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   "source": [
    "# Introduction to Loops"
   ]
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    "import numpy as np"
   ]
  },
  {
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   "source": [
    "## Problem 1\n",
    "\n",
    "**Write code to print the numbers 0 to 4 on each line below.**"
   ]
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   "source": [
    "### Solution"
   ]
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   "source": [
    "### Discussion\n",
    "* Okay, let's keep going to 10000...(!?)"
   ]
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   "source": [
    "## Loops\n",
    "\n",
    "* Loops in programing are used to simplify repetative work.\n",
    "* Above, each line is a print statement.\n",
    "    * The only thing different on each line is the number we are printing.\n",
    "* We can write a loop statment to do the above problem as follows"
   ]
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    "### Discussion\n",
    "* What are the parts of this: list them?\n",
    "* ```i``` can be any variable, not just \"i\"\n",
    "* Is there anything weird or unexpected here?"
   ]
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   "source": [
    "## Problem 2\n",
    "* We can put **any** code in the for loop\n",
    "    * Variables, functions, expressions, math, etc.\n",
    "* We can put variables outside of the loop that are available inside of the loop\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
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   "source": [
    "For the ideal gas law, pressure is given by\n",
    "$$P = nRT/V$$\n",
    "* n = 1 kmol\n",
    "* R = 8314 J/kmol*K\n",
    "* T = 300 K\n",
    "\n",
    "**For V = 2, 3, 4, 5, ... 20, use a loop to find and print the values of P for each V.**\n",
    "    "
   ]
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  {
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   "source": [
    "### Solution"
   ]
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   "source": [
    "### Discussion\n",
    "* What if you wanted V to not step by integers?\n",
    "    * Step by 2? Step by 5?\n",
    "    * Step by 0.2?"
   ]
  },
  {
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   "source": [
    "## Problem 3 \n",
    "\n",
    "**Add all the numbers from 0 to 100**\n",
    "* Before you start the code, think about what you would do if solving this in your head?\n",
    "    * How do you start?\n",
    "    * What things do you keep track of?\n",
    "    * What variables to use?\n",
    "    * What is the loop syntax?\n",
    "    * How do variables interact with the loop?\n",
    "* Think through the process of <font color='red'>the problem</font> to <font color='blue'> the solution</font>\n",
    "* Talk to your neighbor\n",
    "* If you've solved it, help your neighbor."
   ]
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   "source": [
    "### Solution"
   ]
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   "source": [
    "## Nested loops\n",
    "\n",
    "* We can put **any** code in a ```for``` loop. Even another loop.\n",
    "* Look at the code below. \n",
    "* **Before running it, think of what you expect to see.**\n",
    "    * Now run it and compare. Did you get what you thought you would?"
   ]
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    "for i in range(3):\n",
    "    print(\"i =\", i)\n",
    "    for j in range(4):\n",
    "        print(\"   j =\", j)"
   ]
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    "* Note, each time through the top (outer) loop, **everything** inside is done.\n",
    "* This means the **whole** inner loop is computed for each outer loop.\n",
    "* **Think of an actual application for a nested loop.**"
   ]
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   "source": [
    "## Problem 4a\n",
    "\n",
    "* Given the following function\n",
    "$$f(x) = x^2 - 3,$$\n",
    "we can use Newton's method to find $x$ for $f(x)=0$. \n",
    "\n",
    "* Recall, Newton recurrence formula:\n",
    "$$x_{new} = x_{old} - \\frac{f(x_{old})}{f^\\prime(x_{old})}$$\n",
    "\n",
    "### We solved this previously in Excel\n",
    "* **Open Excel and solve this problem.**\n",
    "* Think about the steps involved, how we set it up.\n",
    "* **Use the simplest method possible.**\n"
   ]
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   "source": [
    "### Solution-Excel\n",
    "\n",
    "<img src='https://ignite.byu.edu/cbe263/lectures/loops_fig_1.png' width=300>\n",
    "\n",
    "\n",
    "| Excel | Python|\n",
    "|:-----:|:-----:|\n",
    "|cell for guess |  ?? |\n",
    "|Newton equation in terms of x cell <br> A2 - (A2^2-3)/(2*A2)| ?? |\n",
    "| repeat: fill down | ?? |\n"
   ]
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   "source": [
    "## Problem 4b\n",
    "### Now solve in Python using loops\n",
    "\n",
    "* **Try to write the simplest solution you can think of with as few lines of code as possible.**\n",
    "* **We want 3 decimal places of accuracy.**\n",
    "* Consider: \n",
    "    * What variables are needed?\n",
    "    * How to setup and use the loop.\n",
    "    * What to print and where?"
   ]
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  {
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   },
   "source": [
    "### Solution-Python"
   ]
  },
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   },
   "source": [
    "### Discussion"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
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    }
   },
   "source": [
    "###  Explicitly write out the four steps\n",
    "* We don't have to code it from top to bottom...\n",
    "* Compare to Excel\n",
    "\n",
    "| Excel | Python|\n",
    "|:-----:|:-----:|\n",
    "|cell for guess |  variable for x |\n",
    "|Newton equation in terms of x cell <br> A2 - (A2^2-3)/(2*A2)| Newton equation in terms of x <br> x = x - (x\\*\\*2-3)/(2*x)|\n",
    "| repeat: fill down | repeat: wrap in loop |"
   ]
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   "source": [
    "# step 1:\n",
    "\n",
    "# step 2:\n",
    "\n",
    "# step 3:\n",
    "\n",
    "# step 4:\n"
   ]
  },
  {
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   "source": [
    "## Problem 4c\n",
    "\n",
    "**Redo Solution 1 using functions for $f(x)$ and $f^\\prime(x)$.**\n",
    "* Recall: we discussed functions two classes back."
   ]
  },
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   "source": [
    "## Problem 4d\n",
    "\n",
    "* Now, to get to three decimal places we only needed 3 iterations.\n",
    "* But we didn't know this at the beginning, so we guessed the number of iterations.\n",
    "    * If 6 wasn't enough, we would do more.\n",
    "* This guess and check is not very efficient.\n",
    "\n",
    "**Modify the code above to do just the right number of iterations.**\n",
    "* What Python features do you need to use?\n",
    "* How do you decide when you are done?\n",
    "* **Talk with your neighbor.**"
   ]
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   "source": [
    "### Solution"
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    "### Discussion\n",
    "<!--\n",
    "* Keep track of old and new x: ```xold``` and ```xnew```.\n",
    "    * Need to reset ```xold=xnew``` at the end to get set for next interation.\n",
    "* New keyword\n",
    "```python\n",
    "break\n",
    "```\n",
    "* Define the error and compare it to the desired accuracy.\n",
    "    * Use absolute value: ```np.abs()```\n",
    "-->"
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