Chapter 4: Problem 82
If \(f^{\prime \prime}\) is continuous, show that $$\quad \lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}}=f^{\prime \prime}(x)$$
Short Answer
Expert verified
The limit expression equals \( f''(x) \).
Step by step solution
01
Recognize the Formula
The expression \( \frac{f(x+h)-2f(x)+f(x-h)}{h^{2}} \) is a symmetric finite difference and is used to approximate the second derivative \( f''(x) \) for small \( h \). Our goal is to show that as \( h \to 0 \), this expression approaches \( f''(x) \).
02
Apply Taylor Series Expansion
Apply the Taylor series expansion for \( f(x+h) \) and \( f(x-h) \) around \( x \), assuming \( f'' \) is continuous. We have:\[ f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x) + O(h^4)\]\[ f(x-h) = f(x) - hf'(x) + \frac{h^2}{2}f''(x) - \frac{h^3}{6}f'''(x) + O(h^4)\]
03
Substitute and Simplify
Substitute the series expansions of \( f(x+h) \) and \( f(x-h) \) into the given limit expression:\[\frac{(f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x) + O(h^4)) - 2f(x) + (f(x) - hf'(x) + \frac{h^2}{2}f''(x) - \frac{h^3}{6}f'''(x) + O(h^4))}{h^2}\]Simplifying, the terms \( hf'(x) \) cancel out and we get:\[\frac{h^2f''(x) + O(h^4)}{h^2} = f''(x) + O(h^2)\]
04
Evaluate the Limit
Taking the limit as \( h \rightarrow 0 \) of the expression \( f''(x) + O(h^2) \) gives the result \( f''(x) \) because the error term \( O(h^2) \) vanishes:\[\lim_{h \to 0}(f''(x) + O(h^2)) = f''(x).\]
05
Conclusion: Confirm the Derivative
Thus, we have shown that the expression converges to the second derivative. Therefore, it is verified that:\[\lim_{h \to 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series Expansion
The Taylor Series Expansion is a powerful tool used in calculus to express functions as infinite sums of terms calculated from the values of its derivatives at a single point. For the purposes of approximating functions or derivatives, it provides a means to express a function near a specific point. This expansion around a point, such as \( x \), forms the basis for approximating the function locally with polynomial terms.When dealing with the expression \( f(x+h) \), the Taylor series expansion is applied:
- \( f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x) + O(h^4) \)
- Similarly, \( f(x-h) = f(x) - hf'(x) + \frac{h^2}{2}f''(x) - \frac{h^3}{6}f'''(x) + O(h^4) \)
Symmetric Finite Difference
The symmetric finite difference is a technique to approximate the derivatives of a function using values of the function at certain points. It is especially useful for approximating the second derivative. The formula provided is:\[ \frac{f(x+h)-2f(x)+f(x-h)}{h^2} \]This formula is designed to provide an approximation of the second derivative, \( f''(x) \). The choice of a symmetric pattern is key:
- It utilizes function values at \( x+h \) and \( x-h \), which surround the point \( x \).
- The central position at \( x \) helps in achieving a higher accuracy compared to one-sided differences.
Limit as h approaches zero
Limits are foundational in calculus, particularly when analyzing the behavior of functions as they approach certain values. The process of taking the limit as \( h \) approaches zero plays a critical role in differentiating functions, as seen in this problem.As \( h \to 0 \), the expression containing the finite difference approaches the actual second derivative, \( f''(x) \). The process follows these steps:
- The initial approximation \( \frac{f(x+h)-2f(x)+f(x-h)}{h^2} \) includes a minor error term \( O(h^2) \).
- As \( h \) diminishes, the error term becomes negligible, simplifying to just \( f''(x) \).