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Various question on power series 

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If $a_n \\\\neq 0$  for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n}$ $\\\\to$ $\\\\ell$ with $|\\\\ell | <1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ converges.

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If $a_n \\\\geq \\\\dfrac{1}{n^2}$ for all $n \\\\in \\\\mathbb{N}$, then $\\\\Sigma a_n$ converges.

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If a power series $\\\\Sigma a_n x^n$ has radius of convergence $R$ with $0 < R <\\\\infty$, then the power series converges for all $x$ with $|x|<R$. 

\"", "name": "tr13", "description": ""}, "tr3": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\not\\\\to 0$ as $n \\\\to \\\\infty$, then the series $\\\\Sigma a_n$ diverges.

\"", "name": "tr3", "description": ""}, "f4": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If the series $\\\\Sigma a_n$ diverges, then $a_n \\\\not\\\\to 0$ as $n \\\\to \\\\infty$.

\"", "name": "f4", "description": ""}, "f3": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If  $a_n \\\\to 0$ as $n \\\\to \\\\infty$, then the series $\\\\Sigma a_n$ converges.

\"", "name": "f3", "description": ""}, "tr2": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\geq \\\\dfrac{1}{n}$ for all $n \\\\in \\\\mathbb{N}$, then $\\\\Sigma a_n$ diverges.

\"", "name": "tr2", "description": ""}, "f15": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If a power series $\\\\Sigma a_n x^n$ has radius of convergence $R$ with $0<R<\\\\infty$, then $a_n \\\\neq 0$ and $\\\\dfrac{a_{n+1}}{a_n}$ tends to a limit as $n \\\\to \\\\infty$.

\"", "name": "f15", "description": ""}, "f9": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $\\\\Sigma a_n$ is convergent then it is absolutely convergent.

\"", "name": "f9", "description": ""}, "tr11": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n = (-1)^{n-1} u_n$ with $u_n >0$ and if $\\{u_n\\}$ does not converge to $0$, then $\\\\Sigma a_n$ diverges.

\"", "name": "tr11", "description": ""}, "f10": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $\\\\Sigma a_n$ is not divergent then it is absolutely convergent.

\"", "name": "f10", "description": ""}, "tr6": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n>0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n} \\\\to \\\\ell >1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ diverges.

\"", "name": "tr6", "description": ""}, "tr16": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If a power series $\\\\Sigma a_n x^n$ diverges for some value $x=X \\\\neq 0$, then the radius of convergence $R$ satisfies $R \\\\leq |X|$

\"", "name": "tr16", "description": ""}, "ch8": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(x=1,f13,if(x=2,f14,if(f=x,f15,f16)))", "name": "ch8", "description": ""}, "f13": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If a power series $\\\\Sigma a_n x^n$ has radius of convergence $R$ with $0 < R <\\\\infty$, then the power series converges for $x=R$. 

\"", "name": "f13", "description": ""}, "f14": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If a power series $\\\\Sigma a_n x^n$ has radius of convergence $R$ with $0 < R <\\\\infty$, then the power series diverges for $x=R$

\"", "name": "f14", "description": ""}, "tr1": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\geq 0$ for all $n \\\\in \\\\mathbb{N}$ and if $a_n \\\\leq \\\\dfrac{1}{n^2}$ for all $n \\\\in \\\\mathbb{N}$, then $\\\\Sigma a_n$ converges.

\"", "name": "tr1", "description": ""}, "g": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "g", "description": ""}, "u": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "u", "description": ""}, "w": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "w", "description": ""}, "f11": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n = (-1)^{n-1} u_n$ with $u_n >0$ and if $\\{u_n\\}$ is not decreasing, then $\\\\Sigma a_n$ diverges.

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If a power series $\\\\Sigma a_n x^n$ has radius of convergence $R$ with $0 < R <\\\\infty$, then the power series diverges for all $x$ with $|x|>R$. 

\"", "name": "tr14", "description": ""}, "ch1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(t=1,tr1,if(t=2,tr2,if(t=3,tr3,tr4)))", "name": "ch1", "description": ""}, "tr10": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $\\\\Sigma a_n$ is divergent then it is not absolutely convergent.

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If $a_n = (-1)^{n-1} u_n$ with $u_n >0$ and if $\\{u_n\\}$ is increasing, then $\\\\Sigma a_n$ diverges.

\"", "name": "tr12", "description": ""}, "tr20": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $\\\\Sigma a_n$ is divergent then it is not absolutely convergent.

