 # Use a linear approximation (or differentials) to estimate the given number. (1.999)^5

Linear Approximation or Linearization is a method used to approximate or estimate the value of a given function at a particular point using a line expression in terms of a single real variable. The Linear Approximation is represented by L(x).

As per Taylor’s theorem for the case involving $n=1$, we know that a function $f$ of one real number that is differentiated is represented as follows:

$f(x)\ =\ f(a)\ +\ f^\prime(a)(x-a)\ +\ R$

Here, $R$ is defined as the remainder term. For Linear approximation, we do not consider the remainder term $R$. Hence, the Linear Approximation of a single real variable is expressed as follows:

$L(x)\ \approx\ f(a)\ +\ f^\prime(a)(x\ -\ a)$

Given Term is: $=\ {(1.999)}^5$

Let:

$f(x)\ =\ {(1.999)}^5$

And:

$x\ =\ 1.999$

So:

$f(x)\ =\ x^5$

The nearest whole number $a$ to the given value of $x$ will be $2$. Hence:

$a\ =\ 2$

If we approximate $x\approx a$, then:

$f(x)\ \approx\ f(a)$

$f(a)\ =\ a^5$

Since $a=2$, so:

$f(2)\ =\ 2^5$

$f(2)\ =\ 32$

Now we will find the first derivative of $f(a)$ with respect to $a$ as follows:

$f^\prime(a)\ =\ \frac{d}{da}{\ (a)}^5$

$f^\prime(a)\ =\ 5a^4$

Substituting the value for $a=2$, we get:

$f^\prime(2)\ =\ 5{(2)}^4$

$f^\prime(2)\ =\ 80$

As per the expression for Linear Approximation, we know that:

$f(x)\ \approx\ f(a)\ +\ f^\prime(a)(x\ -\ a)$

Substituting the value in the above expression:

$f(1.999)\ \approx\ f(2)\ +\ f^\prime(2)(1.999\ -\ 2)$

Substituting the values for $f(2)$ and $f^\prime(2)$, we get:

$L(1.999)\ \approx\ 32\ +\ (80)(1.999\ -\ 2)$

$L(1.999)\ \approx\ 32\ +\ (80)(-0.001)$

$L(1.999)\ \approx\ 32\ -\ 0.08$

$L(1.999)\ \approx\ 31.92$

## Numerical Result

As per Linear Approximation, the estimated value for $({1.999)}^5$ is $31.92$.

$({1.999)}^5\ =\ 31.92$

## Example

Use a linear approximation (or differentials) to estimate the given number. $({3.001)}^4$

Solution

Given Term is: $=\ {(3.001)}^4$

Let:

$f(x)\ =\ {(3.001)}^4$

And:

$x\ =\ 3.001$

So:

$f(x)\ =\ x^4$

The nearest whole number $a$ to the given value of $x$ will be $3$. Hence:

$a\ =\ 3$

If we approximate $x\approx a$, then:

$f(x)\ \approx\ f(a)$

$f(a)\ =\ a^4$

Since $a=3$, so:

$f(3)\ =\ 3^4$

$f(3)\ =\ 81$

Now we will find the first derivative of $f(a)$ with respect to $a$ as follows:

$f^\prime(a)\ =\ \frac{d}{da}{\ (a)}^4$

$f^\prime(a)\ =\ 4a^3$

Substituting the value for $a=3$, we get:

$f^\prime(3)\ =\ 4{(3)}^3$

$f^\prime(3)\ =\ 108$

As per the expression for Linear Approximation, we know that:

$f(x)\ \approx\ f(a)\ +\ f^\prime(a)(x\ -\ a)$

Substituting the value in above expression:

$f(3.001)\ \approx\ f(3)\ +\ f^\prime(3)(3.001\ -\ 3)$

Substituting the values for $f(2)$ and $f^\prime(2)$, we get:

$L(3.001)\ \approx\ 81\ +\ (108)(3.001\ -\ 3)$

$L(3.001)\ \approx\ 81\ +\ (108)(0.001)$

$L(3.001)\ \approx\ 81\ +\ 0.108$

$L(3.001)\ \approx\ 81.108$

So, as per Linear Approximation, the estimated value for $({3.001)}^4$ is $81.108$.

$({3.001)}^4\ =\ 81.108$