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 Newton’s method – Process, Approximation, and Example
Newton’s method – Process, Approximation, and Example
Newtonâ€™s method is an important application of differential calculus. This method is also known as the NewtonRaphson method, as this is a technique developed by Sir Isaac Newton and Joseph Raphson. Some equations will be challenging to solve despite our efforts and knowledge of algebraic manipulations, and this is where approximations and algorithms like Newtonâ€™s method enter. Newtonâ€™s method (or NewtonRaphson method) is used to solve the approximate roots of a function using the functionâ€™s first derivative. In this article, weâ€™ll sort out the equations that will benefit from this method, and of course, our goal is to make sure that we apply this method properly to approximate the roots of a given function. Weâ€™ll also try out some equations to compare the resulting roots derived by algebraic manipulation and Newtonâ€™s method to show you how close these values are. Letâ€™s go ahead and begin by understanding what makes this method special! What is Newtonâ€™s method?Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Newtonâ€™s method is a technique we use to solve complex equations of the form, $f(x) = 0$, using successive approximations. This numerical method relies on the geometry of curves and the tangent line/s that pass through it. This means that knowledge of derivatives and their common rules is a prerequisite to appreciating this numerical method better. The graph shown above is a great visualization of how Newtonâ€™s method works. This numerical method relies on the tangent line (see the red line passing through $x_2$) that passes through the curve, $f(x)$. The idea is to begin with an initial value say, $x_0$, where $(x_0, f(x_0))$ is a point along the curve of $f(x)$. We can then find a tangent line that passes through the curve and another point, $(x_1, 0)$. This point must be near $x_1$ and consequently, the actual root of $f(x)$. We stop if $f(x_1) = 0$. Otherwise, weâ€™ll continue by finding a tangent line that passes tough $x_2$, $x_3$, and so on. This process continues until we have $f(x_{n 1}) = 0$ or approximately close to $0$. This method will be a great help, especially when itâ€™s difficult to find the root of $f(x)$ using traditional methods.
Traditional Method 
Newtonâ€™s Method 


Helpful Derivative Rules 

\begin{aligned}\dfrac{d}{dx} c = 0\end{aligned}  \begin{aligned}\dfrac{d}{dx} f(g(x))= fâ€™(g(x)) gâ€™(x)\end{aligned} 
\begin{aligned}\dfrac{d}{dx} x^n = nx^{n 1}\end{aligned}  \begin{aligned}\dfrac{d}{dx} e^x = e^x \end{aligned} 
\begin{aligned}\dfrac{d}{dx} c\cdot f(x) = c \cdot fâ€™(x)\end{aligned}  \begin{aligned}\dfrac{d}{dx} a^x = a^x \ln a \end{aligned} 
\begin{aligned}\dfrac{d}{dx} f(x) \pm g(x) = fâ€™(x) \pm gâ€™(x)\end{aligned}  \begin{aligned}\dfrac{d}{dx} \sin x = \cos x\end{aligned} 
\begin{aligned}\dfrac{d}{dx} [f(x) \cdot g(x)] = fâ€™(x) \cdot g(x) + gâ€™(x) \cdot f(x)\end{aligned}  \begin{aligned}\dfrac{d}{dx} \cos x = \sin x\end{aligned} 
\begin{aligned}\dfrac{d}{dx} \left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)fâ€™(x) â€“ f(x) gâ€™(x)}{[g(x)]^2}\end{aligned}  \begin{aligned}\dfrac{d}{dx} \tan x =\sec^2 x\end{aligned} 
 If itâ€™s not yet given, find the expression for $fâ€™(x)$ using the different derivative rules.
 Begin with an initial value, $x_0$ that might be close to the actual root.
 Determine the first approximate using the formula shown below. \begin{aligned}x_1 &= x_0 – \dfrac{f(x_0)}{fâ€™(x_0)}\end{aligned}
 Repeat the iterative process using the formula shown below until we find the root within our allowed accuracy. \begin{aligned}x_{n+1} &= x_n – \dfrac{f(x_n)}{fâ€™(x_n)}\end{aligned}
 Find the values of $f(x_0)$ and $x_0$.
