C program that computes the value of ex using a formula

[SilviaJamieson]

Help please with program, I seem to have blocks when it comes to anything that has an equation in it!

Write a C program that computes the value of ex by using the following formula.

e^x = 1 + x/(1!) + x^2/(2!) + x^3/(3!)+…x^n/(n!)

– The program should ask user to input value of x and number of terms (n) to approximate value of ex
– Then it calculates and displays the approximated value of the result
– Compare the approximated value of ex with the C library function exp() and find the absolute [use fabs()]difference between the two.

An example of input/output dialog is shown below:-

Enter value of x : 2
Enter number of terms (n) : 18
The approximated value of ex : 7.389056098885
The value of ex using C library function : 7.389056098931
The absolute difference between the two value :
0.000000000046

Many thanks, daisy…..

[GWILouisaxwzkla]

try:

<br /> /****************************************************************<br /> * File Name : c:programstempCG.cpp<br /> * Date : December,15,2008<br /> * Comments : new project<br /> * Compiler/Assembler :<br /> * Modifications :<br /> *<br /> *<br /> *<br /> *<br /> *<br /> * Program Shell Generated At: 3:08:27 p.m.<br /> =-=-=-=-=-=-=-=--=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=*/<br /> <br /> <br /> #include < iostream ><br /> //#include < string.h ><br /> //#include < conio.h ><br /> #include < math.h ><br /> //#include < iomanip ><br /> //#include < ctype.h ><br /> <br /> using namespace std;<br /> <br /> //main function ******************************<br /> <br /> int main ( )<br /> {<br /> <br /> int terms;<br /> double powerOfX;<br /> double e = 1;<br /> double i = 1;<br /> double j ;<br /> double k = 2;<br /> <br /> printf ( "Enter value of x : " );<br /> scanf ( "%lf" , &#038; powerOfX );<br /> printf ( "Enter the number of terms ( greater or equal to one ) : " );<br /> scanf ( "%i" , &#038; terms );<br /> j = powerOfX;<br /> <br /> <br /> <br /> while ( terms > 1 )<br /> {<br /> e = e + ( j * ( 1 / i ) );<br /> terms --;<br /> i = i * k;<br /> j = j * powerOfX;<br /> k ++;<br /> <br /> }<br /> printf ( "The approximated value of ex : %.12lf n" , e );<br /> printf ( "The value of ex using C library function %.12lf: n" ,exp( powerOfX ) );<br /> printf ( "The absolute difference between the two value : " );<br /> printf ( "%.12lf n" , fabs ( exp ( powerOfX ) - e ) );<br /> return 0 ;<br /> }<br />

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<br /> /****************************************************************<br /> * File Name : c:programstempCG.cpp<br /> * Date : December,15,2008<br /> * Comments : new project<br /> * Compiler/Assembler :<br /> * Modifications :<br /> *<br /> *<br /> *<br /> *<br /> *<br /> * Program Shell Generated At: 3:08:27 p.m.<br /> =-=-=-=-=-=-=-=--=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=*/<br /> <br /> <br /> #include < iostream ><br /> //#include < string.h ><br /> //#include < conio.h ><br /> #include < math.h ><br /> //#include < iomanip ><br /> //#include < ctype.h ><br /> <br /> using namespace std;<br /> <br /> //main function ******************************<br /> <br /> int main ( )<br /> {<br /> <br /> int terms;<br /> double powerOfX;<br /> double e = 1;<br /> double i = 1;<br /> double j ;<br /> double k = 2;<br /> <br /> printf ( "Enter value of x : " );<br /> scanf ( "%lf" , &#038; powerOfX );<br /> printf ( "Enter the number of terms ( greater or equal to one  ) : " );<br /> scanf ( "%i" , &#038; terms );<br /> j = powerOfX;<br /> <br /> <br /> <br /> while ( terms > 1 )<br /> {<br /> e = e +  ( j * ( 1 / i ) );<br /> terms --;<br /> i = i * k;<br /> j = j * powerOfX;<br /> k ++;<br /> <br /> }<br /> printf ( "The approximated value of ex : %.12lf n" , e );<br /> printf ( "The value of ex using C library function %.12lf: n" ,exp( powerOfX ) );<br /> printf ( "The absolute difference between the two value : " );<br /> printf ( "%.12lf n" , fabs ( exp ( powerOfX ) - e ) );<br /> return 0 ;<br /> }<br />

I didn’t know how many digit to output so I chose 12 like the example. You can change this by changing the number listed after the ‘.’ in the last printf() statements.

