The aim of this article is to find the value of a given number raised to a degree.

The basic concept behind this article is the use of **Linear Approximation** or **Differential** to calculate the value of a given **function** or a **number**.

**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 **r****eal 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:

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

## Expert Answer

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 $81.108$.

\[({3.001)}^4\ =\ 81.108\]