What is the role of Laurent series in solving limits at infinity?

In summary, the conversation discusses using Wolfram Alpha to calculate the integral of log(Gamma(x)) from 0 to 1 and the final step to solve the problem. It also mentions using series expansions at n=infinity and the Laurent series to solve limits. The conversation concludes by clarifying the correct expansions for 1/z and discussing the oddness of letting the expansion point go to infinity.
  • #1
MAGNIBORO
106
26
hi, I try to calculate the integral
$$\int_{0}^{1}log(\Gamma (x))dx$$

and the last step To solve the problem is:
$$1 -\frac{\gamma }{2} + \lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$
and wolfram alpha tells me something about series expansion at ##n=\infty## of laurent series
http://www.wolframalpha.com/input/?...rmassumption={"C",+"limit"}+->+{"Calculator"}

I know a little about series of laurent, but I do not understand how they serve to solve limits
and expansion at ##n=\infty##.
 
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  • #2
If the Laurent series does not include positive powers of n (as in this problem), then the absolute term is your limit (the negative powers vanish for n->infinity).

If it has a finite number of positive powers of n, then your expression does not have a limit. If none of those cases apply, it gets complicated.
 
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  • #3
mfb said:
If the Laurent series does not include positive powers of n (as in this problem), then the absolute term is your limit (the negative powers vanish for n->infinity).

If it has a finite number of positive powers of n, then your expression does not have a limit. If none of those cases apply, it gets complicated.
I do understand that, but what does it mean "expansion at ##n=\infty##". and how wolfram alpha obtained for complicated expresions like
$$\lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$
 
  • #4
It is the same as an expansion around u=0 for u=1/n.

How: Sum all terms of the expansion of the summands in the usual way.
 
  • #5
mfb said:
It is the same as an expansion around u=0 for u=1/n.

How: Sum all terms of the expansion of the summands in the usual way.

but expancion at u=0 of n=1/u Is not it just 1/u?
like expancion at z=0 of y= z^2 - 3z +1/z is not just z^2 - 3z +1/z?

and other question:
the expancion of 1/z at z=n is
$$\frac{1}{z}=\frac{1}{u+n}=\frac{1}{n}\, \frac{1}{1+\frac{u}{n}}=\sum_{k=0}^{\infty }\frac{(-1)^{k}\, (z-n)^{k}}{n^{k+1}}$$

for n tends to infinity Would be
$$\lim_{n\rightarrow \infty }+\frac{1}{n}-\frac{z-n}{n^{2}}+... = 0$$

so... what mistake I have?
 
  • #6
The limit of 1/z for z->infinity is 0. Where is the problem?
If you expand around n, letting the expansion point go to infinity looks odd.
MAGNIBORO said:
but expancion at u=0 of n=1/u Is not it just 1/u?
like expancion at z=0 of y= z^2 - 3z +1/z is not just z^2 - 3z +1/z?
Those are the correct expansions.
 
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  • #7
mfb said:
The limit of 1/z for z->infinity is 0. Where is the problem?
If you expand around n, letting the expansion point go to infinity looks odd.
Those are the correct expansions.
thanks =D
 

Related to What is the role of Laurent series in solving limits at infinity?

1. What are limits and how are they used in mathematics?

Limits are a fundamental concept in mathematics that describe the behavior of a function as the input approaches a certain value. They are used to determine the value of a function at a specific point, as well as to analyze the behavior of a function near a particular point. In other words, limits help us understand how functions behave as we get closer and closer to a certain value.

2. What is a Laurent series and why is it important?

A Laurent series is a representation of a complex function as an infinite sum of powers of the variable, both positive and negative. It is important because it allows us to extend the concept of a power series to functions that are not analytic at all points. This is particularly useful in areas such as physics and engineering, where functions may have singularities or branch cuts.

3. How do you find the Laurent series of a function?

To find the Laurent series of a function, we first need to determine the singularities of the function. These are the points where the function is not analytic. Then, we use the Maclaurin series (or Taylor series if the function is analytic at 0) to find the coefficients of the positive powers of the variable. Finally, we use the Cauchy integral formula to find the coefficients of the negative powers of the variable.

4. Can a function have more than one Laurent series?

Yes, a function can have multiple Laurent series expansions. This is because the coefficients of the series depend on the path chosen for the integral in the Cauchy integral formula. If different paths are chosen, the coefficients will be different, resulting in different Laurent series. However, all of these series will converge to the same function within their respective domains of convergence.

5. What is the difference between a power series and a Laurent series?

The main difference between a power series and a Laurent series is the range of values for the variable. A power series only contains positive powers of the variable, while a Laurent series includes both positive and negative powers. Additionally, a power series has a single center of convergence, while a Laurent series can have multiple singularities and thus multiple regions of convergence.

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