Math: why is the derivative a linear operator?

[Anonymous No.10941557]

I know the standard algebraic definition of a linear transformation / linear operator. But every single source I checked simply states that "the derivative is a linear operator since it satisfies the definition of the linear operator". Really? But "linear" means it is a line. Where is the line? Whats the geometric interpretation? Why is it never ever mentioned in any books? For example ln'(x) = 1/x. Why is this a linear operation? Is it because a derivative at every point is the tangent line ? So the derivative is an operator that maps ever point to the slope of the tangent line at the point? That's the "line" in "linear"? A local linear approximation?

[Anonymous No.10941559]

Why are you using so many big words? You can phrase this question much simpler.

[Anonymous No.10941571]

>>10941559
Not big words. Just trying to be as specific as possible to avoid ambiguity. The simplicity of the definition makes it difficult for me to make sense of it.

[Anonymous No.10941601]

>>10941557
A linear operator maps elements from one space onto another, in general terms it does not mean necessarily it will pop out the equation of a line.
Linear algebra does not have to be only straight lines, for example

[Anonymous No.10941607]

, when using "linear operator" instead of "linear map", mathematicians often mean actions on vector spaces of functions, which also preserve other properties, such as continuity. For example, differentiation and indefinite integration are linear operators; operators that are built from them are called differential operators, integral operators or integro-differential operators.

[Anonymous No.10941628]

>>10941601
>linear algebra does not have to be only straight lines,
Hmm I guess you can transform a non-linear function but the transformation is linear since you define a transformation matrix such a rotation matrix and perform matrix multiplication which is a linear operation since it is multiplication and addition. So you will rotate your input function or a vector.

[Anonymous No.10941697]

>>10941557
Linear operators are defined on LINEAR spaces. First prove that the set X of all differentiable functions is a linear space, where by (f+g) and a*f is pointwise. Then check that the operator A:X->Y defined as (Af)(x)=f'(x), Y where Y is the space of all functions.

[Anonymous No.10941709]

>>10941557
undergrad shitter here.
i thought it's because
a) the sum of the derivatives equals the derivative of the sums, and
b) the derivative of the scaled function equals the scaled derivative of the function

[Anonymous No.10941716]

>>10941557
Consider any function f(x). We know that f'(x) is defined by f(x+dx)-f(x)/dx when dx -> 0. This is equivalent to f'(x)dx = f(x+dx)-f(x). Now set for x any number. f'(a)dx=f'(a+dx)-f(a). If you consider dx to be an independent variable, you get a linear approximation of the actual function f(x) around the point a by virtue of f'(a)*x+f(a) =f(a+x)
where f'(a) and f(a) are constants. This function is nothing else than b*x+c = f(a+x) for b = f'(a), c = f(a). You may set a = 0 to simplify this expression: f'(0)*x+f(0) = f(x)

[Anonymous No.10941739]

>>10941557
>I know the standard algebraic definition of a >linear operator.
I think that is exactly your problem, you do not know or understand.
Try to proof the linearity and you will see its trivial, you can use your own screenshot for that.

[Anonymous No.10941772]

>>10941557
Think of a linear operator as a homomorphism from on vector space to another.

[Anonymous No.10941805]

>>10941709
yeah that's the correct definition i read all the time. i guess it is worth memorizing to pass a test. but i am trying to make sense of it from the geometrical perspective. "linear" as in there must be a line somewhere. i can't find it yet. but i need to read other responses.

[Anonymous No.10941860]

>>10941716
Yes I understand all of this. I just wanted someone confirm that the derivative is a linear operation because it is a pointwise linear approximation, i.e it maps every point of a function to the slope of a tangent line at that point.

[Anonymous No.10941896]

>>10941557
it’s because any LINEAR combination of differentiable functions is also differentiable.
>now he asks why it’s called a linear combination

[Anonymous No.10941900]

>>10941860
this is a complete nonsense, like literally the whole post

[Anonymous No.10941906]

>>10941860
It has nothing to do with the "linear approximation" aspect.
The second derivative operator is also a linear operator.

The operator that evaluates a function at p is linear.
P[f] = f(p)
P[c*f] = c*f(p) = c*P[f]
P[f+g] = f(p) + g(p) = P[f] + P[g]

>>10941772 is probably the best way to think of a linear operator.

