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Re:Calculus for Beginners and Artists (2003) by kstenerud on Sunday July 14, 2019 @12:39PM (#20434346)
http://www-math.mit.edu/~djk/calculus_beginners/chapter03/se...

We can determine the linear function which takes value f(a) at a and f(b) at b by the following formula:

                x-b        x-a
    f(x) = f(a) --- + f(b) ---
                a-b        b-a
The first term is 0 when x is b and is f(a) when x is a, while the second term is 0 when x is a and is f(b) when x is b. The sum of the two is therefore f(a) when x is a and f(b) when x is b. And it is a linear function. Linear functions have a term that is x multiplied by some constant, and may also have a constant term as well.

Umm... whut?

This has me completely lost. I don't see how this can be geared towards beginners and artists.

Re:Calculus for Beginners and Artists (2003) by mkl on Sunday July 14, 2019 @10:06PM (#20437606)
Try this idea:

  g(t) = (1-t) f(a) + t f(b)
where t goes from 0 to 1. At the end points, we get

  g(0) = (1) f(a) + (0) f(b) = f(a)
  g(1) = (0) f(a) + (1) f(b) = f(b).
This can be used to linearly blend any two things, as for t in between 0 and 1, we'll get part of f(a) and part of f(b).

Now, if we want to use x instead of t, where x goes from a to b, we need to convert:

  x = a  <=>  t = 0
  x = b  <=>  t = 1
The distance that x has travelled from a is (x-a). The total distance from a to b is (b-a). t is the proportion of that total distance, so t is the distance travelled over the total distance:

      x-a
  t = ---
      b-a
Now (renaming g to f as we're changing parameterisation* )

  f(x) = (1 - (x-a)/(b-a)) f(a) + ((x-a)/(b-a)) f(b)
       = (1 + (x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b)             //since -(b-a) = a-b
       = ((a-b)/(a-b) + (x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b)   //since 1 = (a-b)/(a-b)
       = ((a-b+x-a)/(a-b)) f(a) + ((x-a)/(b-a)) f(b)             //combining terms with common denominator
       = ((x-b)/(a-b)) f(a) + ((x-a)/(b-a)) f(b)
Does that explain it?

* g(x) is something different: g(x) = (1-x) f(a) + x f(b), just by substituting in.

Re:Calculus for Beginners and Artists (2003) by baby on Sunday July 14, 2019 @12:51PM (#20434405)
1. X-b befomes 0 if x=b right?

2. This means anything multiplying this x-b becomes 0 as well when x=b right?

His definition of linear is also wrong in my book.

Linear: f(x) = a x

Affine: f(x) = a x + b

(For some a, b)

Re:Calculus for Beginners and Artists (2003) by cornstalks on Sunday July 14, 2019 @01:10PM (#20434492)
> His definition of linear is also wrong in my book.

There are two variations of "linear function". To quote https://en.wikipedia.org/wiki/Linear_function:

> In mathematics, the term linear function refers to two distinct but related notions:

> - In calculus and related areas, a linear function is [...] a polynomial function of degree one or zero.

> - In linear algebra, mathematical analysis, and functional analysis, a linear function is a linear map.

I personally prefer the terminology "linear function" for the former definition (polynomials of degree 1 or 0), and "linear transform" for the latter definition (a linear map).

Re:Calculus for Beginners and Artists (2003) by jacobolus on Sunday July 14, 2019 @07:12PM (#20436879)
What you are calling [“linear”, “affine”] are in some contexts called [“homogeneous linear”, “linear”] or the like.

This is not “wrong”, but just a different convention.

Re:Calculus for Beginners and Artists (2003) by kstenerud on Sunday July 14, 2019 @01:03PM (#20434458)
Is that comment for me or about the referenced page? If it's for me, I'm not sure what you're saying.
Re:Calculus for Beginners and Artists (2003) by baby on Sunday July 14, 2019 @01:24PM (#20434576)
I'm trying to explain what he meant. Which step don't you understand specifically in my comment?
Re:Calculus for Beginners and Artists (2003) by kstenerud on Sunday July 14, 2019 @01:46PM (#20434707)
I'm trying to understand why there's f(a) and f(b), why is b being subtracted from x, why it's divided by a-b, why there's a similar thing for f(a), why the two are being added, why the zero values are important, and what the whole thing actually means.
Re:Calculus for Beginners and Artists (2003) by jacobolus on Sunday July 14, 2019 @07:06PM (#20436845)
It’s simpler if you use 0 and 1 instead of a and b. If we want a linear function such that

f(0) = A, f(1) = B,

then the unique solution is

f(t) = (1-t)A + tB

The more general version with a and b is just composing this with a transformation of [a, b] into [0, 1]

Relevant keywords: linear interpolation, affine combination, barycentric coordinates.

Re:Calculus for Beginners and Artists (2003) by llamaz on Sunday July 14, 2019 @02:21PM (#20434925)
It's demonstrating how to find the line that passes through two points x=a, x=b on the function y=f(x).

Without having read the actual page, I assume he's going to move a and b close to eachother to approximate the tangent of the function at a point.

edit: In higher level math, you use circles/spheres or parabolas/paraboloids to approximate functions, but in high school level calculus you stick to using a straight line to approximate a function

Re:Calculus for Beginners and Artists (2003) by kstenerud on Sunday July 14, 2019 @02:54PM (#20435171)
Unfortunately, he doesn't do that. There's an even more complicated formula under that which is supposed to be related to the slope of some line and a ratio between b and a (I think?) and then I'm guessing he reduces it, but doesn't explain how he did it.

I've tried learning calculus 3-4 times during my life, using materials for an "absolute beginner", and this has always been my experience, as if one were teaching programming by going from "This is a variable. It can store data." directly to "A monad is a monoid in the category of endofunctors." To this day I have no idea what calculus even is, or what it's for.

Re:Calculus for Beginners and Artists (2003) by Xeanort on Sunday July 14, 2019 @05:32PM (#20436208)
You want a linear function f(x) such that f(a)=A and f(b)=B given a, b, A, B. A valid f(x) (actually it's the only one) is defined as

             x-b     x-a
    f(x) = A --- + B ---
             a-b     b-a
You can calculate f(a) and f(b) to confirm that it works.

However, I don't find this kind of guide any good for learning math...