N:th derivative of exp(x)/x type function?

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SUMMARY

The discussion centers on deriving the k-th derivative of the function f(x) = exp(x) * ∏(x - y_j)^(-z_j). A conjectured formula for the k-th derivative is presented, involving a sum over k-length vectors of positive integers. The Leibniz rule for derivatives of products is referenced as a method to approach the proof, suggesting that the derivative of the function g(x) must be determined. The conversation highlights the complexity of transitioning from the derived sum to the conjectured formula.

PREREQUISITES
  • Understanding of the Leibniz rule for derivatives of products
  • Familiarity with exponential functions and their derivatives
  • Knowledge of combinatorial mathematics, specifically k-length vectors
  • Experience with product and chain rule applications in calculus
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  • Research the general Leibniz rule for higher-order derivatives
  • Explore combinatorial techniques for deriving sums over integer partitions
  • Study the properties of exponential functions in calculus
  • Investigate existing literature on derivatives of products involving exponential functions
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Mathematicians, calculus students, and researchers in mathematical analysis who are interested in advanced derivative techniques and combinatorial applications in calculus.

winterfors
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I have a function
[tex]f(x) = \exp (x)\prod\limits_{j = 1}^n {{{(x - {y_j})}^{ - {z_j}}}}[/tex]
of which I need to find an expression of the k:th derivative with respct to x:
[tex]\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x)[/tex] .

I have been able to make a conjecture that seems to be correct
[tex]\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x) = f(x)k!\sum\limits_{{\bf{l}} \in {\Lambda _k}} {\frac{1}{{{l_1}!}}{{\left( {1 + \sum\limits_{j = 1}^n {\frac{{{z_j}}}{{{y_i} - x}}} } \right)}^{{l_1}}}\prod\limits_{a = 2}^k {\frac{1}{{{l_a}!}}{{\left( {\frac{1}{a}\sum\limits_{j = 1}^n {\frac{{{z_j}}}{{{{({y_i} - x)}^a}}}} } \right)}^{{l_a}}}} }[/tex]

where [itex]{\Lambda _k}[/itex] is the set of k-length vectors of positive integers so that [itex]{\bf{l}} \cdot {[1,2,...,k]^{\rm{T}}} = k[/itex] [itex]\forall {\bf{l}} \in {\Lambda _k}[/itex].

I have no idea of how to prove it, but I'm sure someome must have looked at such a simple case before. Does anyone have an idea of how to prove it, or know of a reference for this kind of formula?
 
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The Liebniz rule deals with derivatives of products. Writing your function as [itex]f(x) = \exp(x)g(x)[/itex], the rule gives

[tex]f^{(k)}(x) = \exp(x) \sum_{\ell=0}^k \frac{k!}{\ell!(k-\ell)!}g^{(\ell)}(x),[/tex]

so you need to figure out the derivative of the g(x) function, which you can try to do using the Liebniz rule again. You can probably help churn out your formula this way.
 
Mute said:
The Liebniz rule deals with derivatives of products. Writing your function as [itex]f(x) = \exp(x)g(x)[/itex], the rule gives

[tex]f^{(k)}(x) = \exp(x) \sum_{\ell=0}^k \frac{k!}{\ell!(k-\ell)!}g^{(\ell)}(x),[/tex]

so you need to figure out the derivative of the g(x) function, which you can try to do using the Liebniz rule again. You can probably help churn out your formula this way.

Well, using the general Liebniz rule you can quite easily get to
[tex]\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x) = f(x)k!\sum\limits_{{\bf{m}} \in \Theta _k^{n + 1}} {\frac{1}{{\prod\limits_{j = 1}^{n + 1} {{m_j}!} }}} \prod\limits_{j = 1}^n {\frac{{({z_j} - 1 + {m_j})!}}{{({z_j} - 1)!}}} {({y_j} - x)^{ - {m_j}}}[/tex]

where [itex]\Theta _k^{n + 1}[/itex] is the set of all n+1 length vectors of positive integers such that [itex]{m_1} + {m_2} + \cdots + {m_{n + 1}} = k[/itex].

How to get from this sum to the conjecture above seems a bit more tricky...
 

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