Problem 19 - Advanced Calculus

Question

For a natural number \(n,\) the Geometric Sum Formula asserts that $$1+x+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x} \quad \text { if } x \neq 1$$ By differentiating, find a formula for $$1+x+2 x^{2}+\cdots+n x^{n}$$ and then for $$1^{2}+2^{2} x+\cdots+n^{2} x^{n-1}$$.

Step-by-Step Solution

Answer

After differentiating the geometric sum formula once, we find that \[1 + 2x + 3x^2 + \cdots + nx^{n-1} = \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2},\] provided \(x eq 1\). Differentiating this result again yields our second formula: \[1^2 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1} = \frac{2 - (n-1)n x^{n-1} + n(n+1) x^n}{(1-x)^3}.\]

Step 1: Differentiate the Geometric Sum

Using the formula for the geometric sum, differentiate both sides with respect to x: \[\frac{d}{dx}(1 + x + x^2 + \cdots + x^n) = \frac{d}{dx}\left(\frac{1-x^{n+1}}{1-x}\right).\]

Step 2: Apply the Power Rule

Differentiate the left-hand side term by term to obtain \[\frac{d}{dx}x^k = kx^{k-1},\] for each power of x. This gives us \[1 + 2x + 3x^2 + \cdots + nx^{n-1}.\]

Step 3: Differentiate the Right-hand Side

Use the quotient rule, \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2},\] where \[u = 1-x^{n+1}\] and \[v = 1-x.\]

Step 4: Simplify the Result

Apply the quotient rule to differentiate the right-hand side, and then simplify the result. After differentiating and rearranging, you get the formula for \[1 + 2x + 3x^2 + \cdots + nx^{n-1}.\]

Step 5: Differentiate the New Formula

To find a formula for \[1^2+2^2x+\cdots+n^2x^{n-1},\] differentiate \[1 + 2x + 3x^2 + \cdots + nx^{n-1}\] once more with respect to x.

Step 6: Apply the Power Rule Again

Apply the power rule to each term of \[1 + 2x + 3x^2 + \cdots + nx^{n-1}\] to obtain \[2^1 + 2\cdot 3x^{1} + 3\cdot 4x^{2} + \cdots + n(n-1)x^{n-2}.\]

Step 7: Factor Out the Constants

Factor out constants to get the final formula in the form of \[1^2 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1}.\] This involves recognizing the pattern in the coefficients of each term.

Step 8: Write the Final Formulas

Conclude with the formulas for \[1 + 2x + 3x^2 + \cdots + nx^{n-1}\] and \[1^2+2^2x+\cdots+n^2x^{n-1}\] derived from differentiating the original geometric sum formula twice.

Key Concepts

Power Rule Differentiation

Power rule differentiation is a fundamental concept in calculus used to find the derivative of a power function, which is a function where the variable, typically denoted as 'x', is raised to a constant power. Here's the rule: when differentiating an expression of the form \(x^k\), where \(k\) is any real number, the derivative is \(kx^{k-1}\).

For example, if you have \(x^3\), applying the power rule will give you its derivative, \(3x^2\). This plays a crucial role in solving the exercise at hand for finding a formula for \(1 + x + 2x^2 + \textellipsis + nx^n\) by taking the derivative of each term in the series separately.

Geometric Series

A geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The sum of the first \(n\) terms of a geometric series can be expressed as \(1 + x + x^2 + \textellipsis + x^n\), where \(x\) is the common ratio. The formula for the sum of a geometric series is \(\frac{1-x^{n+1}}{1-x}\), assuming that \(xeq1\).

In our exercise, this sum formula is used as the starting point. Understanding how the formula is derived is crucial as it represents the series in a more manageable form which we can then differentiate to find new, related formulas.

Quotient Rule Differentiation

The quotient rule is a technique for differentiating expressions where one function is divided by another. It's stated as \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\), where \(u\) and \(v\) are functions of \(x\).

Applying the quotient rule helps manage complex expressions, such as when differentiating the right-hand side of the geometric sum formula, \(\frac{1-x^{n+1}}{1-x}\). The key is to differentiate the numerator and denominator separately and then combine them according to the rule. Getting comfortable with this rule is vital for solving calculus problems, especially when dealing with ratios of functions.

Mathematical Induction

Mathematical induction is a proof technique used to establish the validity of a statement for all natural numbers. It involves two steps: first, you show that the statement holds for the first natural number (usually 1, called the base case). Then, you assume it holds for an arbitrary natural number \(k\), and prove it for \(k+1\), called the induction step.

Although induction isn't used directly in the provided exercise, knowing this method can be very helpful for verifying the formulas you obtain after differentiating. It ensures that the patterns you've identified and the conclusions you've drawn are indeed correct for all terms in the series, providing a sound mathematical foundation to your solution.