Add the solution file. (#332)

_assessment/index.md

@@ -13,7 +13,7 @@ A mock coursework is available:
 
 The topics for the individual coursework for the past few years have been:
 
-- 2023-2024 [here]({{site.baseurl}}/assets/assessment/2023-2024/assignment.ipynb) ([cy]({{site.baseurl}}/assets/assessment/2023-2024/cy/assignment.ipynb)):
+- 2023-2024 [here]({{site.baseurl}}/assets/assessment/2023-2024/ind/assignment.ipynb) ([cy]({{site.baseurl}}/assets/assessment/2023-2024/ind/cy/assignment.ipynb)) ([solution]({{site.baseurl}}/assets/assessment/2023-2024/ind/assignment.ipynb))
   1. [Differential equations]({{ site.baseurl }}/topics/differential-equations.html)
   2. [Algebra]({{ site.baseurl }}/topics/algebra.html)
   3. [Sequences]({{ site.baseurl }}/topics/sequences.html)

_posts/2024-01-03-general-feedback-for-individual-coursework.md

@@ -8,6 +8,8 @@ tags:
 
 Thanks all for your efforts in doing the individual coursework!
 
+You can find the solution [here]({{site.baseurl}}/assets/assessment/2023-2024/ind/solution.ipynb).
+
 The performance was really good, below are some summary statistics of your
 marks:
 

