Why is gamma(z+1) = z * gamma(z)?

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In summary, the conversation is about understanding and proving the equation gamma ( z+ 1) = z * gamma(z) using the integral definition of gamma function. The conversation discusses using integration by parts and taking limits as x goes to zero and infinity to prove the equation. The conversation also brings up the question of why exponential terms dominate algebraic terms in limits and suggests using l'Hôpital's rule to prove it.
  • #1
shnaiwer
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hi to all ...
it is just an attempt to understand ...
the problem was to prove that
gamma ( z+ 1) = z * gamma(z) using the integral definition of gamma function ...
when i used the integration by parts i get the following :

gamma ( z+ 1 ) = ( x^z) * e^(-x) (0 ,∞) + ⌠ z x^(z-1) e^(-x ) dx
where the limits of integration ( 0 ,∞)

the integration already gives z * gamma (z) , then ( x^z) * e^(-x) │( 0 ,∞) = 0

i don't have the evidense that this term is zero ...

have u ?
 
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  • #2
What does the " | " actually denote? What values do you get when you sub in 0 and take the limit as x goes to infinity ? It should be quite straight forward. Remember exponential terms dominate algebraic terms in limits.
 
  • #3
Gib Z said:
What does the " | " actually denote? What values do you get when you sub in 0 and take the limit as x goes to infinity ? It should be quite straight forward. Remember exponential terms dominate algebraic terms in limits.

thank u ... " | " denote nothing other than ( when or at ) , we can neglect it..
when we sub in zero we get zero , since
( 0^z) * e^(-0) = 0 * 1 = 0

but the problem still hold when we try that limit as x goes to infinity , where it gives
( ∞ / ∞ ) or ( 0 / 0) if we try l'Hôpital's rule , in fact i want to understand or get the answer of the question :
why do exponential terms dominate algebraic terms in limits ?
i think that you put the exact description of what my question originally is , when you said : Remember exponential terms dominate algebraic terms in limits
 
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  • #4
Take the log of L=x^z*e^(-x), log(L)=z*log(x)-x. x grows faster than z*log(x). Use l'Hopital to prove it. So the limit of the log(L) goes to negative infinity. That means L goes to zero.
 

1. Why is gamma(z+1) = z * gamma(z)?

The identity gamma(z+1) = z * gamma(z) is known as the recurrence relation for the gamma function. It is a result of the definition of the gamma function and the properties of the factorial function.

2. What is the gamma function?

The gamma function is a mathematical function that generalizes the factorial function to non-integer values. It is defined as the integral of the function e^-x over the interval [0, ∞), and is denoted by the symbol Γ(z).

3. How is the gamma function related to the factorial function?

The gamma function is closely related to the factorial function. When z is a positive integer, gamma(z) is equal to (z-1)!, which is the factorial of z-1. This relationship is extended to non-integer values through the recurrence relation gamma(z+1) = z * gamma(z).

4. What are the applications of the gamma function?

The gamma function has various applications in mathematics, physics, and statistics. It is used to calculate probabilities in the gamma distribution, to define the beta function, and to solve differential equations in physics.

5. Are there any special values of the gamma function?

Yes, there are several special values of the gamma function. When z is a positive integer, gamma(z) takes on the values of the factorial function. Additionally, gamma(1/2) = √π, gamma(1) = 1, and gamma(2) = 1. These values have important applications in various mathematical and scientific fields.

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