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POSimple iteration algorithm
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copied!<p>If we are given with an array of non-linear equation coefficients and some range, how can we find that equation's root within the range given?</p> <p>E.g.: the equation is </p> <p><img src="https://i.stack.imgur.com/kzQ3Y.gif" alt="enter image description here"> </p> <p>So coefficient array will be the array of <em>a</em>'s. Let's say the equation is </p> <p><img src="https://i.stack.imgur.com/wHf1X.gif" alt="enter image description here"> </p> <p>Then the coefficient array is <code>{ 1, -5, -9, 16 }</code>.</p> <p>As Google says, first we need to morph function given (the equation actually) to some other function. E.g. if the given equation is <code>y = f(x)</code>, we should define other function, <code>x = g(x)</code> and then do the algorithm:</p> <pre><code>while (fabs(f(x)) > etha) x = g(x); </code></pre> <p>To find out the root.</p> <p>The question is: how to define that <code>g(x)</code> using coefficient array and the range given only?</p> <p>The problem is: when i define <code>g(x)</code> like this </p> <p><img src="https://i.stack.imgur.com/efmVu.gif" alt="enter image description here"></p> <p>or </p> <p><img src="https://i.stack.imgur.com/tis1c.gif" alt="enter image description here"> </p> <p>for the equation given, any start value for <code>x</code> will lead me to the second equation's root. And no one of 'em would give me the other two (roots are <code>{ -2.5, 1.18, 6.05 }</code> and my code gives <code>1.18</code> only).</p> <p>My code is something like this:</p> <pre><code>float a[] = { 1.f, -5.f, -9.f, 16.f }, etha = 0.001f; float f(float x) { return (a[0] * x * x * x) + (a[1] * x * x) + (a[2] * x) + a[3]; } float phi(float x) { return (a[3] * -1.f) / ((a[0] * x * x) + (a[1] * x) + a[2]); } float iterationMethod(float a, float b) { float x = (a + b) / 2.f; while (fabs(f(x)) > etha) { x = phi(x); } return x; } </code></pre> <p>So, calling the <code>iterationMethod()</code> passing ranges <code>{ -3, 0 }</code>, <code>{ 0, 3 }</code> and <code>{ 3, 10 }</code> will provide <code>1.18</code> number three times along.</p> <p>Where am i wrong and how should i act to get it work right?</p> <p><strong>UPD1</strong>: i do not need any third-party libraries.</p> <p><strong>UPD2</strong>: i need "Simple Iteration" algorithm exactly.</p>

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