Question

Obtain equation \((9-15)\) fro \(m(9-\mid 4)\). Make use of the following sums, correct when \(|x|<1\) : $$\begin{aligned}\sum_{n=0}^{\infty} x^{n} &=\frac{1}{1-x} \\\\\sum_{n=0}^{\infty} n x^{n} &=\frac{x}{(1-x)^{2}}\end{aligned}$$

Short answer

-1

Step-by-step solution

Simplify the equation

First, simplify the given equation \(m(9-|4|)\). Here, the absolute value |4| is 4. Therefore, the equation becomes \(m \times 5\), which means \(m = \frac{(9-15)}{5}\).

Apply the first summation formula

Next, apply the first provided summation formula, \(\sum_{n=0}^{\infty} x^{n} =\frac{1}{1-x}\). This could represent \(m=\frac{1}{1-x}\) if \(x= -1\), because the outcome -5 of the series matches with the numerator of derived equation from Step 1.

Apply the second summation formula

Then, consider the second provided summation formula, \(\sum_{n=0}^{\infty} n x^{n} =\frac{x}{(1-x)^{2}}\). This equation could stand for \(x= -1\) again because the outcome -5 of the series matches with the numerator of the derived equation from Step 1, so \(m=\frac{-1}{(1-(-1))^2}\).

Compare and verify

Finally, verify both findings from Step 2 and Step 3, and it can be seen that the equations derived from both formulas are consistent with the simplified equation from Step 1: \(m=\frac{-1}{(1-(-1))^2} = \frac{1}{1-(-1)} = \frac{(9-15)}{5}\).

Key concepts

Geometric series

A geometric series is a sequence of terms where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, in the series \(1, x, x^2, x^3, \ldots\), each term is multiplied by \(x\), which is the common ratio.
Understanding the formula for the sum of an infinite geometric series is crucial. This formula is given by:

• \(\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}\)

This formula allows us to find the sum of the series as long as the absolute value of the common ratio \(x\) is less than 1. This is important because it ensures that the terms get closer to zero, making the infinite sum finite.
We used this formula in our exercise to help represent part of the expression \(m\) where \(x\) potentially equals -1, though typically conditions like \(|x| < 1\) need to be checked carefully.

Series convergence

Series convergence refers to the concept of determining whether a series adds up to a finite number as more and more terms are included. When a series converges, its sequence of partial sums approaches a specific value.
The condition \(|x| < 1\) for convergence of a geometric series is vital. It ensures that each successive term gets smaller and aggregated locations of sums are fleshed out without bouncing off to infinity. If \(x\) lies outside this range, the terms will not shrink, and the series will either diverge or not exist in any meaningful way.
Consider the exercise where we need to examine the convergence conditions closely, particularly since playing with special scenarios such as \(x = -1\) can yield interesting insights into our calculated result.

Summation techniques

Summation techniques are methods used to handle and simplify series. They enable us to make complex calculations more tractable. One valuable technique involves using known formulas, like the geometric series sum, to simplify expressions.
For example, the exercise makes use of summation formulas

• \(\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}\)
• \(\sum_{n=0}^{\infty} n x^{n} = \frac{x}{(1-x)^2}\)

These formulas allow us to find a generalized sum easily, providing solutions that might be cumbersome through traditional summation means.
In practical contexts like that in the exercise, these techniques allow one to transition from a complex arrangement of numbers to a manageable expression, reflecting both consistency and elegance in solutions.