Show that this definition of derivative implies the other one.

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I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] to mathbbR$ be a function. Let $x in [a,b]$.

Let $phi: (a,b) setminus xtomathbbR: t mapsto fracf(t)-
f(x)t-x$

Then we define $f'(x) = lim_t to x phi(t)$, provided that the
limit exists.

Here's the second one:

Let $f: E subseteq mathbbR to mathbbR$ be a function, $x in
E$ and $x$ a limit point of $E$. Put

$phi: E setminus x to mathbbR: t mapsto fracf(t)-
f(x)t-x $.

Then we define $f'(x) = lim_x to t phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.

So, in particular, I want to show that for every $x in [a,b]$

$lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$ exists $iff$ $lim_t to x, t in [a,b] setminus x fracf(t)-f(x)t-x$

and if one of the two exists, both are equal.

Clearly, $Leftarrow$ is satisfied (immediately from the limit definition).

For the other direction, put $q:= lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$. If $x in (a,b)$, then it is an interior point and the limit is equal to the other one.

WLOG, assume $x = a$. Let $epsilon > 0$. Choose $delta > 0$ such that $|q - fracf(t)-f(a)t-a| < epsilon$ for all $t in (a,b)$ with $0 < |t-x|< delta$.

Then, if $t in [a,b] setminus a$ with $0 < |t-a| < min $, then $t neq b, t neq a$ and we can make the quantity smaller than $epsilon$.

Does this seem correct? Are there any arguments against using the
second definition over the first one, as it seems more general?

asked 23 hours ago

Math_QED

6,29931344

1

Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other.
â€“Â Kavi Rama Murthy
21 hours ago

Thanks . And what is that reason if I may ask?
â€“Â Math_QED
21 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] to mathbbR$ be a function. Let $x in [a,b]$.

Let $phi: (a,b) setminus xtomathbbR: t mapsto fracf(t)-
f(x)t-x$

Then we define $f'(x) = lim_t to x phi(t)$, provided that the
limit exists.

Here's the second one:

Let $f: E subseteq mathbbR to mathbbR$ be a function, $x in
E$ and $x$ a limit point of $E$. Put

$phi: E setminus x to mathbbR: t mapsto fracf(t)-
f(x)t-x $.

Then we define $f'(x) = lim_x to t phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.

So, in particular, I want to show that for every $x in [a,b]$

$lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$ exists $iff$ $lim_t to x, t in [a,b] setminus x fracf(t)-f(x)t-x$

and if one of the two exists, both are equal.

Clearly, $Leftarrow$ is satisfied (immediately from the limit definition).

For the other direction, put $q:= lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$. If $x in (a,b)$, then it is an interior point and the limit is equal to the other one.

WLOG, assume $x = a$. Let $epsilon > 0$. Choose $delta > 0$ such that $|q - fracf(t)-f(a)t-a| < epsilon$ for all $t in (a,b)$ with $0 < |t-x|< delta$.

Then, if $t in [a,b] setminus a$ with $0 < |t-a| < min $, then $t neq b, t neq a$ and we can make the quantity smaller than $epsilon$.

Does this seem correct? Are there any arguments against using the
second definition over the first one, as it seems more general?

asked 23 hours ago

Math_QED

6,29931344

1

Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other.
â€“Â Kavi Rama Murthy
21 hours ago

Thanks . And what is that reason if I may ask?
â€“Â Math_QED
21 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] to mathbbR$ be a function. Let $x in [a,b]$.

Let $phi: (a,b) setminus xtomathbbR: t mapsto fracf(t)-
f(x)t-x$

Then we define $f'(x) = lim_t to x phi(t)$, provided that the
limit exists.

Here's the second one:

Let $f: E subseteq mathbbR to mathbbR$ be a function, $x in
E$ and $x$ a limit point of $E$. Put

$phi: E setminus x to mathbbR: t mapsto fracf(t)-
f(x)t-x $.

Then we define $f'(x) = lim_x to t phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.

So, in particular, I want to show that for every $x in [a,b]$

$lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$ exists $iff$ $lim_t to x, t in [a,b] setminus x fracf(t)-f(x)t-x$

and if one of the two exists, both are equal.

Clearly, $Leftarrow$ is satisfied (immediately from the limit definition).

For the other direction, put $q:= lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$. If $x in (a,b)$, then it is an interior point and the limit is equal to the other one.

WLOG, assume $x = a$. Let $epsilon > 0$. Choose $delta > 0$ such that $|q - fracf(t)-f(a)t-a| < epsilon$ for all $t in (a,b)$ with $0 < |t-x|< delta$.

