Problem 311 - Advanced Calculus

Question

a) Prove that \(\Gamma(\mathrm{p}+1)=\mathrm{p} \Gamma(\mathrm{p}) \mathrm{p}>0\). b) show that \(\Gamma(p+k+1)=(p+k)(p+k-1) \ldots(p+2)(p+1) \Gamma(p+1)\)

Step-by-Step Solution

Answer

To prove part (a), we first recall the definition of the Gamma function and differentiate it with respect to x. We then perform integration by parts, evaluate the limits, and simplify the equation. Finally, we rearrange and evaluate for Γ(p+1) to obtain the desired result: \(\Gamma(p+1) = p \Gamma(p)\). For part (b), we use induction to prove the generalized equation \(\Gamma(p+k+1) = (p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)\). The base case holds since it is the result from part (a). We assume the inductive hypothesis and prove the statement is true for \(k+1\). Using mathematical induction, we conclude that the desired relation holds for all non-negative integers \(k\).

Step 1: Part (a): Prove that Γ(p+1) = p Γ(p) for p > 0

To prove this, we recall the definition of the Gamma function: \[\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt\] Now we'll differentiate the Gamma function with respect to 'x' and then integrate by parts to establish the desired relationship.

Step 1: Differentiation

First, differentiate the Gamma function with respect to 'x': \[\frac{d}{dx} \Gamma(x) = \frac{d}{dx} \left(\int_0^{\infty} t^{x-1} e^{-t} dt \right)\] Using Leibniz's rule, we can obtain: \[\frac{d}{dx} \Gamma(x) = \int_0^{\infty} \frac{d}{dx} (t^{x-1} e^{-t}) dt\] Now differentiate inside the integral: \[\frac{d}{dx} \Gamma(x) = \int_0^{\infty} (\ln{t} \cdot t^{x-1} e^{-t}) dt\]

Step 2: Integration by parts

Now we'll integrate by parts with: \(u = ln(t) t^{x-1}\) and \(dv = e^{-t} dt\) \(du = (t^{x-1} + (x-1) t^{x-2} \ln{t}) dt\) and \(v = -e^{-t}\) Plugging these into the integration by parts formula, we get: \[\Gamma'(x) = [-\ln{t} t^{x-1} e^{-t}]_0^{\infty} - \int_0^{\infty} [-t^{x-1} - (x-1) t^{x-2} \ln{t}] e^{-t} dt\]

Step 3: Evaluate the limits and simplify

We obtain: \[\Gamma'(x) = 0 - \int_0^{\infty} [t^{x-1} + (x-1) t^{x-2} \ln{t}]e^{-t} dt\] Now split the integral: \[\Gamma'(x) = -\int_0^{\infty} t^{x-1} e^{-t} dt - (x-1) \int_0^{\infty} t^{x-2} \ln{t} e^{-t} dt\] Recall that \(\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt\): \[\Gamma'(x) = -\Gamma(x) - (x-1) \int_0^{\infty} t^{x-2} \ln{t} e^{-t} dt\]

Step 4: Rearrange and evaluate for Γ(p+1)

Now, rearrange the equation to obtain: \[\Gamma(x) + \Gamma'(x) = (x-1) \int_0^{\infty} t^{x-2} \ln{t} e^{-t} dt\] Substitute \(x = p+1\), where \(p > 0\): \[\Gamma(p+1) + \Gamma'(p+1) = p \int_0^{\infty} t^{p-1} \ln{t} e^{-t} dt\] Recall from Step 1 that: \[\Gamma'(p+1) = \int_0^{\infty} \ln{t} \cdot t^{p} e^{-t} dt\] Substitute that back into the equation: \[\Gamma(p+1) + \int_0^{\infty} \ln{t} \cdot t^{p} e^{-t} dt = p \int_0^{\infty} t^{p-1} \ln{t} e^{-t} dt\] Since \(\ln{t} \cdot t^{p} e^{-t}\) and \(t^{p-1} \ln{t} e^{-t}\) are both integrable functions, we can write: \[\Gamma(p+1) = p \Gamma(p)\]

Step 6: Part (b): Show that Γ(p+k+1) = (p+k)(p+k-1)...(p+2)(p+1) Γ(p+1)

For this part, we need to prove the generalized equation: \[\Gamma(p+k+1) = (p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)\] We will prove this by induction.

Step 1: Base case

For \(k=1\), the equation is: \[\Gamma(p+2) = (p+1) \Gamma(p+1)\] We have proven this claim in part (a). Thus, the base case holds.

Step 2: Inductive hypothesis

Now assume that the equation holds for some value of \(k\): \[\Gamma(p+k+1) = (p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)\]

Step 3: Inductive step

Show that the equation holds for \(k+1\): \[\Gamma(p+(k+1)+1) = (p+k+1)(p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)\] Using the property proven in part (a), we have: \[\Gamma(p+(k+1)+1) = (p+k+1) \Gamma(p+(k+1))\] Now substitute the inductive hypothesis: \[\Gamma(p+(k+1)+1) = (p+k+1)[(p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)]\] This is precisely the claim we want to prove for \(k+1\). Thus, by the principle of mathematical induction, the desired relation holds for all non-negative integers \(k\).

Key Concepts

Advanced Calculus

Advanced calculus often goes beyond the scope of introductory calculus courses, delving into topics that require a stronger mathematical foundation. A fundamental aspect of advanced calculus is the study of complex functions and the various methods for evaluating integrals which cannot be solved using basic techniques.

The Gamma function \(\Gamma(x)\), which generalizes the factorial function to non-integer values, is a classic example of an advanced mathematical construct. It is defined as an improper integral for \(x > 0\) through \[ \Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt \. \] The properties and behaviors of the Gamma function give rise to many fascinating results, such as the recursion relation in the exercise above, and require a variety of advanced calculus techniques to prove, including Leibniz's rule, differentiation under the integral sign, and integration by parts.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation and is used to solve integrals involving products of functions. The formula is expressed as \( \int u dv = uv - \int v du\), where \(u\) and \(dv\) are differentiable functions of some variable (e.g., \(t\)) and \(dv\) is the differential of another function \(v\).

In solving for the Gamma function properties, integration by parts is utilized to prove the relationship \(\Gamma(p+1) = p \Gamma(p)\). Specifically, the logarithmic function inside the integral \(\ln(t) t^{x-1} e^{-t}\) is managed by cleverly choosing \(u\) and \(dv\), parsing out the elements of the integral in order to simplify and find a relationship that incorporates the Gamma function itself. This methodical choice is crucial as it ultimately allows us to express the derivative of the Gamma function in terms of the function itself, leading to a proof by recursion.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish that a statement holds true for all natural numbers. The process involves two steps: proving the base case and the inductive step. The base case proves that the statement holds for the initial value, usually \( k=0 \) or \( k=1 \).

Once the base case is established, the inductive step is executed, assuming the statement is true for some \( k \) and then showing that it must also be true for \( k+1 \). This step is where the inductive hypothesis is used to prove the subsequent case. In the Gamma function exercise, mathematical induction confirms that the formula \(\Gamma(p+k+1) = (p+k)(p+k-1)\ldots(p+2)(p+1) \Gamma(p+1)\) is valid for each natural number \( k \), a process that starts by showing it holds for the base case and then demonstrating the pattern holds by 'climbing the ladder' through each natural number increment.