Question

Let \(\mathrm{p}, \mathrm{q}\) be real numbers such that \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=1\) and \(0

Short answer

Option 4: \( p - \frac{1}{2} p^2 + \frac{1}{3} p^3 - \frac{1}{4} p^4 + \ldots \infty = q + \frac{q^2}{2!} + \frac{q^3}{3!} + \frac{q^4}{4!} + \ldots \infty \)

Step-by-step solution

Determine the relationship between p and q

Given the equation \( \frac{1}{p} - \frac{1}{q} = 1 \), solve for \( q \).

Manipulate the equation to solve for q

Rearrange the equation to find a common denominator and isolate \( q \): \[ \frac{1}{p} - \frac{1}{q} = 1 \implies \frac{q - p}{pq} = 1 \implies q - p = pq \implies q(p - 1) = p \implies q = \frac{p}{p-1} \]

Simplify the expression for q

Since \( 0 < p \leq \frac{1}{2} \), we conclude \( q = \frac{p}{p-1} = \frac{p}{-1(1-p)} = \frac{-p}{1-p} \).

Verify expressions given in options for series equivalence

Substitute \( q \) into each given series option and verify equality.

Evaluate series for option 1 and 2

Both option 1 and 2 are alternating series, so test the equality by substituting values. These won't match due to sign changes in terms.

Evaluate series for options 3 and 4

For option 3, substituting \( q = -\frac{p}{p-1} \) results in unequal series due to factorial weights in both. Option 3 series matches the definition of \( e^x \) expansions. Option 4 holds true because both sides represent the entire series.

Key concepts

Series Expansion

Understanding series expansion is crucial for solving this problem. Series expansion involves expressing a function as a sum of terms derived from the function’s derivatives at a single point. Common types of series expansions include Taylor series and Maclaurin series. For example, the expansion of \(\frac{1}{1-x}\) is given by \(1 + x + x^2 + x^3 + ...\). These expansions help in breaking down complex functions into simpler terms, making calculations and comparisons easier. In the exercise, various series expansions are provided, and understanding how they are derived and related helps in finding the equality between different expressions.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate variables or simplify expressions. It’s an essential skill for solving equations and verifying solutions. In the solution provided, algebraic manipulation is used to express \(q\) in terms of \(p\) by rearranging \(\frac{1}{p} - \frac{1}{q} = 1\) to \(q = \frac{p}{p-1}\). Steps involved include finding a common denominator and solving for the variable. Mastering these manipulations allows you to handle complex equations more efficiently. This process is particularly important as it enables the identification of equivalent expressions between different series.

Equivalent Expressions

Equivalent expressions are different mathematical expressions that represent the same value. Recognizing equivalent expressions is vital in simplifying problems and verifying solutions. In this exercise, we compare series to determine whether they are equivalent by substituting the value of \(q = \frac{p}{p-1}\) into the provided series. If the transformations and terms match, the expressions are equivalent. Practicing this concept improves the ability to spot and prove equivalence between different mathematical forms, providing deeper insight into solving complex problems.

Alternating Series

An alternating series is one whose terms alternate in sign. Examples include \(p - \frac{1}{2} p^2 + \frac{1}{3} p^3 - \frac{1}{4} p^4 + ...\). For alternating series, it is important to check convergence and the behavior of the terms. In this exercise, options 1 and 2 involve alternating series. When dealing with alternating series, note how the signs and magnitudes of terms affect the overall sum and verify the equivalence. By substituting the derived \(q\) value into the series and observing how the series terms change, we can determine if two given series are indeed equal, leading to valid solutions.