tux4life
Nearly a Posting Maven
2,350 posts since Feb 2009
Reputation Points: 2,134
Solved Threads: 243
Have something to say? Contribute New Article Reply to this Article
Help with series
Hello,I need help with the following series (sorry I forgot what its called..)
1/1! - 2/2! + 3/3! -4/4!....upto n terms
Here's my program :
#include<conio.h>
#include<stdio.h>
#include<float.h>
void main()
{
float com=0,j=0,j1=0,ans=0; //variable decl of float type
int tillw,i,no=0,fac1=1,fac=1,f1,f2; //variable decl of int type
clrscr();
printf("\n\n\t\t *********************************");
printf("\n\t\t ---------------------------------");
printf("\n\n\t\t SERIES");
printf("\n\n\t\t ---------------------------------");
printf("\n\t\t *********************************");
printf("\n\n\n\n\t\t Enter till where you want : ");
scanf("%d",&tillw);
flushall(); //accept no. from user
for(i=1;i<=tillw;i++)
{
no=no+1;
ans=no%2;
if(ans!=0)
{
//addition part of series
for(f1=1;f1<=no;f1++)
{
fac=fac*f1;
} //find factorial
j=no/fac;
com=com+j;
}
else
{
//subtraction part of series
for(f2=1;f2<=no;f2++)
{
fac1=fac1*f2;
} //find factorial
j1=no/fac1;
com=com-j1;
}
//printf("\n\n\t\t The answer is : %0.2f",com);(for testing answer in each step)
} //main for-loop ends..
printf("\n\n\t\t The answer is : %f",com); //prints final answer
getch();
} //void-main ending..So now my problem is,whatever input I give the answer is always 0.00
PS:Realized that it gives correct answer till 2,but after that whatever the input it gives 0.00 as the answer
See: you have fac > no for all no values because fac = n!. Therefore no/fac == 0 (integer division) and com=com+j is the same as com += 0...
1. Use double type for math calculations (not float).
2. Use floating-point division if you want to get floating-point result, for example:
com += (double)n/fac;3. You will get (unsignalled) integer overflow in factorial calculation: 13! is greater than max integer value.
Better search DaniWeb forseries: there are lots of threads on this topic. It's a bad approach to calculate factorial for every series member...
I changed void main to int main, also using double for calculations..
Still the same error :(
ArkM,what exactly you mean by saying tht it is bad to calculate factorial for each no of series.I did not understand
And also can you plz explain
See: you have fac > no for all no values because fac = n!. Therefore no/fac == 0 (integer division) and com=com+j is the same as com += 0...
I could not understand understand..sorry if i am becoming a headache..
soryy for the double post,how to delete my post btw?
I changed void main to int main, also using double for calculations..
Still the same error :(
ArkM,what exactly you mean by saying that it is bad to calculate factorial for each no. of series.I did not understand
And also can you plz explain ...
See: you have fac > no for all no values because fac = n!. Therefore no/fac == 0 (integer division) and com=com+j is the same as com += 0...
I could not understand...sorry if i am becoming too much of a nuisance .
you dont need to recalculate each and every factorial. because if you've already calculated 9! (for example), then 10! is really just 9! * 10.
and basically it's this: you can't do integer division. because all your quotients will be integer zeros.
3/6 = 0. 4/24 = 0. 5/120 = 0.
you have to use the "double" data type.
and as he already mentioned, it will crap out at 13!, because the maximum unsigned integer 2^32 = ~4 billion.
look into "long long int" data types. and "big number" libraries if you really need to get deep.
power series are not a trival matter in C.