Theres an error in the code I posted for the approximation of e = 1 + 1/1! + 1/2! . Heres the proper code:

<br /> <br /> /****************************************************************<br /> * File Name : c:programstempCG.cpp<br /> * Date : December,15,2008<br /> * Comments : new project<br /> * Compiler/Assembler :<br /> * Modifications :<br /> *<br /> *<br /> *<br /> *<br /> *<br /> * Program Shell Generated At: 3:08:27 p.m.<br /> =-=-=-=-=-=-=-=--=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=*/<br /> <br /> <br /> #include < iostream ><br /> //#include < string.h ><br /> //#include < conio.h ><br /> #include < math.h ><br /> //#include < iomanip ><br /> //#include < ctype.h ><br /> <br /> using namespace std;<br /> <br /> //main function ******************************<br /> <br /> int main ( )<br /> {<br /> <br /> int terms;<br /> double powerOfX;<br /> double e = 1;<br /> double i = 1;<br /> double j ;<br /> double k = 2;<br /> <br /> <br /> printf ( "Enter the number of terms ( greater or equal to one ) : " );<br /> scanf ( "%i" , &#038; terms );<br /> <br /> while ( terms > 1 )<br /> {<br /> e = e + ( 1 / i ) ;<br /> terms --;<br /> i = i * k;<br /> k ++;<br /> <br /> }<br /> printf ( "The approximated value of e : %.12lf n" , e );<br /> printf ( "The given mathematical constant e = 2.718281828459 n");<br /> printf ( "The absolute difference between the two value : " );<br /> printf ( "%.12lf n" , fabs ( 2.718281828459 - e ) );<br /> <br /> return 0 ;<br /> }<br />

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<br /> <br /> /****************************************************************<br /> * File Name : c:programstempCG.cpp<br /> * Date : December,15,2008<br /> * Comments : new project<br /> * Compiler/Assembler :<br /> * Modifications :<br /> *<br /> *<br /> *<br /> *<br /> *<br /> * Program Shell Generated At: 3:08:27 p.m.<br /> =-=-=-=-=-=-=-=--=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=*/<br /> <br /> <br /> #include < iostream ><br /> //#include < string.h ><br /> //#include < conio.h ><br /> #include < math.h ><br /> //#include < iomanip ><br /> //#include < ctype.h ><br /> <br /> using namespace std;<br /> <br /> //main function ******************************<br /> <br /> int main ( )<br /> {<br /> <br /> int terms;<br /> double powerOfX;<br /> double e = 1;<br /> double i = 1;<br /> double j ;<br /> double k = 2;<br /> <br /> <br /> printf ( "Enter the number of terms ( greater or equal to one  ) : " );<br /> scanf ( "%i" , &#038; terms );<br /> <br /> while ( terms > 1 )<br /> {<br /> e = e +  ( 1 / i ) ;<br /> terms --;<br /> i = i * k;<br /> k ++;<br /> <br /> }<br /> printf ( "The approximated value of e : %.12lf n" , e );<br /> printf ( "The given mathematical constant e = 2.718281828459 n");<br /> printf ( "The absolute difference between the two value : " );<br /> printf ( "%.12lf n" , fabs (  2.718281828459 - e ) );<br /> <br /> return 0 ;<br /> }<br />

[SilviaJamieson]

Thanks dman, and for the correction to the other code. I am trying to learn computer programming at home and you are making it so much easier to understand, for that i am very grateful.

daisy….

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