\"", "name": "tr20", "description": ""}, "f7": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n > 0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n}$ $\\\\to$  $\\\\ell >0$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ converges.

\"", "name": "f7", "description": ""}, "tr7": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\neq 0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n}$ $\\\\to$ $\\\\infty$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ diverges.

\"", "name": "tr7", "description": ""}, "tr9": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $\\\\Sigma a_n$ is absolutely convergent then it is convergent.

\"", "name": "tr9", "description": ""}, "h": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "h", "description": ""}, "f2": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\geq 0$ for all $n \\\\in \\\\mathbb{N}$ and if $a_n \\\\leq \\\\dfrac{1}{n}$ for all $n \\\\in \\\\mathbb{N}$, then $\\\\Sigma a_n$ converges.

\"", "name": "f2", "description": ""}, "tr15": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If a power series $\\\\Sigma a_n x^n$ converges for some value $x=X \\\\neq 0$, then the radius of convergence $R$ satisfies $R \\\\geq |X|$. 

\"", "name": "tr15", "description": ""}, "ch6": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(g=1,f5,if(g=6,f2,if(g=3,f7,f8)))", "name": "ch6", "description": ""}, "tr5": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n > 0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n} \\\\to \\\\ell <1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ converges.

\"", "name": "tr5", "description": ""}, "f5": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n > 0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n} \\\\to \\\\ell =1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ converges.

\"", "name": "f5", "description": ""}, "f": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "name": "f", "description": ""}, "ch7": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(h=1,f9,if(h=2,f10,if(h=3,f11,f12)))", "name": "ch7", "description": ""}, "f8": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n \\\\neq 0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n}$ $\\\\to$ $\\\\ell<1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ converges.

\"", "name": "f8", "description": ""}, "f12": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n = (-1)^{n-1} u_n$ with $u_n \\\\geq 0$ and if $\\{u_n\\}$ is increasing, then $\\\\Sigma a_n$ diverges.

\"", "name": "f12", "description": ""}, "f6": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If $a_n>0$ for all $n \\\\in \\\\mathbb{N}$ and if $\\\\dfrac{a_{n+1}}{a_n}$ $\\\\to$ $\\\\ell=1$ as $n \\\\to \\\\infty$, then $\\\\Sigma a_n$ diverges.

\"", "name": "f6", "description": ""}, "ch3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(v=1,tr9,if(v=2,tr10,if(v=3,tr11,tr12)))", "name": "ch3", "description": ""}, "ch4": {"templateType": "anything", "group": "Ungrouped variables", "definition": "if(w=1,tr13,if(w=2,tr14,if(w=3,tr15,tr16)))", "name": "ch4", "description": ""}, "tr4": {"templateType": "long string", "group": "Ungrouped variables", "definition": "\"

If the series $\\\\Sigma a_n$ converges, then $a_n \\\\to 0$ as $n \\\\to \\\\infty$.

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If a power series $\\\\Sigma a_n x^n$ with $a_n \\\\neq 0$ for all $n$ has radius of convergence $R$ with $R<\\\\infty$, then $\\\\dfrac{a_{n+1}}{a_n}$ tends to a limit as $n \\\\to \\\\infty$.

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{Ch1}

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{Ch2}

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{Ch3}

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{Ch4}

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{Ch5}

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{Ch6}

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{Ch7}

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{Ch8}

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[[0]]

\n \n \n \n", "showCorrectAnswer": true, "marks": 0}], "statement": "

Answer the following question on series. Note that every correct answer is worth 1 mark, but every wrong answer loses a mark.

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17/04/2015:

\n

(OK) new question based on a similar style question on sequences. Changed the statements to long text to enable better mathematical expressions. Encountered problems when editing (math expressions not recognised).

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Multiple response question (4 correct out of 8) covering properties of convergent and divergent series and including questions on power series. Selection of questions from a pool.

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You should be able to work out the correct answers from your notes.

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questions on radius of convergence

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Find the radius of convergence of the following power series:

", "advice": "

Recall that for a given power series
\\[\\displaystyle\\sum_{n=1}^\\infty a_n,\\]

\n

one can use
\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right|\\] 

\n

to find its radius of convergence.

\n

\n

a) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{(-1)^{n+1}}{n}}{\\frac{(-1)^{n+2}}{n+1}}\\right|
= \\lim_{n\\to\\infty}\\frac{\\frac{1}{n}}{\\frac{1}{n+1}} = 1.\\]

\n

\n

So the power series is convergent when $|x|<1$. Next we need to analyse the behaviour at the end points to determine the convergence. 

