Mixing
Prove this formula follows from a function being continuously differentiable?
$begingroup$
I'm studying for an exam in an electrical engineering course (stochastic process in dynamic systems), though this section is strictly on the math. A given practice problem (with no solution given, of course) is:
"Show that if $f: mathbb{R}^n rightarrow mathbb{R}$ is continuously differentiable around $x_0$, then $$f(x) = f(x_0) + A(x-x_0)+g(x)$$ where $$lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
Note that in the first equation, A is the matrix of partial derivatives of $f$ at $x_0$
The approach I've taken, which I don't think is correct, is as follows:
$f$ is continuously differentiable, then $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||}=0$$
So we have $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||} = lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
$$frac{lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0)}{lim_{xto x_0}||x-x_0||} = frac{lim_{xto x_0} ||g(x)||}{lim_{xto x_0} ||x-x_0||}$$
$$lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0) = lim_{xto x_0} ||g(x)||$$
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+||g(x)||$$
Since the codomain of $f(x)$ is $mathbb{R}$, we can assume the codomain of $g(x)$ is also $mathbb{R}$, meaning $||g(x)|| = sqrt{g(x)^2} = g(x)$, which leaves us with
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+g(x)$$
Here, I'm tempted to just drop the limits, leaving the equation which was to be shown. But I feel like this is wrong, because it seems strange to equate two functions because their limits are the same. For example, if $r(x)=x$ and $s(x)=sin(x)$, $$lim_{xto 0}r(x) = 0 = lim_{xto 0}s(x)$$
But certainly $xneq sin(x)$
Can anyone help me solve this?
limits multivariable-calculus derivatives
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
$endgroup$
|
show 1 more comment
$begingroup$
I'm studying for an exam in an electrical engineering course (stochastic process in dynamic systems), though this section is strictly on the math. A given practice problem (with no solution given, of course) is:
"Show that if $f: mathbb{R}^n rightarrow mathbb{R}$ is continuously differentiable around $x_0$, then $$f(x) = f(x_0) + A(x-x_0)+g(x)$$ where $$lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
Note that in the first equation, A is the matrix of partial derivatives of $f$ at $x_0$
The approach I've taken, which I don't think is correct, is as follows:
$f$ is continuously differentiable, then $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||}=0$$
So we have $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||} = lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
$$frac{lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0)}{lim_{xto x_0}||x-x_0||} = frac{lim_{xto x_0} ||g(x)||}{lim_{xto x_0} ||x-x_0||}$$
$$lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0) = lim_{xto x_0} ||g(x)||$$
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+||g(x)||$$
Since the codomain of $f(x)$ is $mathbb{R}$, we can assume the codomain of $g(x)$ is also $mathbb{R}$, meaning $||g(x)|| = sqrt{g(x)^2} = g(x)$, which leaves us with
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+g(x)$$
Here, I'm tempted to just drop the limits, leaving the equation which was to be shown. But I feel like this is wrong, because it seems strange to equate two functions because their limits are the same. For example, if $r(x)=x$ and $s(x)=sin(x)$, $$lim_{xto 0}r(x) = 0 = lim_{xto 0}s(x)$$
But certainly $xneq sin(x)$
Can anyone help me solve this?
