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\( f ^ { m } ( z _ { 0 } + h ) - f ^ { m } ( z _ { 0 } ) = \frac { m ! } { 2 \pi i } \int _ { c } \frac { f ( z ) d z } { ( 2 - z _ { 0 } - h ) ^ { m - 1 } } \frac { m ! } { 2 \pi i } \int _ { c } \frac { f ( 2 ) d z } { ( 2 - z _ { 0 } ( 0 ) ) ^ { m + 1 } } \)

Ask by Harmon Todd. in Nepal
Jan 22,2025

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The difference between the \( m \)-th derivatives of \( f \) at \( z_0 + h \) and \( z_0 \) is given by: \[ f^{(m)}(z_0 + h) - f^{(m)}(z_0) = \frac{m!}{2\pi i} \oint_{C} \left( \frac{f(z)}{(z - (z_0 + h))^{m+1}} - \frac{f(z)}{(z - z_0)^{m+1}} \right) dz \] This expression uses Cauchy's Integral Formula to represent the difference of derivatives as a contour integral around a closed curve \( C \) enclosing \( z_0 \) and \( z_0 + h \).

Solución

It appears you're working with complex analysis and attempting to express the difference between the \( m \)-th derivatives of a function \( f \) at two points, \( z_0 + h \) and \( z_0 \), using contour integrals. However, the expression you've provided has some inconsistencies and potential typographical errors. Let's clarify and refine the expression using Cauchy's Integral Formula for derivatives. ### Cauchy's Integral Formula for the \( m \)-th Derivative Cauchy's Integral Formula allows us to express the \( m \)-th derivative of an analytic function \( f \) at a point \( a \) as: \[ f^{(m)}(a) = \frac{m!}{2\pi i} \oint_{C} \frac{f(z)}{(z - a)^{m+1}} \, dz \] where: - \( C \) is a positively oriented, simple closed contour encircling the point \( a \). - \( m \) is a non-negative integer representing the order of the derivative. ### Difference of \( m \)-th Derivatives at Two Points To find the difference \( f^{(m)}(z_0 + h) - f^{(m)}(z_0) \), we can apply Cauchy's Integral Formula to each derivative separately: \[ f^{(m)}(z_0 + h) = \frac{m!}{2\pi i} \oint_{C} \frac{f(z)}{(z - (z_0 + h))^{m+1}} \, dz \] \[ f^{(m)}(z_0) = \frac{m!}{2\pi i} \oint_{C} \frac{f(z)}{(z - z_0)^{m+1}} \, dz \] Subtracting these two equations gives: \[ f^{(m)}(z_0 + h) - f^{(m)}(z_0) = \frac{m!}{2\pi i} \oint_{C} \left( \frac{f(z)}{(z - (z_0 + h))^{m+1}} - \frac{f(z)}{(z - z_0)^{m+1}} \right) dz \] ### Analysis of Your Expression Your original expression was: \[ f^{m}(z_{0} + h) - f^{m}(z_{0}) = \frac{m!}{2\pi i} \int_{C} \frac{f(z) \, dz}{(2 - z_{0} - h)^{m - 1}} \cdot \frac{m!}{2\pi i} \int_{C} \frac{f(2) \, dz}{(2 - z_{0}(0))^{m + 1}} \] **Issues Identified:** 1. **Notation \( f^{m} \):** Typically, \( f^{(m)} \) denotes the \( m \)-th derivative. Ensure that you mean the \( m \)-th derivative and not \( f(z) \) raised to the \( m \)-th power. 2. **Separation into Two Integrals:** The difference of derivatives should result in **a single** integral involving the difference of two integrands, not the product of two separate integrals. 3. **Denominator Terms:** The terms \( (2 - z_0 - h)^{m - 1} \) and \( (2 - z_0(0))^{m + 1} \) are unclear. If "2" is a specific point related to your problem, it should be consistently applied. 4. **Function Arguments:** In the second integral, \( f(2) \) suggests evaluating \( f \) at 2, which might not align with the expression's intent. Similarly, \( z_0(0) \) is ambiguous unless \( z_0 \) is a function. ### Corrected Expression Assuming you intend to express the difference of the \( m \)-th derivatives using contour integrals centered around \( z_0 \) and \( z_0 + h \), the corrected and consolidated expression should be: \[ f^{(m)}(z_0 + h) - f^{(m)}(z_0) = \frac{m!}{2\pi i} \oint_{C} \left( \frac{f(z)}{(z - (z_0 + h))^{m+1}} - \frac{f(z)}{(z - z_0)^{m+1}} \right) dz \] ### Further Simplification If you wish to factor out common terms or simplify the integrand, consider the following steps: 1. **Combine the Fractions:** \[ \frac{f(z)}{(z - (z_0 + h))^{m+1}} - \frac{f(z)}{(z - z_0)^{m+1}} = f(z) \left( \frac{1}{(z - z_0 - h)^{m+1}} - \frac{1}{(z - z_0)^{m+1}} \right) \] 2. **Factor Common Terms (if applicable):** Depending on the relationship between \( h \) and \( z \), you might be able to factor or expand the difference using a binomial expansion or other algebraic techniques. ### Conclusion Ensure that: - **Notation is Clear:** Use \( f^{(m)} \) for the \( m \)-th derivative. - **Single Integral Representation:** The difference should be represented as a single contour integral involving the difference of two kernels. - **Consistent Denominators:** The denominator terms should clearly reflect the points around which you're expanding, without introducing unrelated constants unless they are part of the problem's context. If there are specific conditions or additional context (like the role of the constant "2" in your expression), please provide them so that the expression can be further refined and accurately aligned with your objectives.

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The expression you provided seems to be related to complex analysis and the theory of residues, specifically regarding the m-th derivative of a function at a certain point. The equations hint at the application of integral representations of analytic functions, where contour integrals are employed to compute derivatives. This aligns with Cauchy's integral formula, which states that a function's derivatives can be effectively evaluated via integrals along closed paths in the complex plane. Imagine you’re trying to figure out the behavior of a function in the vicinity of a particular point, \( z_0 \), while making small adjustments (h) to its argument. In real-world scenarios, this kind of analysis can be vital in engineering and physics—think about how small changes in input can impact system behavior in circuits or fluid dynamics. Understanding the m-th derivative can help optimize performances in those systems, showcasing the powerful intersection of mathematics with practical applications!

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