[Anonymous No.10941930]

>>10941860
>>10941900
A map [math]\varphi:V \to W[/math] between vector spaces is linear if [math]\varphi(au+bv) = a\varphi(u)+b\varphi(v)[/math]. It's called linear because it preserves linear combinations (the term in the bracket). A corollary of this is that it maps lines to lines. A line is something like [math]p+tv[/math], apply [math]\varphi[/math] to see that the image is also a line.

In our case, [math]V = W = C^{\infty}(\mathbb{R})[/math], the set of all smooth functions on the real line, [math]\varphi = \tfrac{d}{dx}[/math]. The map eats a function and gives back another function. It's linear because we know that [math]\tfrac{d}{dx}(af+bg) = a\tfrac{df}{dx}+b\tfrac{dg}{dx}[/math] from calculus, that's literally the fact that derivative preserves linear combinations of functions.

Linearity of the derivative has absolutely nothing to do with the fact that the derivative is related to the slope or whatever. Integral is also a linear operator and integral is about areas and not about lines.

[Anonymous No.10941949]

>>10941900
>>10941906
>>10941930
>nothing to do with linear approximation/slope/whatever
OK thank you anons I think I got it. I think I confused two unrelated terms that both refer to linearity.

[Anonymous No.10941960]

>>10941557
[eqn] \frac{d}{dx} f(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} [/eqn]

[eqn] \frac{d}{dx} a \cdot f(x) + b \cdot g(x) = \lim\limits_{h \to 0} \frac{a \cdot f(x+h) + b \cdot g(x+h) - a \cdot f(x) - b \cdot g(x)}{h} = \lim\limits_{h \to 0 } \frac{a \cdot f(x+h) - a \cdot f(x)}{h} + \lim\limits_{h \to 0} \frac{b \cdot g(x+h) - b \cdot g(x)}{h} = a \cdots \lim\limits_{h \to 0 } \frac{f(x+h) - f(x)}{h} + b \cdots \lim\limits_{h \to 0 } \frac{g(x+h) - g(x)}{h} = a \cdot \frac{d}{dx} f(x) + b \cdot \frac{d}{dx} g(x) [/eqn]

[Anonymous No.10941963]

>>10941557
Read shlomo

[Anonymous No.10941964]

>>10941963
http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf

[Anonymous No.10941968]

>>10941557
Linear operators, O, obey this rule:
O[ a*f(x) + b*g(x) ] = a*O[f(x)] + b*O[g(x)].

It's just called a linear operator for "historical reasons". Don't confuse it with lines.

[Anonymous No.10941988]

>>10941963
>>10941964
based shlomo. thanks anon. searching for shlomo on libgen returns about 300 hits. awesome

>>10941968

i think my problem is i insisted on a geometric interpretation but there is not much to interpret geometrically other than the properties of a linear transformation. i need to learn more about this. i got too fixated on searching for a 'line' and suspected it was a tangent line, a totally unrelated thing. i am overthinking everything.

[Anonymous No.10942688]

What is this about? Partial derivatives or total?

Yes, it does have to do with "lines". They are basis vectors of the tangential space.

[Anonymous No.10942697]

>>10941559
You have small brain.

[Anonymous No.10943340]

>>10942688
This is about a geometrical proof that taking the derivative is a linear operation. Not partial, just a simple derivative in one variable.

>>10941960
Nice latex

[Anonymous No.10943353]

>>10941557
on a scale from air to uranium, why are you as dense as a neutron star?

if I differentiate af+bg (with a and b real numbers and f,g as differentiable functions), I get af' + bg' which is exactly the definition of a linear operator.

[Anonymous No.10943365]

>>10943353
Then yours must be the density of a black hole. He's asking about geometry.

Also OP, your answer is here, under "Properties":

https://en.m.wikipedia.org/wiki/Tangent_space

[Anonymous No.10943371]

>>10943365
there is no geometry in infinite dimensional spaces.

[Anonymous No.10943384]

>>10943365
Thanks

>>10943371
Well what about a simple function of one variable y=f(x). No f and g, just f. One single function. y= ln(x) and take the derivative and get 1/x. Thats what made me think about linearity. But I just need to accept that since (f + g)' = f' + g' then it is a linear operator. And then there are various proofs like this one: >>10941960
They are cool and valid but there is something unsatisfactory about them.

[Anonymous No.10943390]

>>10941607
Isn't the word operator mainly used to denote maps from a space to itself? Am I missing something?

