Find the least integer $N_1$ such that
\n\\[\\left|\\simplify[std]{x_n -({a} / {c})}\\right| \\le 10 ^ { -\\var{r1}}, \\textrm{ for } n \\geq N_1\\]
\nLeast $N_1=$ [[0]]
", "marks": 0}, {"scripts": {}, "gaps": [{"correctAnswerFraction": false, "showPrecisionHint": false, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{N_2}", "minValue": "{N_2}", "showCorrectAnswer": true, "marks": 1}], "type": "gapfill", "showCorrectAnswer": true, "prompt": "Find the least integer $N_2$ such that
\n\\[\\left|\\simplify[std]{x_n -({a} / {c})}\\right| \\le 10 ^ {- \\var{r2}}, \\textrm{ for } n \\geq N_2\\]
\nLeast $N_2 = $ [[0]]
", "marks": 0}], "statement": "Let \\[x_n=\\simplify[std]{({a}n+{b})/({c}n+{d})}, \\quad n=1,\\; 2,\\; 3 \\ldots \\]
", "tags": ["checked2015", "convergence of a sequence", "limit", "limit of a sequence", "limits", "MAS1601", "query", "sequences", "taking the limit", "tested1", "udf"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "4/07/2012:
\nChecked calculations.
\nSmall changes to Advice display.
\nLeft inequalities as $\\lt$.
\n21/07/2012:
\nAdded description.
\nAdded function chcop(a, b) to create coprime pairs - better display of solution.
\nThis finds an integer coprime to a in the range 1..20. b is set to a random value in the range.
\nChanged definition of variables a, b, c, d.
\n27/7/2012:
\nAdded tags.
\nQuestion appears to be working correctly.
\n24/12/2012:
\nChanged $\\lt$ to $\\le$ and $\\gt$ t $\\ge$ throughout in order to be consistent with question 2 in this assignment. Added query tag to check on this.
\nChecked calculations, OK. Added tested1 tag.
\nAdded more information about the function chcop. Added udf tag.
\n", "licence": "Creative Commons Attribution 4.0 International", "description": "
$x_n=\\frac{an+b}{cn+d}$. Find the least integer $N$ such that $\\left|x_n -\\frac{a}{c}\\right| \\le 10 ^{-r},\\;n\\geq N$, $2\\leq r \\leq 6$.
"}, "variablesTest": {"condition": "gcd(a,c)=1 and gcd(c,d)=1", "maxRuns": 100}, "advice": "To find the least $N_1$ such that all terms from the the $N_1$th are within $10^{\\var{-r1}}$ of the limit we proceed as follows:
\n\\begin{align}
\\left| \\simplify[std]{x_n -({a} / {c})} \\right| \\le 10^{ -\\var{r1}} &\\iff \\left| \\simplify[std]{({a}n+{b})/({c}n+{d})-{a}/{c}}\\right| \\le 10 ^ { -\\var{r1}} \\\\
&\\iff \\simplify[std]{abs({b*c-a*d})/({c^2}n+{c*d})}\\le 10 ^ { -\\var{r1}}
\\end{align}
(We can get rid of the absolute value in the denominator as $\\simplify[std]{{c^2}n+{c*d}} \\gt 0$, $\\forall n=1,\\; 2,\\; 3 \\ldots$)
\nRearranging this last inequality by multiplying both sides by $(\\simplify[std]{{c^2}n+{c*d}}) \\times 10^{\\var{r1}}$ (this is positive and so the inequality does not reverse) we get:
\n\\[ \\simplify[std]{{c^2}n+{c*d}} \\ge \\var{10^r1*abs(b*c-a*d)} \\iff n \\ge \\frac{1}{\\var{c^2}}\\left(\\simplify[std]{{10^r1*abs(b*c-a*d)}-{c*d}}\\right) = \\var{tval} \\]
\nHence the least integer value is given by $N_1=\\var{N_1}$.
\nUsing the same method you should obtain $N_2=\\var{N_2}$.
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}