 Use these values to find $x_1$ using the formula, $x_1 = x_0 – \dfrac{f(x_0)}{fâ€™(x_0)}$.
\begin{aligned}\boldsymbol{n}\end{aligned}  \begin{aligned}\boldsymbol{x_n}\end{aligned}  \begin{aligned}\boldsymbol{f(x_n)}\end{aligned}  \begin{aligned}\boldsymbol{fâ€™(x_n)}\end{aligned}  \begin{aligned}\boldsymbol{x_{n+1} = x_n – \dfrac{f(x_n)}{fâ€™(x_n)}}\end{aligned} 
\begin{aligned}0\end{aligned}  \begin{aligned}1\end{aligned}  \begin{aligned}1\end{aligned}  \begin{aligned}2\end{aligned}  \begin{aligned}1.5\end{aligned} 
\begin{aligned}1\end{aligned}  \begin{aligned}1.5\end{aligned}  \begin{aligned}0.875\end{aligned}  \begin{aligned}5.75\end{aligned}  \begin{aligned}1.347826\end{aligned} 
\begin{aligned}2\end{aligned}  \begin{aligned}1.347826\end{aligned}  \begin{aligned}0.1006822\end{aligned}  \begin{aligned}4.4499055\end{aligned}  \begin{aligned}1.325200\end{aligned} 
\begin{aligned}3\end{aligned}  \begin{aligned}1.325200\end{aligned}  \begin{aligned}0.0.002058\end{aligned}  \begin{aligned}4.2684683\end{aligned}  \begin{aligned}1.324718\end{aligned} 
\begin{aligned}4\end{aligned}  \begin{aligned}1.324718\end{aligned}  \begin{aligned}0.0000009\end{aligned}  \begin{aligned}4.2646347\end{aligned}  \begin{aligned}1.324718\end{aligned} 
\begin{aligned}\boldsymbol{n}\end{aligned}  \begin{aligned}\boldsymbol{x_n}\end{aligned}  \begin{aligned}\boldsymbol{x_{n+1} = x_n – \dfrac{f(x_n)}{fâ€™(x_n)}}\end{aligned} 
\begin{aligned}0\end{aligned}  \begin{aligned}1\end{aligned}  \begin{aligned}0.3195\end{aligned} 
\begin{aligned}1\end{aligned}  \begin{aligned}0.3195\end{aligned}  \begin{aligned}0.4208\end{aligned} 
\begin{aligned}2\end{aligned}  \begin{aligned}0.4208\end{aligned}  \begin{aligned}0.4263\end{aligned} 
\begin{aligned}3\end{aligned}  \begin{aligned}0.4263\end{aligned}  \begin{aligned}0.4263\end{aligned} 
\begin{aligned}\boldsymbol{n}\end{aligned}  \begin{aligned}\boldsymbol{x_n}\end{aligned}  \begin{aligned}\boldsymbol{x_{n+1} = x_n – \dfrac{f(x_n)}{fâ€™(x_n)}}\end{aligned} 
\begin{aligned}0\end{aligned}  \begin{aligned}1\end{aligned}  \begin{aligned}0.891396\end{aligned} 
\begin{aligned}1\end{aligned}  \begin{aligned}0.891396\end{aligned}  \begin{aligned}0.876985\end{aligned} 
\begin{aligned}2\end{aligned}  \begin{aligned}0.876985\end{aligned}  \begin{aligned}0.876726\end{aligned} 
\begin{aligned}3\end{aligned}  \begin{aligned}0.876726\end{aligned}  \begin{aligned}0.876726\end{aligned} 
\begin{aligned}\boldsymbol{n}\end{aligned}  \begin{aligned}\boldsymbol{x_n}\end{aligned}  \begin{aligned}\boldsymbol{x_{n+1} = x_n – \dfrac{f(x_n)}{fâ€™(x_n)}}\end{aligned} 
\begin{aligned}0\end{aligned}  \begin{aligned}3\end{aligned}  \begin{aligned}2.64691358\end{aligned} 
\begin{aligned}1\end{aligned}  \begin{aligned}2.64691358 \end{aligned}  \begin{aligned}2.52497770\end{aligned} 
\begin{aligned}2\end{aligned}  \begin{aligned}2.52497770\end{aligned}  \begin{aligned}2.51202148 \end{aligned} 
\begin{aligned}3\end{aligned}  \begin{aligned}2.51202148 \end{aligned}  \begin{aligned}2.51188645\end{aligned} 
\begin{aligned}4\end{aligned}  \begin{aligned}2.51188645 \end{aligned}  \begin{aligned}2.51188643\end{aligned} 
\begin{aligned}5\end{aligned}  \begin{aligned}2.51188643\end{aligned}  \begin{aligned}2.51188643\end{aligned} 
