diff --git a/0.3 Some Exercises.ipynb b/0.3 Some Exercises.ipynb
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--- a/0.3 Some Exercises.ipynb	
+++ b/0.3 Some Exercises.ipynb	
@@ -1,420 +1,420 @@
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     "#  Some exercises\n",
     "\n",
     "**Course**: [_Analyse Numérique pour SV_](https://moodle.epfl.ch/course/info.php?id=) (MATH-2xx)\n",
     "\n",
     "Original by prof. Fabio Nobile\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "- Before starting to work on this exercise session, you should review the material in the short `Python tutorial` available on Moodle."
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 0 -- Learning Python using Python."
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "Type the following commands and look at the results. Begin by importing numpy and matplotlib.pyplot  as "
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "import numpy as np\n",
     "import matplotlib.pyplot as plt"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "These are the libraries that python relies upon for scientific computing and plotting. "
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "| No. | Code | Instruction |\n",
     "| --- | --- | --- |\n",
     "| 1 |  `a=7` | Define $a$ as a scalar variable which has type `int` |\n",
     "| 2 |  `a=7.1` | Define $a$ as a scalar variable which has type `float`   |\n",
     "| 3 |  `b=np.array([1,2,3])` | Define $b\\in\\mathbb{R}^{1 \\times 3}$. What happens if you replace `,` by `''` or `;`?   |\n",
     "| 4 |  `g=b[2]` | Return the **third** element of  $b$. Note: Indexation starts at 0.   |\n",
     "| 5 |  `E=np.array([[1, 2, 3],[4, 5, 6]])` | Create matrix $E\\in\\mathbb{R}^{2\\times3}$. Note: Python is case sensitive.   |\n",
-    "| 6 |  `E[0,:]` | Return first column of $E;$ $(1,4)^T$   |\n",
+    "| 6 |  `E[0,:]` | Return first row of $E;$ $(1,4)^T$   |\n",
     "| 7 |  `h=E[1,2]` | Return the element $E_{2,3}$ of the matrix $E$ ($E_{2,3}=6$).   |\n",
     "| 8 |  `%whos` | Fetch information about the used variables.    |\n",
     "| 9 |  `help(np.sin)` | Fetch help for the command ` np.sin`. Same as `? np.sin` in iPython.   |\n",
     "| 10 |  `A=np.eye(3)` | Return the identity matrix $3\\times 3$    |\n",
     "| 11 |  `I=np.ones((4,4))` | Return matrix $4\\times 4$ for which each value is 1   |\n",
     "| 12 |  `F=np.zeros((2,3))` | Return $F\\in\\mathbb{R}^{2\\times 3}$ for which each value is 0   |\n",
     "| 13 |  `F.T` | Perform transpose of $F$   |\n",
     "| 14 |  `x=np.linspace(0,1,3)` | Return a vector of length $3$ whose values are equally distributed between $0$ and $1$ |\n",
     "| 15 |  `D=np.diag(b)` | Return diagonal matrix with diagonal given by vector b.   |\n",
     "| 16 |  `np.shape(F)` | Return the number of lines and columns of $F$ in a line vector.   |\n",
     "| 17 |  `A@D` | Return product of two matrices   |\n",
     "| 18 |  `E@x` | Return matrix vector product   |\n",
     "| 19 |  `A[-1,1]` | Return element of $A$ which is on the last line, second column   |\n",
     "| 20 |  `H=np.array([[1, 3],[9, 11]])；np.linalg.solve(H,E)` | Compute $H^{-1}E$. Note: Never use `np.linalg.inv(H)@E`    |\n",
     "| 21 |  `np.linalg.det(H)` | Compute determinant of $H$   |"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "## type here and execute the 21 commands above\"\n",
     "#1\n",
     "a=7\n",
     "print('#1',a),\n",
     "#2 continue by yourself"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 1\n",
     "\n",
     "Fibonacci's series is defined as follows:\n",
     "$$\\begin{align*}\n",
     "F_0&=1\\\\\n",
     "F_1&=1\\\\\n",
     "F_{n}&=F_{n-1}+F_{n-2}, \\quad \\forall n \\geq 2\n",
     "\\end{align*}$$\n",
     "* Complete the `Python` function below  which computes the $30^{\\text{th}}$ Fibonacci's number.\n",
     "\n",
     "* Write a function which takes as an argument a natural number $n$ and returns the $n^{\\text{th}}$ Fibonacci's number.\n"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {
     "jupyter": {
      "source_hidden": true
     }
    },
    "outputs": [],
    "source": [
     "#COMPLETE THIS CODE\n",
     "def Fibonacci(n):\n",
     "    # n should be a positive integer\n",
     "    # Complete this code\n",
     "    Fn=None #Change this value\n",
     "    # ...\n",
     "    return Fn\n",
     "\n",
     "F30=Fibonacci(30)\n",
     "print(F30)"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 2\n",
     "Let us consider the matrices \n",
     "\\begin{align*}\n",
     "A=\n",
     "\\begin{bmatrix}\n",
     "5 & 3 & 0\\\\\n",
     "1 & 1 & -4\\\\\n",
     "3 & 0 & 0\n",
     "\\end{bmatrix}, \\quad\n",
     "B=\n",
     "\\begin{bmatrix}\n",
     "4 & 3 & 2\\\\\n",
     "0 & 1 & 0\\\\\n",
     "5 & 0 & 1/2\n",
     "\\end{bmatrix}.\n",
     "\\end{align*}\n",
     "Compute (without using any loops), the matrix $C=AB$ (matrix product) and the matrix $D$ which\n",
     "has as elements $D_{ij}=A_{ij}B_{ij}$ (element-wise product)."