limits multivariable-calculus derivatives
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
$endgroup$
$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
1
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50
|
show 1 more comment
$begingroup$
I'm studying for an exam in an electrical engineering course (stochastic process in dynamic systems), though this section is strictly on the math. A given practice problem (with no solution given, of course) is:
"Show that if $f: mathbb{R}^n rightarrow mathbb{R}$ is continuously differentiable around $x_0$, then $$f(x) = f(x_0) + A(x-x_0)+g(x)$$ where $$lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
Note that in the first equation, A is the matrix of partial derivatives of $f$ at $x_0$
The approach I've taken, which I don't think is correct, is as follows:
$f$ is continuously differentiable, then $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||}=0$$
So we have $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||} = lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
$$frac{lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0)}{lim_{xto x_0}||x-x_0||} = frac{lim_{xto x_0} ||g(x)||}{lim_{xto x_0} ||x-x_0||}$$
$$lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0) = lim_{xto x_0} ||g(x)||$$
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+||g(x)||$$
Since the codomain of $f(x)$ is $mathbb{R}$, we can assume the codomain of $g(x)$ is also $mathbb{R}$, meaning $||g(x)|| = sqrt{g(x)^2} = g(x)$, which leaves us with
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+g(x)$$
Here, I'm tempted to just drop the limits, leaving the equation which was to be shown. But I feel like this is wrong, because it seems strange to equate two functions because their limits are the same. For example, if $r(x)=x$ and $s(x)=sin(x)$, $$lim_{xto 0}r(x) = 0 = lim_{xto 0}s(x)$$
But certainly $xneq sin(x)$
Can anyone help me solve this?
limits multivariable-calculus derivatives
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
$endgroup$
I'm studying for an exam in an electrical engineering course (stochastic process in dynamic systems), though this section is strictly on the math. A given practice problem (with no solution given, of course) is:
"Show that if $f: mathbb{R}^n rightarrow mathbb{R}$ is continuously differentiable around $x_0$, then $$f(x) = f(x_0) + A(x-x_0)+g(x)$$ where $$lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
Note that in the first equation, A is the matrix of partial derivatives of $f$ at $x_0$
The approach I've taken, which I don't think is correct, is as follows:
$f$ is continuously differentiable, then $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||}=0$$
So we have $$lim_{xto x_0}frac{f(x)-f(x_0)-Adot(x-x_0)}{||x-x_0||} = lim_{xto x_0} frac{||g(x)||}{||x-x_0||}=0$$
$$frac{lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0)}{lim_{xto x_0}||x-x_0||} = frac{lim_{xto x_0} ||g(x)||}{lim_{xto x_0} ||x-x_0||}$$
$$lim_{xto x_0}f(x)-f(x_0)-Adot(x-x_0) = lim_{xto x_0} ||g(x)||$$
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+||g(x)||$$
Since the codomain of $f(x)$ is $mathbb{R}$, we can assume the codomain of $g(x)$ is also $mathbb{R}$, meaning $||g(x)|| = sqrt{g(x)^2} = g(x)$, which leaves us with
$$lim_{xto x_0}f(x) = lim_{xto x_0} f(x_0)+Adot(x-x_0)+g(x)$$
Here, I'm tempted to just drop the limits, leaving the equation which was to be shown. But I feel like this is wrong, because it seems strange to equate two functions because their limits are the same. For example, if $r(x)=x$ and $s(x)=sin(x)$, $$lim_{xto 0}r(x) = 0 = lim_{xto 0}s(x)$$
But certainly $xneq sin(x)$
Can anyone help me solve this?
limits multivariable-calculus derivatives
limits multivariable-calculus derivatives
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
edited Jan 24 at 4:59
W. MacTurk
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
asked Jan 23 at 23:31
W. MacTurkW. MacTurk
135
135
$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
1
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50
|
show 1 more comment
$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
1
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50
$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
1
1
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50
|
show 1 more comment
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$begingroup$
$g(x)=sin(x)$ does not fulfill $lim_{xto 0} frac{||g(x)||}{||x-0||}=0$
$endgroup$
– B.Swan
Jan 23 at 23:54
$begingroup$
Your argument is completely faulty. You have zeros in the denominator at many places.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:01
$begingroup$
I think the hint that will help you solve this is that if $f$ is continuously differentiable, then $A=df$ is continuous.
$endgroup$
– B.Swan
Jan 24 at 0:08
1
$begingroup$
Looks like just an unpacking of the definition of derivative.
$endgroup$
– Matematleta
Jan 24 at 0:20
$begingroup$
B.swan: sorry for the confusing notation. I was just providing an example of a case where two functions whose limits are equal are not themselves equal. g(x)=sin(x) is a separate function, not meant to satisfy the actual problem given.
$endgroup$
– W. MacTurk
Jan 24 at 4:50