9.3 Quadratic approximation

If $Q(x)$ is the quadratic approximation of $f$ about $3$, then $Q(3)=f(3)$, $Q'(3)=f'(3)$, and $Q''(3)=f''(3)$.

$f(3)=9$, $f'(3)=0$, $f''(3)=-2$; there is not enough information to know $f'''(3)$.

If $Q(x)$ is the quadratic approximation of $f$ about $3$, then $Q(3)=f(3)$, $Q'(3)=f'(3)$, and $Q''(3)=f''(3)$. There is no guarantee that $f(x)$ and $Q(x)$ share the same third derivative, though, so we do not have enough information to know $f'''(3)$.

It is a very good approximation.

$f(x) \approx 2x+5$

The quadratic approximation of $f(x)$ about $x=a$ is

We subsitute $f(a)=2a+5$, $f'(a)=2$, and $f''(a)=0$:

So, our approximation is $f(x) \approx 2x+5$.

Remark: Our approximation is exact for every value of $x$. This will always happen with a quadratic approximation of a function that is quadratic, linear, or constant.

Approximate $f(x)=\log x$.

$\log(0.93) \approx -0.07245$

We approximate the function $f(x)=\log x$ about the point $x=1$. We choose 1 because it is close to $0.93$, and we can evaluate $f(x)$ and its first two derivatives at $x=1$.

So,

When $x=0.93$:

We estimate $\log(0.93) \approx -0.07245$.

Remark: a calculator approximates $\log(0.93) \approx -0.07257$. We're pretty close.

You'll probably want to centre your approximation about $x=0$.

$\cos\left(\dfrac{1}{15}\right )\approx \dfrac{449}{450}$

We approximate the function $f(x)=\cos x$. We can easily evaluate $\cos x$ and $\sin x$ ($\sin x$ will appear in the first derivative) at $x=0$, and 0 is quite close to $\dfrac{1}{15}$, so we centre our approximation about $x=0$.

Using the quadratic approximation $f(x) \approx f(0)+f'(0)(x-0)+\frac{1}{2}f''(0)(x-0)^2$:

We approximate $\cos\left(\dfrac{1}{15}\right)\approx\dfrac{449}{450}$.

Remark: $\dfrac{449}{450}= 0.99\overline{77}$, while a calculator gives $\cos\left(\frac{1}{15}\right )\approx 0.9977786$. Our approximation has an error of about $0.000001$.

The quadratic approximation of a function $f(x)$ about $x=a$ is

$e^{2x}\approx 1+2x+2x^2$

The quadratic approximation of a function $f(x)$ about $x=a$ is

We compute derivatives.

Substituting:

One way to go about this is to approximate the function $f(x) = 5 \cdot x^{1/3}$ , because then $5^{4/3} = 5 \cdot 5^{1/3} =f(5)$.

One approximation: $e^{\tfrac{4}{3}}\approx\dfrac{275}{32}$

There are a few functions we could choose to approximate. For example:

There isn't much difference between the first and third bullets. We'll go with $f(x)=5\sqrt[3]{x}$, centred about $x=8$.

Using the quadratic approximation $f(x) \approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2$:

We estimate $5^{4/3}\approx\dfrac{275}{32}$

Remark: $\dfrac{275}{32} = 8.59375$, and a calculator gives $5^{4/3}\approx 8.5499$. Although 5 and 8 are somewhat far apart, our estimate is only off by about $0.04$.

For (c), look for cancellations.

(a) $26$ (b) $16$ (c) $\dfrac{10}{11}$ (d) $\dfrac{75}{64}$

Compare (c) to (b).
Compare (e) and (f) to (d).
To get an alternating sign, consider powers of $(-1)$.

For each of these, there are many solutions. We provide some below.

For each of these, there are many solutions. We provide some below.

You can evaluate $f(1)$ exactly.
Recall $\ds\diff{}{x}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}$.

$f(1) \approx 2$, $f(1)=\pi$

Let's start by taking the first two derivative of $f(x)$.

Now, we can find the quadratic approximation about $x=0$.

Our quadratic approximation gives $2\arcsin(1) \approx 2$. However, $2\arcsin(1)$ is exactly equal to $2\left(\dfrac{\pi}{2}\right)=\pi$. We've just made the rather unfortunate approximation $\pi \approx 2$.

Let $f(x)=e^x$, and use the quadratic approximation of $f(x)$ about $x=0$ (given in your text, or you can reproduce it) to approximate $f(1)$.

$e\approx 2.5$

From the text, the quadratic approximation of $e^x$ about $x=0$ is

So,

We estimate $e \approx 2.5$.

Remark: actually, $e \approx 2.718$.

Be wary of indices: for example $\ds\sum_{n=1}^3 n = \ds\sum_{n=5}^7 (n-4)$.

$\{(\text{a}),(\text{d}), (\text{e})\}$, $\{(\text{b}),(\text{g})\}$, $\{(\text{c}),(\text{f})\}$

We group our expressions in to collections of equivalent expressions as follows:
$\{(\text{a}),(\text{d}), (\text{e})\}$, $\{(\text{b}),(\text{g})\}$, $\{(\text{c}),(\text{f})\}$