The recurrence relation for the Gamma function states:

\[
\Gamma(n+1) = n \times \Gamma(n)
\]

This relation is very similar to how factorials work, where \( (n+1)! = (n+1) \times n! \).

Let's prove this recurrence relation step by step.

### Step 1: Start with the definition of the Gamma function
The Gamma function is defined as:

\[
\Gamma(n) = \int_0^\infty x^{n-1} e^{-x} \, dx
\]

### Step 2: Express \( \Gamma(n+1) \) using the definition
By substituting \( n+1 \) into the definition, we get:

\[
\Gamma(n+1) = \int_0^\infty x^n e^{-x} \, dx
\]

Now, compare this with the original definition \( \Gamma(n) \). The only difference is that the exponent of \( x \) has increased from \( n-1 \) to \( n \).

### Step 3: Use integration by parts
To prove the recurrence relation, we'll apply the technique of **integration by parts**. Integration by parts is a method that comes from the product rule of differentiation and is given by:

\[
\int u \, dv = uv - \int v \, du
\]

For this proof, let's choose:

- \( u = x^n \)
- \( dv = e^{-x} dx \)

Now, we need to find \( du \) and \( v \):

- Differentiating \( u = x^n \), we get \( du = n x^{n-1} dx \)
- Integrating \( dv = e^{-x} dx \), we get \( v = -e^{-x} \)

### Step 4: Apply integration by parts
Now apply the formula for integration by parts:

\[
\Gamma(n+1) = \int_0^\infty x^n e^{-x} \, dx = \left[ -x^n e^{-x} \right]_0^\infty + \int_0^\infty n x^{n-1} e^{-x} \, dx
\]

Let’s break this into parts:

1. **Boundary term**: \( \left[ -x^n e^{-x} \right]_0^\infty \)

   - When \( x \to \infty \), \( x^n e^{-x} \) goes to 0 because \( e^{-x} \) decreases much faster than \( x^n \) increases.
   - When \( x = 0 \), \( x^n e^{-x} \) also becomes 0 since \( x^n = 0 \) at \( x = 0 \).

   So, the boundary term equals 0.

2. **Remaining integral**: 

   \[
   \int_0^\infty n x^{n-1} e^{-x} \, dx
   \]

   Notice that this integral is just \( n \times \Gamma(n) \), because:

   \[
   \Gamma(n) = \int_0^\infty x^{n-1} e^{-x} \, dx
   \]

### Step 5: Combine the results
So, we are left with:

\[
\Gamma(n+1) = 0 + n \times \Gamma(n)
\]

This simplifies to the desired recurrence relation:

\[
\Gamma(n+1) = n \times \Gamma(n)
\]

### Conclusion
We've now proven that the Gamma function follows the recurrence relation \( \Gamma(n+1) = n \times \Gamma(n) \).

### Do You Have Any Questions?

Here are 5 related questions:

1. How does the recurrence relation help in calculating Gamma functions for large numbers?
2. Can you use the recurrence relation to find \( \Gamma(4) \) if you know \( \Gamma(3) \)?
3. What are some practical applications of the Gamma function's recurrence relation?
4. How does the recurrence relation connect to the factorial function for integers?
5. What is the significance of the boundary term becoming 0 in the proof?

### Tip:
When solving problems involving the Gamma function, always remember to use the recurrence relation to simplify your calculations, especially when dealing with large numbers!

Explore the proof of the recurrence relation for the Gamma function, Gamma(n+1) = n * Gamma(n). Step-by-step explanation using the Gamma function definition and integration by parts method. Learn practical applications and connections to factorial functions.

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