What is the error bound derived from the scaling $n^2-3 \ge 2n^2/3$ for $n \ge 3$ ?

Q: What is the error bound derived from the scaling $n^2-3 \ge 2n^2/3$ for $n \ge 3$?

Using the inequality $n^2 - 3 \ge \frac{2n^2}{3}$ for $n \ge 3$, the error term $\frac{9}{n^2-3}$ is bounded above by $\frac{9}{2n^2/3} = \frac{27}{2n^2} = \frac{13.5}{n^2}$. This provides a tighter upper bound than the $18/n^2$ estimate, potentially leading to a smaller valid integer cutoff $N$.

Using the inequality

n^2 - 3 \ge \frac{2n^2}{3}

for

n \ge 3

, the error term

\frac{9}{n^2-3}

is bounded above by

\frac{9}{2n^2/3} = \frac{27}{2n^2} = \frac{13.5}{n^2}

. This provides a tighter upper bound than the

18/n^2

estimate, potentially leading to a smaller valid integer cutoff

N

Conditions

$n \ge 3$ .
Error term is $\frac{9}{n^2-3}$ .
Scaling inequality is $n^2 - 3 \ge \frac{2n^2}{3}$ .

Reasoning, step by step

Start with the error expression $\frac{9}{n^2-3}$ .
Apply the lower bound for the denominator: $n^2 - 3 \ge \frac{2n^2}{3}$ .
Invert the inequality for the fraction: $\frac{1}{n^2-3} \le \frac{1}{2n^2/3} = \frac{3}{2n^2}$ .
Multiply by the numerator 9: $\frac{9}{n^2-3} \le 9 \cdot \frac{3}{2n^2} = \frac{27}{2n^2} = \frac{13.5}{n^2}$ .
Conclude that the error is at most $\frac{13.5}{n^2}$ .
Set $\frac{13.5}{n^2} < \varepsilon$ to find the required $n > \sqrt{\frac{13.5}{\varepsilon}}$ .
Verify that the chosen $N$ satisfies $n \ge 3$ .

Example

The script states: 'For $n\ge 3$ , the sharper bound n²− $3\ge 2n$ ²/3 gives error at most $13.5/n$ ². Different bounds can therefore yield different integer cutoffs.'

Common misconceptions

Calculating the coefficient incorrectly as 13.5 instead of $27/2$ or vice versa.
Ignoring the condition $n \ge 3$ which validates the inequality.
Assuming this bound is always tighter than the direct algebraic solution (it is an approximation, not the exact error).

Watch the explanation

BilibiliWhy the cutoff N in a limit proof is not unique

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Connected concepts

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What is the error bound derived from the scaling $n^2-3 \ge 2n^2/3$ for $n \ge 3$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Conditions

Reasoning, step by step

Example

Common misconceptions

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What is the error bound derived from the scaling $n^2-3 \ge 2n^2/3$ for $n \ge 3$ ?