[Anonymous No.10943405]

>>10943365
>In a sense, the derivative is the best linear approximation to phi near x, etc
So that entire section talks about the linearity of the derivative in the context of local linear approximation. But I was told itt that this is utter nonsense and it has nothing to do with that and has everything to do with the superposition f(x+y) = f(x) + f(y) and af(x) = f(ax).

[Anonymous No.10943412]

>>10943405
I know. Probably undergrads. Pretty much everything I know has a geometrical interpretation behind it. Claiming the derivative doesn't is just nonsense.

[Anonymous No.10943457]

>>10943412
Found this book:
Advanced Calculus: A geometric view.

My understanding is in case of the derivative, the superposition f(x+y) = f(x) + f(y) and af(x) = f(ax) that is used as a general definition of a linear operator is the *consequence* of treating derivative as a local linear approximation.

[Anonymous No.10943527]

>>10943384
you can't geometrically represent it because to represent linearity you need at least 4 dimensions.

You can see part of the linearity this way :

imagine if you could order all the multiples of a differentiable function f along an axis. 0*f, 0.1*f, 0.2*f...10*f, 100*f and so on. Call that the "f axis"

and imagine you could order all the multiples of its derivative f' along another axis.
0*f', 0.1*f', 0.2*f'...10*f', 100*f' and so on. Call that the "f' axis"
These axes form a plane.

Then the "projection" of the graph of the derivative operator on this plane is a goddamn line.

[Anonymous No.10944181]

>>10943457
Interesting source. Thanks Anon!

[Anonymous No.10944496]

>>10943527
I can see linearity right here in this proof: >>10941960
for example. You can think of a limit as a linear operator. Because these are all linear operations: addition and multiplication. These fractions are essentially the slope so f(a+dx)=f(a) + f'(a)dx + o(1) . As you can see with the vanishing o(1) it is simply the differential or in other words an equation of the line in the slope-intercept form. So that is the "line" in the "linear transformation".

>>10941906
Yes I know now that it has everything to do with the linear approximation: it is right there in the definition of the derivative via limits.

>superposition formulas
well that's the consequence of the fact that derivative is a linear approximation. dy = mdx is the differential which is a line.

>>10941930
>Linearity of the derivative has absolutely nothing to do with the fact that the derivative is related to the slope or whatever
again, thats the consequence of the fact that the derivative is a local linear approximation so you can (f+g)' = f' + g'.

>Integral is also a linear operator and integral is about areas and not about lines.
Well, again you can see that the Riemann sum is a linear operation since it is just addition and multiplication. That's the reason the superposition formulas hold, not the other way around (it is not defined via the superposition requirement).

>>10941968
>O[ a*f(x) + b*g(x) ] = a*O[f(x)] + b*O[g(x)].
That's a general rule for all linear operators. They may be linear for different reasons but since they are linear that rule holds. But we can see why they are linear on a case per case basis. For example a matrix is a linear operation since matrix multiplication is linear. The derivative is a linear operator because it is local linear approximation due to the definition of the limit and the differential etc. Integral is a linear operator because the riemann sums are linear (multiplication + addition). etc. This is my take anyway.

[Anonymous No.10944502]

Linear operator doesn't have anything to do with lines. It means it satisfies some conditions such as preservation of addition and scalar multiplication. This is from the linear algebra class I took last winter which I forget now.

[Anonymous No.10944505]

>>10941559
Quantum eraser

[Anonymous No.10944524]

>>10944502
Yes you can just accept that definition and run with it. But you can also ask yourself why is a *particular* linear operator linear? What you saw in the linear algebra class was most likely matrices and transformations and a matrix as a linear operator. But why is it linear? Because matrix multiplication is linear. That's why it satisfies these superposition rules, i,e, addition and scalar multiplication. Likewise the derivative can also be defined using these rules. But it doesn't explain *why* these rules hold true. And the "line" is right there. It y=kx+b. You can see it in the definition of the derivative as a limit.

[Anonymous No.10944657]

>>10941557
Derivative of a function at a point can be defined as the linear map which best approximates the function at a given point.

[Anonymous No.10945023]

A linear operator is just a name for a map between vector spaces (sometimes called linear spaces) that respects the addition and scalar multiplication, there is no geometric reason.

[Anonymous No.10945261]

>>10943527
Nice post and quality thread