Tolong dibantu no.4 saja, jangan asal pliss

Berikut ini adalah pertanyaan dari sigup pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\displaystyle \begin{array}{|c|c|c|c|c|}\hline x&1&2&3&4\\ \hline f(x)&1&1.414&1.732&2\\ \hline \end{array}$

$f(2.25) =y_{\text{eksak}}= 1.5$

Rumus Interpolasi Lagrange:

$\displaystyle \boxed{f_n(x) = \sum_{i=0}^n L_i(x)f(x_i),\quad L_i(x)=\prod_{\begin{array}{c} j=0\\j\neq i \end{array}}^n \frac{x-x_j}{x_i-x_j}}$

4.1 Interpolasi Lagrange orde ke-1

$\displaystyle \begin{array}{|c|c|c|}\hline x_i&x_0=2&x_1=3\\ \hline f(x_i)&1.414&1.732\\ \hline\end{array}$

$\displaystyle \begin{aligned} L_0(x) &= \frac{x-x_1}{x_0-x_1}\\ L_0(2.25) &= \frac{2.25-3}{2-3} = 0.75 \\ L_1(x) &= \frac{x-x_0}{x_1-x_0}\\ L_1(2.25) &= \frac{2.25-2}{3-2} = 0.25 \end{aligned}$

$\displaystyle f_1(2.25) = 1.414\cdot 0.75+1.732\cdot 0.25 = 1.4935$

$\displaystyle \varepsilon_r = \left| \frac{1.5-1.4935}{1.5} \right| \times 100\% \approx \boxed{\bold{\underline{\underline{0.4333\% }}}}$

4.2 Interpolasi Lagrange orde ke-2

$\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&x_0=1&x_1=2&x_2=3 \\ \hline f(x_i)&1&1.414&1.732\\ \hline\end{array}$

$\displaystyle \begin{aligned} L_0(x) &= \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\\ L_0(2.25) &= \frac{(2.25-2)(2.25-3)}{(1-2)(1-3)} = -\frac{3}{32} \\ L_1(x) &= \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\\ L_1(2.25) &= \frac{(2.25-1)(2.25-3)}{(2-1)(2-3)} = \frac{15}{16} \\ L_2(x) &= \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\\ L_2(2.25) &= \frac{(2.25-1)(2.25-2)}{(3-1)(3-2)} = \frac{5}{32}\end{aligned}$

$\displaystyle f_2(2.25) = 1\cdot\left(-\frac{3}{32}\right)+1.414\cdot \frac{15}{16}+1.732\cdot \frac{5}{32} = 1.5025$

$\displaystyle \varepsilon_r = \left| \frac{1.5-1.5025}{1.5} \right| \times 100\% \approx \boxed{\bold{\underline{\underline{0.1667\% }}}}$

4.3 Interpolasi Lagrange orde ke-3

$\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&x_0=1&x_1=2&x_2=3&x_3=4 \\ \hline f(x_i)&1&1.414&1.732&2\\ \hline\end{array}$

$\displaystyle \begin{aligned} L_0(x) &= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\\ L_0(2.25) &= \frac{(2.25-2)(2.25-3)(2.25-4)}{(1-2)(1-3)(1-4)} = -\frac{7}{128} \\ L_1(x) &= \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\\ L_1(2.25) &= \frac{(2.25-1)(2.25-3)(2.25-4)}{(2-1)(2-3)(2-4)} = \frac{105}{128} \\ L_2(x) &= \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\\ L_2(2.25) &= \frac{(2.25-1)(2.25-2)(2.25-4)}{(3-1)(3-2)(3-4)} = \frac{35}{128} \\ L_3(x) &= \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\\ L_3(2.25) &= \frac{(2.25-1)(2.25-2)(2.25-3)}{(4-1)(4-2)(4-3)} = -\frac{5}{128}\end{aligned}$

$\displaystyle f_3(2.25) = 1\cdot\left(-\frac{7}{128}\right)+1.414\cdot \frac{105}{128}+1.732\cdot \frac{35}{128}+2\cdot\left(-\frac{5}{128}\right) \approx 1.5007$