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Serge Lang: Multivariable Derivative Definition?

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1

1

$begingroup$

I'm confused about how the absolute value appears in the following formula (**) in Serge Lang's "Multivariable Calculus of Several Variables".

Context: He has already defined the single-variable derivative as the limit of the difference quotient. He proceeds to define the following function for a fixed x:

$g(h) = fracf(x+h)-f(x)h -f'(x)$

This is then rewritten as:

$f(x+h)-f(x) = f'(x)h + hg(h)$

After a comment on how this is defined for all values of $h$, since the definition works for $hnot = 0$ and we specifically define $g(0) = 0$, we get this paragraph:

"Furthermore, we can replace $h$ by $-h$ if we replace $g$ by $-g$. Thus we have shown that if $f$ is differentiable, there exists a function $g$ such that:

(**) $f(x+h)-f(x) = f'(x)h + |h|g(h)$

$lim_hrightarrow 0 g(h) = 0$

Question: Where does the absolute value come from? Also, is what the point of replacing $h$ and $g$ with their negatives.

I've tried calculating $-g(h)$ from the original formula, and I get:

$-g(h) = fracf(x+h)-f(x)-h +f'(x)$

which leads back to the original rewrite:

$f(x+h)-f(x) = f'(x)h + hg(h)$

So I get that a change to $-g$ corresponds with a change to $-h$.

But the right-hand side of (**) is: $f'(x)h + |h|g(h)$

Why is there absolute value in the second term, but not the first?

multivariable-calculus derivatives definition

share|cite|improve this question

asked Mar 12 at 21:17

Cassius12Cassius12

13812

$endgroup$

add a comment | 

1

1

$begingroup$

I'm confused about how the absolute value appears in the following formula (**) in Serge Lang's "Multivariable Calculus of Several Variables".

Context: He has already defined the single-variable derivative as the limit of the difference quotient. He proceeds to define the following function for a fixed x:

$g(h) = fracf(x+h)-f(x)h -f'(x)$

This is then rewritten as:

$f(x+h)-f(x) = f'(x)h + hg(h)$

After a comment on how this is defined for all values of $h$, since the definition works for $hnot = 0$ and we specifically define $g(0) = 0$, we get this paragraph:

"Furthermore, we can replace $h$ by $-h$ if we replace $g$ by $-g$. Thus we have shown that if $f$ is differentiable, there exists a function $g$ such that:

(**) $f(x+h)-f(x) = f'(x)h + |h|g(h)$

$lim_hrightarrow 0 g(h) = 0$

Question: Where does the absolute value come from? Also, is what the point of replacing $h$ and $g$ with their negatives.

I've tried calculating $-g(h)$ from the original formula, and I get:

$-g(h) = fracf(x+h)-f(x)-h +f'(x)$

which leads back to the original rewrite:

$f(x+h)-f(x) = f'(x)h + hg(h)$

So I get that a change to $-g$ corresponds with a change to $-h$.

But the right-hand side of (**) is: $f'(x)h + |h|g(h)$

Why is there absolute value in the second term, but not the first?

multivariable-calculus derivatives definition

share|cite|improve this question

asked Mar 12 at 21:17

Cassius12Cassius12

13812

$endgroup$

add a comment | 

$begingroup$

I'm confused about how the absolute value appears in the following formula (**) in Serge Lang's "Multivariable Calculus of Several Variables".

Context: He has already defined the single-variable derivative as the limit of the difference quotient. He proceeds to define the following function for a fixed x:

$g(h) = fracf(x+h)-f(x)h -f'(x)$

This is then rewritten as:

$f(x+h)-f(x) = f'(x)h + hg(h)$

After a comment on how this is defined for all values of $h$, since the definition works for $hnot = 0$ and we specifically define $g(0) = 0$, we get this paragraph:

"Furthermore, we can replace $h$ by $-h$ if we replace $g$ by $-g$. Thus we have shown that if $f$ is differentiable, there exists a function $g$ such that:

(**) $f(x+h)-f(x) = f'(x)h + |h|g(h)$

$lim_hrightarrow 0 g(h) = 0$

Question: Where does the absolute value come from? Also, is what the point of replacing $h$ and $g$ with their negatives.

I've tried calculating $-g(h)$ from the original formula, and I get:

$-g(h) = fracf(x+h)-f(x)-h +f'(x)$

which leads back to the original rewrite:

$f(x+h)-f(x) = f'(x)h + hg(h)$

So I get that a change to $-g$ corresponds with a change to $-h$.

But the right-hand side of (**) is: $f'(x)h + |h|g(h)$

Why is there absolute value in the second term, but not the first?

multivariable-calculus derivatives definition

share|cite|improve this question

asked Mar 12 at 21:17

Cassius12Cassius12

13812

$endgroup$

share|cite|improve this question

asked Mar 12 at 21:17

Cassius12Cassius12

13812

share|cite|improve this question

asked Mar 12 at 21:17

Cassius12Cassius12

13812

asked Mar 12 at 21:17

Cassius12Cassius12

13812

asked Mar 12 at 21:17

Cassius12Cassius12

13812

1 Answer
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0

$begingroup$

By definition of $g$, relationship

$$ f(x+h)-f(x) = f'(x)h + hg(h)$$

holds for any $h$.

