4.2.4 Approximating Values

Approximating values is a common technique in mathematics, used when exact values are difficult or unnecessary to calculate. Approximations can help simplify calculations, estimate answers, or work with complex functions. There are several methods and concepts for approximating values, each suited to different situations.

1. Rounding

Rounding is one of the simplest forms of approximation. It involves adjusting a number to a nearby value, often to make calculations easier.

To the nearest integer:
- $3.6$ rounds to $4$
- $3.4$ rounds to $3$
To a specific decimal place:
- $5.6789$ rounded to two decimal places is $5.68$
- $5.6789$ rounded to one decimal place is $5.7$

2. Truncation

Truncation involves cutting off digits beyond a certain point without rounding up. This is different from rounding because it simply drops the digits rather than rounding them.

Truncating $5.6789$ to two decimal places gives $5.67$

3. Using Significant Figures

Significant figures are used to indicate the precision of a number. This method is common in science and engineering.

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Example:
$0.004567$ to three significant figures is $0.00457$
$123,456$ to three significant figures is $123,000$

4. Linear Approximation (Tangent Line Approximation)

Linear approximation uses the tangent line at a point on a curve to approximate the value of the function near that point. It's often used for functions that are difficult to evaluate exactly.

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Formula: If you have a function $f(x)$ and you want to approximate it near a point $a$ , you use:

$f(x) \approx f(a) + f'(a)(x - a)$

where $f'(a)$ is the derivative of $f(x)$ at $x = a$

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Example: Approximating $\sqrt{4.1}$ :
$f(x) = \sqrt{x}$ , and we know $f(4) = 2$ and $f'(x) = \frac{1}{2\sqrt{x}}$
So, $f'(4) = \frac{1}{4}$
$\sqrt{4.1} \approx 2 + \frac{1}{4} \times 0.1 = 2 + 0.025 = :success[2.025]$

5. Binomial Approximation

When dealing with expressions like $(1 + x)^n$ , where $x$ is small, you can use the binomial approximation: $(1 + x)^n \approx 1 + nx$

This is the first term in the binomial series expansion and works well when $x$ is small.

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Example: Approximate $(1.02)^5$ :

6. Taylor Series Approximation

Taylor series approximation expands a function into an infinite sum of terms calculated from the values of its derivatives at a single point. For small values of $x$ , the first few terms can provide a good approximation.

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Example: Approximate $e^x$ near $x = 0$ :

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

For $x = 0.1$ :

$e^{0.1} \approx 1 + 0.1 + \frac{(0.1)^2}{2} = 1 + 0.1 + 0.005 = :success[1.105]$

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Example

Question

Expand $(1 - 2x)^{\frac{1}{2}}, |x| < \frac{1}{2}$ , in ascending powers of $x$ up to and including the term in $x^3$ .
By substituting a suitable value of $x$ in your expansion, find an estimate for $\sqrt{0.98}$ .
Show that $\sqrt{0.98} = \frac{7}{10} \sqrt{2}$ and hence find the value of $\sqrt{2}$ correct to 8 significant figures.

Solution:

a. Expand $(1 - 2x)^{\frac{1}{2}}, |x| < \frac{1}{2}$ , in ascending powers of $x$ up to and including the term in $x^3$ .

1 + \frac{(0.5)(-2x)}{1!} + \frac{(0.5)(-0.5)(-2x)^2}{2!} + \frac{(0.5)(-0.5)(-1.5)(-2x)^3}{3!} + \ldots

= 1 - x - \frac{x^2}{2} - \frac{x^3}{2} + \ldots

b. By substituting a suitable value of $x$ in your expansion, find an estimate for $\sqrt{0.98}$ .

Notice $(1-2x)^{\frac{1}{2}} \equiv \sqrt{1-2x}$ , so think about what x should be to get 0.98 under the square root.

Let\ \ x = :highlight[0.01] \Rightarrow \sqrt{1-2(0.01)} = \sqrt{0.98} \quad (LHS)

RHS: \quad 1 - 0.01 - \frac{(0.01)^2}{2} - \frac{(0.01)^3}{2} = :success[0.9899495]

Checking this answer using a calculator is very close, therefore this gives a good approximation.

c. Show that $\sqrt{0.98} = \frac{7}{10} \sqrt{2}$ and hence find the value of $\sqrt{2}$ correct to 8 significant figures.

\sqrt {0.98} = \sqrt {\frac{98}{100}} = \frac{\sqrt 98}{\sqrt 100} = \frac{\sqrt{49} \times \sqrt{2}}{10} = \frac{7\sqrt{2}}{10}

From $(b): \sqrt{0.98} \approx 0.9899495$

\frac{7\sqrt{2}}{10} \approx 0.9899495 \Rightarrow \sqrt{2} = \frac{0.9899495 \times 10}{7}

\approx :success[1.4142136]

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Summary:

Rounding: Adjusting to a nearby value.
Truncation: Cutting off digits without rounding.
Significant Figures: Keeping a specified number of meaningful digits.
Linear Approximation: Using the tangent line at a point to approximate the function.
Binomial Approximation: Simplifying expressions like $(1 + x)^n$ for small $x$
Taylor Series: Expanding a function into a series to approximate its value.

Approximating Values Simplified Revision Notes for A-Level AQA Maths Pure

4.2.4 Approximating Values

1. Rounding

2. Truncation

3. Using Significant Figures

4. Linear Approximation (Tangent Line Approximation)

5. Binomial Approximation

6. Taylor Series Approximation

Example

Solution:

Summary:

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