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

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 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:

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

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 $81.108$.

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

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