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		<title>Metode Numerik : SOLUSI PERSAMAAN NIRLANJAR (II)</title>
		<link>https://ramzilhuda.com/metode-numerik-solusi-persamaan-nirlanjar-ii/</link>
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		<dc:creator><![CDATA[ramzilhuda]]></dc:creator>
		<pubDate>Thu, 05 Nov 2020 00:16:05 +0000</pubDate>
				<category><![CDATA[Metode Numerik]]></category>
		<category><![CDATA[konvergen]]></category>
		<category><![CDATA[metode]]></category>
		<category><![CDATA[Metode Fixed-Point]]></category>
		<category><![CDATA[Metode Newton-Raphson]]></category>
		<category><![CDATA[Metode Secant]]></category>
		<category><![CDATA[metode terbuka]]></category>
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					<description><![CDATA[SOLUSI PERSAMAAN NIRLANJAR (II) Metode Fixed-Point Metode Newton-Raphson Metode Secant  METODE TERBUKA Metode Lelaran Titik Tetap (fixed-point iteration) &#8211; Susunlah persamaan f(x) = 0 menjadi bentuk x = g(x) &#8211; Bentuk prosedur lelaran  xr+1 = g(xr) &#8211; Tentukan nilai awal x0  lalu hitung nilai x1 , x2 , dst yang mudah2an  konvergen ke akar sejati [&#8230;]]]></description>
										<content:encoded><![CDATA[<h1><b>SOLUSI PERSAMAAN NIRLANJAR (II)</b></h1>
<ul>
<li><b>Metode Fixed-Point</b></li>
<li><b>Metode Newton-Raphson</b></li>
<li><b>Metode Secant</b></li>
</ul>
<h2><b> METODE TERBUKA</b></h2>
<p><b>Metode Lelaran Titik Tetap (</b><b><i>fixed-point</i></b> <b><i>iteration</i></b><b>)</b></p>
<p>&#8211; Susunlah persamaan <i>f(x) </i>= 0 menjadi bentuk <i>x</i> = <i>g(x)</i></p>
<p>&#8211; Bentuk prosedur lelaran  <i>x</i><i>r+1</i> = <i>g(x</i><i>r</i><i>)</i></p>
<p>&#8211; Tentukan nilai awal <i>x</i><i>0 </i><i> </i>lalu hitung nilai <i>x</i><i>1</i><i> , x</i><i>2</i><i> , </i>dst yang mudah2an  <b>konvergen</b> ke akar sejati</p>
<p>&#8211; Kondisi berhenti dinyatakan bila : | <i>x</i><i>r+1</i><i> – x</i><i>r</i> | &lt; ε</p>
<h2><b>Kriteria Konvergensi </b></h2>
<p><b>Teorema </b>: Misalkan <i>g(x)</i> dan  <i>g'(x)</i> menerus di dalam selang [<i>a,b</i>] = [ <i>s-h, s+h</i>] yang mengandung titik tetap  <i>s </i>dan nilai awal  <i>x</i><i>0</i> dipilih dalam selang tersebut.</p>
<p>Jika  |<i>g'(x)</i>|&lt; 1 untuk semua   <i>x</i>  Î [<i>a,  b</i>] maka lelaran  <i>x</i><i>r+1</i> =  <i>g(x</i><i>r</i><i>)</i> akan konvergen ke <i>s</i>.  Pada kasus ini <i>s</i> disebut juga  titik atraktif .</p>
<p>Jika  |<i>g'(x)</i>| &gt; 1 untuk semua <i>x  </i>Î [<i>a, b</i>] maka lelaran <i>x</i><i>r+1</i> =  <i>g(x</i><i>r</i><i>)</i> akan divergen dari  <i>s</i>.</p>
<p>Berdasarkan teorema tersebut, dapat disimpulkan :</p>
<p>Didalam selang <i>I</i> =[ <i>s-h, s+h </i>] dengan <i>s </i> titik tetap ,</p>
<p>-Jika 0 &lt; <i>g’(x) </i>&lt; 1 untuk setiap <i>x ∈</i> <i>I </i>, maka lelarannya <b>konvergen monoton</b></p>
<p>-Jika -1 &lt; <i>g’(x) </i>&lt; 0 untuk setiap <i>x ∈</i> <i>I </i>, maka lelarannya <b>konvergen berosilasi</b></p>
<p>-Jika  <i>g’(x) </i>&gt; 1 untuk setiap <i>x ∈</i> <i>I </i>, maka lelarannya <b>divergen monoton</b></p>
<p>-Jika  <i>g’(x) </i>&lt; -1 untuk setiap <i>x ∈</i> <i>I </i>, maka lelarannya <b>divergen berosilasi</b></p>
<p>&nbsp;</p>
<h2><strong>Contoh  1 :</strong></h2>
<p>Carilah akar persamaan <i>f(x) </i>= <i>x</i><i>2</i> – 2<i>x</i> – 3 = 0,  gunakan  ε = 0.000001</p>
<p>Penyelesaian :</p>
<p>Terdapat beberapa kemungkinan prosedur     lelaran yang dapat dibentuk</p>
<ol>
<li><i>x</i><i>2</i> – 2<i>x</i> – 3 = 0 -&gt; <i>x</i><i>2</i> = 2<i>x </i>– 3</li>
</ol>
<p><i>x</i> = √(2<i>x</i> + 3) -&gt; <i>g(x)</i></p>
