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

In summary: Well, using the general Liebniz rule you can quite easily get to \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
  • #1
winterfors
71
0
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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  • #2
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.
 
  • #3
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...
 

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

1. What is the general formula for the N:th derivative of exp(x)/x type function?

The general formula for the N:th derivative of exp(x)/x type function is given by:

f(N)(x) = (-1)N N! exp(x)/x(N+1)

2. How do you find the N:th derivative of exp(x)/x type function?

To find the N:th derivative of exp(x)/x type function, you can use either the product rule or the quotient rule. The process involves taking the N:th derivative of each individual term in the function, and then combining them using the appropriate rule.

3. Can the N:th derivative of exp(x)/x type function be simplified?

Yes, the N:th derivative of exp(x)/x type function can be simplified by factoring out common terms and using properties of exponents. The resulting expression may not always be in the simplest form, but it can help in further analysis of the function.

4. What is the behavior of the N:th derivative of exp(x)/x type function as N increases?

As N increases, the N:th derivative of exp(x)/x type function becomes more oscillatory in nature. This means that the function will have more extreme values and will oscillate more rapidly as N increases.

5. What is the significance of the N:th derivative of exp(x)/x type function?

The N:th derivative of exp(x)/x type function is significant in many applications, including physics, engineering, and economics. It can be used to model the behavior of various systems and to make predictions about their future behavior. It also has important connections to other mathematical concepts, such as Taylor series and Laplace transforms.

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