Aproximação constante por zeros da função

Let's say we want to approximate the function $f(x)=e^x$. A constant approximation at a point $x=c$ (where $c$ is a constant) would be the horizontal line $y=e^c$.

For example, here is a constant approximation of $f(x)=e^x$ about point $0$:

As you can see, constant approximations are pretty boring, and extremely erroneous. However, we will definitely get on to more exciting and accurate approximations in later topics!

Error analysis for the zeroth order approximation is very simple, and as noted in the previous article, the error is very significant. Consider the function $f(x)= e^x$, whose approximation is the horizontal line $y=e^c$, as we found in the previous article. The graph of both functions is shown below.

The error function, denoted E(x), in this case would be defined as $E(x) = |e^x - e^c|$. To see why this is true, notice our function measures the distance between our function and the approximation. In other words, it measures how far off our approximation is from our actual function. E(x), for $f(x) = e^x$ is graphed below.

As you can see, the error increases significantly as x gets large. Therefore, our zeroth order approximation is very inaccurate. This motivates a desire for higher degree approximations, which will be explored in another topic.

Finally, the general definition for the error function of zeroth degree must be stated. For a function f(x), and a constant approximation of f(c), the error function E(x) is given by: $\boxed {E(x) = |f(x) - f(c)|}$.

Now for a question:

Now that you have seemed to learn this topic, move on to quadratic approximations, which while slightly harder, are much, much more accurate.

Let's consider an infinitely differentiable function $f$ that we'd like to approximate close to a point $a$. That is, let's try to find a function $g$ that is easy to calculate such that if $x$ is close to $a$, then $g(x)$ is close to $f(x)$.

The simplest approximation to $f$ near $a$, which we will call $g_0$, should require no knowledge but the value of $f$ at $a$. Namely: $g_0(x) = f(a)$. One might complain that this is not a very good approximation at all - and in fact it isn't! But extremely close to $a$, it does okay. Let's look at $f(x) = e^x$, $a = c$ for some number $c$.

Then $f(c) = e^c$, so we have $g_0(x) = e^c$. As you can see in the plot below, this isn't very good - so we'd like to be able to do better.

In the next section, we'll analyze the error of this approximation. Then, we'll find a better one!