Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a mapping. This exercise explores how we can consider $f$ a dynamical system.
", "advice": "The exercise should be somewhat self-explanatory. Ask me if you are in doubt.
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\\[ f^{\\circ n}(x)= \\begin{cases} x & \\text{when } n = 0 ; \\\\ f(f^{\\circ (n-1)}(x)) & \\text{when } n =1,2,3,\\ldots \\end{cases} \\]
When $f$ is invertible, when can also define $f^{\\circ n}$ for $n = -1, -2, \\ldots$, by letting $f^{\\circ -k}$ denote the $k$'th iterate of the inverse map, when $k \\in \\mathbb{N}$.
\nWe can think of $f$ as defining a flow, by letting $\\phi_n(x)=f^{\\circ n}(x)$. Notice, that in this setting \"time\" n is discrete, $n \\in \\mathbb{N}$ (or $n \\in \\mathbb{Z}$ in the invertible case).
\nGiven $x_0$, we can let $x_1 = f(x_0)$, $x_2 = f(x_1)$, ... Then $x_n = f^{\\circ n}(x_0)$. The points $z_0, z_1, z_2, \\ldots,$ form what is called the forward orbit of $z_0$.
\nA point $x^*$ where $f(x^*) = x^*$ is called a fixed point.
\nHow many fixed points do the mapping $x \\mapsto x^3$ have: [[0]].
\nLet the diagonal in $\\mathbb{R}^2$ be the points $\\{(x,x):x\\in\\mathbb{R}\\}$. Consider the following statement:
\nThe fixed points of f are the values of x where the graph of f intersects the diagonal.
\nIs this statement true? [[1]]
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\nDetermine a condition that is equivalent (meaning sufficient and necessary) to the following:
For any $x_0 \\in \\mathbb{R}$ it holds that $f^{\\circ n}(x_0) \\to 0$ when $n\\to \\infty$.
Suppose $f : \\mathbb{R} \\to \\mathbb{R}$ is differentiable at the point $x^*$, and that $f(x^*)=x^*$.
\nLet $\\lambda = f'(x^*)$, and suppose $\\lvert \\lambda \\rvert < 1$.
\nProve that there exists an interval open $I = (x^*-\\delta, x^*+\\delta)$, so that when $x_0 \\in I$, it holds that $f^{\\circ n}(x_0) \\to x^*$, when $n \\to \\infty$.
\nHint: Given any $\\kappa$ with $\\kappa > \\lvert \\lambda \\rvert$, there exists an open interval around $x^*$ where $\\lvert f(x)-f(x^*)\\rvert \\leq \\kappa \\lvert x-x^* \\rvert$.
\n"}, {"type": "information", "useCustomName": true, "customName": "solution", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "conclusion", "rawLabel": "", "otherPart": 4, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "showPenaltyHint": true, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "We can assume $x^*=0$ with no loss of generality.
\nPick $\\kappa$ with $\\lvert \\lambda \\rvert < \\kappa < 1$.
\nSince $f'(0)=\\lambda$, we know $\\frac{f(x-0)}{x-0}=\\frac{f(x)}{x}\\to \\lambda$ when $x\\to 0$.
\nIt follows that $\\left\\lvert \\frac{f}{x}\\right\\rvert \\to \\lvert \\lambda\\rvert$ when $x\\to 0$. It particular, there exists $\\delta > 0$ such that $\\lvert \\frac{f(x)}{x} \\rvert \\leq \\kappa$ when $0 < \\lvert x \\rvert <\\delta$.
\nWe deduce that $\\lvert f(x) \\rvert \\leq \\kappa \\lvert x\\rvert$, when $x \\in I =(-\\delta, \\delta)$.
\nBy induction, $\\lvert f^{\\circ n}(x) \\rvert \\leq \\kappa^n \\lvert x \\rvert \\to 0$ when $n \\to \\infty$, for any $x \\in I$.
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\nEnd of exercise.
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