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Show, using the definition of derivative, that $f:mathbbR^2rightarrowmathbbR$ defined by $f(x,y)=xy$ is differentiable everywhere.
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This question has the following hint:
Find a linear function $L(x,y)$ such that $f(x+h,y+k)-f(x,y)-L(h,k)=hk$. Note that $|hk|leq l^2$ where $l:=maxk$. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an $epsilon-delta$ proof.
My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued.
Let $h,kinmathbbR$ such that $|hk|leq l^2$ where $l:=mboxmaxk$, and suppose $L(h,k)=kx+hy$. Thus,
$$frac(x+h)(y+k)-xy-(kx+hy)sqrth^2+k^2=fracsqrth^2+k^2.$$
Then, note that
$$|h|^2leq h^2+k^2Rightarrow|h|leqsqrth^2+k^2Rightarrowfracsqrth^2+k^2leq1Rightarrowfracsqrth^2+k^2leq|k|.$$
Hence,
$$frac=fracsqrth^2+k^2leq|k|.$$
So,
$$lim_(h,k)rightarrow(0,0)frac=0.$$
Therefore, by the definition of derivative, $f$ is differentiable everywhere.
It seems to me that an $epsilon-delta$ proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved?
real-analysis multivariable-calculus proof-verification proof-writing
asked Jul 26 at 23:44
Atsina
513113
add a comment |Â
up vote
0
down vote
favorite
This question has the following hint:
Find a linear function $L(x,y)$ such that $f(x+h,y+k)-f(x,y)-L(h,k)=hk$. Note that $|hk|leq l^2$ where $l:=maxk$. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an $epsilon-delta$ proof.
My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued.
Let $h,kinmathbbR$ such that $|hk|leq l^2$ where $l:=mboxmaxk$, and suppose $L(h,k)=kx+hy$. Thus,
$$frac(x+h)(y+k)-xy-(kx+hy)sqrth^2+k^2=fracsqrth^2+k^2.$$
Then, note that
$$|h|^2leq h^2+k^2Rightarrow|h|leqsqrth^2+k^2Rightarrowfracsqrth^2+k^2leq1Rightarrowfracsqrth^2+k^2leq|k|.$$
Hence,
$$frac=fracsqrth^2+k^2leq|k|.$$
So,
$$lim_(h,k)rightarrow(0,0)frac=0.$$
Therefore, by the definition of derivative, $f$ is differentiable everywhere.
It seems to me that an $epsilon-delta$ proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved?
real-analysis multivariable-calculus proof-verification proof-writing
asked Jul 26 at 23:44
Atsina
513113
I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question has the following hint:
Find a linear function $L(x,y)$ such that $f(x+h,y+k)-f(x,y)-L(h,k)=hk$. Note that $|hk|leq l^2$ where $l:=maxk$. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an $epsilon-delta$ proof.
My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued.
Let $h,kinmathbbR$ such that $|hk|leq l^2$ where $l:=mboxmaxk$, and suppose $L(h,k)=kx+hy$. Thus,
$$frac(x+h)(y+k)-xy-(kx+hy)sqrth^2+k^2=fracsqrth^2+k^2.$$
Then, note that
$$|h|^2leq h^2+k^2Rightarrow|h|leqsqrth^2+k^2Rightarrowfracsqrth^2+k^2leq1Rightarrowfracsqrth^2+k^2leq|k|.$$
Hence,
$$frac=fracsqrth^2+k^2leq|k|.$$
So,
$$lim_(h,k)rightarrow(0,0)frac=0.$$
Therefore, by the definition of derivative, $f$ is differentiable everywhere.
It seems to me that an $epsilon-delta$ proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved?
real-analysis multivariable-calculus proof-verification proof-writing
asked Jul 26 at 23:44
Atsina
513113
This question has the following hint:
Find a linear function $L(x,y)$ such that $f(x+h,y+k)-f(x,y)-L(h,k)=hk$. Note that $|hk|leq l^2$ where $l:=maxk$. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an $epsilon-delta$ proof.
My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued.
Let $h,kinmathbbR$ such that $|hk|leq l^2$ where $l:=mboxmaxk$, and suppose $L(h,k)=kx+hy$. Thus,
$$frac(x+h)(y+k)-xy-(kx+hy)sqrth^2+k^2=fracsqrth^2+k^2.$$
Then, note that
$$|h|^2leq h^2+k^2Rightarrow|h|leqsqrth^2+k^2Rightarrowfracsqrth^2+k^2leq1Rightarrowfracsqrth^2+k^2leq|k|.$$
Hence,
$$frac=fracsqrth^2+k^2leq|k|.$$
So,
$$lim_(h,k)rightarrow(0,0)frac=0.$$
Therefore, by the definition of derivative, $f$ is differentiable everywhere.
It seems to me that an $epsilon-delta$ proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved?
real-analysis multivariable-calculus proof-verification proof-writing
asked Jul 26 at 23:44
Atsina
513113
edited Jul 27 at 4:36
asked Jul 26 at 23:44
Atsina
513113
asked Jul 26 at 23:44
Atsina
513113
asked Jul 26 at 23:44
Atsina
513113
513113
I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31
add a comment |Â
I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31
I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31
add a comment |Â
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I would write $L_x,y(h,k)$ for clarity as the linear function must depend on $x$ and $y$. Check that $L_x,y(h,k)=kx+hy$ works.
– Batominovski
Jul 26 at 23:48
In the second link you posted, the linear function and the full proof that the limit is zero is written. If your question is how to synthesize that formula, the idea is to assume that the limit is zero, and move the increment $(h,k)$ along the axes towards $(0,0)$. Moving along one axis gives you one of the coefficients, moving along the other will give you the other coefficient. Not only for this problem, but in general this shows that when there is differentiability the coefficients should be equal to the partial derivatives.
– user578878
Jul 26 at 23:52
@nextpuzzle I should have been more explicit...my question was in fact regarding the intuition for the choice in linear function. Your comment helped. I'll whip up a solution, but this honestly should be marked as a duplicate of that second link.
– Atsina
Jul 27 at 0:03
Well, that sequence of inequalities makes up an $epsilon$-$delta$ proof. It shows that one can choose $delta=epsilon$ since $|k|leq |(h,k)|<delta=epsilon$.
– user578878
Jul 27 at 0:31