Problem 68 - Calculus Single Variable

Question

Involve the factorial numbers \(n !\), which were introduced in Section \(1.4 .\) They can be defined by \(0 !=1\) and, for a positive integer \(n, n !=n(n-1) \ldots 3 \ldots 2 \cdot 1\). Let \(n\) be a positive integer. If \(k\) is a positive integer no greater than \(n\), then the expression $$\left(\begin{array}{l}n \\\k\end{array}\right)=\frac{n !}{k !(n-k) !}=\frac{n(n-1)(n-2) \cdots(n-k+1)}{k \cdot(k-1) \cdots \cdot 3 \cdot 2 \cdot 1}$$ is called a binomial coefficient. For example, if \(n=7,\) and \(k=3,\) then \(n-k+1=5,\) and $$\left(\begin{array}{l}7 \\\3\end{array}\right)=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}=35$$ By definition \(\left(\begin{array}{l}n \\ 0\end{array}\right)=1 .\) Leibniz's Rule states that \((f \cdot g)^{(n)}(c)\) is the sum of the \(n+1\) expressions that are obtained by substituting \(k=0,1,2, \ldots, n\) in the formula \(\left(\begin{array}{c}n \\ k\end{array}\right) \cdot f^{(n-k)}(c) \cdot g^{(k)}(c) .\) Explicitly write out Leibniz's Rule for \(n=1,2,\) and \(3 .\) Observe that the \(n=1\) case is the Product Rule, the \(n=2\) case is equation (3.7.2). Verify Leibniz's Rule for \(n=2\) .

Step-by-Step Solution

Answer

Leibniz's Rule for \(n=2\): \((f \cdot g)''(c) = f''(c)g(c) + 2f'(c)g'(c) + f(c)g''(c)\). This verifies the rule.

Step 1: Leibniz's Rule for n=1

For \(n=1\), Leibniz's Rule states:\[(f \cdot g)'(c) = \binom{1}{0} f'(c)g(c) + \binom{1}{1} f(c)g'(c)\] Calculating the binomial coefficients, we have:\[ \binom{1}{0} = 1, \quad \binom{1}{1} = 1 \] Thus, Leibniz's Rule for \(n=1\) is:\[(f \cdot g)'(c) = f'(c)g(c) + f(c)g'(c)\]This is the classic Product Rule for differentiation.

Step 2: Leibniz's Rule for n=2

For \(n=2\), Leibniz's Rule can be expressed as:\[(f \cdot g)''(c) = \binom{2}{0} f''(c)g(c) + \binom{2}{1} f'(c)g'(c) + \binom{2}{2} f(c)g''(c)\]Calculating the binomial coefficients, we get:\[ \binom{2}{0} = 1, \quad \binom{2}{1} = 2, \quad \binom{2}{2} = 1 \]Thus, Leibniz's Rule for \(n=2\) becomes:\[(f \cdot g)''(c) = f''(c)g(c) + 2f'(c)g'(c) + f(c)g''(c)\]

Step 3: Leibniz's Rule for n=3

For \(n=3\), Leibniz's Rule is:\[(f \cdot g)^{(3)}(c) = \binom{3}{0} f^{(3)}(c)g(c) + \binom{3}{1} f''(c)g'(c) + \binom{3}{2} f'(c)g''(c) + \binom{3}{3} f(c)g^{(3)}(c)\]Calculating the binomial coefficients, we have:\[ \binom{3}{0} = 1, \quad \binom{3}{1} = 3, \quad \binom{3}{2} = 3, \quad \binom{3}{3} = 1 \]Thus, the rule for \(n=3\) can be written as:\[(f \cdot g)^{(3)}(c) = f^{(3)}(c)g(c) + 3f''(c)g'(c) + 3f'(c)g''(c) + f(c)g^{(3)}(c)\]

Step 4: Verifying Leibniz's Rule for n=2

To verify Leibniz's Rule for \(n=2\), consider the binomial expansions calculated in Step 2.Rewriting, we have:\[(f \cdot g)''(c) = f''(c)g(c) + 2f'(c)g'(c) + f(c)g''(c)\]This expression matches the product of individual derivatives for \(n=2\), confirming that Leibniz's Rule is applicable and correct.

Key Concepts

Factorial Numbers

Factorial numbers are a fundamental concept in mathematics, especially in counting and combinatorics. If you're working with permutations or combinations, you've already encountered factorial numbers. A factorial is denoted by the symbol "!" and refers to the product of all positive integers up to a specific number. For example, the factorial of 5, written as 5!, is calculated as follows:

• 5! = 5 × 4 × 3 × 2 × 1 = 120

It provides the number of ways to arrange n objects into a sequence, making it crucial in probability and statistics. And it's not just limited to positive integers. By definition, the factorial of 0 is 1, which is important for simplifying many mathematical expressions involving factorials. This definition helps keep formulas and functions consistent and avoids division by zero in certain cases.

Binomial Coefficients

Binomial coefficients arise in the binomial theorem and are used to describe the number of ways to choose k items from a set of n items without regard to order. The notation is expressed as \(\binom{n}{k}\), and it can be computed using the factorials as follows:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

These coefficients are the building blocks of Pascal's triangle, where each entry is the sum of the two directly above it. For example, to find \(\binom{7}{3}\), you calculate:

• \(\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)

Binomial coefficients form the basis for expanding expressions raised to a power in the binomial theorem. They have applications in various computational problems and probabilities, where understanding combinations is essential.

Product Rule for Differentiation

The product rule is a fundamental technique used in calculus to differentiate the product of two functions. It is a special case of Leibniz's Rule for differentiation. If you have two functions, \(f(x)\) and \(g(x)\), and you want to differentiate \(f(x) \cdot g(x)\), the product rule tells us how to do it:

• The derivative is \((f \cdot g)' = f'(x) \cdot g(x) + f(x) \cdot g'(x)\).

This formula effectively states that you take the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function.
The product rule becomes especially useful when dealing with functions that are complex and cannot be simplified into simpler terms before taking the derivative. It ensures that each part of the product is accounted for in the derivative, maintaining accuracy in calculus.