Knowlet

Unit 5: Beta & Gamma Functions and Numerical Techniques

Beta and Gamma Functions: Definitions

Gamma Function (Γ(n))

The Gamma function is an extension of the factorial function to complex and real numbers. It is defined by a definite integral:

Γ(n) = ∫0 e-x xn-1 dx (for n > 0)

Beta Function (B(m, n))

The Beta function is a related function defined by a different integral:

B(m, n) = ∫01 xm-1 (1-x)n-1 dx (for m > 0, n > 0)

Properties and Relation between Beta and Gamma

Properties of Gamma Function

  • Recurrence Relation: This is the most important property.
    Γ(n+1) = n Γ(n)
  • Relation to Factorial: If 'n' is a positive integer:
    Γ(n+1) = n!
    (e.g., Γ(4) = 3! = 6; Γ(1) = 0! = 1)
  • Important Values:
    • Γ(1) = 1
    • Γ(1/2) = √π (This is extremely useful to memorize)
    • Using the recurrence relation, we can find other half-integer values:
      Γ(3/2) = (1/2) Γ(1/2) = (1/2)√π
      Γ(5/2) = (3/2) Γ(3/2) = (3/2)(1/2)√π = (3/4)√π

Properties of Beta Function

  • Symmetry:
    B(m, n) = B(n, m)
  • Trigonometric Form: By substituting x = sin²(θ):
    B(m, n) = 2 ∫0π/2 (sinθ)2m-1 (cosθ)2n-1

Relation Between Beta and Gamma Functions

This is the key identity that connects the two functions and allows for the easy evaluation of many integrals.

B(m, n) = [ Γ(m) Γ(n) ] / Γ(m+n)

Expression of Integrals in terms of Gamma Functions

We can use the relation `B(m, n) = [ Γ(m) Γ(n) ] / Γ(m+n)` and the trigonometric form of the Beta function to solve a very common class of definite integrals.

We know:

0π/2 (sinθ)p (cosθ)q dθ = (1/2) * B( (p+1)/2, (q+1)/2 )

...by comparing `p = 2m-1` and `q = 2n-1`.

Combining these, we get the W_all_is Integral Formula:

0π/2 (sinθ)p (cosθ)q dθ = [ Γ((p+1)/2) * Γ((q+1)/2) ] / [ 2 * Γ((p+q+2)/2) ]
Example Problem: Evaluate I = ∫0 e-x² dx (Gaussian Integral)
  1. Let t = x², so x = t1/2. Then dx = (1/2)t-1/2 dt.
  2. As x → 0, t → 0. As x → ∞, t → ∞.
  3. Substitute: I = ∫0 e-t (1/2)t-1/2 dt = (1/2) ∫0 e-t t(1/2) - 1 dt
  4. This is in the form of the Gamma function: (1/2) * Γ(1/2).
  5. Since Γ(1/2) = √π, the integral is √π / 2.

Numerical Integration

These techniques are used to find the approximate value of a definite integral `I = ∫[a, b] f(x) dx` when `f(x)` is too difficult or impossible to integrate analytically. The method is to divide the interval `[a, b]` into `n` equal subintervals of width `h`.

h = (b - a) / n

Let `y₀ = f(x₀)`, `y₁ = f(x₁)`, ..., `yₙ = f(xₙ)`, where `x₀ = a` and `xₙ = b`.

Trapezoidal Rule

This method approximates the area under the curve in each subinterval as a trapezoid.

I ≈ (h/2) [ (y₀ + yₙ) + 2(y₁ + y₂ + ... + yn-1) ]

In words: (h/2) * [ (first + last ordinates) + 2 * (sum of all other ordinates) ]

Simpson's 1/3 Rule

This method is more accurate. It approximates the function over two intervals at a time using a parabola. It requires `n` to be an even number.

