:57
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 Holo, Community♦ Jul 17 at 17:57
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.
2
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
1
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
2
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
1
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
2
2
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
1
1
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
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accepted
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.
answered Jul 17 at 15:47
gimusi
65.4k73584
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.
answered Jul 17 at 15:47
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
accepted
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.
answered Jul 17 at 15:47
gimusi
65.4k73584
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.
answered Jul 17 at 15:47
gimusi
65.4k73584
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.
answered Jul 17 at 15:47
gimusi
65.4k73584
answered Jul 17 at 15:47
gimusi
65.4k73584
answered Jul 17 at 15:47
gimusi
65.4k73584
answered Jul 17 at 15:47
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
2
This is known as Clairaut's Theorem. Googling "proof of Clairaut's Theorem" yields many results including this one: math.ubc.ca/~feldman/m105/mixedPartials.pdf
– Umberto P.
Jul 17 at 15:41
1
For weaker result from the link there is this pdf
– Holo
Jul 17 at 16:03
FYI, the act of taking a derivative is called "differentiation," not "derivation."
– Tiwa Aina
Jul 17 at 16:25
@SantiMontouliu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05