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Y=mx+b & linearity

  1. Jul 22, 2012 #1
    I understand that a linear relation needs to satisfy both the property of superposition and homogeneity. Y(x)=mx+b does not satisfy both property at the same time yet any equations in this form are called a "linear function" and it is used in linear approximation.

    For example, sin(x), which is a non-linear function, can be approximated as sin(x0)+cos(x0)(x-x0), where b=sin(x0), m=cos(x0) and x=(x-x0), using Taylor series expansion with the operating point x0. This expression fails the linearity test, yet it is refer as a "linear" approximation of non-linear function sin(x).

    So if the form y=mx+b fails the superposition & homogeneity test, why do we consider it as a "linear function"? What I am missing here? Thank you so much!
     
  2. jcsd
  3. Jul 22, 2012 #2

    Mark44

    Staff: Mentor

    We consider y = mx + b to be a linear function because its graph is a straight line. The term "linear" when used in the context of transformations has a different meaning, as you know, with T(u + v) = T(u) + T(v), and so on.

    For your example of the linear approximation of the sine function, "linear" means that the graph of the approximation is a straight line that is tangent to the sine function at x0 and has the same slope there.

    It's confusing, but the context usually indicates whether what's being described is a straight line or a transformation from one vector space to another.
     
  4. Jul 22, 2012 #3
    To add to what Mark44 said, observe that the derivative of a function at a point, [itex]c[/itex] is a linear appoximation of [itex]f(x) - f(c)[/itex]. For example, let's take the function [itex]f(x) = x^2[/itex]. Then let [itex]c=2[/itex] and note that [itex]f(x) - f(2) = x^2 - 4[/itex]. Now, [itex]f'(2) = 4[/itex], and we see that [itex]f'(2) * (x-2) = 4x - 8[/itex]. Now, let's take some number close to [itex]2[/itex]. Let's say [itex]x_0=1.98[/itex]. Then, [itex]f(1.98) - f(2) = 1.98^2 - 4 = -.0796[/itex]. On the other hand [itex]f'(2) * (1.98 - 2) = -.08[/itex].

    OK, so how does this relate to your question? Well, we know that [itex]f(x) - f(c) \simeq f'(c) * (x-c)[/itex]. Therefore, we know that [itex]f(x) \simeq f'(c) * (x-c) + f(c)[/itex]. That is, this is a linear approximation to the value of [itex]f[/itex] near [itex]x=c[/itex]. So, the way I view the statement "linear approimation to [itex]f[/itex]" is that the derivative is a linear function (in this case, it is just a function from R to R, that is, it is just a number that multiles other numbers) approximation to [itex]f(x) - f(c)[/itex]

    For example, let's keep [itex]f(x)=x^2[/itex]. We know that [itex]f(1.98)=3.9204[/itex]. But, using this approximation, we have [itex]f(1.98) = 4 * (1.98 - 2) + 4 = 3.92[/itex].

    Now, I don't know if this is what you were looking for, but it is VERY important to realise that the derivaitve of a function at a point [itex]c[/itex] is the linear apporiximation of [itex]f(x) - f(x)[/itex], so, I decided to explain it.
     
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