Are Derivatives Merely Approximations?

[Femme_physics]

According to wiki

"The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. "

Is this to say that that derivatives are not 100% accurate? They're linear approximation? Which is confusing to me, with math being such an exact science an all, how can this be?

 

[hotvette]

The derivative is accurate. They are talking about one practical application of derivatives (i.e. forming a linear approximation of a function at a point).

 

[wisvuze]

continuous functions are locally approximately ( to any practical degree) constant
differentiable functions are locally approximately linear

 

[Mute]

What the wiki article means is that for a differentiable function f(x), at some point x = a,

$$f(x) \approx f(a) + f'(a)x$$
where f'(x) is the derivative of f(x) (which is exact) and f'(a) is the derivative of f(x) evaluated at the point a. This approximation holds for values of x "close to" x=a.

Basically, the derivative of f(x) is the slope of f(x) at a given point x, so f'(a) is the slope of f(x) at the point x=a. The tangent line through x=a is the best linear approximation of the curve f(x) at the point x=a.

(note that this all assumes things like f'(a) is non-zero, f'(x) exists at x=a, etc).

 

[lurflurf]

Who told you math was exact? Are you aware of approximation theory? Is it not math? If you want to be pedantic, it can be said that an approximation to the object x is an (exact) object y related in some way to x. That is we require x and y to have some properties in common, but allow or require that some of their properties differ. Often we desire a "best" approximation, that is an approximation that is (in some sense) as good as possible. Other times we may (due to the diffuculty of finding or marginal benifit of knowing the best approximation) be satisfied by any "good" (in some sense) approximation. For example we might say 1 is the best approximation in the uniform sense by a constant function to the function x/x (which is undefined when x=0). Another example we might say 7 is the best positive integer multiple of 7 approximation of the real number pi in the sense of smallest difference.

 

[Fredrik]

According to wiki

"The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. "

Is this to say that that derivatives are not 100% accurate? They're linear approximation? Which is confusing to me, with math being such an exact science an all, how can this be?

They're referring to the fact that f'(x)h is the first-order approximation of how much f(x) changes when you change x by h. More precisely,

$$f(x+h)-f(x)=f'(x)h+\mathcal O(h^2)$$

as h→0. (The "O" is explained here). There's nothing inexact about the derivative itself.

 

[mjpam]

They're referring to the fact that f'(x)h is the first-order approximation of how much f(x) changes when you change x by h. More precisely,

$$f(x+h)-f(x)=f'(x)h+\mathcal O(h^2)$$

as h→0. (The "O" is explained here). There's nothing inexact about the derivative itself.

Is this a restatement of the Fundamental Theorem of Calculus?

 

[Hurkyl]

Is this a restatement of the Fundamental Theorem of Calculus?

Nope. However

$$f(a+h) - f(a) = L h + o(h)$$​

which is a restatement of "f is differentiable at a, and has derivative L". (note the little-oh instead of the big-oh)

What he said is true if f is also twice-differentiable, but I'm not sure it's equivalent to twice-differentiability.

 

[Fredrik]

I meant the big O, but Hurkyl's statement is good too. His statement with the little o is equivalent to "f is differentiable", while my statement with the big O is just implied by it. (Hm, maybe f actually has to be twice differentiable for my statement to be valid. I don't have time to think about that now). If f is smooth (differentiable infinitely many times), then my O(h²) represents a power series such that each term contains a factor of hⁿ with n≥2.

 

[Hurkyl]

(Hm, maybe f actually has to be twice differentiable for my statement to be valid. I don't have time to think about that now).

Try f(x) = x^3/2

If f is smooth (differentiable infinitely many times), then my O(h²) represents a power series such that each term contains a factor of hⁿ with n≥2.

The truncated form of Taylor series let's you prove it with twice-differentiability -- the error term for the degree-1 Taylor polynomial involves the second derivative of f.

 

[lavinia]

According to wiki

"The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. "

Is this to say that that derivatives are not 100% accurate? They're linear approximation? Which is confusing to me, with math being such an exact science an all, how can this be?

The derivative is the best linear approximation to the change in a function for small changes. The key to calculus is that differentiable functions are nearly linear in the small.

 

[HallsofIvy]

According to wiki

"The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. "

Is this to say that that derivatives are not 100% accurate? They're linear approximation? Which is confusing to me, with math being such an exact science an all, how can this be?

This has been answered by a number of people but I want to point out that they are not saying that the derivative is being approximated. It is saying that the derivative is itslelf an approximation (the best possible linear approximation) to the function.

If we use the derivative to approximate the function itself, for example, if we use the function f(x)= x to approximate sin(x) (sin(0)= 0 and its derivative, cos(x), is equal to 1 at x= 0) we are using the tangent line to approximate sin(x)- for small values of x, $x\approx sin(x)$.