\n\n

Therefore the interval of convergence is $[-1,1]$ (or $-1\\leq x \\leq 1$). 

\n

\n

b) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{1}{(2n)!}}{\\frac{1}{(2n+2)!}}\\right|
= \\lim_{n\\to\\infty} (2n+2)(2n+1) = \\infty.\\]

\n

\n

So the power series is convergent for all $x\\in\\mathbb R$. 

\n

\n

c) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{(-1)^{n+1}}{(-1)^{n+2}}\\right|
=  1.\\]

\n

\n

So the power series is convergent when $|x|<1$. Next we need to analyse the behaviour at the end points to determine the convergence. 

\n\n

Therefore the interval of convergence is $(-1,1)$ (or $-1< x < 1$).

\n

\n

\n

d) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{n}{5^{n-1}}}{\\frac{(n+1)}{5^{n}}}\\right|
= 5.\\]

\n

\n

So the power series is convergent when $|x+2|<5$; equivalently when $-7 < x < 3$. Next we need to analyse the behaviour at the end points to determine the convergence. 

\n\n

Therefore the interval of convergence is $(-7,3)$ (or $-7< x < 3$).

\n

\n

e) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{2^n}{3^n}}{\\frac{2^{n+1}}{3^{n+1}}}\\right|
= \\frac{3}{2}.\\]

\n

\n

So the power series is convergent when $|x|<\\frac{3}{2}$. Next we need to analyse the behaviour at the end points to determine the convergence. 

\n\n

Therefore the interval of convergence is $(-3/2,3/2)$ (or $-3/2< x < 3/2$). 

\n

\n

f) The radius of convergence is 

\n

\\[\\displaystyle\\lim_{n\\to\\infty}\\left| \\frac{a_n}{a_{n+1}}\\right| = \\lim_{n\\to\\infty}\\left| \\frac{\\frac{(-1)^n}{n^2+1}}{\\frac{(-1)^{n+1}}{(n+1)^2+1}}\\right|
= \\lim_{n\\to\\infty}\\frac{(n+1)^2+1}{n^2+1} = 1.\\]

\n

\n

So the power series is convergent when $|x + 2| < 1$, i.e. $-3< x< -1$. Next we need to analyse the behaviour at the end points to determine the convergence. 

\n\n

Therefore the interval of convergence is $[-3,-1]$ (or $-3\\leq x \\leq -1$).

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The power series $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}x^n}{n}$ has radius of convergence [[0]]

\n

The interval of convergence is [[1]]

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The power series $\\displaystyle\\sum_{n=1}^\\infty (-1)^{n+1}x^{2n}$ has radius of convergence [[0]]

\n

The interval of convergence is [[1]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "1", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "-1The power series $\\displaystyle\\sum_{n=1}^\\infty \\frac{n(x+2)^n}{5^{n-1}}$ has radius of convergence [[0]] 

\n

The interval of convergence is [[1]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "5", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "-7The power series $\\displaystyle\\sum_{n=1}^\\infty \\frac{(2x)^n}{3^{n}}$ has radius of convergence [[0]].

\n

The interval of convergence is [[1]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "3/2", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "-1The power series $\\displaystyle\\sum_{n=1}^\\infty \\frac{(-1)^n (x+2)^n}{n^2 + 1}$ has radius of convergence [[0]].

\n

The interval of convergence is [[1]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "1", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "-3 <= x <= -1", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "power series sine", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}], "tags": [], "metadata": {"description": "

power series expansions for sine

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Find the power series expansion of $\\sin x$ by cnsidering the following.

", "advice": "

a) We know that $\\sin x$ has infitely many zeros. On the other hand a polynomial of degree $n$ can have at most $n$ roots. Hence $\\sin x$ cannot be a polynomial. (If we try to express $sin x $ as a polynomial, we see that the degree of the polynomial must be infinite. In other words, we need a power series!)

\n

\n

b) Suppose we have a polynomial 

\n

\\[y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \\ldots + a_nx^n\\] 

\n

which is a solution to the given differential equation. 

