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<h2 class="hd hd-2 unit-title">1. Motivation</h2>
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<h3 class="hd hd-2">Newton's Method</h3>
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<h2 class="hd hd-2 unit-title">2. Newton's Method</h2>
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<p><b class="bf"><span style="color:#99182C">Iterative numeric method</span></b> for finding <b class="bf"><span style="color:#99182C"> roots</span></b> of functions. </p><p><b class="bfseries">Objectives</b></p><p>
At the end of this sequence, and after some practice, you should be able to: </p><ul class="itemize"><li><p>
Apply <b class="bf"><span style="color:#27408C">Newton's Method</span></b> to find roots of functions. </p></li><li><p>
Describe Newton's Method <b class="bf"><span style="color:#27408C"> geometrically</span></b>. </p></li><li><p>
Determine conditions <b class="bf"><span style="color:#27408C"> where Newton's Method will fail</span></b> to find a root. </p></li></ul><p><b class="bfseries">Contents: 10 pages</b></p><p>
5 videos (17 minutes at 1x speed) 10 questions </p>
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<h2 class="hd hd-2 unit-title">3. Roots and Newton's Method</h2>
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Review roots of polynomials
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Given a quadratic polynomial [mathjaxinline]\ f(x) = ax^2 + bx + c[/mathjaxinline], choose all statements below that describe what a <b class="bf">root</b> of this polynomial is. </p>
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<text>Points where the tangent line slope [mathjaxinline]\ f'(x) = 0[/mathjaxinline]</text>
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<text>[mathjaxinline]\displaystyle {x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}}[/mathjaxinline]</text>
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<text>[mathjaxinline]x=0[/mathjaxinline]</text>
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The value [mathjaxinline]\sqrt {2}[/mathjaxinline] is a root of the function [mathjaxinline]\ f(x) = 2-x^2[/mathjaxinline]. If you had to guess the value of [mathjaxinline]\sqrt {2}[/mathjaxinline], which of the following integer values is the closest to the the root? </p>
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<text> 2</text>
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What is the equation for the tangent line [mathjaxinline]L(x)[/mathjaxinline] to [mathjaxinline]\ f(x) = 2-x^2[/mathjaxinline] at [mathjaxinline]x_0=1[/mathjaxinline]. Then find the value of [mathjaxinline]x_1[/mathjaxinline] where this tangent line crosses the [mathjaxinline]x[/mathjaxinline]-axis. </p>
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<p style="display:inline">[mathjaxinline]y=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]x_1=[/mathjaxinline]</p>
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Is the value [mathjaxinline]x_1[/mathjaxinline] closer to or further from the value of the root [mathjaxinline]\sqrt {2}[/mathjaxinline] as compared to the initial guess [mathjaxinline]x_0[/mathjaxinline]? </p>
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<h2 class="hd hd-2 unit-title">5. Newton's Method</h2>
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Given a function [mathjaxinline]\ f(x)[/mathjaxinline], find [mathjaxinline]x[/mathjaxinline] such that [mathjaxinline]\ f(x) = 0[/mathjaxinline]. </p><ol class="enumerate"><li value="1"><p>
Make a good guess [mathjaxinline]x_0[/mathjaxinline]. </p></li><li value="2"><p>
Call [mathjaxinline]x_1[/mathjaxinline] the [mathjaxinline]x[/mathjaxinline]-intercept of the tangent line through [mathjaxinline](x_0, f(x_0))[/mathjaxinline]. It has the formula </p><table id="a0000000585" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\boxed { \displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}.}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(5.57)</td></tr></table></li><li value="3"><p>
Repeat. The general formula is </p><table id="a0000000586" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\boxed { \displaystyle x_{n+1} = x_ n - \frac{f(x_ n)}{f'(x_ n)}}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(5.58)</td></tr></table><p>
for [mathjaxinline]n=0,1,2, \cdots[/mathjaxinline]. </p></li></ol>
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Suppose that a function [mathjaxinline]\ f(x)[/mathjaxinline] has the tangent line [mathjaxinline]y = 4x-1[/mathjaxinline] at the point [mathjaxinline]x=1[/mathjaxinline]. If your first approximation in the Newton Method for finding the roots of [mathjaxinline]\ f(x)[/mathjaxinline] is [mathjaxinline]x=1[/mathjaxinline], what is your second approximation? </p>
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<p style="display:inline">[mathjaxinline]x_1 =[/mathjaxinline]</p>
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Square root of 5
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Let us use [mathjaxinline]\ f(x) = x^2 - 5[/mathjaxinline] to find the numerical value for [mathjaxinline]\sqrt {5}[/mathjaxinline]. Enter your guess in the first box, and apply 2 iterations of Newton's Method to your guess. </p>
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(Use the calculator tab on the lower right hand part of the screen. Enter a formula. You can copy and paste the answer into the answer box. Copy and paste the result of each iteration back into the formula bar in the calculator to find the next entry. Enter results with at least 4 decimal places of accuracy. DO NOT ENTER FRACTIONS.) </p>
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<h2 class="hd hd-2 unit-title">6. Convergence Rate</h2>