\begin{aligned}\boldsymbol{n}\end{aligned}  \begin{aligned}\boldsymbol{x_n}\end{aligned}  \begin{aligned}\boldsymbol{x_{n+1} = x_n – \dfrac{f(x_n)}{fâ€™(x_n)}}\end{aligned} 
\begin{aligned}0\end{aligned}  \begin{aligned}2\end{aligned}  \begin{aligned}1.42834\end{aligned} 
\begin{aligned}1\end{aligned}  \begin{aligned}1.42834 \end{aligned}  \begin{aligned}1.49476\end{aligned} 
\begin{aligned}2\end{aligned}  \begin{aligned}1.49476\end{aligned}  \begin{aligned}1.52342\end{aligned} 
\begin{aligned}3\end{aligned}  \begin{aligned}1.52342 \end{aligned}  \begin{aligned}1.53700\end{aligned} 
\begin{aligned}4\end{aligned}  \begin{aligned}1.53700\end{aligned}  \begin{aligned}1.54365\end{aligned} 
\begin{aligned}5\end{aligned}  \begin{aligned}1.54365\end{aligned}  \begin{aligned}1.54693\end{aligned} 
\begin{aligned}6\end{aligned}  \begin{aligned}1.54693\end{aligned}  \begin{aligned}1.54857\end{aligned} 
\begin{aligned}7\end{aligned}  \begin{aligned}1.54857\end{aligned}  \begin{aligned}1.54938\end{aligned} 
\begin{aligned}8\end{aligned}  \begin{aligned}1.54938\end{aligned}  \begin{aligned}1.54979 \end{aligned} 
\begin{aligned}9\end{aligned}  \begin{aligned}1.54979 \end{aligned}  \begin{aligned}1.54999\end{aligned} 
\begin{aligned}10\end{aligned}  \begin{aligned}1.54999\end{aligned}  \begin{aligned}1.55009\end{aligned} 
\begin{aligned}11\end{aligned}  \begin{aligned}1.55009\end{aligned}  \begin{aligned}1.55014\end{aligned} 
\begin{aligned}12\end{aligned}  \begin{aligned}1.55014\end{aligned}  \begin{aligned}1.55017\end{aligned} 
\begin{aligned}13\end{aligned}  \begin{aligned}1.55017\end{aligned}  \begin{aligned}1.55018\end{aligned} 
\begin{aligned}14\end{aligned}  \begin{aligned}1.55018\end{aligned}  \begin{aligned}1.55019\end{aligned} 
\begin{aligned}15\end{aligned}  \begin{aligned}1.55019\end{aligned}  \begin{aligned}1.55019\end{aligned} 
\begin{aligned}\boldsymbol{x_0}\end{aligned}  \begin{aligned}\boldsymbol{x}\end{aligned} 
\begin{aligned}x_0 = 2.5\end{aligned}  \begin{aligned}x \approx 2.39749\end{aligned} 
\begin{aligned}x_0 = 3\end{aligned}  \begin{aligned}x \approx 2.84948\end{aligned} 
\begin{aligned}x_0 = 3.2\end{aligned}  \begin{aligned}x \approx 3.15963\end{aligned} 
Practice Questions
1. Find the solution to the equation, $x^3 + 2x^2 + 1= 0$ using the Newtonâ€™s method with an accuracy of $5$Â decimal places. Use $x_0 = 2$ as your initial value. 2. Find the solution to the equation, $2x+ e^{3x} = 0$ using the Newtonâ€™s method with an accuracy of $5$Â decimal places. Use $x_0 = –1$ as your initial value. 3. Find the solution to the equation, $\cos x = 2x^2$ using Newtonâ€™s method with an accuracy of $6$Â decimal places. Use $x_0 = 1$ as your initial value. 4. Estimate the seventh root of $32$ to eight decimal places using the NewtonRaphson method. 5.Determine one approximate solution for the equation,$\cos (e^2x) = 1$. Estimate the root to the nearest five decimal places and make sure that the solution is within the interval, $\left[0, \dfrac{\pi}{2}\right]$.Answer Key
1. $x\approx 2.20557$ 2. $x\approx 0.24195$ 3. $x \approx 0.634560$ 4. $\sqrt[7]{32} \approx 1.64067071$ 5. $x \approx 0.91894$, $x \approx 1.26552$, or $x \approx 1.46825$ Images/mathematical drawings are created with GeoGebra.
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