αေးαားαော αုံαွα် $x$ αα် $f(x)$ ၏ root αြα ်αα်αို αိαိုα်αါαα်။ $f(x)=0$ ၏ root αို αα်္αျာ αုα်αုံးαျား αြα့် α‘αွα်ααူαှာ၍αααα့် α‘αြေα‘αေαိုαါα ို့။ αို့αြောα့် αြိုαα်αα့်αှα်းαားαော $x_0$ αို $f(x)$ root αု αူααα်။ Figure 2 αို αြα့်αါ။
$f(x_0)\ne 0$ αြα ်αောαြာα့် $x_0$ αα် $f(x)$ ၏ root ααုα်αြောα်းαိαိုα်αα်။ Graph αေါ်αှိ $(x_0,f(x_0))$ α‘αှα်αှ tangent αျα်းαα ်αြောα်းαွဲαိုα်αα်။ α‘αိုαါ tangent αျα်းαα် $x$ αα်αိုးαို αြα်αွားαောαေαာ $x_1$ αα် $f(x)=0$ ၏ root αှα့် αီးα α်αေαα်αို αွေ့ααα်။ αို့αြောα့် $x_1$ αို $f(x)=0$ ၏ approximate root αု αူα၍ $x_1$ αို αှာαါαα်။
$x_1$ αα် tangent line α $x$-axis αို αြα်αောαေαာ αြα ်αောαြောα့် tangent line equation αို α¦းα ွာαှာαα်။ tangent line ၏ gradient αှာ $f'(x_0)$ αြα ်αြီး $(x_0,f(x_0))$ α‘αှα်αα် tangent line αေါ်αွα်αှိαα်။ αို့αြောα့် tangent line ၏ equation αှာ ...
$y=f'(x_0)(x-x_0)+f(x_0)$
αြα ်αါαα်။ $x=x_1$ αြα ်αျှα် $y=0$ αြα ်αα်။ αို့αြောα့်
$$\begin{aligned} &0=f'(x_0)(x_1-x_0)+f(x_0)$\\\\ &f'(x_0)(x_1-x_0) = -f(x_0)\\\\ &x_1-x_0=-\frac{f(x_0)}{f'(x_0)}\\\\ \therefore\ & x_1=x_0-\frac{f(x_0)}{f'(x_0)}\\\\ \end{aligned}$$αူ၍ $x_1$ αို αှာαိုα်αါαα်။
αို့αာαွα် $x_1$ αα်αα်း $f(x)$ ၏ root ααုα်αေးαါ။ $f(x)$ ၏ root αှα့် αီးα α်αာαα့် α‘αြေα‘αေαွα်αာ αှိαေးαα်။ αို့αြောα့် α‘αα်αွα်αှα်းαα်းαဲ့αော αုα်αောα်αျα်α‘αိုα်း approximate root $(x_2)$ αို αα်αံαှာαူαါαα်။ Figure 3 αို αြα့်αါ။
α‘αα်αွα် αော်αြαဲ့αော αုα်αောα်αျα်α‘αိုα်း $(x_1,f(x_1))$ α‘αှα်αှိ tangent line αှ $x$-axis αို αြα်αောα‘αှα် $x_2$ αို αα်αα်αှာαူαα်။ αိုα‘αါ $$x_2=x_1-\frac{f(x_1)}{f'(x_1)}$$ αြα ်αα်။ ၎α်းαα်းαα်းα‘αိုα်း αα်αါ αα်αါ αှာαူαြα်းαြα့် $f(x)=0$ ၏ root αှα့် αα်αူαျαုαီးαါး αြα ်αာαော root αα်αိုးαို ααှိαါαα်။ Figure 4 αို αြα့်αါ။
αော်αြαါ αα့်αဲαှာαူαြα်းαα်းαα်းαြα့် $f(x)=0$ ၏ root αို αှာαူαိုα်αော αုံαေαα်း $x_{n+1}$ αို α‘ောα်αါ α‘αိုα်းαှα်αားαိုα်αα်။
$$\begin{array}{|c|} \hline x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\\ \hline \end{array}$$α‘αα်αော်αြαါ αုံαေαα်းαြα့် root αှာαူαြα်း process αို Newton-Raphson Iteration Method αုαေါ်αα်။ αα်αွေ့αုα ္αာαြေαှα်းαာαွα် $x_{n+1}\approx x_n$ αြα ်αα့်α‘αါ Iteration Process αြီးαုံးαြီαု αှα်αူαိုα်αα်။
α‘ောα်αါ α‘αြေαေαို့αွα် Newton-Raphson method αို αုံး၍ αααိုα်αါ။
Case I
Starting point $x_0$ αα် α‘αှα်αြα ်αော root αှα့် αေးαေαောα‘αါ tangent αα် root αှα့် αို၍αို၍ αေးαော $x$ coordinate αို αောα်αှိαွားαြီး Newton-Raphson Formula αα် divergent αြα ်αွားαα်။
Case II
$f'(x_0)=0$ αြα ်αျှα် tangent αα် horizontal αြα ်αွားαα်။ $(x_0, f(x_0))$ stationary point αြα ်αေαα်။ αိုα‘αါ tangent αα် $x$-axis αို αα်αော့αျှ ααြα်αော့αါ။ αိုα‘αြေα‘αေαွα် $ x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$ αα် undefined αြα ်αွား၍ Newton-Raphson Formula αို αုံး၍αααိုα်αါ။
Example (1)
Using Newton-Raphson method, find the approximate value of $\sqrt[3]{19}$ in four decimal places.
