The derivative of f(x) = cos(x) can be estimated using the approximation \(f'\left( x \right) = \frac{{f\left( {x + h} \right) - f\left( {x - h} \right)}}{{2h}}\) . The percentage error is calculated as \(\left( {\frac{{Exact\;value - Approximate\;value}}{{Exact\;value}}} \right) \times 100\). The percentage error in the derivative of f(x) at x = π/6 radian, choosing h = 0.1 radian, is
> 0.1 % and < 1 %
This problem asks us to calculate the percentage error when approximating the derivative of a function \(f(x) = \cos(x)\) using a specific numerical method, known as the central difference approximation. We are given the function, the approximation formula, the point of evaluation, and the step size, along with the formula for percentage error.
First, let's find the exact derivative of the given function \(f(x) = \cos(x)\).
Now, we need to evaluate the exact derivative at the given point \(x = \frac{\pi}{6}\) radian.
The problem provides an approximation formula for the derivative:
\[f'\left( x \right) = \frac{{f\left( {x + h} \right) - f\left( {x - h} \right)}}{{2h}}\]
To find the percentage error, we need to understand the relationship between the exact value and this approximation. We can use Taylor series expansions to analyze the error associated with this central difference formula.
Let's expand \(f(x+h)\) and \(f(x-h)\) around \(x\) using Taylor series:
Now, subtract the expansion of \(f(x-h)\) from \(f(x+h)\):
\[f(x+h) - f(x-h) = (f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x) + \dots) - (f(x) - hf'(x) + \frac{h^2}{2}f''(x) - \frac{h^3}{6}f'''(x) + \dots)\]
\[f(x+h) - f(x-h) = 2hf'(x) + 2\frac{h^3}{6}f'''(x) + O(h^5)\]
\[f(x+h) - f(x-h) = 2hf'(x) + \frac{h^3}{3}f'''(x) + O(h^5)\]
Divide by \(2h\) to get the approximate derivative:
\[\text{Approximate value} = \frac{f(x+h) - f(x-h)}{2h} = f'(x) + \frac{h^2}{6}f'''(x) + O(h^4)\]
This shows that the approximation directly includes the exact derivative \(f'(x)\) and an error term.
The error in the approximation is given by:
\[\text{Error} = \text{Exact value} - \text{Approximate value}\]
\[\text{Error} = f'(x) - \left( f'(x) + \frac{h^2}{6}f'''(x) \right)\]
\[\text{Error} = -\frac{h^2}{6}f'''(x)\]
Now, we need to find \(f'''(x)\) for \(f(x) = \cos(x)\):
Substitute \(f'''(x) = \sin(x)\) into the error formula:
\[\text{Error} = -\frac{h^2}{6}\sin(x)\]
At \(x = \frac{\pi}{6}\) and \(h = 0.1\):
So, \[\text{Error} = -\frac{0.01}{6} \times 0.5 = -\frac{0.005}{6}\]
The percentage error formula is given as:
\[\text{Percentage Error} = \left( {\frac{{\text{Exact value} - \text{Approximate value}}}{{\text{Exact value}}}} \right) \times 100\]
We know that \(\text{Exact value} - \text{Approximate value} = \text{Error}\). So:
\[\text{Percentage Error} = \left( {\frac{{\text{Error}}}{{\text{Exact value}}}} \right) \times 100\]
Substitute the expressions for Error and Exact Value:
\[\text{Percentage Error} = \left( {\frac{{-\frac{h^2}{6}\sin(x)}}{{-\sin(x)}}} \right) \times 100\]
Notice that \((-\sin(x))\) cancels out from the numerator and denominator, assuming \(\sin(x) \neq 0\). At \(x = \frac{\pi}{6}\), \(\sin(\frac{\pi}{6}) = 0.5 \neq 0\).
\[\text{Percentage Error} = \left( {\frac{{h^2}}{6}} \right) \times 100\]
Now, substitute the value of \(h = 0.1\):
\[\text{Percentage Error} = \left( {\frac{{(0.1)^2}}{6}} \right) \times 100\]
\[\text{Percentage Error} = \left( {\frac{{0.01}}{6}} \right) \times 100\]
\[\text{Percentage Error} = \frac{1}{6}\%\]
\[\text{Percentage Error} \approx 0.1666\dots \%\]
Let's check where our calculated percentage error \(0.1666\dots \%\) falls among the given options:
The calculated percentage error falls within the range of "> 0.1 % and < 1 %".
The final answer is \(\text{Percentage Error} \approx 0.1666\dots \% \), which is in the range > 0.1 % and < 1 %.
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