Gamma Function

Worked Example

Using the recurrence relation Γ(z+1) = z Γ(z) and the known value Γ(1/2) = √π, we can determine the value of the Gamma function at half-integers. Let's calculate the exact result for Γ(5/2).

Calculation trace:

1. Reduce the argument Γ(5/2) = (3/2) * Γ(3/2)
2. Reduce the argument again Γ(3/2) = (1/2) * Γ(1/2)
3. Substitute known base value Γ(1/2) = √π
4. Combine to find the expected result: (3/2) * (1/2) * √π = (3/4) * √π ≈ 1.329340388

Therefore, the expected result for Γ(5/2) is exactly (3/4)√π.

Interactive Experiment

Explore the properties of the Gamma function by using the interactive Gamma Function module. Adjust the controls to see how the recurrence relation connects points on the curve. Notice how the function grows for positive values and alternates signs with simple poles for negative integers.

1. Set the "Argument x" to 0.5 to view the finite result 1.77245...
2. Change the "Argument x" to 0 and observe the result transition to a pole.

Limitations

The Gamma function is not defined at nonpositive integers (0, -1, -2, ...), which are simple poles. At these points, the function approaches positive or negative infinity depending on the direction of approach. Therefore, evaluation exactly at these inputs is structurally undefined. The visualization skips drawing continuous curves precisely across these poles to accurately reflect the distinct vertical asymptotes.

References

• NIST Digital Library of Mathematical Functions: 5.2 The Gamma Function
• NIST Digital Library of Mathematical Functions: 5.5 Recurrence Relations

The NIST DLMF references above provide comprehensive and rigorous documentation of the properties, recurrence relations, and approximations of the Gamma function and related mathematical functions.