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Let $ \sqrt[3]{19}=x$. $\begin{aligned} &\\ \therefore \ & x^3 = 19\\\\ &x^3 - 19=0\\\\ \end{aligned}$ Hence, $ \sqrt[3]{19}$ is the root of the equation $x^3 - 19=0$. $\begin{aligned} &\\ &\text{Let } f(x)=x^3-19\\\\ \therefore \ & f'(x) = 3x^2\\\\ &x_{n+1}=x_n-\frac{{x_n}^3-19}{3{x_n}^2}\\\\ \end{aligned}$ Since $2^3=8$ and $3^3=27$, $ \sqrt[3]{19}$ is between $2$ and $3$. Let $x_0=3$. $\begin{array}{|l|l|c|} \hline \quad x_n & x_{n+1} & \text{Remark}\\ \hline 3 & 2.703704 & x_{n+1}\ne x_n \\ \hline 2.703704 & 2.668861 & x_{n+1}\ne x_n \\ \hline 2.668861 & 2.668402 & x_{n+1}\ne x_n \\ \hline 2.668402 & 2.668402 & x_{n+1}\approx x_n \\ \hline \end{array}$ $\therefore\ \sqrt[3]{19}=2.6684$ |
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Example (2)
Show that equation $e^x=3x+1$ has a root $\alpha=0$. Show by calculation that this equation also has a root, $\beta$, such that $1 < \beta < 2 $. Hence using Newton-Raphson method, find the $\beta$ correct to 4 decimal places.
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$\begin{aligned}
&e^x=3x+1\\\\
\therefore\ & e^x-3x-1=0\\\\
&\text{Let } f(x)=e^x-3x-1\\\\
&f(0)=e^0-3(0))-1\\\\
&f(0)=1-0)-1=0\\\\
\therefore\ & x=0 \text{ is a root of } f(x).\\\\
&f(1)=e^1-3-1=-1.28172 < 0\\\\
&f(2)=e^2-3(2)-1=0.389056 >0\\\\
\therefore\ & 1 < \beta < 2\\\\
&f'(x)=e^x-3\\\\
\therefore\ & x_{n+1}=x_n-\frac{e^{x_n}-3{x_n}-1}{e^{x_n}-3}\\\\
\end{aligned}$ Let $x_0=2$. $\begin{array}{|l|l|c|} \hline \quad x_n & x_{n+1} & \text{Remark}\\ \hline 2 & 1.911358 & x_{n+1}\ne x_n \\ \hline 1.911358 & 1.911358 & x_{n+1}\approx x_n \\ \hline \end{array}$ $\therefore\ \beta=1.9114$ |
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Example (3)
Given that $f(x)=x^3 - x - 1$. Show that $f(x)=0$ has a root between $1$ and $2$. Taking $x_0=1.5$, use Newton-Raphson method to find the root of $f(x)=0$ correct to $3$ decimal places.
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$\begin{aligned}
&f(x)=x^3 - x - 1\\\\
&f(1)=1-1-1=-1< 0\\\\
&f(2)=8-2-1=5 > 0\\\\
\therefore\ & f(x)=0 \text{ has a root between } 1 \text{ and } 2. \\\\
&f'(x)=3x^2-1\\\\
\therefore\ & x_{n+1}=x_n-\frac{{x_n}^3 - {x_n} - 1}{3{x_n}^2-1}\\\\
\end{aligned}$ Taking $x_0=1.5$, $\begin{array}{|l|l|c|} \hline \quad x_n & x_{n+1} & \text{Remark}\\ \hline 1.5 & 1.34783 & x_{n+1}\ne x_n \\ \hline 1.34783 & 1.32520 & x_{n+1}\ne x_n \\ \hline 1.32520 & 1.32472 & x_{n+1}\approx x_n \\ \hline 1.32472 & 1.32472 & x_{n+1}\approx x_n \\ \hline \end{array}$ $\therefore\ $ The required root is $x=1.325$. |
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