<p>Jika terkaan awal <i>x</i><i>0 </i>= 4</p>
<p>Untuk <i>g(x) </i>= √(2<i>x</i> + 3)</p>
<p><img data-recalc-dims="1" decoding="async" class="aligncenter size-full wp-image-330" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-1.jpg?resize=252%2C194&#038;ssl=1" alt="" width="252" height="194" /></p>
<p>Grafik fungsi <i>f(x) </i>= <i>x</i><i>2</i> – 2<i>x</i> – 3, pada selang [-5, 5]</p>
<p><img data-recalc-dims="1" fetchpriority="high" decoding="async" class="aligncenter size-full wp-image-331" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-2.jpg?resize=400%2C247&#038;ssl=1" alt="" width="400" height="247" srcset="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-2.jpg?w=400&amp;ssl=1 400w, https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-2.jpg?resize=300%2C185&amp;ssl=1 300w" sizes="(max-width: 400px) 100vw, 400px" /></p>
<p><img data-recalc-dims="1" decoding="async" class="aligncenter size-full wp-image-332" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-3.jpg?resize=648%2C469&#038;ssl=1" alt="" width="648" height="469" srcset="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-3.jpg?w=648&amp;ssl=1 648w, https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-3.jpg?resize=300%2C217&amp;ssl=1 300w" sizes="(max-width: 648px) 100vw, 648px" /></p>
<p><img data-recalc-dims="1" loading="lazy" decoding="async" class="aligncenter size-full wp-image-333" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-4.jpg?resize=650%2C496&#038;ssl=1" alt="" width="650" height="496" srcset="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-4.jpg?w=650&amp;ssl=1 650w, https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-4.jpg?resize=300%2C229&amp;ssl=1 300w" sizes="auto, (max-width: 650px) 100vw, 650px" /></p>
<p><img data-recalc-dims="1" loading="lazy" decoding="async" class="aligncenter size-full wp-image-334" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-5.jpg?resize=628%2C442&#038;ssl=1" alt="" width="628" height="442" srcset="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-5.jpg?w=628&amp;ssl=1 628w, https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-5.jpg?resize=300%2C211&amp;ssl=1 300w" sizes="auto, (max-width: 628px) 100vw, 628px" /></p>
<p>Grafik fungsi <i>f(x) </i>= <i>x</i><i>2</i> – 2<i>x</i> – 3 pada selang [-2, 5]</p>
<p><img data-recalc-dims="1" loading="lazy" decoding="async" class="aligncenter size-full wp-image-335" src="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-6.jpg?resize=614%2C375&#038;ssl=1" alt="" width="614" height="375" srcset="https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-6.jpg?w=614&amp;ssl=1 614w, https://i0.wp.com/ramzilhuda.com/wp-content/uploads/2020/11/Metode-Numerik-Fixed-Point-6.jpg?resize=300%2C183&amp;ssl=1 300w" sizes="auto, (max-width: 614px) 100vw, 614px" /></p>
<p><b>Latihan :</b></p>
<p>Temukan akar persamaan dari <i>f</i>(<i>x</i>) = 2<i>x</i>2 + 10<i>x</i> -20, dengan menggunakan metode Lelaran Titik Tetap (<i>Fixed Point</i>) dengan <i>x</i><i>0</i> = 1 dan ε = 0.00001</p>
<p>Bagi teman &#8211; teman yang sudah mempelajari solusi persamaan nirlanjar II dapat melanjutkan dengan <a href="https://ramzilhuda.com/metode-numerik-metode-eliminasi-gauss-pivoting-penskalaan/">Metode Numerik : Metode Eliminasi Gauss Pivoting Penskalaan.</a></p>
<p>Silahkan bagi teman &#8211; teman yang belum memahami tentang solusi persamaan nirlanjar II dapat juga menonton video berikut ini <a href="https://www.youtube.com/watch?v=3dbNxj5g6kc">Solusi persamaan nirlanjar</a></p>
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