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As everyone knows, for a function $ f: \mathbb{R} \to \mathbb{R} $ and a point $ a \in \mathbb{R} $, we say that the derivative of $ f $ at $ a $ equals $ L $ if and only if $$ \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = L. $$

Now, most textbooks on particle mechanics offer the following recipe for defining the instantaneous velocity of a particle traveling along a straight line.

  • Let $ f: \mathbb{R} \to \mathbb{R} $ denote the displacement function of a particle traveling along the $ x $-axis, with respect to time. The goal is to define the instantaneous velocity of the particle at time $ t_{0} $.

  • Pick a sequence of non-degenerate bounded closed intervals $ ([a_{n},b_{n}])_{n \in \mathbb{N}} $ such that $ t_{0} \in (a_{n},b_{n}) $ for all $ n \in \mathbb{N} $ and $ \displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0 $.

  • Compute the sequence of average velocities $ (v_{\text{ave},n})_{n \in \mathbb{N}} $ of the particle over these closed intervals, i.e., $$ (v_{\text{ave},n})_{n \in \mathbb{N}} \stackrel{\text{def}}{=} \left( \frac{f(a_{n}) - f(b_{n})}{a_{n} - b_{n}} \right)_{n \in \mathbb{N}}. $$

  • Define the instantaneous velocity of the particle at time $ t_{0} $ as $ \displaystyle \lim_{n \to \infty} v_{\text{ave},n} $, if it exists.

This recipe is somehow proposing a new definition of the derivative of a function. Let me describe it in precise mathematical terms.

Consider $ f: \mathbb{R} \to \mathbb{R} $. Let $ a \in \mathbb{R} $, and let $ \Lambda $ be the set of non-degenerate bounded closed intervals $ I $ such that $ a \in I^{\circ} $. Define a partial ordering $ \preceq $ on $ \Lambda $ by $ I \preceq J \iff I \supseteq J $ for all $ I,J \in \Lambda $. One can easily verify that $ (\Lambda,\preceq) $ is a directed set. Next, define a net $ \lambda: \Lambda \to \mathbb{R} $ by $$ \forall I \in \Lambda: \quad \lambda(I) \stackrel{\text{def}}{=} \frac{f({\frak{l}}(I)) - f({\frak{r}}(I))}{{\frak{l}}(I) - {\frak{r}}(I)}, $$ where $ {\frak{l}}(I) $ and $ {\frak{r}}(I) $ denote the left- and right-endpoints of $ I $ respectively. Then define $$ f'(a) \stackrel{\text{def}}{=} \lim_{I \in \Lambda} \lambda(I), $$ if it exists.

Question: How can we show that this new definition of the derivative agrees with the usual one?

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I'm not familiar with nets. Is your proposal equivalent to defining $f'(a)=\lim\big(f(b_2)-f(b_1)\big)/(b_2-b_1)$ for $(b_1,b_2)\to(a,a), b_1<a<b_2$? –  Rahul Feb 2 '13 at 7:35
    
@AndréNicolas, that condition is implicit in the definition of limit of a net. –  Mariano Suárez-Alvarez Feb 2 '13 at 8:09
    
Sorry, I read "the set" as "a set." –  André Nicolas Feb 2 '13 at 8:14
    
@$ \mathbb{R}^{\text{n}} $: Yes. I’m sorry about my usage of nets, which is just a distraction. –  Haskell Curry Feb 3 '13 at 21:25

2 Answers 2

up vote 4 down vote accepted

In the following it is assumed that $a<0<b$ whenever the letters $a$ and $b$ appear. Then one has the following

Proposition. Let $f:\ U\to{\mathbb R}$ be defined in a neighborhood of $0$. If the limit $$\lim_{b-a\to0}{f(b)-f(a)\over b-a}=:p$$ exists then the one-sided limits $\lim_{x\to0-} f(x)$ and $\lim_{x\to0+} f(x)$ exist and are equal. If $f(0)$ equals this common limit then $f'(0)=p$.

Proof. Subtracting a linear function from $f$ we may assume $p=0$. Let an $\epsilon>0$ be given. Then by assumption there is a $\delta\in\ \bigl]0,{1\over4}\bigr]$ such that for $$-\delta < a < 0 < b <\delta$$ one has $$|f(b)-f(a)|<\epsilon(b-a)<{\epsilon\over2}\ .\tag{1}$$ It follows that for arbitrary $x$, $x'\in\ ]0,\delta[\ $ and $a:=-{\delta\over2}$ we are sure that $$|f(x)-f(x')\leq |f(x)-f(a)|+|f(a)-f(x')|<\epsilon\ .$$ This implies by Cauchy's criterion the existence of the $\lim_{x\to0+} f(x)$. Similarly for $\lim_{x\to0-} f(x)$, and then the equality of the two limits is obvious.

