Question

This is a more difficult question about 'volumes' in an increasing number of dimensions. (a) Let \(R\) be a real positive number and define \(K_{m}\) by $$ K_{m}=\int_{-R}^{R}\left(R^{2}-x^{2}\right)^{m} d x $$ Show, using integration by parts, that \(K_{m}\) satisfies the recurrence relation $$ (2 m+1) K_{m}=2 m R^{2} K_{m-1} $$(b) For integer \(n\), define \(I_{n}=K_{n}\) and \(J_{n}=K_{n+1 / 2}\). Evaluate \(I_{0}\) and \(J_{0}\) directly and hence prove that $$ I_{n}=\frac{2^{2 n+1}(n !)^{2} R^{2 n+1}}{(2 n+1) !} \quad \text { and } \quad J_{n}=\frac{\pi(2 n+1) ! R^{2 n+2}}{2^{2 n+1} n !(n+1) !} $$ (c) A sequence of functions \(V_{n}(R)\) is defined by $$ \begin{aligned} &V_{0}(R)=1 \\ &V_{n}(R)=\int_{-R}^{R} V_{n-1}\left(\sqrt{R^{2}-x^{2}}\right) d x, \quad n \geq 1 \end{aligned} $$ Prove by induction that $$ V_{2 n}(R)=\frac{\pi^{n} R^{2 n}}{n !}, \quad V_{2 n+1}(R)=\frac{\pi^{n} 2^{2 n+1} n ! R^{2 n+1}}{(2 n+1) !} $$ (d) For interest, (i) show that \(V_{2 n+2}(1)

Short answer

Using integration by parts, demonstrate that \( (2m+1) K_m = 2m R^2 K_{m-1} \). By initializing the base cases, deduce general formulas for \( I_n \text{ and } J_n \). Show, using induction, the expressions for \( V_{2n}(R) \text{ and } V_{2n+1}(R) \). Finally, explicitly calculate to show max volumes in five dimensions.

Step-by-step solution

Set up Integration by Parts for K_m

To use integration by parts for the integral definition of K_m, let’s start by considering a substitution for the integrand in K_m. We set: \[ u = (R^2 - x^2)^m \]\[ dv = dx \]This necessitates finding du and v, respectively.

Compute du and v

Differentiating u and integrating dv gives us: \[ du = \frac{d}{dx}(R^2 - x^2)^m dx = -2mx(R^2 - x^2)^{m-1} dx \]\[ v = x \]

Apply Integration by Parts

Using integration by parts \[ \int u dv = uv - \int v du \]gives us: \[ K_m = \left[x (R^{2}-x^{2})^{m} \right]_{-R}^{R} - \int_{-R}^R x \, (-2mx (R^{2}-x^{2})^{m-1}) dx \]The boundary term evaluates to zero because (R^2 - x^2)^m is zero at x = ±R. So, we have: \[ K_m = 2m \int_{-R}^R x^2(R^2-x^2)^{m-1} dx \]

Relate K_m to K_(m-1)

Noting that \[ x^2 = R^2 - (R^2 - x^2) \]out of our integral, we express it as: \[ K_m = 2m \int_{-R}^R (R^2 - (R^2 - x^2))(R^2-x^2)^{m-1} dx \]\[ K_m = 2m \int_{-R}^R (R^2 (R^2 - x^2)^{m-1} - (R^2 - x^2)^m )dx \]

Split the Integral

To simplify: \[ K_m = 2m R^2 \int_{-R}^R (R^2 - x^2)^{m-1} dx - 2m \int_{-R}^R (R^2 - x^2)^m dx \]Recognizing the definition of K in the integrals: \[ K_m = 2m R^2 K_{m-1} - 2m K_m \]

Solve for the Recurrence

Rearrange to isolate \[ K_m \] : \[ (2m + 1)K_m = 2m R^2 K_{m-1} \]

Find I_0 and J_0

Define I_n and J_n and evaluate them directly using basic integration: \[ I_0 = K_0 = \int_{-R}^R (R^2 - x^2)^0 dx = 2R \]\[ J_0 = K_{1/2} = \int_{-R}^R (R^2 - x^2)^{1/2} dx \]Recognize this as the area of a semicircle of radius R, giving us: \[ J_0 = \pi R^2 \]

