The value ex can be approximated by the sum
1 + x + x2/2! + x3/3! + … + xn/n!
Write a program that takes a value x as input and outputs this sum for n taken
to be each of the values 1 to 100. The program should also output ex
calculated using the predefined function exp. The function exp is a predefined
function such that exp(x) returns an approximation to the value ex. The
function exp is in the library with the header file cmath. Your program should
repeat the calculation for new values of x until the user says she or he is
through.

This post has been edited by madawa: 13 Mar, 2007 - 09:51 PM

Then where is the problem?
you need to write a method to calculate power of x.
and need to write a method to calulate e^x.

just try to code the steps which we follow while calculating it manually.
and still if there is any problem the put the code here.
we will try to solve it.

i tried i dont get it am trying for da 3rd day nw

Please post the code you have written, and we'd be happy to provide some guidance.

The nice thing about this problem is there is a simple constant time relationship between each term, therefore, you don't need nor want a power function or a factorial function. Also this will eliminate problems caused by the fact the factorial function and the pow function are difficult to calculate for large values.

T0 = 1
T1 = x
T(n + 1) = Tn * x / n

and exp(x) = T0 + T1 + ... + Tn + ...

This post has been edited by msg555: 14 Mar, 2007 - 06:01 AM

And take care not to exceed a certain value for finding factorials.
Finding factorials of numbers beyond the range of int of long data types, results in garbage values which may mess up the final result.

So don't keep the loop exit condition too large.

Why don't you build an interface that calls a function such as:

answer = calculate( x );

at least.

The function calculate(x)
could just be:

long calculate(int x) {
return 10;
}

for now.

Then you can add the math latter.
The last line describes the interface.

I would also do the math out on paper to see if there are any paterns that can be found.

#include<iostream>
#include<cmath>

using namespace std;

double fact( int x)
{
if ( x == 1 )
return 1;
else
return x * fact( x - 1) ;
}

int main()

{
char ans;
do{
double sum =0;
system("color 2b");
double Entry;
cout<<endl;
system("color 1b");
cout<<"please enter two values x :"<<endl;
cout<<endl;
cin>>Entry;
sum = sum + 1 + Entry ;
double facto;

for ( int i = 2 ; i < 101; i++)
{
facto = fact(i);
system("color 2d");
sum += pow(Entry,i) / facto;
cout<<"\tSUM="<<sum;
system("color 2a");
if( i % 3 == 0)
cout<<"\n"<<endl;
system("color 1d");

}

cout<<endl;
cout<<endl;
system("color 2a");
cout<<endl;
cout<<"Do you want to continue?(y/n)" ;
cout<<endl;
system("color 1a");
cout<<endl;
cin>>ans;
cout<<endl;
}while(ans == 'y');
system("color 1e");

system("pause");
system("pause");
system("pause");
cout<<endl;





system("color 2b");
system("color 2c");
return 0;

}

IS THIS CORRECT ???????

assuming that x^2 = pow(x,2) then is this the summ you are trying to computer?

F(x,n) = 1 + x + (x^2)/2 + (x^3)/3! + (x^4)/4! + ... + (x^n)/n!

or

G(x,n) = 1 + x + (x*2)/2 + (x*3)3! + (x*4)/4! + ... + (x*n)/n!

The original post makes it look like you are trying for the second, but your code looks like the first.

You ask for two values, but only read in 1 (entry) that causes the "DO you wish to continue..." part to be passed up.

other than that it seems to preform F(x, n) as defined above. You could improve the program with a little math and or precalculation. For example if you were to allocate an array large enough to hold all n numbers, then you can calculate out n! while saving all of the intermediate values in the table.


The above code calculates a table containing all of the factorial values between n=1 and n=100. It does so all at once so you don't waste so much time calculateing the save values over and over. (5!=5*4!, 4!=4*3! so if you have a loop from 1 to 5, you calcuate 1! 5 time, 2! 4 times, 3! 3 times, 4! twice, and 5! once... the above program calculates all factorials once).

There is a neat little algorithm for calculating polynomials

say p(n) = a(0) + a(1)* x +a(2)*x^2 +a(3)*x^3 +... + a(n)*x^n

for the computer to calculate x^n it takes n multiplications, then x^(n-1) is n-1 multiplications... so this formula has n + (n-1) + (n-2)... +1 multiplications to preform.

We can speed things along by re-writing this as:

p(n) = a(0) + x*(a(1) + x*(a(2) + x*(a(3) + ... + x*(a(n)) ... )))

This preforms the same calculation in only n multiplications. That is significantly faster!

If you combine the two ideas you can make the calculation much faster... of course you need to really think about what they are doing, else when your teacher says, "oh very good! now why did you do it this way?" you will not look bad when you explain.

tks for the support ,, i need to see the following code matches for this question .. and any way i can improve its accuricy


An approximate value of pi can be calculated using the series given below:
Pi = 4 [ 1 – 1/3 + 1/5 – 1/7 + 1/9 … + ((-1)n)/(2n+1)]
Write a C++ program to calculate the approximate value of pi using this
series. The program takes an input n that determines the number of terms in
the approximation of the value of pi and outputs the approximation. Includes a
loop that allows the user to repeat this calculation for new values n until the
user says she or he wants to end the program.



#include<iostream>
#include<cmath>

using namespace std;

int main()
{
double firstFactor;
double pi ;
double determinated =-1;
int numberOfInputs;
system("color 7f");
cout<<endl;
system("color 7a");
cout<<endl;
system("color 7c");
cout<<"please input a number "<<endl;
system("color 2f");
cout<<endl;
system("color 7b");
cout<<endl;
system("color 5a");
cin>>numberOfInputs;
for(int i =0; i<= numberOfInputs; i++)
{

firstFactor += pow(determinated,i) / (2*i +1);
}

pi= 4 * firstFactor;
cout<<endl;
cout<<pi;
cout<<endl;

cout<<endl;
system("pause");
system("color 2a");
system("color 2b");
system("color 2e");
cout<<endl;
system("pause");

cout<<endl;
return 0;

not sure how you can improve accuracy on the one, I know that there is no reason to use the pow() function which will increase the speed.

basicly you need to alternate between two values +1 and -1, if i is even, then (-1)^i is +1 else it is -1 so using (i % 2) ? -1 : 1 would be faster than pow(). If the conditional operator is too confusing you can also do it with:Which would basicly be the same thing as the conditional.

I just thought I throw this out there. This is the way I would calculate exp(x).