Answered # solving a series

Rashakil Fol 978 pspwxp fan 1 Freaky_Chris 299 Rashakil Fol 978 Discussion Starter trinity_neo Lerner 582 mvmalderen 2,072 Discussion Starter trinity_neo siddhant3s 1,429 Write a C program that should create a 10 element array of random integers (0 to 9). The program should total all of the numbers in the odd positions of the array and compare them with the total of the numbers in the even positions of the array and indicate ...

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Well it's almost like the series for exp(x), except the signs are alternating and it's missing the x^1 term and the x^0 term has the wrong sign. If consider the series you get for exp(-x), you should be close to the answer :)

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I'm a newbie here but as much as i know, shouldnt this thread not be in C++? Its math, for heavens sake. Also, its wierd that someone has come asking for help with random math equations. Like a binomial theorem that goes on forever :P.

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Perhaps his real question is how can he compute the value of that series to a given figure amount using C++?

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In that case, the right solution would still be to use the closed form expression :)

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I missed a term in series.Really sorry. the actual series is

E(x)=1-x/1!+x^2/2!-x^3/3!+........................x^n/n!

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This should get you pretty close. Each term has the form (-1^n)*((x^n)/n!) where 0! is defined as 1 and x^0 is also defined as 1. E(x) is the sum of the number of terms you want to use. You calculate the sum using a running total within a loop adding each term as it is calculated. Calculate each term by changing the form to: a *(b/c) and then calculate a, b and c before calculating the term. a, b, and c can each be calculated using loops. As an alternative a and b could be calculated using pow(). As a further alternative a could be calculated using an if statement and n % 2.

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I wrote two simple C++-functions:

-> One wich raises a number x to the power of y ( `apow(x, y);`

)

-> One which calculates the faculty of number x ( `faculty(x);`

)

```
long long faculty(long x)
{
long long y = 1;
y = 1;
for(int i = 1; i < (x+1); i++)
y = y * i;
return y;
}
long double apow(float x, int y)
{
long double result = 1;
if(y == 0)
return result;
if(y < 0)
{
y = -y;
for(int i = 0; i < y; i++)
result = result * x;
return 1/result;
}
for(int i = 0; i < y; i++)
result = result * x;
return result;
}
```

And here's a full example where you can see these functions at work:

```
#include <iostream>
long long faculty(long x);
long double apow(float x, int y);
using namespace std;
int main(void)
{
long double answer;
int x = 1;
cout << "6! = " << faculty(6) << endl;
cout << "10^2 = " << apow(10, 2) << endl;
cout << "10^-2 = " << apow(10, -2) << endl;
cout << "10^0 = " << apow(10, 0) << endl;
return 0;
}
long long faculty(long x)
{
long long y = 1;
y = 1;
for(int i = 1; i < (x+1); i++)
y = y * i;
return y;
}
long double apow(float x, int y)
{
long double result = 1;
if(y == 0)
return result;
if(y < 0)
{
y = -y;
for(int i = 0; i < y; i++)
result = result * x;
return 1/result;
}
for(int i = 0; i < y; i++)
result = result * x;
return result;
}
```

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You want to calculate the n-term series and not the infinite series?

YES

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First convert your equation to a closed form. The close form clearly is [tex]\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}[/tex]

Now construct a loop and evaluate it to desired **n**

You would need a function to calculate factorial and powers.

This question has already been answered. Start a new discussion instead.

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