I ≈ (h/3) [ (y₀ + yₙ) + 4(y₁ + y₃ + y₅ + ...) + 2(y₂ + y₄ + y₆ + ...) ]

In words: (h/3) * [ (first + last) + 4*(sum of odd ordinates) + 2*(sum of even ordinates) ]

Simpson's 3/8 Rule

This method approximates the function over three intervals at a time using a cubic. It requires `n` to be a multiple of 3.

I ≈ (3h/8) [ (y₀ + yₙ) + 3(y₁ + y₂ + y₄ + y₅ + ...) + 2(y₃ + y₆ + y₉ + ...) ]

In words: (3h/8) * [ (first + last) + 3*(sum of non-multiple-of-3 ordinates) + 2*(sum of multiple-of-3 ordinates) ]

Key Point: Simpson's 1/3 rule is generally more accurate and more commonly used than the Trapezoidal rule. The 3/8 rule is for specific cases where `n` is a multiple of 3.

Solution of Equations (Root Finding)

(Note: The syllabus has a typo "Solution of linear equations...". Bisection and Newton-Raphson are for non-linear equations of the form `f(x) = 0`.)

Bisection Method

This is a "bracketing" method that is slow but guaranteed to work.

  1. Requirement: Find two points, `a` and `b`, such that `f(a)` and `f(b)` have opposite signs. This guarantees a root exists between them.
  2. Step 1: Calculate the midpoint `c = (a + b) / 2`.
  3. Step 2: Evaluate `f(c)`.
  4. Step 3 (Update):
    • If `f(a)` and `f(c)` have opposite signs, the root is in `[a, c]`. Set `b = c`.
    • If `f(b)` and `f(c)` have opposite signs, the root is in `[c, b]`. Set `a = c`.
  5. Step 4 (Repeat): Go back to Step 1 with the new, smaller interval. Repeat until the interval `[a, b]` is sufficiently small.

Newton-Raphson Method

This is an "open" method that is much faster but can sometimes fail to converge. It uses the tangent line to approximate the root.

  1. Requirement: You need the function `f(x)` and its derivative `f'(x)`.
  2. Step 1: Start with an initial "guess" `x₀` that is close to the root.
  3. Step 2 (Iterate): Generate the next, better guess `x₁` using the formula:
    xn+1 = xn - f(xn) / f'(xn)
  4. Step 3 (Repeat): Use `x₁` to find `x₂`, `x₂` to find `x₃`, and so on.
  5. Step 4 (Stop): Stop when the difference between `x_{n+1}` and `xₙ` is very small.
Comparison:
  • Bisection: Slow, reliable, guaranteed convergence (linear).
  • Newton-Raphson: Very fast (quadratic convergence), but can diverge if the initial guess is bad or if `f'(x)` is close to zero near the root.

Interpolation

Interpolation is the process of estimating the value of a function `y` at a point `x` that lies between known data points `(x₀, y₀), (x₁, y₁), ...`.

These formulas are for equally spaced data, where `h = x₁ - x₀ = x₂ - x₁`.

Newton-Gregory Forward Difference Formula

This formula is used to interpolate values of `y` near the beginning of the dataset (near `x₀`).

It uses a forward difference table.
First difference: Δy₀ = y₁ - y₀
Second difference: Δ²y₀ = Δy₁ - Δy₀
...and so on.

The formula is:

y(x) = y₀ + p Δy₀ + [p(p-1)/2!] Δ²y₀ + [p(p-1)(p-2)/3!] Δ³y₀ + ...

where p = (x - x₀) / h. (`x` is the point you are interpolating at).

Newton-Gregory Backward Difference Formula

This formula is used to interpolate values of `y` near the end of the dataset (near `xₙ`).

It uses a backward difference table.
First difference: ∇yₙ = yₙ - yn-1
Second difference: ∇²yₙ = ∇yₙ - ∇yn-1
...and so on.

The formula is:

y(x) = yₙ + p ∇yₙ + [p(p+1)/2!] ∇²yₙ + [p(p+1)(p+2)/3!] ∇³yₙ + ...

where p = (x - xₙ) / h. (Note the different definition of `p`!)

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