Question

Explain, in terms of linear approximations or differentials, why the approximation is reasonable.37. \(\ln 1.04 \approx 0.04\)

Short answer

The approximation \(\ln 1.04 \approx 0.04\) is reasonable.

Step-by-step solution

Linear Approximation

The Linear approximation is basically an expanded representation of differentials in calculus that can be expressed as:

\(L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\)

Examining the function using approximation method:

Let we have:

\(y = f\left( x \right) = \ln x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{say}}} \right)\)

On differentiating, we have:

\(f'\left( x \right) = \frac{1}{x}\)

Now, solving at\(x = 1\):

\(\begin{aligned}{l}f\left( 1 \right) &= 0\\f'\left( 1 \right) &= 1\end{aligned}\)

So, the linear approximation at\(x = 1\)will be:

\(\begin{aligned}{c}f\left( {1.04} \right) &= f\left( 1 \right) + f'\left( 1 \right)\left( {x - 1} \right)\\ &= 0 + \left( {1.04 - 1} \right)\\ &= 1.04 - 1\\ &= 0.4\end{aligned}\)

Hence, the approximation \(\ln 1.04 \approx 0.04\)is reasonable.