Geometric Series - replace pow

[avels]

I have a problem to replace "pow" in my program below (infinite geometric series).If the input a=2 and r=.99999 the program will get the answer so slow. Otherwise the program output is correct.

Could anyone help me to replace the pow with something else.



#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
int n, i=1, x=0, b=i-1;
double a, r, number, suma, sumb=0;

cout << "Please enter a starting number for your infinite geometric series\n";
cin >> a;

cout << "Enter the number between -1 and 1 for the series to increment by\n";
cin >> r;

while (r>1 && r<-1)
{
cout << "The increment must be between -1 and 1";
cout << endl;
cout << "Enter the number between -1 and 1 for the series to increment by\n";
cin >> r;
}

suma = a /(1-r); //compute the sum by the formula

//check the condition for both of sum
while (fabs (suma-sumb)> fabs (suma*0.0000001))
{
number = a*(pow(r,b)); //THIS THE PART THAT I HAVE TO REPLACE
sumb = sumb + number; //compute the sum by add the number
i++;
}

cout << endl;
cout.precision (15);
cout << scientific << "Sum by formula S= a/(1-r) is " << suma;
cout << endl;
cout << scientific << "Sum by add the series numbers is " << sumb;
cout << endl;

return 0;
}

[melkiy]

In the cycle

while(fabs(suma-sumb) > .... )
{
....
}

the number = a*pow(r,b) does not change, so you must calculate it once before the cycle, and only use it inside the cycle.
But i'm not sure of the correctness of the logics...

[avels]

Melkiy,

the problem is in the "pow" I have to change with something else. Example for the area of circle (phi*pow(radius,2)), I have to change to (phi*radius*radius).

The result with pow is too slow for some numbers input.

[firedraco]

Is it just slow? Or is it an infinite loop?

[melkiy]

Avels,
You can quicken your program manyfold by rewriting the main cycle as

number = a*(pow(r,b));
while (fabs (suma-sumb)> fabs (suma*0.0000001)) 
{
   sumb = sumb + number;
   i++;
}

because the pruduction of a*pow(r,b) does not change from iteration to iteration.
But i guess the algorithm is not correct, and to optimize a senseless program is senseless.

[avels]

@firedrace

sorry.....the program at the top have a little bit mistake, I pulled the declare the b at the top. Program at the bottom suppose work correctly.

I should make a program that read 2 input a (the initial of the geometric series) and r (the increment, the value is -1<r<1). Then calculate the sum of the geometric series in 2 different ways, first by formula sum=a/(1-r) and the other is by adding every individual terms until the answer match for 7 digits. For the last sum, two real values have match with this condition |suma-sumb|<|suma*10^-n|. The program below is correct. For an example you can put a = 1 and r =0.1. Then the sum is 1.1111111 (the result for both ways is the same pattern for 7 digits). You can try for different number also. But for some number. example if the a=2 and the r=0.9999, it will take a long time for the computer give the result. The problem is in the pow. I should not use "pow" but something else.

@Melkiy, you still use pow. I have to avoid pow in my program.

Here the program that I should replace the "pow"

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
int n, i=1, x=0;
double a, r, number, suma, sumb=0;

cout << "Please enter a starting number for your infinite geometric series\n";
cin >> a;

cout << "Enter the number between -1 and 1 for the series to increment by\n";
cin >> r;

while (r>1 && r<-1)
{
cout << "The increment must be between -1 and 1";
cout << endl;
cout << "Enter the number between -1 and 1 for the series to increment by\n";
cin >> r;
}

suma = a /(1-r); //compute the sum by the formula

//check the condition for both of sum
while (fabs (suma-sumb)> fabs (a*0.0000001))
{
int b=i-1;
number = a*(pow(r,b));
sumb = sumb + number; //compute the sum by add the number
i++;
}
cout << endl;
cout.precision (15);
cout << scientific << "Sum by formula S= a/(1-r) is " << suma;
cout << endl;
cout << scientific << "Sum by add the series numbers is " << sumb;
cout << endl;

return 0;
}

[melkiy]

It's a different pair of shoes!

Notice that the next member or the series is r times "greater" then the previous. So replace
number = a*(pow(r,b));
with
number = prev_number*r;

Of course you must update number and prev_number velues timely.