For the last statement we now assume $f(0)=\lim_{x\to0} f(x)$. Letting $a\to0-$ in $(1)$ we see that $$|f(b)-f(0)|\leq\epsilon b\qquad(0<b<\delta)\ ,$$ and as $\epsilon>0$ was arbitrary this implies $$\lim_{b\to0+}{f(b)-f(0)\over b}=0\ .$$ Arguing similarly about the lefthand limit we are done.$\qquad\square$

The converse of this proposition has been dealt with by coffeemath.

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The crucial thing is that you are proving the two one-sided limits must be the same if they exist. Then if $f(0)$ is defined as their common value, it works. What I noted in an example of the EDIT is that initially $f(0)$ might not have been that common value, and the "new" definition could say $f$ diff. at 0, however the difference quotient for this "alternate" derivative definiton does not refer to $f(0)$ anywhere, as the intervals about $0$ must contain $0$ in their interiors. So it looks like we have equivalence of the two defs, if one initially assumes $f$ is continuous at $0$. +1 –  coffeemath Feb 4 '13 at 13:36

The two versions can be compared using the equation (where $h,k>0$) $$\frac{f(x+h)-f(x-k)}{h+k}=\frac{h}{h+k}\frac{f(x+h)-f(x)}{h}+\frac{k}{h+k}\frac{f(x)-f(x-k)}{k}.$$

If $f$ is differentiable in the usual sense, then the two difference quotients on the right agree and have common value $f'(x)$ as $h,k \to 0.$ Then in the limit the right side is $f'(x)$ in the usual definition, since the right is an affine combination.

On the other hand, using the "new" derivative definition, one is allowed to let either of $h$ or $k$ approach zero first, and then have the other approach zero. Assuming $f$ is continuous, if we first let $h\to 0$ we end up with the usual left derivative, and if we first let $k$ go to zero we end up with the usual right derivative, so that these match and $f$ is differentiable in the usual sense with value as in the limit of the "new" definition.

EDIT: Haskell Curry has pointed out that one cannot let one of $h,k$ approach zero first, and then the other. In other words, during the approach of each to zero, both of $h,k$ must remain positive. Given this, the "new" definition will say that the function $f(x)=0$ for $x\ne 0$ and $f(0)=1$ has the derivative $0$ at $x=0$, but this function $f$ is not differentiable in the usual sense at $x=0$ (it's not even continuous there).

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Dear coffeemath, could you elaborate on the second paragraph? I’m a little doubtful about the legitimacy of letting one of the variables tend to $ 0 $ first. In my opinion, both variables should tend to $ 0 $ simultaneously. In other words, the convergence of both variables to $ (0,0) $ should be in the $ \| \cdot \|_{2} $-sense. –  Haskell Curry Feb 3 '13 at 21:22
    
@HaskellCurry Yes there is some delicacy I didn't discuss. But I think if say one wants first $h \to 0$ one can rig up an argument where first $h$ goes very near zero, then $k$ catches up some, then $h$ goes very very near zero, and so on. The idea would be to choose the simultaneous approach in such a way that $h$ went to zero at a sufficiently faster rate than did $k$, so that one could show one of the two sided derivatives of $f$ existed. Then the reverse with $k$ going to zero faster for the other one sided derivative of $f$. Vague I agree.. –  coffeemath Feb 4 '13 at 0:22
    
@HaskellCurry: If during the approach both of $h,k$ must remain positive, the "new" and usual derivative notions are not the same. See the "EDIT" in my answer above. –  coffeemath Feb 4 '13 at 8:37
    
The "affine combination" argument is not really convincing, since you're letting the variables $h$ and $k$ tend to zero in the difference quotients but leaving them as they are in the factors $h/(h+k)$ and $k/(h+k)$; this is not a valid operation. I think you'll need to write $\frac{1}{h}(f(x+h)-f(x)) = f'(x) + \rho_1(h)$, where $\rho_1(h) \to 0$ as $h \to 0$, and similarly for the other one, so that you can form the affine combination before going to the limit, and make some estimates to show that the remainder tends to zero. –  Hans Lundmark Feb 4 '13 at 19:02

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