Error bounds in linear approximation
- Author:
- Joseph Lo
Here we want to use linear approximation to find an approximate value of \((2.4)^5\).
Consider the function \(f(x) = x^5\) near a convenient point. Since \(2^5\) is easy to calculate and 2 is "close" to 2.4, we use the linear approximation of \(f(x)\) at \(x = 2\) to approximate \(f(2.4) = (2.4)^5\).
Therefore, \[f(x) \approx L(x) = f(a) + f'(a)(x-a),\] where \(a = 2\), \(f(a) = 32\) and \(f'(a) = 80\). It follows that \[x^5 \approx 32 + 80(x-a)\] and \[(2.4)^5 \approx 32 + 80(2.4-2) = 64\]
First of all, since \(f''(x) = 20x^3 > 0\) when \(x \in [2,2.4]\), the function is concave up and the tangent line is below the curve. It means that the linear approximation \((2.4)^5 \approx 64\) is an underestimate.
To find the error bound, we look at the upper bound of the magnitude (or absolute value) of the second derivative, i.e. \(|f''(x)| = 20|x|^3\) for \(x \in [2, 2.4]\).
\[\text{upper bound of } |f''(x)| = 20\cdot(2.4)^3 = 276.48 =M\]
Therefore,
\[| \text{error in linear approximation} | \le \dfrac{1}{2}M|x-a|^2\]
or
\[ |\text{error in linear approximation}| \le \frac{1}{2}\cdot 276.48 \cdot|2.4-2|^2 = 22.118\]
Recall that the linear approximation \((2.4)^5 \approx 64\) is an underestimate. As a result, the true value of \((2.4)^5\) is somewhere between \(64\) and \(64 + 22.118\), or in another word, the true value is bounded by \[64 \le (2.4)^5 \le 86.118\]
In the diagram, this range where the true value lies is shown by the blue bracket. The blue dash curve corresponds to the upper bound and the red dash curve corresponds to the lower bound of the estimate (in this case, the linear approximation). Note that the actual curve \(y = f(x)\) must lie between the blue and the red curves.
You can move the slider to change the value of \(x\).