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "#### Solution:\n",
     "\n",
     "We can use the Python commands `@` for matrix multiplication and `*` for element-wise multiplication, as shown below:\n"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "A=np.array([ [5,3,0],\n",
     "             [1,1,-4],\n",
     "            [3,0,0]])\n",
     "B=np.array([ [4,3,2],\n",
     "             [0,1,0],\n",
     "            [5,0,1/2]])\n",
     "\n",
     "\n",
     "# Complete this code\n",
     "C=None #Change this value\n",
     "D=None #Change this value\n",
     "\n",
     "\n",
     "print('C matrix')\n",
     "print(C)\n",
     "\n",
     "print('D matrix')\n",
     "print(D)\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 3\n",
     "\n",
     "Define (without using any loops) the\n",
     "bidiagonal matrix of size $n=5$ whose main diagonal is a vector\n",
     "of equally distributed points between $3$ and $6$, i.e. $$D=(3, 3.75, 4.5, 5.25, 6)^T,$$\n",
     "\n",
     "and the sub-diagonal is the vector of equally distributed points between $2$ and $3.5$, i.e. $$S=(2, 2.5, 3, 3.5)^T.$$\n",
     "\n",
     "\n",
     "Tip: See ``https://numpy.org/doc/stable/reference/generated/numpy.diag.html``\n",
     "\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "#### Solution\n",
     "\n",
     "Following the reference for the`diag` method of numpy we obtain:"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "# Complete this code\n",
     "D=None #Change this value\n",
     "S=None #Change this value\n",
     "M=None #Change this value\n",
     "\n",
     "print('The matrix M is')\n",
     "print(M)"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "Remember that you can also obtain help from a python function using the command \n",
     "`help(NAME OF FUNCTION)`. In this case: `help(np.diag)`"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "help(np.diag)"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 4\n",
     "Let us consider the vectors $x=(1,4,7,2,1,2)$ and $y=(0,9,1,4,3,0)$. \n",
     "Compute (without using any loops for points a and b):\n",
     "1. the product, component by component, between two vectors $x$ and $y$ (tip: use the operator `*`)\n",
     "2. the scalar product between the same vectors $x$ and $y$ (tip: use the operator `@`)\n",
     "3. a vector whose elements are defined by:\n",
     "$v_1=x_1\\,y_n, \\quad v_2=x_2\\, y_{n-1},\\quad \\dots, \\quad v_{n-1}=x_{n-1}\\, y_2,\\quad v_n=x_n\\,y_1$.\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "#### Solution"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "x=np.array([1,4,7,2,1,2])\n",
     "y=np.array([0,9,1,4,3,0])\n",
     "\n",
     "# Complete this code\n",
     "v1=None #Change this value\n",
     "v2=None #Change this value\n",
     "v3=None #Change this value\n",
     "\n",
     "print('v1 = '+str(v1))\n",
     "print('v2 = '+str(v2))\n",
     "print('v3 = '+str(v3))"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 5\n",
     "\n",
     "Plot the functions\n",
     "\\begin{align*}\n",
     "f(x)=x^3 , \\quad x \\in [1,10]\n",
     "\\end{align*}\n",
     "\\begin{align*}\n",
     "g(x)=\\exp(4x) , \\quad x \\in [1,10]\n",
     "\\end{align*}\n",
     "\n",
     "in linear scale (i.e. using the function `plt.plot()` and logarithmic scale (i.e. using the function `plt.semilogy()` and `plt.loglog()`. We will use the definition interval to plot these graphs considering 200 equally distributed points.\n",
     "\n",
     "\n",
     "\n",
     "\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "#### Solution \n",
     "\n"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "# Complete this code\n",
     "f=None #Change this value to define inline function f\n",
     "g=None #Change this value to define inline function g \n",
     "\n",
     "x=np.linspace(1,10,200)\n",
     "\n",
     "\n",
     "#plot ins plot plot \n",
     "plt.title(' Normal scale')\n",
     "plt.plot(x,f(x),label=r'$f(x)$')\n",
     "# Complete this code\n",
     "# add a plot here for g\n",
     "plt.legend()\n",
     "plt.show()\n",
     "\n",
     "#plot in semilogy \n",
     "\n",
     "\n",
     "# Complete this code\n",
     "\n",
     "\n",
     "\n",
     "\n",
     "#plot in log-log\n",
     "\n",
     "# Complete this code\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "### Exercise 6\n",
     "\n",
     "1. Given \n",
     "\\begin{align*}\n",
     "f(x)=\\frac{x^2}{2}\\sin(x) , \\quad x \\in [1,20]\n",
     "\\end{align*}\n",
     "plot the function $f$ using 10, 20 and 100 equidistant points in the given interval.\n",
     "Plot the three graphs on the same figure with three\n",
     "different colors. Which one gives the best representation of $f$?\n",
     "\n",
     "2. Do the same for the functions:\n",
     "\\begin{align*}\n",
     "g(x) &= \\frac{x^3}{6}\\cos(\\sin(x))\\exp(-x)+\\left(\\frac{1}{1+x}\\right)^2 , \\quad x \\in [1,20] \\\\\n",
     "h(x) &= x(1-x)+\\frac{\\sin(x)\\cos(x)}{x^3}, \\quad x \\in [1,20].\n",
     "\\end{align*}\n"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "#### solution"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "# complete the code here for question 1"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
    "metadata": {},
    "outputs": [],
    "source": [
     "# complete the code here for question 2\n"
    ]
   }
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