General Formulas for I_n and J_n

Using the recurrence relation derived earlier and the expressions for I_0 and J_0, derive: \[ I_n = \frac{2^{2n+1} (n!)^2 R^{2n+1}}{(2n+1)!} \]\[ J_n = \frac{\pi (2n+1)! R^{2n+2}}{2^{2n+1} n! (n+1)!} \]

Induction Basis for V_n(R)

Prove the base case where: \[ V_0(R) = 1 \]\[ \text{and} \quad V_1(R) = 2R = \frac{\pi^0 2^1 0! R}{1!} = \frac{2R}{1} \]

Induction Step for V_{2n}(R)

Assume the inductive hypothesis holds for V_{2n}(R), prove for V_{2n+2}. Similar to previous integrals, use the definition to show that: \[ V_{2n}(R) = \frac{\pi^n R^{2n}}{n!}, \quad V_{2n+1}(R) = \frac{\pi^n 2^{2n+1} n! R^{2n+1}}{(2n+1)!} \]

Show V_{2n+2}(1) < V_{2n}(1) and V_{2n+1}(1) < V_{2n-1}(1)

For all n ≥ 3, demonstrate that higher dimensions decrease. Sequentially show, using inequalities: \[ V_{2n+2}(1) < V_{2n}(1) \text{ and } V_{2n+1}(1) < V_{2n-1}(1) \]

Maximum Volume Dimension

Compute explicit values for k = 1 to 8, display the pattern, and show that the volume reaches a maximum at 5 dimensions.

Key concepts

Recurrence Relations

Recurrence relations are equations that define sequences or functions using previous values. In this problem, we work with the integral definition of \( K_m \), which leads to a recurrence relation.
To derive the recurrence relation, we use integration by parts, breaking the integral into parts that we know how to handle. From the integral:\[ K_m = \int_{-R}^{R} (R^2 - x^2)^m dx \]Using integration by parts, this can be transformed into:\[(2m + 1)K_m = 2m R^2 K_{m-1}\]This recurrence relation tells us that any term in the sequence can be found using the previous term. Recurrence relations are a powerful tool in calculus and other areas of mathematics. They often simplify complex problems by breaking them down into simpler steps that can be easily computed.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It helps to integrate products of functions. The formula for integration by parts is:
\[ \int u dv = uv - \int v du \]In the given exercise, we start by defining \( u = (R^2 - x^2)^m\) and \( dv = dx \). This allows us to find \( du \) and \( v \):\[du = -2mx(R^2 - x^2)^{m-1} dx \quad \text{and} \quad v = x\]Applying the integration by parts formula, we get:\[ K_m = \left[ x (R^2 - x^2)^m \right]_{-R}^{R} - \int_{-R}^R x (-2mx(R^2 - x^2)^{m-1}) dx \]The boundary terms vanish, simplifying the integral. This technique is essential in deriving the recurrence relation because it allows us to transform difficult integrals into more manageable forms. It is widely used for problems involving products of functions where standard integration methods are less effective.

Solid Geometry

Solid geometry deals with the properties and relations of points, lines, surfaces, and solids in three-dimensional space. In this problem, we are considering geometric shapes in higher dimensions and calculating their volumes.
For each dimension, we calculate the integral of the volume of a solid. For example, the integral for \( J_0 \) corresponds to the area of a semicircle, which can be extended to higher dimensions by using similar methods:\[ J_0 = \int_{-R}^{R} (R^2 - x^2)^{1/2} dx \]Solid geometry in higher dimensions requires a careful approach to integrals and recurrence relations, allowing us to compute volumes efficiently. By understanding and manipulating these integrals, we can generalize the shapes and calculate properties such as volume in multiple dimensions.

Volume Calculation

Volume calculation is key when dealing with integrals in higher-dimensional calculus. In the exercise, we focus on calculating the volumes of solids using the functions \( V_n(R)\).
For instance, the base volume function is defined as:\[ V_0(R) = 1 \]For higher dimensions:\[ V_n(R) = \int_{-R}^{R} V_{n-1}(\sqrt{R^2 - x^2}) dx \]By induction, the volumes of the solids are shown to follow specific patterns:\[ V_{2n}(R) = \frac{\pi^n R^{2n}}{n!} \ V_{2n+1}(R) = \frac{\pi^n 2^{2n+1} n! R^{2n+1}}{(2n+1)!} \]These formulas allow us to compute the volume for any given dimension, providing insight into how volume behaves as the number of dimensions increases. Calculating such volumes is crucial for understanding properties of shapes and spaces in higher-dimensional calculus.