CALCULATION OF APPROXIMATIONS USING THE DIFFERENTIAL - MATHS - 2025

An approximation in mathematics is a number that is not the exact value of something, but is so close to it that it is considered as useful as that exact value.

When approximations are made in mathematics, it is because manually it is difficult (or sometimes impossible) to know the precise value of what you want.

The main tool when working with approximations is the differential of a function.

The differential of a function f, denoted by Δf (x), is nothing more than the derivative of the function f times the change in the independent variable, that is, Δf (x) = f '(x) * Δx.

Sometimes df and dx are used instead of Δf and Δx.

Approximations using the differential

The formula that is applied to carry out an approximation through the differential arises precisely from the definition of the derivative of a function as a limit.

This formula is given by:

f (x) ≈ f (x0) + f '(x0) * (x-x0) = f (x0) + f' (x0) * Δx.

Here it is understood that Δx = x-x0, therefore x = x0 + Δx. Using this the formula can be rewritten as

f (x0 + Δx) ≈ f (x0) + f '(x0) * Δx.

It should be noted that "x0" is not an arbitrary value, but a value such that f (x0) is easily known; moreover, "f (x)" is just the value we want to approximate.

Are there better approximations?

The answer is yes. The above is the simplest of the approximations called "linear approximation".

For better quality approximations (the error made is less), polynomials with more derivatives called "Taylor polynomials" are used, as well as other numerical methods such as the Newton-Raphson method among others.

Strategy

The strategy to follow is:

- Choose a suitable function f to carry out the approximation and the value «x» such that f (x) is the value to be approximated.

- Choose a value "x0", close to "x", such that f (x0) is easy to calculate.

- Calculate Δx = x-x0.

- Calculate the derivative of the function y f '(x0).

- Substitute the data in the formula.

Solved approximation exercises

In what continues there are a series of exercises where approximations are made using the differential.

First exercise

Approximately √3.

Solution

Following the strategy, a suitable function must be chosen. In this case, it can be seen that the function to choose must be f (x) = √x and the value to be approximated is f (3) = √3.

Now we must choose a value "x0" close to "3" such that f (x0) is easy to calculate. If "x0 = 2" is chosen, then "x0" is close to "3" but f (x0) = f (2) = √2 is not easy to calculate.

The appropriate value of "x0" is "4", since "4" is close to "3" and also f (x0) = f (4) = √4 = 2.

If "x = 3" and "x0 = 4", then Δx = 3-4 = -1. Now we proceed to calculate the derivative of f. That is, f '(x) = 1/2 * √x, so f' (4) = 1 / 2√4 = 1/2 * 2 = 1/4.

Substituting all the values ​​in the formula you get:

√3 = f (3) ≈ 2 + (1/4) * (- 1) = 2 - 1/4 = 7/4 = 1.75.

If you use a calculator you get that √3≈1.73205… This shows that the previous result is a good approximation of the real value.

Second exercise

Approximately √10.

Solution

As before, f (x) = √xy is chosen as a function, in this case x = 10.

The value of x0 to choose this time is "x0 = 9". We then have that Δx = 10-9 = 1, f (9) = 3 and f '(9) = 1 / 2√9 = 1/2 * 3 = 1/6.

When evaluating in the formula it is obtained that

√10 = f (10) ≈ 3 + 1 * 1/6 = 3 + 1/6 = 19/6 = 3.1666…

Using a calculator it is obtained that √10 ≈ 3.1622776… Here it can also be seen that a good approximation was obtained before.

Third exercise

Approximate ³√10, where ³√ denotes the cube root.

Solution

Clearly the function to be used in this exercise is f (x) = ³√x and the value of "x" must be "10".

A value close to "10" such that its cube root is known is "x0 = 8". Then we have Δx = 10-8 = 2 and f (x0) = f (8) = 2. We also have f '(x) = 1/3 * ³√x², and consequently f' (8) = 1/3 * ³√8² = 1/3 * ³√64 = 1/3 * 4 = 1/12.

Substituting the data in the formula, it is obtained that:

³√10 = f (10) ≈ 2 + (1/12) * 2 = 2 + 1/6 = 13/6 = 2.166666….

The calculator says that ³√10 ≈ 2.15443469… Therefore, the approximation found is good.

Fourth exercise

Approximate ln (1.3), where "ln" denotes the natural logarithm function.

Solution

First we choose as a function f (x) = ln (x) and the value of "x" is 1.3. Now, knowing a little about the logarithm function, we can know that ln (1) = 0, and furthermore "1" is close to "1.3". Therefore, "x0 = 1" is chosen and thus Δx = 1.3 - 1 = 0.3.

On the other hand f '(x) = 1 / x, so that f' (1) = 1. When evaluating in the given formula we have:

ln (1.3) = f (1.3) ≈ 0 + 1 * 0.3 = 0.3.

Using a calculator we have that ln (1.3) ≈ 0.262364… So the approximation made is good.

References

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