assets/assessment/2023-2024/ind/solution.ipynb

@@ -0,0 +1 @@
+{"cells": [{"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Computing for Mathematics - 2023/2024 individual coursework\n\n**Important** Do not delete the cells containing: \n\n```\n### BEGIN SOLUTION\n\n\n### END SOLUTION\n```\n\nwrite your solution attempts in those cells.\n\nTo submit this notebook:\n\n- Change the name of the notebook from `main` to: `<student_number>`. For example, if your student number is `c1234567` then change the name of the notebook to `c1234567`.\n- **Write all your solution attempts in the correct locations**;\n- **Do not delete** any code that is already in the cells;\n- Save the notebook (`File>Save As`);\n- Follow the instructions given to submit."}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "#### Question 1 (30 marks)\n\n(__Hint__: This question is similar to the [this section of the python for mathematics text](https://vknight.org/pfm/tools-for-mathematics/09-differential-equations/how/main.html#how-to-obtain-the-general-solution-of-a-differential-equation).)\n\nFor each of the differential equations output the general solution(s).\n\na. \\\\(\\frac{dy}{dx}=x\\\\)\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 1, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-a"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle y{\\left(x \\right)} = C_{1} + \\frac{x^{2}}{2}$", "text/plain": "Eq(y(x), C1 + x**2/2)"}, "execution_count": 1, "metadata": {}, "output_type": "execute_result"}], "source": "import sympy as sym\n\n### BEGIN SOLUTION\ny = sym.Function(\"y\")\nx = sym.Symbol(\"x\")\n\nlhs = sym.diff(y(x), x)\nrhs = x\nequation = sym.Eq(lhs, rhs)\nsym.dsolve(equation, y(x))\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. \\\\(\\frac{dy}{dx}=y\\\\)\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 5, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-b"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle y{\\left(x \\right)} = C_{1} e^{x}$", "text/plain": "Eq(y(x), C1*exp(x))"}, "execution_count": 5, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\nlhs = sym.diff(y(x), x)\nrhs = y(x)\nequation = sym.Eq(lhs, rhs)\nsym.dsolve(equation, y(x))\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. \\\\(\\frac{dy}{dx}=\\cos(y) \\sin(x)\\\\)\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 9, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-c"]}, "outputs": [{"data": {"text/plain": "[Eq(y(x), pi - asin((C1 + exp(2*cos(x)))/(C1 - exp(2*cos(x))))),\n Eq(y(x), asin((C1 + exp(2*cos(x)))/(C1 - exp(2*cos(x)))))]"}, "execution_count": 9, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\nlhs = sym.diff(y(x), x)\nrhs = sym.cos(y(x)) * sym.sin(x)\nequation = sym.Eq(lhs, rhs)\nsym.dsolve(equation, y(x))\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "d. \\\\(\\frac{dy}{dx}=\\frac{e^{y}}{x}\\\\)\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 13, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-d"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle y{\\left(x \\right)} = \\log{\\left(- \\frac{1}{C_{1} + \\log{\\left(x \\right)}} \\right)}$", "text/plain": "Eq(y(x), log(-1/(C1 + log(x))))"}, "execution_count": 13, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\nlhs = sym.diff(y(x), x)\nrhs = sym.exp(y(x)) / x\nequation = sym.Eq(lhs, rhs)\nsym.dsolve(equation, y(x))\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "e. \\\\(\\frac{dy}{dx}=x ^ 3 - \\cos(x) + \\tan(x)\\\\)\n\nAvailable marks: 6"}, {"cell_type": "code", "execution_count": 17, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q1-e"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle y{\\left(x \\right)} = C_{1} + \\frac{x^{4}}{4} - \\log{\\left(\\cos{\\left(x \\right)} \\right)} - \\sin{\\left(x \\right)}$", "text/plain": "Eq(y(x), C1 + x**4/4 - log(cos(x)) - sin(x))"}, "execution_count": 17, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\nlhs = sym.diff(y(x), x)\nrhs = x ** 3 - sym.cos(x) + sym.tan(x)\nequation = sym.Eq(lhs, rhs)\nsym.dsolve(equation, y(x))\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 2 (18 marks)\n\n\n(__Hint__: This question is similar to the [third exercise of the Algebra chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/02-algebra/solutions/main.html#question-3))\n\nConsider the equation $\\alpha ^ 2 - \\beta / \\alpha = 5 + \\alpha$\n\na. Create a variable `general_solution` which has value the set of solutions to this equation for $\\alpha$ (in other words: what values can $\\alpha$ take as function of $\\beta$).\n\navailable marks: 12"}, {"cell_type": "code", "execution_count": 21, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-a"]}, "outputs": [], "source": "alpha = sym.Symbol(\"alpha\")\nbeta = sym.Symbol(\"beta\")\n\n### BEGIN SOLUTION\nlhs = alpha ** 2 - beta / alpha\nrhs = 5 + alpha\nequation = sym.Eq(lhs=lhs, rhs=rhs)\ngeneral_solution = sym.solveset(equation, alpha)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Output the set of solution for the case when $\\beta=0$.\n\navailable marks: 6"}, {"cell_type": "code", "execution_count": 25, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q2-b"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle \\left\\{\\frac{1}{3} - \\frac{16}{3 \\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}} \\left(- \\frac{1}{2} - \\frac{\\sqrt{3} i}{2}\\right)} - \\frac{\\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}} \\left(- \\frac{1}{2} - \\frac{\\sqrt{3} i}{2}\\right)}{3}, \\frac{1}{3} - \\frac{\\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}} \\left(- \\frac{1}{2} + \\frac{\\sqrt{3} i}{2}\\right)}{3} - \\frac{16}{3 \\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}} \\left(- \\frac{1}{2} + \\frac{\\sqrt{3} i}{2}\\right)}, \\frac{1}{3} - \\frac{\\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}}}{3} - \\frac{16}{3 \\sqrt[3]{- \\frac{47}{2} + \\frac{45 \\sqrt{7} i}{2}}}\\right\\} \\setminus \\left\\{0\\right\\}$", "text/plain": "Complement({1/3 - 16/(3*(-47/2 + 45*sqrt(7)*I/2)**(1/3)*(-1/2 - sqrt(3)*I/2)) - (-47/2 + 45*sqrt(7)*I/2)**(1/3)*(-1/2 - sqrt(3)*I/2)/3, 1/3 - (-47/2 + 45*sqrt(7)*I/2)**(1/3)*(-1/2 + sqrt(3)*I/2)/3 - 16/(3*(-47/2 + 45*sqrt(7)*I/2)**(1/3)*(-1/2 + sqrt(3)*I/2)), 1/3 - (-47/2 + 45*sqrt(7)*I/2)**(1/3)/3 - 16/(3*(-47/2 + 45*sqrt(7)*I/2)**(1/3))}, {0})"}, "execution_count": 25, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\ngeneral_solution.subs({beta: 0})\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 3 (24 marks)\n\n(__Hint__: This question is similar to the [fourth exercise of the Sequences chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/07-sequences/solutions/main.html#question-4).)