Taylor approximations

Chapter 14 Taylor approximations

We started the course (back in Preface ) with the idea of approximating a function \(f(x)\) by its tangent line. The problem of actually finding the tangent line is what motivated us to think hard about limits and how they work, then to apply that idea to secant lines.

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Here’s another definition of the tangent line (in addition to the ones we saw in Unit 3.2):

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Definition 14.1. alternative definition of the tangent line.

The tangent line to the graph of \(f(x)\) at \(x=c\) is the graph of the linear function \(L(x)=mx+b\) which is as good an approximation to \(f(x)\) nearby to \(x=c\) as any linear function can be.

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This definition might make you a bit uncomfortable. How would you use this definition to actually find a tangent line?

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We’ll use the notion of asymptotic equivalence from Unit 6.3. What we’d like, first of all, is that \(f(x)\sim L(x)\) as \(x\to c\text{.}\) In other words, we want

\begin{equation*} \displaystyle\lim_{x\to c}\frac{f(x)}{L(x)}=1\ \ . \end{equation*}

Now as long as \(f(x)\) and \(L(x)\) are both continuous, and \(L(c)\neq 0\text{,}\) the rules of limits tell us that \(\displaystyle\lim_{x\to c}\frac{f(x)}{L(x)}=\frac{f(c)}{L(c)}\text{.}\) So we need \(\frac{f(c)}{L(c)}=1\text{,}\) which is just the requirement that \(L(c)=f(c)\text{.}\) For reference later, we’ll say that the value \(f(c)\) approximates the function \(f(x)\) to zeroth order.

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A line is determined by two pieces of information:

a slope \(m\) and an intercept \(b\)

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a slope \(m\) and a point \((x_0,y_0)\) on the line

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two points \((x_0,y_0)\) and \((x_1,y_1)\) on the line

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We just learned that the point \((c,f(c))\) has to be on the line we’re looking for, so let’s work with the second option and determine \(L\) by finding its slope.

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Imagine a competition among all lines which pass through the point \((c,f(c))\text{,}\) for which one is "most like" \(f(x)\) nearby \(x=c\text{.}\) Using the point-slope form of the equation for a line, we know that any such competitor has to look like \(L(x)=f(c)+m(x-c)\text{.}\) Now:

\begin{align*} &\text{We want }f(x)\sim f(c)+m(x-c)\text{ as }x\to c\\ &\text{That's the same as demanding }f(x)-f(c)\sim m(x-c)\text{ as }x\to c \end{align*}

All \(f(x)-f(c)\sim m(x-c)\text{ as }x\to c\) means is that

\begin{equation*} \displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{m(x-c)}=1 \end{equation*}

which can only happen if \(m=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f'(c)\text{.}\)

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In other words, this approach (finding an asymptotically-equivalent line) gives the same formula for the slope of the tangent line that we know and love.

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You might ask: why all the work just to get the same formula? Because the understanding of the tangent line as "the line that matches \(f(x)\) near \(x=c\) as well as any line can" will generalize to questions like:

What parabola matches the graph of \(f(x)\) near \(x=c\) as well as any parabola can?

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What cubic function approximates \(f(x)\) near \(x=c\) the best among all cubic functions?

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etc.

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These polynomial approximations are known as the Taylor polynomials for \(f(x)\text{.}\) They allow you to use your knowledge of polynomials -- which you’ve no doubt spent many years honing -- to understand how nonpolynomial functions behave.

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