# In derivative as linear approximation why does E(h)/h->0 as h->0

1. Jun 15, 2013

### Melac12

When I started learning Multivariable calc first we went back and developed a new notion of derivative as a linear approximation. And what we came up with was

F(a+h)=F(a)+mh+E(h) *

where m is the derivative. Basically the function minus the line tangent to point a. However there is a requirement that E(h)/h->0 as h->0. In the book and my proff say that we want E(h) to go faster to 0 then h, but that's not very mathematical! I asked my proff and he said that we don't have to devide by h it just makes approx better. The reason I am heaving trouble is because if I were to discovering the derivative for the 1st time using line approx (without knowing limit way) the condition for E(h)/h->0 as h->0 would not be obvious. It seems out of no wear. Why cant
E(h)->0 as h->0 on its own without heaving to worry what happens when to E(h)/h as h->0. Could some guide me through the logic for the requirement of E(h)/h->0 as h->0 Thank you.

PS: I know that if we rearrange * then E(h)/h->0 as h->0 works out but im asking if we just focus in this linear approx way only!

2. Jun 15, 2013

### SammyS

Staff Emeritus
I suppose E(h) is meant to be the error term.

Solve
F(a+h)=F(a)+mh+E(h)​
for
(F(a+h) - F(a))/h​
and take the limit as h → 0 .

3. Jun 15, 2013

### bolbteppa

Taking the single variable definition of a derivative as our motivation we have that

$$f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$$

We can remove the limits from this definition (explicitly) if we re-write the above as

$$f'(a) \ = \ \frac{f(a + h) - f(a)}{h} \ - \ e(h)$$

$$f'(a)h \ = \ f(a + h) - \ f(a) \ - \ e(h)h$$

$$f(a + h) \ = \ f(a) \ + \ f'(a)h \ + \ e(h)h$$

Where $$\lim_{h \rightarrow 0} e(h) = 0$$

thus you see the idea of an error term with a limit going to zero is a natural consequence of the classical definition of a derivative. Note the minus beside the error was only included to make things look nice on re-arrangement. Note that we could have called the error term in the last line something else, i.e.

$$f(a + h) \ = \ f(a) \ + \ f'(a)h \ + \ \nu (h)$$

so long as we specified that $$\lim_{h \rightarrow 0} \frac{\nu (h)}{h} = 0$$

Something we need to do because we want to be able to re-derive the classical form of the derivative as we'd originally defined it.

Now, for the multi-variable case you are following the same principles, you want to be able to re-derive the difference quotient idea so that you can define the directional derivative.

If we defined the derivative as

$$f(a + h) = f(a) + f'(a)h + E(h)$$

where E(h) goes to zero as h goes to zero then if you tried to derive the directional derivative using h = tv you'd end up with an error term E(tv)/t which makes no sense as far as I can see, whereas if you mimic the classical theory you naturally end up with the directional derivative as a theorem you can prove. In fact, some books (such as Apostol's Mathematical Analysis) start by defining the derivative as the directional derivative, which is a very very natural extension of the single variable derivative (probably the first thing you'd define yourself if you tried to define derivatives of several variables) & then he shows how this definition leads to the possibility of directional derivatives existing in every direction while the function not being continuous, in other words differentiability implies continuity does not hold in the multivariable case (the reason being the error term E(h) going to zero slower than the h in E(h)/h !) thus if you want to extend other basic single-variable results to multiple dimensions you need to look at derivatives from a different perspective, namely as a linear approximation & the above counter-example needs to be incorporated in whatever you do via the linear approximation methods!

4. Jun 16, 2013

### Melac12

Ok the explanation is good and I do realize that this linear approx. works really well with the original definition of derivative. But I am asking if we were formulating the derivative for the first time. If we were just discovering the derivative using line approximation, We have our

F(a)=L(a) L(x)=mx+b and since b=F(a)-ma so L(x)=F(a)+m(x-a)

now take the function and minus the line F(x)-L(x)=F(x)-F(a)-m(x-a) make x-a=h

so F(x)-L(x)=F(a+h)-F(a)-m(h) and we say that F(x)-L(x) is a function of h E(h) so

E(h)=F(a+h)-F(a)-m(h) so F(a+h)=F(a)+m(h)+E(h) is the derivative as the line

approximates the function. Wouldn't E(h)->0 as h->0 be enough. what would be my train of thought from here that would logically lead to hey we need to make sure E(h)/h->0 as h->0.

5. Jun 17, 2013

### bolbteppa

Ah I see, well if you send h to zero in your second last line:

"E(h)=F(a+h)-F(a)-m(h) so F(a+h)=F(a)+m(h)+E(h) is the derivative as the line"

what you get is:

0 = lim E(h) = lim [F(a+h)-F(a)-m(h)] = lim F(a + h) - lim F(a) - lim m(h) = F(a) - F(a) - m(0) = m(0)

Thus all this sentence amounts to is showing that m(0) = 0, it doesn't give us any information about what m actually is when h goes to zero (i.e. what m does at the point a) so we need a way to explicitly give more information about m at a. Therefore if you divide by h & take a limit (where E(h)/h goes to zero) you have

0 = lim [E(h)/h] = lim [F(a + h) - F(a)]/h - lim m(h)/h = lim [F(a + h) - F(a)]/h - lim m(h)/h = lim [F(a + h) - F(a)]/h - lim m

Thus m = lim [F(a + h) - F(a)]/h

This way we can actually establish what m is at the point a. Obviously if lim [E(h)/h] =/= 0 then you can't end up with m = lim [F(a + h) - F(a)]/h or a uniquely defined derivative. Hopefully that's more helpful!

6. Jun 17, 2013

### Melac12

I think that's very nice. So your saying that we make the E(h)/h->0 as h->0 rule because it actually helps us see what the derivative has to be. My proff said that its just make approximation even better then if its just E(h)->0 as h->0. I think this is what he meant and extra condition that helps us actually see what the derivative is dong and makes approximation better.
Thank you very much that question bothered, but I think I get it now.

7. Jun 17, 2013

### HallsofIvy

Staff Emeritus
If we do NOT require that, the "derivative" would not be unique.