\n\na. Create a function `generate_sequence` which returns the $n$ term of the following sequence:\n\n\n$$\na_n =\n\\begin{cases}\n    n &n \\in \\{1, 2, 3\\}\\\\\n    a_{n-1} - a_{n-3}&\\text{if }n\\text{ odd}\\\\\n    p a_{n-2}&\\text{otherwise}\n\\end{cases}\n$$\n\nFor some parameter $p$.\n\n_Available marks: 7_"}, {"cell_type": "code", "execution_count": 27, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-a"]}, "outputs": [], "source": "p = sym.Symbol(\"p\")\n\ndef generate_sequence(n, p=p):\n### BEGIN SOLUTION\n    \"\"\"\n    Returns the sequence corresponding to question 4.\n    \"\"\"\n    if n in {1, 2, 3}:\n        return n\n    if n % 2 == 1:\n        return generate_sequence(n=n - 1, p=p) - generate_sequence(n=n - 3, p=p)\n    return p * generate_sequence(n=n - 2, p=p)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Output $a_5$ in terms of $p$\n\n_Available marks: 2_"}, {"cell_type": "code", "execution_count": 31, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-b"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle 2 p - 2$", "text/plain": "2*p - 2"}, "execution_count": 31, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\ngenerate_sequence(n=5, p=p)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. Output $a_6$ for $p=\\frac{1}{2}$\n\n_Available marks: 6_"}, {"cell_type": "code", "execution_count": 33, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-c"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle \\frac{1}{2}$", "text/plain": "1/2"}, "execution_count": 33, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\ngenerate_sequence(n=6, p=sym.S(1) / 2)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["description:3c-value-is-approximately-correct"]}, "source": "d. Output the set of values that $p$ can take which ensures that $a_{10}=a_{11}$.\n\n_Available marks: 9_"}, {"cell_type": "code", "execution_count": 36, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q3-d"]}, "outputs": [{"data": {"text/latex": "$\\displaystyle \\left\\{0\\right\\}$", "text/plain": "{0}"}, "execution_count": 36, "metadata": {}, "output_type": "execute_result"}], "source": "### BEGIN SOLUTION\nlhs = generate_sequence(n=10, p=p)\nrhs = generate_sequence(n=11, p=p)\nequation = sym.Eq(lhs, rhs)\nsym.solveset(equation, p)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "### Question 4 (28 marks)\n\n(__Hint__: This question uses aspects from [the Matrices chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/04-matrices/) as well as [the statistics chapter of Python for mathematics](https://vknight.org/pfm/tools-for-mathematics/08-statistics/))\n\na. Create variable `determinants` which has value a list with the determinants of each of the matrices in the tuple `matrices`:\n\n_Available marks: 10_"}, {"cell_type": "code", "execution_count": 39, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-a"]}, "outputs": [], "source": "matrices = (\n    sym.Matrix((1,)),\n    sym.Matrix(((1, 2), (2, 1))),\n    sym.Matrix(((3, 0), (0, 1))),\n    sym.Matrix(((2, 1, 1), (3, 2, 4), (6, 1, -1))),\n    sym.Matrix(((2, 1, 3), (3, 2, 4), (6, 1, -1))),\n    sym.Matrix(((6, 2, 3), (3, -2, -4), (6, 1, -1))),\n    sym.Matrix(((1, 2, 3, 1), (1, 3, -2, -4), (5, 6, 1, -1), (1, 2, -1, -4))),\n    sym.Matrix(((-4, 3), (1, 1))),\n    sym.Matrix(((3, 12), (0, 1))),\n)\n\n### BEGIN SOLUTION\ndeterminants = [matrix.det() for matrix in matrices]\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "b. Create variable `number_of_rows` which has value a list with the number of rows of each of the matrices in the tuple `matrices`:\n\n_Available marks: 12_"}, {"cell_type": "code", "execution_count": 43, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-b"]}, "outputs": [], "source": "### BEGIN SOLUTION\nnumber_of_rows = [matrix.rows for matrix in matrices]\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "raw_mimetype": "", "slideshow": {"slide_type": ""}, "tags": []}, "source": "c. Output the pearson correlation coefficient between the number of rows and the determinants.\n\n_Available marks: 3_"}, {"cell_type": "code", "execution_count": 47, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-c"]}, "outputs": [{"data": {"text/plain": "0.6804510993672779"}, "execution_count": 47, "metadata": {}, "output_type": "execute_result"}], "source": "import statistics as st\n\n### BEGIN SOLUTION\nst.correlation(number_of_rows, determinants)\n### END SOLUTION"}, {"cell_type": "markdown", "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": []}, "source": "d. Create two variables: `slope` and `intercept` which have values the slope ($a$) and the intercept ($b$) for a line of best fit estimating the determinants as a function of the number of rows\n\n_Available marks: 3_"}, {"cell_type": "code", "execution_count": 51, "metadata": {"editable": true, "slideshow": {"slide_type": ""}, "tags": ["answer:q4-d"]}, "outputs": [], "source": "### BEGIN SOLUTION\nslope, intercept = st.linear_regression(number_of_rows, determinants)\n### END SOLUTION"}], "metadata": {"celltoolbar": "Tags", "kernelspec": {"display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3"}, "language_info": {"codemirror_mode": {"name": "ipython", "version": 3}, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.11.4"}}, "nbformat": 4, "nbformat_minor": 4}