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Magazine Chapter 2: Problem 146
Assertion: If \(f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}(a>0)\), then \(\sum_{r=1}^{2 n-1} 2 f\left(\frac{r}{2 n}\right)=2 n-1\) Reason: \(f(x)+f(1-x)=1 \forall x\)
Short Answer
Expert verified
The sum equals \( 2n - 1 \), confirming that the assertion is correct.
Step by step solution
01
Understanding the assertion
The problem provides a function \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \) where \( a > 0 \). The task is to evaluate the sum \( \sum_{r=1}^{2n-1} 2 f\left(\frac{r}{2n}\right) \) and show that it equals \( 2n-1 \).
02
Understanding the reason
The reason given is \( f(x) + f(1-x) = 1 \) for all \( x \). This property of the function can be useful in simplifying the sum for our assertion.
03
Simplifying the sum
Consider the transformation using the property \( f\left(\frac{r}{2n}\right) + f\left(\frac{2n-r}{2n}\right) = 1 \). Tranform the summation into pairs: \( r + (2n-r) = 2n \), which leads to this identity \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \) for \( r = 1, 2, \ldots, n-1 \).
04
Evaluating the sum using pair property
The pairs for \( r \) and \( 2n-r \) exist for each \( r \) from 1 to \( n-1 \). For these \( n-1 \) distinct pairs, the contributions sum up to \( n-1 \) because each pair sums to 2 due to the doubling factor in the sum: \( 2(f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right)) = 2\).
05
Consider the midpoint and edge case
When \( r = n \), \( f\left(\frac{n}{2n}\right) = f\left(\frac{1}{2}\right) \). For the middle term, the contribution is naturally 1 due to symmetry, as \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \), implying \( 2f\left(\frac{1}{2}\right) = 1 \), giving an extra contribution of 1.
06
Final result
Thus, summing the pairs contributes \( n-1 \times 2 = 2(n-1) \) and adding the middle term contributes additional 1, yielding \( 2(n-1) + 1 = 2n-1 \).
07
Final Calculation
Calculate: \(2(n-1) + 1 = 2n - 1\) confirming that the assertion is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Real Functions
In mathematics, a **real function** is a function that associates a real number to each element in its domain, which is also a set of real numbers. One way to represent such a function is through its formula, as we see in the problem where \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \). Here, \( a \) is a positive real number, and \( x \) can take any real value.
Real functions are fundamental because they help us model and solve real-world problems where inputs and outputs are both real numbers.
**Key Properties of Real Functions:**
Real functions are fundamental because they help us model and solve real-world problems where inputs and outputs are both real numbers.
**Key Properties of Real Functions:**
- **Continuity:** A real function can be continuous, meaning that small changes in the input \( x \) result in small changes in the output \( f(x) \).
- **Domain and Range:** For the given function, the domain is \( \mathbb{R} \), the set of all real numbers, and the range, intuitively, is between 0 and 1 since both the numerator and denominator are positive, and the denominator is greater than the numerator.
- **Symmetric Properties:** For some functions, special properties such as \( f(x) + f(1-x) = 1 \) can simplify calculations.
Summation
**Summation** refers to the addition of a sequence of numbers, and it is denoted by the sigma (\( \Sigma \)) symbol. In this exercise, we deal with a specific kind of summation over terms derived from a real function.
The summation given is \( \sum_{r=1}^{2n-1} 2 f\left(\frac{r}{2n}\right) \). This expression requires us to evaluate the function repeatedly at different points determined by \( \frac{r}{2n} \) for \( r = 1, 2, ..., 2n-1 \).
**Strategies for Simplifying Summations:**
The summation given is \( \sum_{r=1}^{2n-1} 2 f\left(\frac{r}{2n}\right) \). This expression requires us to evaluate the function repeatedly at different points determined by \( \frac{r}{2n} \) for \( r = 1, 2, ..., 2n-1 \).
**Strategies for Simplifying Summations:**
- **Symmetrical Pairing:** The problem exploits a unique symmetry in the function values such that \( f\left(\frac{r}{2n}\right) + f\left(1-\frac{r}{2n}\right) = 1 \).
- **Pairwise Addition:** By grouping terms symmetrically, we get consistent results, simplifying the summation significantly. Each pair sums to 2, reducing the need to calculate each term individually.
- **Handling Edge Cases:** Special attention is given to the middle term \( r = n \). Evaluating the middle separately ensures we account for every term correctly.
Symmetry in Mathematics
**Symmetry** in mathematics conveys a form of balance and predictability in structures. In this exercise, symmetry plays a crucial role in simplifying the summation.
The function \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \) possesses a symmetric property: \( f(x) + f(1-x) = 1 \). This is known as **reflective symmetry** about the line \( x = \frac{1}{2} \).
**Benefits of Symmetry in Mathematical Problems:**
The function \( f(x) = \frac{a^x}{a^x + \sqrt{a}} \) possesses a symmetric property: \( f(x) + f(1-x) = 1 \). This is known as **reflective symmetry** about the line \( x = \frac{1}{2} \).
**Benefits of Symmetry in Mathematical Problems:**
- **Reduction of Complexity:** Symmetry allows us to pair terms and simplify calculations, significantly reducing the work needed to compute large sums.
- **Predictable Outcomes:** With symmetry, we can predict that certain properties will hold throughout calculations, giving us confidence in the correctness of results.
- **Optimization:** Symmetry can lead to optimizations in computations, avoiding unnecessary work by grouping similar or equivalent terms.