Replacing $h$ by $-h$, we also have:

$$ f(x-h)-f(x) = f'(x)(-h) + (-h)g(-h),$$

which is equivalent to

$$ f(x-h)-f(x) = f'(x)(-h) + h(-g)(-h).$$

The function that exists and respects Lang's relationship is $alpha$ defined as

$alpha(h)=g(h)$ if $h$ is positive, $alpha(h)=(-g)(h)$ if $h$ is negative, and $alpha(0)=0$. Its limit at $0$ is $0$ because both $g$ and $-g$ have limits $0$ at $0$.

It is this new function (not original $g$) that gives:

$$ f(x+h)-f(x) = f'(x)h + |h|alpha(h)$$

for any $h$.

Note that for any non-zero $h$, we simply have:

$$ alpha(h)=frachg(h).$$

share|cite|improve this answer

edited Mar 13 at 22:43

answered Mar 13 at 21:50

ir7ir7

4,16311115

$endgroup$

add a comment | 

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0

$begingroup$

By definition of $g$, relationship

$$ f(x+h)-f(x) = f'(x)h + hg(h)$$

holds for any $h$.

Replacing $h$ by $-h$, we also have:

$$ f(x-h)-f(x) = f'(x)(-h) + (-h)g(-h),$$

which is equivalent to

$$ f(x-h)-f(x) = f'(x)(-h) + h(-g)(-h).$$

The function that exists and respects Lang's relationship is $alpha$ defined as

$alpha(h)=g(h)$ if $h$ is positive, $alpha(h)=(-g)(h)$ if $h$ is negative, and $alpha(0)=0$. Its limit at $0$ is $0$ because both $g$ and $-g$ have limits $0$ at $0$.

It is this new function (not original $g$) that gives:

$$ f(x+h)-f(x) = f'(x)h + |h|alpha(h)$$

for any $h$.

Note that for any non-zero $h$, we simply have:

$$ alpha(h)=frachg(h).$$

share|cite|improve this answer

edited Mar 13 at 22:43

answered Mar 13 at 21:50

ir7ir7

4,16311115

$endgroup$

add a comment | 

0

$begingroup$

By definition of $g$, relationship

$$ f(x+h)-f(x) = f'(x)h + hg(h)$$

holds for any $h$.

Replacing $h$ by $-h$, we also have:

$$ f(x-h)-f(x) = f'(x)(-h) + (-h)g(-h),$$

which is equivalent to

$$ f(x-h)-f(x) = f'(x)(-h) + h(-g)(-h).$$

The function that exists and respects Lang's relationship is $alpha$ defined as

$alpha(h)=g(h)$ if $h$ is positive, $alpha(h)=(-g)(h)$ if $h$ is negative, and $alpha(0)=0$. Its limit at $0$ is $0$ because both $g$ and $-g$ have limits $0$ at $0$.

It is this new function (not original $g$) that gives:

$$ f(x+h)-f(x) = f'(x)h + |h|alpha(h)$$

for any $h$.

Note that for any non-zero $h$, we simply have:

$$ alpha(h)=frachg(h).$$

share|cite|improve this answer

edited Mar 13 at 22:43

answered Mar 13 at 21:50

ir7ir7

4,16311115

$endgroup$

add a comment | 

$begingroup$

By definition of $g$, relationship

$$ f(x+h)-f(x) = f'(x)h + hg(h)$$

holds for any $h$.

Replacing $h$ by $-h$, we also have:

$$ f(x-h)-f(x) = f'(x)(-h) + (-h)g(-h),$$

which is equivalent to

$$ f(x-h)-f(x) = f'(x)(-h) + h(-g)(-h).$$

The function that exists and respects Lang's relationship is $alpha$ defined as

$alpha(h)=g(h)$ if $h$ is positive, $alpha(h)=(-g)(h)$ if $h$ is negative, and $alpha(0)=0$. Its limit at $0$ is $0$ because both $g$ and $-g$ have limits $0$ at $0$.

It is this new function (not original $g$) that gives:

$$ f(x+h)-f(x) = f'(x)h + |h|alpha(h)$$

for any $h$.

Note that for any non-zero $h$, we simply have:

$$ alpha(h)=frachg(h).$$

share|cite|improve this answer

edited Mar 13 at 22:43

answered Mar 13 at 21:50

ir7ir7

4,16311115

$endgroup$

share|cite|improve this answer

edited Mar 13 at 22:43

answered Mar 13 at 21:50

ir7ir7

4,16311115

answered Mar 13 at 21:50

ir7ir7

4,16311115

answered Mar 13 at 21:50

ir7ir7

4,16311115