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Subsection 3.4.3 Second Approximation — the Quadratic Approximation

We next develop a still better approximation by now allowing the approximating function be to a quadratic function of \(x\text{.}\) That is, we allow \(F(x)\) to be of the form \(A+Bx+Cx^2\text{,}\) for some constants \(A\text{,}\) \(B\) and \(C\text{.}\) To ensure that \(F(x)\) is a good approximation for \(x\) close to \(a\text{,}\) we choose \(A\text{,}\) \(B\) and \(C\) so that

  • \(f(a)=F(a)\) (just as in our zeroth approximation),
  • \(f'(a)=F'(a)\) (just as in our first approximation), and
  • \(f''(a)=F''(a)\) — this is a new condition.

These conditions give us the following equations

\begin{align*} F(x)&=A+Bx+Cx^2 & &\implies & F(a)=A+Ba+\phantom{2}Ca^2&=f(a)\\ F'(x)&=B+2Cx & &\implies & F'(a)=\phantom{A+a}B+2Ca&=f'(a)\\ F''(x)&=2C & &\implies & F''(a)=\phantom{A+aB+a}2C&=f''(a) \end{align*}

Solve these for \(C\) first, then \(B\) and finally \(A\text{.}\)

\begin{align*} C &=\half f''(a) & \text{substitute}\\ B &= f'(a) - 2Ca = f'(a)-af''(a) & \text{substitute again}\\ A &= f(a)-Ba-Ca^2 = f(a)-a[f'(a)-af''(a)]-\half f''(a)a^2\hskip-0.5in \end{align*}

Then put things back together to build up \(F(x)\text{:}\)

\begin{align*} F(x)&=f(a)-f'(a)a+\half f''(a)a^2 & &\text{(this line is $A$)}\cr &\phantom{=f(a)\hskip3pt}+f'(a)\,x\hskip3pt- f''(a)ax & & \text{(this line is $Bx$)}\\ &\phantom{=f(a)-f'(a)a\hskip3.5pt}+\half f''(a)x^2 & &\text{(this line is $Cx^2$)}\\ &=f(a)+f'(a)(x-a)+\half f''(a)(x-a)^2 \end{align*}

Oof! We again write it in this form because we can now clearly see that our second approximation is just an extension of our first approximation.

Our second approximation is called the quadratic approximation:

Here is a figure showing the graphs of a typical \(f(x)\) and approximating function \(F(x)\text{.}\)

This new approximation looks better than both the first and second.

Now there is actually an easier way to derive this approximation, which we show you now. Let us rewrite  5 Any polynomial of degree two can be written in this form. For example, when \(a=1\text{,}\) \(3 + 2x + x^2 = 6 + 4(x-1) + (x-1)^2\text{.}\)

\(F(x)\) so that it is easy to evaluate it and its derivatives at \(x=a\text{:}\)

\begin{align*} F(x) &= \alpha + \beta\cdot (x-a) + \gamma \cdot(x-a)^2 \end{align*}


\begin{align*} F(x) &= \alpha + \beta\cdot (x-a) + \gamma \cdot(x-a)^2 & F(a) &= \alpha = f(a)\\ F'(x) &= \beta + 2\gamma \cdot(x-a) & F'(a)&=\beta = f'(a)\\ F''(x) &= 2\gamma & F''(a) &= 2\gamma = f''(a) \end{align*}

And from these we can clearly read off the values of \(\alpha,\beta\) and \(\gamma\) and so recover our function \(F(x)\text{.}\) Additionally if we write things this way, then it is quite clear how to extend this to a cubic approximation and a quartic approximation and so on.

Return to our example:

Use the quadratic approximation to estimate \(e^{0.1}\text{.}\)

Solution Set \(f(x) = e^x\) and \(a=0\) as before.

  • To form the quadratic approximation we need \(f(a), f'(a)\) and \(f''(a)\text{:}\)
    \begin{align*} f(x) &= e^x & f(0) & = 1\\ f'(x) &= e^x & f'(0) & = 1\\ f''(x) &= e^x & f''(0) & = 1 \end{align*}
  • Then our quadratic approximation is
    \begin{align*} F(x) &= f(0) + x f'(0) + \frac{1}{2} x^2 f''(0) = 1 + x + \frac{x^2}{2}\\ F(0.1) &= 1.105 \end{align*}

Recall that \(e^{0.1} = 1.105170918\dots\text{,}\) so the quadratic approximation is quite accurate with very little effort.

Before we go on, let us first introduce (or revise) some notation that will make our discussion easier.