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POe to the power x in C
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  1. POe to the power x in C
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    copied!<p>I am trying to find e to the power x. Please indicate what could possibly be wrong with my current implementation rather than simply suggesting new/ore efficient solutions (i could find many on the net). There seems to be a logical or run time error, debugging it doesn't show anything. Thanks in advance!</p> <p>cmath library has been included, program compiles fine..stops on run</p> <pre><code>double find_e_power_x(int x){ int i = 1, j = 1, count = 1, accuracy = 15; double xd = static_cast&lt;double&gt;(x); //pow takes double args double jd = static_cast&lt;double&gt;(j); double epx = 1.0; while ( count &lt; accuracy ){ while ( ++i &amp;&amp; ++j) { epx += ( pow(xd,jd) / fact(i) ); //static_cast count++; } } return epx; } </code></pre> <p>In response to the comments (pointing out my infinite inner loop), </p> <p><strong>EDIT:</strong></p> <pre><code>while ( count &lt; accuracy ){ epx += ( pow(xd,jd) / fact(i) ); //static_cast count++; i++; j++; } </code></pre> <p>input =3 answer is incorrect though it does give an output as 3.15485</p> <hr> <p>below is the <strong>final version</strong> works fine </p> <p>i/p = 4 </p> <p>o/p = 54.8278</p> <pre><code>double find_e_power_x(int x){ int i = 1, count = 1, accuracy = 15; double xd = static_cast&lt;double&gt;(x); //pow takes double args double id = static_cast&lt;double&gt;(i); double epx = 1.0; while ( count &lt; accuracy ){ epx += ( pow(xd,id) / fact(i) ); //static_cast ++count; ++id; ++i; } return epx; } </code></pre>
 

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