Approximations: Symbol, Error Calculation & Examples

Approximation is a numerical figure that is intentionally similar but not exactly equal to something else. It is basically a rough estimate or guess at something. 

• Approximations are used when it is difficult to find the exact value of a number. 
• It is used in various fields such as Mathematics, Science, and day-to-day life.
• Approximation Theory is an integral component of Mathematics that relies on the derivatives of the functions.
• The symbol ‘≈’ meaning “almost equal to” is used to represent the approximate values.

Example: The approximate value of π is 3.14, thus, it is represented as π ≈ 3.14.

Read More: NCERT Solutions For Class 12 Mathematics Applications of Derivatives

Key Terms: Approximations, Derivatives, Absolute Error, Differentiation, Differential Calculus, Errors, Functions, Variables

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Approximation in Mathematics

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Approximation is an important theory in Mathematics that is employed when it's difficult to seek out the precise value of any number. 

• The term ‘Approximation’ is derived from Latin word ‘Proximus’ meaning ‘very near’ or ‘nearest’.
• It is the nearest estimate or measure of something. 
• It is also used in order to round off the errors leading to approximation.
• Approximation is used in various disciplines such as mathematics, science, and day-to-day life.

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Approximations Symbol

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Approximation is defined as something which is similar to but not the same as something else. Approximations are generally denoted by the wavy equal “≈” sign which means “almost equal to”.

Example: The approximate value of √2 is 1.414. 

Thus, it can be written as √2 ≈ 1.414.

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Differential Calculus Approximations

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Differential Calculus Approximation is used to make smaller portions of anything to investigate the rate of change. Differentials are used to approximate certain quantities. Let y = f(x) be any function of x. Let Δx be the small change in x and Δy be the corresponding change in y.

Then, Δy = f(x + Δx) – f(x).

Then, dy = f'(x) dx or dy = (dy/dx) * Δx is an approximate value of Δy.

Now with this, we can define the following: 

• The differential of x which is denoted by dx, can be defined by dx = Δx.
• The differential of y which is denoted by dy, can be defined by dy = f’(x) dx or dy = (dy/dx) * Δx

If dx = Δx is relatively small in comparison to x then dy is an approximate value of Δy and dy ≈ Δy.

Here are some of the important points that need to be remembered: 

• Differential of the dependent variable can not be equal to the increment of the variable.
• Differential of the independent variable can be equal to the increment of the variable.
• Absolute Error: Absolute error in x is the change Δx in x.

Read More: Limits And Derivatives

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Approximation Examples

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Here are a few solved examples on Approximations for better concept clarity: 

Example 1: Find the approximate value of √26.

Solution: Here it's very easy to determine the value of the under root if the given number is a perfect square, however, for such numbers we've to use derivatives to seek out the approximate value of the function.

Let the f(x) =√x and therefore the derivative of this is 

\(f'(x) = \frac{1}{2x^{1/2}}\)

According to the Approximation Formula, 

Δy ≈ dy = (\(\frac{dy}{dx}\)) . Δx 

f( x + Δx) – f(x) = f’(x).Δx

f( x + Δx) = f(x) + f’(x).Δx

Here, we'll assume x almost 26 which may be a perfect square.

So we'll assume

x = 25

x₂ – x₁

= 26 – 25

= 1

It tells us the change in x.

Let x = 25 and put the values within the formula.

f( x + Δx) = f(x) + f’(x).Δx

f( 25 + 1) = f(25) + f’(25)

f(26) = \(\sqrt{25} + \frac{1}{2.25 ^{1/2}}1\)

= 5 + \(\frac{1}{10}\)

\(\sqrt{26}\) = 5 + 0.1

= 5.1

Read More: Limits and Derivatives Formulas

Example 2: Determine the approximation of √25.5 using differential. 

Solution: Assume that, y = √x, where, x = 25 and ∆x = 0.5.

Then,

∆y = \(\begin{array}{l}\sqrt{x+\bigtriangleup x} -\sqrt{x}\end{array}\)

∆y = \(\begin{array}{l}\sqrt{25.5} -\sqrt{25}\end{array}\)

∆y = \(\begin{array}{l}\sqrt{25.5} – 5\end{array}\)

\(\begin{array}{l}\sqrt{25.5} = \bigtriangleup y + 5\end{array}\)

As dy is approximately equal to ∆y, thus, 

\(\begin{array}{l}dy = \frac{\mathrm{d} y}{\mathrm{d} x}\bigtriangleup x = \frac{1}{2\sqrt{x}}(0.5) = 0.05\end{array}\)

Hence, the approximate value of √25.5 will be

√25.5 = 5 + 0.05 = 5.05

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Application of Derivatives for Approximation

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Differentiation is used to find the approximate values of certain quantities. If there's a very small change in one variable corresponding to the opposite variable, then we use the differentiation to seek out the approximate value.

• The differentiation of x is represented by dx and is defined by dx = x where dx is that the minor change in x.
• The differential of y is defined by dy = (dy/dx) x and is represented by dy

Application of derivatives as dx is extremely small compared to x, so dy is that the approximation of y (dy = y).

• dy = f’(x)dx
• ∆x = dx
• ∆y = dy

Thus, it can be derived that ∆y = f’(x)∆x.

It can also be written as f(x + ∆x) = f(x) + ∆y = f(x) + f’(x)∆x

This shows that the derivative of the variable isn't equal to the increase of the variable whereas the derivative of the independent variable is equal to the increase of the variable.

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Approximation and Errors

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Approximations of functions are also used to approximate the errors in calculating certain functions if their exact dependence on the independent parameters is known.

Consider a function as 

y = f(a, b, c...)

The error in the calculation of y at points a = a’, b = b’….., owing to the errors in the values of a, b, c…

Where, 

\(\frac{\partial{y}}{\partial{a}}]_{a = a’}\) is the derivative of y with respect to a (at a = a’), where other variables like b, c…. are constant.

The given formula follows directly as a consequence of the method of approximations and the logical assumption that the total error in the calculation of y is the sum of the different errors obtained in y due to the errors in the measurement of independent variables a, b, c…etc.

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Things to Remember

• Approximation is the numerical estimate or rough guess of any number whose precise value is not determined. 
• It is a mathematical tool that is dependent on the derivatives of functions.
• Approximation is denoted by the symbol ‘≈’ meaning “almost equal to”.
• Differentials are used to determine the approximation of certain quantities in Mathematics.
• It is also used to approximate the errors in calculating certain functions. 

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Previous Years’ Questions

1. The maximum area of a rectangle inscribed in the circle… (KCET - 2018)
2. The angle of intersection of the two curves…
3. A wire of length 20 cm is bent in the form of a sector of a… (KCET - 2010)
4. If θ is semi-vertical angle of a cone of maximum volume and… (KEAM)
5. If the tangent at a point P, with parameter t, on the curve… (JEE Main - 2016)
6. A wire of length 2 units is cut into two parts which are bent… (JEE Main - 2016)
7. If f : R → R is a differentiable function such that… (JEE Advanced - 2017)
8. If the area of a circle increases at a uniform rate, then… (COMEDK UGET - 2011)
9. Length of the subtangent at (a, a) on the curve… (COMEDK UGET - 2012)
10. A spherical balloon is being inflated at the rate of… (COMEDK UGET - 2014)