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Chapter 1: Problem 38

Find the least value of \(N\) such that when \(n \geqslant N\), (a) \(n^{2}+2 n>100\) (b) \(\frac{n^{2}}{2^{n}}<\frac{1}{1000}\)(c) \(\frac{1}{n}-\frac{(-1)^{\mu}}{n^{2}}<0.000001\) (d) \(\sqrt{(n+1)-\sqrt{n}<\frac{1}{10}}\) (e) \(\frac{n^{2}+2}{n^{2}-1}-1<0.01\)

Short Answer

Expert verified
N is largest when N = 1000001 (Step c).

Step by step solution

01

Solve (a)

The inequality \(n^2 + 2n > 100\) is given. We need to find the smallest integer \(N\) such that for all \(n \geq N\), the inequality holds.First, treat it as a quadratic equation: \(n^2 + 2n - 100 = 0\). Use the quadratic formula to find its roots: \[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].Here, \(a = 1\), \(b = 2\), and \(c = -100\):\[n = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-100)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 400}}{2}\]\[n = \frac{-2 \pm 20}{2}\]Calculate the two roots: \(n = 9\) and \(n = -11\).Since we're looking for the smallest \(n\) where \(n^2 + 2n > 100\), and the curve is a parabola opening upwards, choose \(n = 10\) as the least integer satisfying this inequality for all \(n \geq 10\).
02

Solve (b)

For the inequality \(\frac{n^2}{2^n} < \frac{1}{1000}\), we look for the smallest integer \(N\) such that this is true for all \(n \geq N\).Test successive values of \(n\) starting from a reasonable point (e.g., \(n = 10\)), calculating \(\frac{n^2}{2^n}\) each time:- For \(n = 10\), calculate \(\frac{100}{1024} \approx 0.0977\).- For \(n = 15\), calculate \(\frac{225}{32768} \approx 0.00687\).- Continue incrementing \(n\) until \(\frac{n^2}{2^n} < 0.001\).Verify that \(n = 19\) satisfies this, as for \(n = 19\), \(\frac{361}{524288} \approx 0.000688\). Hence, \(N = 19\).
03

Solve (c)

Consider the inequality \(\frac{1}{n} - \frac{(-1)^n}{n^2} < 0.000001\). Assume \((-)^{\mu} = 1\) for simplicity since \((-1)^n\) does not affect magnitude.This simplifies to \(\frac{1}{n} - \frac{1}{n^2} < 0.000001\):Test values of \(n\) to find the smallest \(N\).Estimate initially with \(\frac{1}{n} < 0.000001\) when \(n > 1000000\).At \(n = 1000001\), \(\frac{1}{1000001} - \frac{1}{(1000001)^2}\) approximates small and satisfies the inequality.Thus, \(N = 1000001\).
04

Solve (d)

The inequality is given by \(\sqrt{(n+1) - \sqrt{n}} < \frac{1}{10}\). To solve:Square both sides to eliminate the square root: \[n+1 - \sqrt{n} < \frac{1}{100}\].Rearrange to: \[n + 1 < \sqrt{n} + \frac{1}{100}\].Estimate with values of \(n\) by testing intentionally:Calculate for \(n = 0\), \(1\), etc., and determine compliance.Find that for \(n = 10000\), \(\sqrt{10001} < \sqrt{10000} + 0.1\), and the inequality holds.So, \(N = 10000\).
05

Solve (e)

Consider the inequality \(\frac{n^2 + 2}{n^2 - 1} - 1 < 0.01\). Simplify:\[\frac{n^2 + 2}{n^2 - 1} = 1 + \frac{3}{n^2 - 1}\].Set up \(\frac{3}{n^2 - 1} < 0.01\):\[3 < 0.01 \times (n^2 - 1)\] or \[3 < 0.01n^2 - 0.01\].Solve \[0.01n^2 > 3.01\].Thus, \[n^2 > 301\], and \(n > 17.37\).Therefore, choose \(n = 18\) as the least integer to satisfy all \(n \geq 18\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Inequalities
Quadratic inequalities, such as those involving expressions like \( n^2 + 2n > 100 \), often appear in mathematics problems.
Solving these inequalities requires finding values of the variable that satisfy the given condition for all subsequent values.

To tackle a quadratic inequality, you can follow these steps:
  • First, treat it like a quadratic equation by setting it to zero and finding its roots. This is done using the quadratic formula, \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are coefficients from the expression.
  • Identify the roots of the equation. These roots help determine intervals over which the inequality holds true.
  • With the roots found, remember that the parabola opens upwards since the leading coefficient \( a \) is positive. This means the inequality holds true beyond the roots when \( n \geq \) largest root.
In this exercise, the smallest integer \( N \) satisfying \( n^2 + 2n > 100 \) is found to be 10, as it is greater than the larger root found.
Exponential Inequalities
Exponential inequalities involve expressions where the variable is part of an exponent, like \( \frac{n^2}{2^n} < \frac{1}{1000} \).
These can be particularly challenging, as they contain exponential growth or decay.

To solve exponential inequalities, you may follow these steps:
  • Start by testing small values of \( n \) that might satisfy the inequality.
  • With each successive integer \( n \), calculate the value on the left side of the inequality.
  • Once the calculated value becomes smaller than the right side of the inequality, that \( n \) becomes your smallest satisfactory integer \( N \).
For this specific exercise, by testing several values, it was established that \( N = 19 \). At this point, the inequality \( \frac{19^2}{2^{19}} < \frac{1}{1000} \) holds true. Exponential functions grow or shrink rapidly, requiring you to test values incrementally to determine \( N \).
Sequence Inequalities
Sequence inequalities are about finding terms in sequences that satisfy a particular condition.
For the given inequality \( \frac{1}{n} - \frac{(-1)^n}{n^2} < 0.000001 \), these often involve series or alternating terms.

Here is a suggested method to approach sequence inequalities:
  • Simplify the expression as much as possible. Consider the absolute magnitude of alternating terms, focusing on the dominant one.
  • Estimate values for \( n \) such that the dominant term is smaller than a given threshold.
  • Evaluate the simplified expression by plugging in values until it meets the condition.
In this exercise, simplifying results in a finding for \( N = 1000001 \). It's an estimate with large \( n \), where \( \frac{1}{n} \) becomes sufficiently small to satisfy the inequality.
Mathematical Proofs
Mathematical proofs are the backbone of validating any solved inequality or mathematical expression.
Proofs provide a logical foundation that ensures the statement or solution is universally true and free from errors.

To effectively use proofs in solving inequalities, consider the following tips:
  • Understand the initial conditions and premises thoroughly.
  • Use logical arguments and derivations from known mathematical principles to demonstrate the truth.
  • Cross-check the results with multiple approaches or verify by alternative means.
In the given examples, the solutions aren't just about computing, but how you logically arrive at them. Checking endpoints, verifying calculations, or using alternate assumptions helps in proof. Proofs help solidify understanding and make results reliable, as in solving the complex set of inequalities from the exercise.

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Most popular questions from this chapter

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