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POCalculate multiplicative inverse of positive real number
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  1. POCalculate multiplicative inverse of positive real number
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    copied!<p>I wanted to calculate the multiplicative inverse of a positive real number without using division. After reading through various pages on Newton Raphson method, i implemented the following code for calculating the inverse:</p> <pre><code>#define PRECISION 0.1e-5 double inverse(double num) { double ans = num; while (ans * num &gt; 1 + PRECISION || ans * num &lt; 1 - PRECISION) { ans = ans * (2 - num * ans); } return ans; } </code></pre> <p>Now, the problem is, I actually do not get the inverse value. The loop goes on infinitely.</p> <p>So, <strong>1st</strong> thing I noticed was: The value of <code>ans</code> become negative, I added a statement <br><code>if (ans &lt; 0) ans *= -1;</code> <br>so that <code>ans</code> remains positive always.</p> <p><strong>2nd</strong> point to be noticed: If my initialization for <code>ans = 0.5</code> then I get the correct answer for a few values of <code>num</code> = <code>(1, 2, 3, 5)</code>.</p> <p>So, my assumption is, that the implementation isn't working because of inappropriate initialization of the variable <code>ans</code>.</p> <p>So, finally my questions: <br><strong>1</strong>.Can this method actually be used to calculate inverse of all positive real numbers? <br><strong>2</strong>.If yes, then are there any conditions required for the initial value assumed when using Newton Raphson method? <br><strong>3</strong>.Is there any other better method to calculate reciprocal without using division?</p> <h2><strong>EDIT:</strong></h2> <p>Thanks for the answer. So, as mentioned in the answer, I changed my initial value of <code>ans</code> to <code>PRECISION</code> and it works! Also, now as the initial value is good for particular max limit on <code>num</code>, <code>ans</code> never becomes negative, so no need for the negative checking condition that I had added initially.</p> <p>So, this is the working code (atleast works for the input I have given.)</p> <pre><code>double inverse(double num) { // Added to make it valid for all inputs. // For a too large number under the precision constraint, the answer is 0. if (num &gt; 999999) return 0; double ans = PRECISION; while (ans * num &gt; 1 + PRECISION || ans * num &lt; 1 - PRECISION) { ans = ans * (2 - num * ans); } return ans; } </code></pre>
 

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