#include<conio.h>
#include<stdio.h>
#include<float.h>
int main()
{
double com=0,j=0,j1=0,ans=0;
int no=0;
double tillw,i,fac1=1,fac=1,f1,f2;
clrscr();
printf("\n\n\t\t *********************************");
printf("\n\t\t ---------------------------------");
printf("\n\n\t\t SERIES");
printf("\n\n\t\t ---------------------------------");
printf("\n\t\t *********************************");
printf("\n\n\n\n\t\t Enter till where you want : ");
scanf("%d",&tillw);
flushall(); //accept no. from user
for(i=1;i<=tillw;i++)
{
no=no+1;
ans=no%2;
if(ans!=0)
{
//addition part of series
fac=fac*i;
j=(double)no/fac;
com=(double)com+j;
}
else
{
//subtraction part of series
fac=fac*i;
j1=(double)no/fac1;
com=(double)com-j1;
}
//printf("\n\n\t\t The answer is : %0.2f",com);(for testing answer in each step)
} //main for-loop ends..
printf("\n\n\t\t The answer is : %f",com); //prints final answer
getch();
return (0);
} //void-main ending..
I removed the for loops for calculating the factorials,used double and removed void main.STill the same problem :(
Oh i dont need to go to such high numbers.I will be happy if it can calculate upto 5:)
The series breaks down to::
1 - 2 / (2 *1) +3 / (3*2!) -....
i.e 1 - 1 + 1/2! - 1/3!
So the first (major) 2 terms of the series cancel out.
Now, may be you define a variable
int sign_carrier = 1 ;
long fact = 1;
double sum = 0;
for( i = 2;i<=tillw;i++) /*start from 2*/
/*start from 3rd term in the series*/
{
if (i is even) sign_carrier =1;
else sign_carrier =-1;
now get the factorial with the help of i;
sum = sum + (sign_carrier / fact);
}Well, let f(i) is i-th member without sign (it's not C).
f(i) = i/i! = i/(1*2*...*(i-1)*i) =
i/(i*(1*2...*(i-1)) = 1/(i-1)!
f(i+1) = 1/i! = 1/((i-1)!*i) = f(i)/i
i = 1, 2,... (0! = 1 by definition)In other words, if we know i-th member of series then the next member is equal to this member / i (ignore sign).
if we know that i-th member is positive then the next member is negative and vice versa.
We know that 2-nd member is equal to -1.0 (negative). We know that the partial sum of the series is equal to 1.0 at this moment: f(1) == 1/1!...
Now we must define calculation loop condition. Obviously no sense to continue calculations when the next member of series is too small: double type precision is ~16 decimal digits. We want to stop calculations when f(i) < eps, where eps == 1e-16 (for example).
Yet another tip: no need in int variables at all!
double Series()
{
const double eps = 1e-16;
double sum = 1.0; /* sum(1) */
double f, i;
int neg = 1;/* the 2nd member < 0 */
for (f = 1.0, i = 2.0; f > eps; f /= i, i += 1.0) {
/* cout << i << '\t' << f << '\t'; */
sum += (neg? -f: f);
/* cout << setprecision(15) << sum << '\t'
<< (sum?1/sum:0) << '\n'; */
neg ^= 1;
}
/* cout << sum << endl; */
return sum;
}The answer: 0.367879441171442
It's equal to 1/e constant. Exactly.
#include<conio.h>
#include<stdio.h>
int main()
{
int sign_carrier,i,tillw,ans;
long int fac=1;
float sum=0;
clrscr();
printf("\n\n\t\t *********************************");
printf("\n\t\t ---------------------------------");
printf("\n\n\t\t SERIES");
printf("\n\n\t\t ---------------------------------");
printf("\n\t\t *********************************");
printf("\n\n\n\n\t\t Enter till where you want : ");
scanf("%d",&tillw);
flushall(); //accept no. from user
for(i=1;i<=tillw;i++)
{
ans=i%2; //to find odd or even..
if(ans!=0) //odd
{
sign_carrier=1;
}
else //even
{
sign_carrier=-1;
}
fac=fac*i;
sum+=(double)(i*sign_carrier)/fac;
} //main for-loop ends..
printf("\n\n\t\t Answer is : %0.20f",sum);
getch();
return sum;
} //void-main ending..
This works :)
Thank you all for your time.really appreciate it.. :)
I have started it from 1 instead of 2 which you guys had suggested
This question has already been solved
You
© 2012 DaniWeb® LLC