However, the derivative itself is the "exact" derivative- we are NOT approximating that. And it has many applications other than just being used to give a linear approximation.

 

[arildno]

The derivative yields the slope of the best linear approximation at any particular point; the derivative does NOT, all by itself provide us with that linear approximation.

 

[Studiot]

From your other threads you are just starting calculus so some of the comments may seem obscure.

with math being such an exact science an all

What is exact? Consider and compare these statements.

1)The probability of all possible outcomes of some action add up to exactly 1.

2)The birth rate in some country or other is exactly 1.7 births per family.

Arildno made a good comment. The derivative does not approximate the function. It is about (small) changes.

Also talking about small changes we have the exchange between Fredrik and Hurkyl. You will meet this idea many times over in technical maths.

A very small quantity multiplied by another very small quantity is a very very tiny quantity indeed.
Such a tiny quantity can often be considered insignificant compared to other quantities involved and ignored.

 

[HallsofIvy]

The derivative yields the slope of the best linear approximation at any particular point; the derivative does NOT, all by itself provide us with that linear approximation.

That is true if you interpret "derivative" to mean "slope of the tangent line" for a function from R to R. However, the wiki quote refers, I believe, the derivative of a function from Rⁿ to R^m, the linear transformation from Rⁿ to R^m that best approximtes the function at that point. That definition, with m= n= 1, would define the derivative to be the function, y= ax, rather than the number a.

 

[Studiot]

In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity. Conversely, the integral of the object's velocity over time is how much the object's position changes from the time when the integral begins to the time when the integral ends.

The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. In higher dimensions, the derivative of a function at a point is a linear transformation called the linearization

~Since we seem to be discussing the Wiki article perhaps the full text of the section involved is appropriate. I have colour coded my annotations.

1) The above text very clearly states that the derivative is about the change, not the original

2) The above text says 'describes' not is the best linear approximation. In the context and taking the illustrations it is clearly intended to show that the tangent line obtained via the derivative can be locally substitued for the function itself, but it does not say the derivative is the tangent line either.

3)This seems to be what HOI is referring to, but is an aside from the thrust of the article.

 

[lavinia]

In physics dt and ds often stand for small changes in t and s. Given these small changes you can form the ratio. ds/dt is a well defined ratio if you understand it to mean the change in s for a very small change in t. But this way of looking at it is not the same as the derivative. The derivative is a limit of these ratios as dt goes to zero.

Maybe historically people thought of limits of ratios as infinitesimal ratios. I don't know but I can sort of see thinking that way. Nowadays we do not do that. We have the derivative and we have the ratios, ds/dt, of changes in s for small changes in t. The notation is the same because the derivative is assumed to well approximate the ratio.

 

[Landau]

The derivative yields the slope of the best linear approximation at any particular point; the derivative does NOT, all by itself provide us with that linear approximation.

This is the classical view of the derivative of a function R\to R, which is somewhat misleading in that it implicitly leans on the canonical isomorphism
$\mathbb{R}\cong \text{Hom}_\mathbb{R}(\mathbb{R},\mathbb{R})$
$x\mapsto (1\mapsto x).$

If you are doing analysis in $\mathbb{R}^{n\geq 2}$, you'll learn that the derivative IS itself the best linear approximation.

 

[epenguin]

The rigour experts may correct me, but is there any difference, any more doubt, in saying there is a derivative of a smooth function at a point than there is in saying there is a function at a point or that there is a point? If we come to practice we can never exactly [STRIKE]pinpoint[/STRIKE] locate a point exactly in the way we think of it mathematically exactly? Somehow if we go from one place to another we have no difficulty in thinking that in between there are points and we visited all of them, (or if we want to be mathematically minimalist there is a point and we did visit it and this is then true of another point and then another ad infinitum, is that, which sounds like stuff I have read, better mathematical spirit?:shy:) - then if we accelerated smoothly between 0 and 100 kph at some point or instant we were doing 60kph, we do not have to keep going for an hour at the same speed to say we ever did 60 km per hour.

Is lavinia trying to be rigorous about one thing and needs to be equally rigorous about everything, and is it possible to be?

 

[Hurkyl]

If you are doing analysis in $\mathbb{R}^{n\geq 2}$, you'll learn that the derivative IS itself the best linear approximation.

I don't think 3 is a very good linear approximation to 3x.

 

[Fredrik]

This is the classical view of the derivative of a function R\to R, which is somewhat misleading in that it implicitly leans on the canonical isomorphism
$\mathbb{R}\cong \text{Hom}_\mathbb{R}(\mathbb{R},\mathbb{R})$
$x\mapsto (1\mapsto x).$

Not sure what you mean here. $x\mapsto(\text{something})$ looks like a specification of a function with domain ℝ. But $1\mapsto x$? Is it a function with domain {1}? Did you mean $x\mapsto(y\mapsto xy)$? (The real number x is mapped to "the straight line through the origin with slope x").