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<h2 class="hd hd-2 unit-title">7. Failure to converge</h2>
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Consider the function [mathjaxinline]y = \arctan x .[/mathjaxinline] The root of this function is [mathjaxinline]x=0[/mathjaxinline]. Apply Newton's method with an initial guess of [mathjaxinline]x_0=2[/mathjaxinline] and see what happens. </p>
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Making the best guess
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For which of the following initial guesses will Newton's method <b class="bf">fail</b> to converge to the root of the function graphed below that is between [mathjaxinline]0[/mathjaxinline] and [mathjaxinline]0.5[/mathjaxinline]? </p>
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(Choose all that apply.) </p>
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Identify ending root for the various starting points
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Consider the following function which has 3 roots, A, B, and C labeled in the image below. </p>
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<img alt="Function is increasing when x is less than 0, decreasing when x is between 0 and 2, and increasing when x is greater than 2. The point A is at approxiamtely negative 1.2. The point B is at approximately 1.4. The point C is at approximately 3.4. The function is concave down when x is less than 0.2, concave up when x is between 0.2 and 0.9, concave down when x is between 0.9 and 1.6, concave up when x is between 1.6 and 3.4, and concave down when x is greater than 3.4. Function has horizontal tangent lines at x equals 0 and x equals 2." src="/assets/courseware/v1/c698b9fd557425b3a02f4fac7dafc75b/asset-v1:MITx+18.01.1x+2T2019+type@asset+block/images_u3s3_img2.svg" style="margin: 10px 25px 25px 25px" width="450px"/>
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For each initial starting point [mathjaxinline]x_0[/mathjaxinline] described below, choose the root to which Newton's method will converge. </p>
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(Choose None if it will not converge, or is unclear from the image.) </p>
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</div></div>&#8195;&#8195;&#8195;&#8195;<p style="display:inline">[mathjaxinline]x_0=3[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div class="inputtype option-input inline">
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</div></div>&#8195;&#8195;&#8195;&#8195;<p style="display:inline">[mathjaxinline]x_0=4[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 6" role="group"><div class="inputtype option-input inline">
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<h2 class="hd hd-2 unit-title">9. Finding Pi</h2>
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Finding Pi
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Suppose that you want to find a numerical approximation for [mathjaxinline]\pi[/mathjaxinline]. To which of the following functions could you apply Newton's Method to in order to find it? </p>
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<text>[mathjaxinline]\sin (x)[/mathjaxinline]</text>
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<text>[mathjaxinline]\tan (x)[/mathjaxinline]</text>
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<text>[mathjaxinline]\arcsin (x)[/mathjaxinline]</text>
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Newton&#39;s method slow convergence
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Let us use [mathjaxinline]\ f(x) = \cos (x)+1[/mathjaxinline] to find the numerical value for [mathjaxinline]\pi[/mathjaxinline]. This is the wrong function to use to find this root. Newton's method converges very, very slowly because this function has a double root at [mathjaxinline]\pi[/mathjaxinline]. Let's see just how slow it is. (How many steps does it take to get an accurate answer to 4 decimal places? 5 decimal places? Compare this to other examples we've already seen.) </p>
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(Enter your initial guess in the first box, and enter the first two approximations found using Newton's Method in the 2nd and 3rd boxes. Use the calculator button on the bottom right to aid your computation. Submit answers with at least 7 decimal places.) </p>
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<h2 class="hd hd-2 unit-title">10. Summary</h2>
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<p><b class="bfseries">Newton's Method</b></p><p>
Given a function [mathjaxinline]\ f(x)[/mathjaxinline], find [mathjaxinline]x[/mathjaxinline] such that [mathjaxinline]\ f(x) = 0[/mathjaxinline]. </p><ol class="enumerate"><li value="1"><p>
Make a good guess [mathjaxinline]x_0[/mathjaxinline]. </p></li><li value="2"><p>
Call [mathjaxinline]x_1[/mathjaxinline] the [mathjaxinline]x[/mathjaxinline]-intercept of the tangent line through [mathjaxinline](x_0, f(x_0))[/mathjaxinline]. It has the formula </p><table id="a0000000595" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\boxed { \displaystyle x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}.}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(5.59)</td></tr></table></li><li value="3"><p>
Repeat. The general formula is </p><table id="a0000000596" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\boxed { \displaystyle x_{n+1} = x_ n - \frac{f(x_ n)}{f'(x_ n)}}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(5.60)</td></tr></table><p>
for [mathjaxinline]n=0,1,2, \cdots[/mathjaxinline]. </p></li></ol>
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