Then, if $t in [a,b] setminus a$ with $0 < |t-a| < min $, then $t neq b, t neq a$ and we can make the quantity smaller than $epsilon$.

Does this seem correct? Are there any arguments against using the
second definition over the first one, as it seems more general?

asked 23 hours ago

Math_QED

6,29931344

I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] to mathbbR$ be a function. Let $x in [a,b]$.

Let $phi: (a,b) setminus xtomathbbR: t mapsto fracf(t)-
f(x)t-x$

Then we define $f'(x) = lim_t to x phi(t)$, provided that the
limit exists.

Here's the second one:

Let $f: E subseteq mathbbR to mathbbR$ be a function, $x in
E$ and $x$ a limit point of $E$. Put

$phi: E setminus x to mathbbR: t mapsto fracf(t)-
f(x)t-x $.

Then we define $f'(x) = lim_x to t phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.

So, in particular, I want to show that for every $x in [a,b]$

$lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$ exists $iff$ $lim_t to x, t in [a,b] setminus x fracf(t)-f(x)t-x$

and if one of the two exists, both are equal.

Clearly, $Leftarrow$ is satisfied (immediately from the limit definition).

For the other direction, put $q:= lim_t to x, t in (a,b) setminus x fracf(t)-f(x)t-x$. If $x in (a,b)$, then it is an interior point and the limit is equal to the other one.

WLOG, assume $x = a$. Let $epsilon > 0$. Choose $delta > 0$ such that $|q - fracf(t)-f(a)t-a| < epsilon$ for all $t in (a,b)$ with $0 < |t-x|< delta$.

Then, if $t in [a,b] setminus a$ with $0 < |t-a| < min $, then $t neq b, t neq a$ and we can make the quantity smaller than $epsilon$.

Does this seem correct? Are there any arguments against using the
second definition over the first one, as it seems more general?

asked 23 hours ago

Math_QED

6,29931344

asked 23 hours ago

Math_QED

6,29931344

asked 23 hours ago

Math_QED

6,29931344

asked 23 hours ago

Math_QED

6,29931344

1

Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other.
â€“Â Kavi Rama Murthy
21 hours ago

Thanks . And what is that reason if I may ask?
â€“Â Math_QED
21 hours ago

add a commentÂ |Â

1

Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other.
â€“Â Kavi Rama Murthy
21 hours ago

Thanks . And what is that reason if I may ask?
â€“Â Math_QED
21 hours ago

Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other.
â€“Â Kavi Rama Murthy
21 hours ago

Thanks . And what is that reason if I may ask?
â€“Â Math_QED
21 hours ago

add a commentÂ |Â

1 Answer
1

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up vote
1
down vote

accepted

Let us generalize the second definition. For $x in mathbbR$ we denote by $mathfrakN(x)$ the set of all neighborhoods of $x$ in $mathbbR$ (a neighborhood of $x$ is a set $N subset mathbbR$ such that $(x -varepsilon, x +varepsilon ) subset N$ for some $varepsilon > 0$).

Let $x in E subset mathbbR$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N in mathfrakN(x)$ such that $x$ is a limit point of $N cap E$.

(3) For all $N in mathfrakN(x)$, $x$ is a limit point of $N cap E$.

Now let $f: E to mathbbR$ be a function and $x in E$ be a limit point of $E$. For each $N in mathfrakN(x)$ define
$$phi_N: N cap E setminus x to mathbbR: t mapsto fracf(t)-
f(x)t-x .$$
The map $phi$ from the second definition is given as $phi = phi_mathbbR$. If $lim_x to t phi_N (t)$ exists, we denote it by $f'_N(x)$. In case $N = mathbbR$ we simply write $f'(x)$.

The following are obvious:

(1) If $lim_x to t phi (t)$ exists, then $lim_x to t phi_N (t)$ exists for all $N in mathfrakN(x)$ and $f'_N(x) = f'(x)$.

(2) If $lim_x to t phi_N (t)$ exists for some $N in mathfrakN(x)$, then $lim_x to t phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $phi$ from the first definition will be denoted by $Phi$.

For $x in (a,b)$ we have $Phi = phi_(a,b)$, for $x = a$ we have $Phi = phi_(a-1,b)$ and for $x = b$ we have $Phi = phi_(a,b+1)$.

This shows that the first and the second definition are equivalent.

answered 18 hours ago

Paul Frost

3,323320

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

1

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

1

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

1

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Â |Â
show 4 more comments

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Let $x in E subset mathbbR$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N in mathfrakN(x)$ such that $x$ is a limit point of $N cap E$.

(3) For all $N in mathfrakN(x)$, $x$ is a limit point of $N cap E$.

