Question

Carry out derivation and calculations analogous to those in Example \(1.2\), using the expression $$ \frac{f\left(x_{0}+h\right)-f\left(x_{0}-h\right)}{2 h} $$ for approximating the first derivative \(f^{\prime}\left(x_{0}\right)\). Show that the error is \(\mathcal{O}\left(h^{2}\right)\). More precisely, the leading term of the error is \(-\frac{h^{2}}{3} f^{\prime \prime \prime}\left(x_{0}\right)\) when \(f^{\prime \prime \prime}\left(x_{0}\right) \neq 0\).

Short answer

Answer: The error in using the central difference approximation is of order \(h^2\) (\(\mathcal{O}(h^2)\)). The leading term in the error expression when the third derivative is not equal to zero is \(-\frac{h^2}{3}f'''(x_0)\).

Step-by-step solution

Taylor Series Expansion

We will need to expand the function f(x) around the point x_0 using Taylor series. This means we will need to calculate the Taylor series expansions for f(x_0 + h) and f(x_0 - h). f(x_0 + h) can be expanded as follows: $$ f(x_0 + h) = f(x_0) + h f'(x_0) + \frac{h^2}{2}f''(x_0) + \frac{h^3}{6}f'''(x_0) + \dots $$ Similarly, f(x_0 - h) can be expanded as: $$ f(x_0 - h) = f(x_0) - h f'(x_0) + \frac{h^2}{2}f''(x_0) - \frac{h^3}{6}f'''(x_0) + \dots $$ Both expansions represent function f(x) around x_0 with the error term approaching zero as the power of h increases.

Derive the central difference quotient

Now, we need to derive the expression given in the exercise for the central difference quotient. We can do this by subtracting the Taylor series of f(x_0 - h) from f(x_0 + h) and then dividing by 2h: $$ \frac{f(x_0 + h) - f(x_0 - h)}{2h} = \frac{2 h f'(x_0) + \frac{2h^3}{6}f'''(x_0) + \dots}{2h} = f'(x_0) + \frac{h^2}{6}f'''(x_0) + \dots $$

Determine the error term

To find the error term, we need to compare the central difference quotient derived in step 2 with the exact derivative of f(x), which is f'(x_0). Here, we observe that the error term consists of higher-order derivatives of f(x): $$ \text{Error} = \left(f'(x_0) + \frac{h^2}{6}f'''(x_0) + \dots\right) - f'(x_0) = \frac{h^2}{6}f'''(x_0) + \mathcal{O}(h^4) $$ From this expression, we can see that the error is of the order of \(h^2\), denoted as \(\mathcal{O}(h^2)\).

Determine the leading term when f'''(x_0) ≠ 0

Given that the third derivative is not equal to 0 (f'''(x_0) ≠ 0), the leading term in the error expression is: $$ -\frac{h^2}{3}f'''(x_0) $$ Thus, the problem has been solved, and it has been shown that the error in using the central difference quotient to approximate the first derivative of a function f(x) is of order \(\mathcal{O}(h^2)\), and the leading term in the error expression is \(-\frac{h^2}{3}f'''(x_0)\) when the third derivative of the function is not equal to 0.

Key concepts

Taylor Series Expansion

To understand numerical differentiation, it is essential to grasp the concept of Taylor Series Expansion. This expansion allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It's like breaking down a function into a series of simpler polynomial terms.
When we expand a function like \(f(x)\) around a point \(x_0\), with increments by \(h\), the Taylor series expansion helps in approximating the value of the function at nearby points such as \(x_0 + h\) or \(x_0 - h\).
This expansion is expressed in mathematical terms as:
\[f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{h^2}{2}f''(x_0) + \frac{h^3}{6}f'''(x_0) + \dots \]
And similarly,
\[f(x_0 - h) = f(x_0) - hf'(x_0) + \frac{h^2}{2}f''(x_0) - \frac{h^3}{6}f'''(x_0) + \dots \]
As you expand these, notice that every term is just a tiny tweak of the prior one, getting tinier as the powers of \(h\) increase. This gives us a very close approximation of the original function. By using these expansions, you can derive the relationship needed for estimating derivatives numerically.

Central Difference Method

The Central Difference Method is a popular numerical differentiation technique developed using Taylor series expansions. It is used to approximate the derivative of a function. Unlike forward or backward difference methods, the central difference takes an average approach and generally gives more accurate results.
Here's how it works: Using the Taylor expansions previously discussed, if you subtract the expansion of \(f(x_0 - h)\) from \(f(x_0 + h)\), you get
\[f(x_0 + h) - f(x_0 - h) = 2hf'(x_0) + \frac{2h^3}{6}f'''(x_0) + \dots \]
Dividing this difference by \(2h\) yields an estimate of the first derivative \(f'(x_0)\):
\[\frac{f(x_0 + h) - f(x_0 - h)}{2h} = f'(x_0) + \frac{h^2}{6}f'''(x_0) + \dots \]
This method assumes the function is continuous and smooth, allowing the higher-order derivatives to be small so that errors are significantly reduced. It gives the derivative at the midpoint, making it central, hence the name. It's a great balance between accuracy and computational simplicity.

Error Analysis in Numerical Methods

Error Analysis is a vital aspect of numerical methods, as it helps us understand how precise our approximations are. In the Central Difference Method, understanding the error helps gauge the accuracy of the derivative approximation.
Whenever we approximate, like in our expression \(\frac{f(x_0 + h) - f(x_0 - h)}{2h}\), there are inherent errors due to omitted higher-order terms from the Taylor series.
In this context, the error is given by the expression:
\[\text{Error} = \left( f'(x_0) + \frac{h^2}{6}f'''(x_0) + \dots \right) - f'(x_0) = \frac{h^2}{6}f'''(x_0) + \mathcal{O}(h^4) \]
As indicated, the dominant error term here is \(\frac{h^2}{6}f'''(x_0)\). This means the error is proportional to \(h^2\), making it \(\mathcal{O}(h^2)\), which tells us it decreases quadratically as \(h\) gets smaller.
Understanding the size and behavior of the error term helps decide the step size \(h\) you should use to achieve your desired accuracy. Essentially, when the third derivative of the function at that point isn't zero, the error's leading term is \(-\frac{h^2}{3}f'''(x_0)\). Recognizing and calculating these error terms empowers you to make smarter choices in numerical problem solving.