# Homework Help: Y=mx+b & linearity

1. Jul 22, 2012

### dominicfhk

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. Jul 22, 2012

### 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.

3. Jul 22, 2012

### Robert1986

To add to what Mark44 said, observe that the derivative of a function at a point, $c$ is a linear appoximation of $f(x) - f(c)$. For example, let's take the function $f(x) = x^2$. Then let $c=2$ and note that $f(x) - f(2) = x^2 - 4$. Now, $f'(2) = 4$, and we see that $f'(2) * (x-2) = 4x - 8$. Now, let's take some number close to $2$. Let's say $x_0=1.98$. Then, $f(1.98) - f(2) = 1.98^2 - 4 = -.0796$. On the other hand $f'(2) * (1.98 - 2) = -.08$.

OK, so how does this relate to your question? Well, we know that $f(x) - f(c) \simeq f'(c) * (x-c)$. Therefore, we know that $f(x) \simeq f'(c) * (x-c) + f(c)$. That is, this is a linear approximation to the value of $f$ near $x=c$. So, the way I view the statement "linear approimation to $f$" 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 $f(x) - f(c)$

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

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 $c$ is the linear apporiximation of $f(x) - f(x)$, so, I decided to explain it.