Math Problem Statement

Prove that Γ(n+1) = n ⋅ Γ(n) for n > 0. Also compute Γ(5/2) using the recurrence relation of the Gamma function.

Solution

The question asks to:

  1. Prove that Γ(n+1)=nΓ(n)\Gamma(n+1) = n \cdot \Gamma(n) for n>0n > 0, using the recurrence relation of the Gamma function.
  2. Compute Γ(5/2)\Gamma(5/2) using this recurrence relation.

Let’s solve this step-by-step:


Proof of the Recurrence Relation:

The Gamma function is defined as: Γ(x)=0tx1etdt\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt

If we substitute x=n+1x = n+1 into the definition, we have: Γ(n+1)=0tnetdt\Gamma(n+1) = \int_0^\infty t^{n} e^{-t} \, dt

Using integration by parts:

  • Let u=tnu = t^n, so du=ntn1dtdu = n t^{n-1} \, dt
  • Let dv=etdtdv = e^{-t} \, dt, so v=etv = -e^{-t}

Then, tnetdt=tnet0+ntn1etdt\int t^n e^{-t} \, dt = -t^n e^{-t} \Big|_0^\infty + n \int t^{n-1} e^{-t} \, dt

The term tnet0-t^n e^{-t} \Big|_0^\infty vanishes because:

  • As tt \to \infty, ete^{-t} decays faster than tnt^n grows.
  • At t=0t = 0, tn=0t^n = 0.

Thus: Γ(n+1)=n0tn1etdt=nΓ(n)\Gamma(n+1) = n \int_0^\infty t^{n-1} e^{-t} \, dt = n \Gamma(n)

This proves the recurrence relation: Γ(n+1)=nΓ(n)\Gamma(n+1) = n \cdot \Gamma(n)


Compute Γ(5/2)\Gamma(5/2):

Using the recurrence relation Γ(n+1)=nΓ(n)\Gamma(n+1) = n \cdot \Gamma(n), we compute step-by-step:

  1. Start with Γ(5/2)\Gamma(5/2): Using the recurrence, Γ(5/2)=32Γ(3/2)\Gamma(5/2) = \frac{3}{2} \cdot \Gamma(3/2).

  2. Now compute Γ(3/2)\Gamma(3/2): Again using the recurrence, Γ(3/2)=12Γ(1/2)\Gamma(3/2) = \frac{1}{2} \cdot \Gamma(1/2).

  3. Evaluate Γ(1/2)\Gamma(1/2): It is a known fact that Γ(1/2)=π\Gamma(1/2) = \sqrt{\pi}.

  4. Substitute back: Γ(3/2)=12π\Gamma(3/2) = \frac{1}{2} \cdot \sqrt{\pi}

    Γ(5/2)=3212π=34π\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3}{4} \sqrt{\pi}

Thus, Γ(5/2)=34π\Gamma(5/2) = \frac{3}{4} \sqrt{\pi}.


Final Results:

  1. The recurrence relation is proved: Γ(n+1)=nΓ(n)\Gamma(n+1) = n \cdot \Gamma(n).
  2. The value of Γ(5/2)\Gamma(5/2) is 34π\frac{3}{4} \sqrt{\pi}.

Would you like a more detailed explanation or further derivations?
Here are 5 questions to extend your understanding:

  1. What is the significance of the Gamma function in mathematics and physics?
  2. How is Γ(x)\Gamma(x) related to factorials for integer values of xx?
  3. Can you use the recurrence relation to compute Γ(7/2)\Gamma(7/2)?
  4. How would you prove that Γ(1)=1\Gamma(1) = 1 using the definition of Γ(x)\Gamma(x)?
  5. Can you explore the relationship between the Gamma function and the Beta function?

Tip: When working with the Gamma function, remember its symmetry properties and relation to factorials, as these often simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Gamma function
Integration by parts
Recurrence relations

Formulas

Γ(x) = ∫₀⁺∞ t^(x-1) e^(-t) dt
Γ(n+1) = n ⋅ Γ(n)
Γ(1/2) = √π

Theorems

Gamma function recurrence relation

Suitable Grade Level

Undergraduate Mathematics