\n

Then, since $y^{\\prime\\prime} = - y$ we have 

\n

\\[2a_2 + 6a_3x +12a_4x^2+\\ldots + n(n-1)a_nx^{n-2} =  -a_0 - a_1x - a_2x^2 - a_3x^3 - \\ldots - a_nx^n .\\]

\n

Then

\n
    \n
  • $2a_2= -a_0=0$ (since y(0)=0), so $a_2=0$
  • \n
  • $6a_3 = -a_1 = -1$ (since y^prime(0)=1), so $a_3 = \\frac{-1}{3!}$.
  • \n
  • $12a_4 = -a_2 = 0$, so $a_4=0$.
  • \n
  • $20a_5 = -a_3 = \\frac{1}{5!}$. 
  • \n
  • In general $a_n=0$ if $n$ is odd, and if $n$ is odd, $a_n=\\frac{-1}{n!}$ if $n$ or $a_n=\\frac{1}{n!}$ alternatingly 
  • \n
\n

But observe that the degrees of $y^{\\prime\\prime}$  and $-y$ does not match. So, infact $y$ should be a power series:

\n

\\[y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \\ldots + a_nx^n+\\ldots\\]

\n

with the coefficents computed as above. Discarding the zero terms we can rewrite it as 

\n

\\[y(x) = -1+ -\\frac{1}{3!} x^3+ \\frac{1}{5!} x^5 -\\frac{1}{7!}x^7+\\frac{1}{9!}x^9 -\\frac{1}{11!}x^{11} + \\ldots.\\]

\n

\n

c) We know that $sin x$ is the only solution to the differential equation in part b) (why it is the only one?). So we must have 

\n

\\[\\sin x =-1+ -\\frac{1}{3!} x^3+ \\frac{1}{5!} x^5 -\\frac{1}{7!}x^7+\\frac{1}{9!}x^9 -\\frac{1}{11!}x^{11} + + \\ldots.\\]

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Explain why $\\sin x$ is not a polynomial (Hint: Think about how many zeros $sin x$ has). 

"}, {"type": "information", "useCustomName": true, "customName": "b)", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Give an \"infinite polynomial\" solution to 

\n

\\[\\frac{d^2 y}{dx^2} = -y; \\mbox{ and } y(0)=0, y^\\prime(0) = 1.\\]

\n
"}, {"type": "information", "useCustomName": true, "customName": "c)", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Hence express $\\sin x$ as an \"infinite polynomial\".

"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Ugur's copy of Taylor series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}], "tags": ["3 term Taylor series", "approximation", "approximations", "Calculus", "calculus", "checked2015", "function", "functions", "series approximation", "series expansion", "taylor series", "Taylor series", "three term Taylor series"], "metadata": {"description": "

Find the first 3 terms in the Taylor series at $x=c$ for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

find the first 3 terms in the Taylor series around $x=\\var{c}$ for $f(x)=(\\simplify[all]{{a-b*c}+{b}*x})^{1/\\var{n}}$ (i.e. up to terms in $x^2$).

", "advice": "

The first three terms in the Taylor series are given by $\\simplify[all]{a+b(x-{c})+c(x-{c})^2}$ where $\\displaystyle a=f(\\var{c}),\\;\\;b=f'(\\var{c}),\\;\\;c=\\frac{f''(\\var{c})}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a-b*c}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a-b*c}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(\\var{c})=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(\\var{c})=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(\\var{c})}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the Taylor series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2} \\]

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Input the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.

\n


Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.

\n ", "answer": "{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2", "answerSimplification": "all,fractionNumbers,!collectNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 1e-06, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "notallowed": {"strings": ["!", "."], "showStrings": false, "partialCredit": 0, "message": "

Do not input factorials or decimals in the Taylor series.

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Find the first 3 terms in the MacLaurin series for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Find the first 3 terms in the MacLaurin series for $f(x)=(\\simplify[all]{{a}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$.

", "advice": "

The first three terms in the MacLaurin series are given by $a+bx+cx^2$ where $\\displaystyle a=f(0),\\;\\;b=f'(0),\\;\\;c=\\frac{f''(0)}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(0)=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(0)=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(0)}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the MacLaurin series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2} \\]

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First 3 terms = ?
Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.

