# Solving for k With Gamma Function

• Matthollyw00d
In summary, the conversation discusses solving for k in the equation k*Γ((n+1)/2 + 1)=Γ(n/2 + 1) using properties of the Gamma function and the Beta function. The initial solution of k= \frac{\Gamma(\frac{n}{2}+ 1)}{\Gamma(\frac{n-1}{2}+ 1)} is mentioned, but the individual is looking for a simpler solution without the Gamma function. The conversation also mentions using Euler's Reflective formula to work with the value of Γ(-1/2) and clarifies the correct equation to solve for k.
Matthollyw00d
k*Γ((n-1)/2 + 1)=Γ(n/2 + 1)

I need to solve for k, and I'm having some difficulty manipulating the gamma function to obtain my desired result. Any properties, hints or help would be greatly appreciated.

The Gamma function relates to the factoria function for a positive integer by $$\Gamma (n+1) =n!$$ If things like n/2 are a problem, we have $$\Gamma((n+1)/2+1)=(n+1)/2*\Gamma\((n+1)/2)$$

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Matthollyw00d said:
k*Γ((n-1)/2 + 1)=Γ(n/2 + 1)

I need to solve for k, and I'm having some difficulty manipulating the gamma function to obtain my desired result. Any properties, hints or help would be greatly appreciated.
I must be completely misunderstanding the question.
$$k= \frac{\Gamma(\frac{n}{2}+ 1)}{\Gamma(\frac{n-1}{2}+ 1)}$$
What more do you want? To reduce the right side to a single gamma function?

HallsofIvy said:
I must be completely misunderstanding the question.
$$k= \frac{\Gamma(\frac{n}{2}+ 1)}{\Gamma(\frac{n-1}{2}+ 1)}$$
What more do you want? To reduce the right side to a single gamma function?

Yes, sorry. Obviously that could be a solution and I'll live with that solution if it's the best I can get; however, I'm pretty sure k can be reduced to just a simple expression of n without the Gamma function hanging around.

And Robert Ihnot, that's pretty much all I've been using and a bit of the Beta Function, but was unable to get very far last night. I kept getting a Γ(-1/2) and I can't work with that.

You can work with that using Euler's Reflective formula: $$\Gamma(1-z)\Gamma(z)$$
$$=\pi divided by sin(\pi(z))$$

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As it turns out I had made an error early on in the problem and it turns out I need to find k for k*Γ((n+1)/2 + 1)=Γ(n/2 + 1) instead of k*Γ((n-1)/2 + 1)=Γ(n/2 + 1). Which now seems much more promising and I should be able to find a solution with the Beta Function. Now I just need to find the solution to
B((n/2) + 1, 1/2)

## 1. What is the gamma function?

The gamma function is a mathematical function that is used to extend the factorial function to non-integer values. It is denoted by the symbol Γ and is defined as Γ(z) = ∫0 xz-1e-x dx.

## 2. How is the gamma function related to solving for k?

The gamma function is often used in solving for k in a variety of mathematical equations. This is because the gamma function can be used to simplify complex equations involving factorials and powers of variables.

## 3. Can the gamma function be used for negative values of k?

Yes, the gamma function is defined for all complex numbers except for negative integers. This means that it can be used for negative values of k as long as k is not a negative integer.

## 4. What are some other uses for the gamma function?

The gamma function has many applications in mathematics, physics, and engineering. It is used in probability theory, statistics, and the calculation of areas and volumes. It also has applications in solving differential equations and in the field of complex analysis.

## 5. Are there any special properties of the gamma function?

Yes, the gamma function has several special properties that make it a useful tool in mathematical analysis. These include the reflection formula, the duplication formula, and the recurrence relation. It also has connections to other mathematical functions, such as the beta function and the zeta function.

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