 

[arildno]

This is the classical view of the derivative of a function R\to R, which is somewhat misleading in that it implicitly leans on the canonical isomorphism
$\mathbb{R}\cong \text{Hom}_\mathbb{R}(\mathbb{R},\mathbb{R})$
$x\mapsto (1\mapsto x).$

If you are doing analysis in $\mathbb{R}^{n\geq 2}$, you'll learn that the derivative IS itself the best linear approximation.

Thank you for reminding me, sort of slipped my mind.
Possibly because OP seemed to be fairly new to maths, so I assumed he was talking one-dimensionally.
Or possibly because it has, indeed, been some time since I did multi-dimensional analysis.
Or possibly because I'm enjoying a cigarette right now, along with a cup of coffee.

 

[Landau]

Not sure what you mean here. $x\mapsto(\text{something})$ looks like a specification of a function with domain ℝ. But $1\mapsto x$? Is it a function with domain {1}? Did you mean $x\mapsto(y\mapsto xy)$? (The real number x is mapped to "the straight line through the origin with slope x").

It is the $\mathbb{R}$-linear map $\mathbb{R}\to\mathbb{R}$ which sends 1 to x. Recall that a linear map is uniquely specified by its action on basis elements, and of course 1 is a basis element of the 1-dimensional space R. In other words, it is the 1x1-matrix

$$(x)$$

which is indeed 'multiplication by x'.

I don't think 3 is a very good linear approximation to 3x.

I am not sure what you mean? The map $$f:\mathbb{R}\to\mathbb{R}$$ given by f(x)=3x has "classical derivative = slope of tangent line" 3, which under this isomorphism becomes $$1\mapsto 3$$, in other words, $$x\mapsto 3x$$, i.e. f itself. So f, being already linear, is its own derivative.

 

[disregardthat]

What about the function f(x) = 3x+2, Landau? f'(0)=3 is not a linear approximation of f at 0 under any of the two interpretations of the derivative. As arildno says it generates a linear approximation by specific formulas. In the R^n-->R^m case, a linear expression L(x) = Df(x-a)+f(a), which is an approximation to f(x) around a.

 

[Landau]

Sure, formally we should say that the derivative is the "best affine approximation". Again, some sloppiness arose because of an implicit identification: all tangent spaces of R^n are canonically isomorphic; we are forgetting the origin.

 

[disregardthat]

Perhaps it is best not to talk too loudly about approximations then, as you sure see that "approximations" was meant in this context as a linear function yielding approximate numerical values to the function at hand in some sense. By stating that the derivate yield a "canonical affine approximation" all you really say is that the derivative is the slope of the function in a fancy way!

This is not pedantry, it is a matter of fact the the derivative does not contain enough information to numerically approximate a function (In the R^n-->R^m case).

 

[Goongyae]

Take a function like f(x)=x^2

Then evaluate it at x and x+e where e is an infinitesimal number. The line passing between the two points (x,f(x)) and (x+e,f(x+e)) will have slope df/dx = (f(x+e)-f(x))/(x+e-x) = (xx+2xe+ee-xx)/e = 2x + e

The slope at the point x is thus 2x+e, but since e is infinitesimal you can take the "standard part" of the expression and just use the approximation 2x as the slope/derivative.

So the derivative is approximate in the sense that you are discarding infinitesimal contributions to a precise expression.

 

[Studiot]

I've normally seen this as x+h rather than e but no matter.
But I also normally see a limit as h tends to 0,
rather than 'discarding' h as a small quantity.
Again normally we do not discard first powers of small quantities only second and higher.

Edit I wish I could get the symbol I want.

 

[Hurkyl]

Take a function like f(x)=x^2

Then evaluate it at x and x+e where e is an infinitesimal number. The line passing between the two points (x,f(x)) and (x+e,f(x+e)) will have slope df/dx = (f(x+e)-f(x))/(x+e-x) = (xx+2xe+ee-xx)/e = 2x + e

The slope at the point x is thus 2x+e, but since e is infinitesimal you can take the "standard part" of the expression and just use the approximation 2x as the slope/derivative.

So the derivative is approximate in the sense that you are discarding infinitesimal contributions to a precise expression.

What do you mean by "infinitessimal"? By your use of "standard part", it sounds like you are referring to non-standard analysis. In that case, your usage is actually wrong; we have:

df(x) = f'(x) dx​

So in this particular case,

d(x²) = 2 x dx​

Also, you make it sound like f'(x) is imprecise; that is not so. It is a precisely defined function of x. But as you stated, it does differ from the difference quotient by an infinitessimal number.