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Taylor Approximation ofr $\\cos(2x)$

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Fin the fourth order Taylor Approximation of $\\cos(2x)$

", "advice": "

a)

\n

We need the derivatives of $\\cos(2x)$ up to fourth order. They are 

\n

\\begin{align}
\\cos^{(0)}(2x) &= \\cos(2x) \\quad   &\\cos^{(1)}(2x) &= -2\\sin(2x)\\\\
\\cos^{(2)}(2x) &= -4\\cos(2x) \\quad  &\\cos^{(3)}(2x) &= 8\\sin(2x)\\\\
\\cos^{(4)}(2x) &= 16\\cos(2x) \\quad  
\\end{align}

\n

Then the fourth order Taylor Approximation to the given function is 

\n

\\[\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right) -
2\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right) -
\\frac{4\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{2}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^2
+\\frac{8\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3!}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^3 +
\\frac{16\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{4!}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^4 =
\\\\[3mm]
\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right) -
2\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right) -
2\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^2 +
\\frac{4\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^3 +
\\frac{2\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^4
\\]

\n

(Note that in the answer above, the algorithm approximates sine and cosine values that appear in the polynomial we gave here)

\n

\n

b)

\n

We need the fifth derivative for the error term. That is $\\cos^{(5)}(2x) = -32\\sin(2x)$. Then by Taylor's Theorem the error can computed by 

\n

\\[R_5(\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}) = \\frac{(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi - {r}*pi})}{5!}\\cos^{(5)}(2\\eta) =
\\frac{(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi - {r}*pi})}{5!}(-32(\\sin(2\\eta)) \\quad \\\\[1mm] \\mbox{for some } \\eta\\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}\\right] \\quad (\\mbox{or }\\;
\\eta\\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}\\right])\\]

\n

\n

Note that findin what $\\eta$ should is not possible at this stage. However, one can still bound the error. For that you need to find the $\\eta \\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}\\right]$ such that $\\sin(2\\eta)$ is maximal.

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Find the $4$-th order Taylor Approximation to $\\cos x$ around $x=\\simplify[zeroFactor, fractionNumbers]{{r}*pi}$.

\n

[[0]]

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Find an upper bound for the error between the approximation you found above and $\\cos(2x)$ at $x=\\simplify[zeroFactor, unitFactor, fractionNumbers]{{r2}*pi}$.

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Do not enter decimals in your answer.

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Do not enter decimals in your answer.

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In your answer use the symbols a0 and a1 for $a_0$ and $a_1$ respectively.  In addition, do not enter decimals.

\n

$a_2=$ [[0]].

\n

$a_3=$ [[1]].

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{addSumFunction()}

\n

Seek a power series solution, about $x=0$, in the form

\n

\\[y(x)=\\sum_{n=0}^{\\infty}{a_nx^n},\\]

\n

of the differential equation

\n

\\[\\simplify{y''+{a1}*x*y'+{b1}*y}=0.\\]

\n

Take $a_0$ and $a_1$ to be arbitrary constants, and enter the coefficients $a_2$ and $a_3$ as functions of $a_0$ and $a_1$.

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Uses a custom function to allow simplification of a LaTeX sum, in the same manner as e.g. int() or defint().

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Power series solution of $y''+axy'+by=0$ about $x=0$.

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We have

\n

\\[y(x)=\\sum_{n=0}^{\\infty}{a_nx^n},\\]

\n

so

\n

\\[y'(x)=\\sum_{n=1}^{\\infty}{a_nnx^{n-1}},\\]

\n

and

\n

\\[y''(x)=\\sum_{n=2}^{\\infty}{a_nn(n-1)x^{n-2}}.\\]

\n

Substitute these expressions into the original differential equation to obtain

\n

\\[\\simplify{sum(a_n*n*(n-1)x^(n-2),n=2,infty)+{a1}*sum(a_n*n*x^n,n=1,infty)+{b1}*sum(a_n*x^n,n=0,infty)}=0.\\]

\n

Now reset the index $m=n-2$ in the first summation, and $m=n$ in the second and third summations to obtain

\n

\\[\\simplify{sum({amp2}*(m+2)*(m+1)x^m,m=0,infty)+{a1}*sum(a_m*m*x^m,m=1,infty)+{b1}*sum(a_m*x^m,m=0,infty)}=0.\\]

\n

This equation must be valid for all values of $x$, so the coefficients of like powers of $x$ must vanish.  Take $m=0$ to obtain the coefficients of $x^0$, then

\n

\\[\\simplify{2*a2+{b1}*a0}=0,\\]

\n

and so

\n

\\[a_2=\\simplify{{-b1}*a0/2}.\\]

\n

Now take $m=1$ to obtain the coefficients of $x^1$, so

\n

\\[\\simplify{6*a3+{a1}*a1+{b1}*a1}=0,\\]

\n

then

\n

\\[a_3=\\simplify{-{a1+b1}*a1/6}.\\]

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