The following are obvious:

(1) If $lim_x to t phi (t)$ exists, then $lim_x to t phi_N (t)$ exists for all $N in mathfrakN(x)$ and $f'_N(x) = f'(x)$.

(2) If $lim_x to t phi_N (t)$ exists for some $N in mathfrakN(x)$, then $lim_x to t phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $phi$ from the first definition will be denoted by $Phi$.

For $x in (a,b)$ we have $Phi = phi_(a,b)$, for $x = a$ we have $Phi = phi_(a-1,b)$ and for $x = b$ we have $Phi = phi_(a,b+1)$.

This shows that the first and the second definition are equivalent.

answered 18 hours ago

Paul Frost

3,323320

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

1

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

1

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

1

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Â |Â
show 4 more comments

up vote
1
down vote

accepted

Let $x in E subset mathbbR$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N in mathfrakN(x)$ such that $x$ is a limit point of $N cap E$.

(3) For all $N in mathfrakN(x)$, $x$ is a limit point of $N cap E$.

The following are obvious:

(1) If $lim_x to t phi (t)$ exists, then $lim_x to t phi_N (t)$ exists for all $N in mathfrakN(x)$ and $f'_N(x) = f'(x)$.

(2) If $lim_x to t phi_N (t)$ exists for some $N in mathfrakN(x)$, then $lim_x to t phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $phi$ from the first definition will be denoted by $Phi$.

For $x in (a,b)$ we have $Phi = phi_(a,b)$, for $x = a$ we have $Phi = phi_(a-1,b)$ and for $x = b$ we have $Phi = phi_(a,b+1)$.

This shows that the first and the second definition are equivalent.

answered 18 hours ago

Paul Frost

3,323320

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

1

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

1

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

1

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Â |Â
show 4 more comments

up vote
1
down vote

accepted

Let $x in E subset mathbbR$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N in mathfrakN(x)$ such that $x$ is a limit point of $N cap E$.

(3) For all $N in mathfrakN(x)$, $x$ is a limit point of $N cap E$.

The following are obvious:

(1) If $lim_x to t phi (t)$ exists, then $lim_x to t phi_N (t)$ exists for all $N in mathfrakN(x)$ and $f'_N(x) = f'(x)$.

(2) If $lim_x to t phi_N (t)$ exists for some $N in mathfrakN(x)$, then $lim_x to t phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $phi$ from the first definition will be denoted by $Phi$.

For $x in (a,b)$ we have $Phi = phi_(a,b)$, for $x = a$ we have $Phi = phi_(a-1,b)$ and for $x = b$ we have $Phi = phi_(a,b+1)$.

This shows that the first and the second definition are equivalent.

answered 18 hours ago

Paul Frost

3,323320

Let $x in E subset mathbbR$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N in mathfrakN(x)$ such that $x$ is a limit point of $N cap E$.

(3) For all $N in mathfrakN(x)$, $x$ is a limit point of $N cap E$.

The following are obvious:

(1) If $lim_x to t phi (t)$ exists, then $lim_x to t phi_N (t)$ exists for all $N in mathfrakN(x)$ and $f'_N(x) = f'(x)$.

(2) If $lim_x to t phi_N (t)$ exists for some $N in mathfrakN(x)$, then $lim_x to t phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $phi$ from the first definition will be denoted by $Phi$.

For $x in (a,b)$ we have $Phi = phi_(a,b)$, for $x = a$ we have $Phi = phi_(a-1,b)$ and for $x = b$ we have $Phi = phi_(a,b+1)$.

This shows that the first and the second definition are equivalent.

answered 18 hours ago

Paul Frost

3,323320

answered 18 hours ago

Paul Frost

3,323320

answered 18 hours ago

Paul Frost

3,323320

answered 18 hours ago

Paul Frost

3,323320

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

1

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

1

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

1

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Â |Â
show 4 more comments

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

1

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

1

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

1

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Thanks for your detailed answer! I was wondering if my approach would work too?
â€“Â Math_QED
18 hours ago

Yes, of course, your approch works. I only wanted to round off the picture.
â€“Â Paul Frost
18 hours ago

By the way, where did you find the second definition?
â€“Â Paul Frost
18 hours ago

Thank you! I personally prefer this definition, but one hardly ever finds it in the literature.
â€“Â Paul Frost
18 hours ago

I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x in E$ if there exist an open neighborhood $U$ of $x$ in $mathbbR$ and a differentiable function $hat f : U to mathbbR$ such that $hat f mid_U cap E = f mid_U cap E$. In my eyes this is more a theorem than a definition.
â€“Â Paul Frost
18 hours ago

Â |Â
show 4 more comments

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