# Calculating upper and lower limits using C++

I am writing a class that spits out polynomials based on a constants array a and exponent array b, such that this equation is generated:

However, this equation does not have a solution for f(0), but it can be calculated using the limits from both sides (given that they are equal). How could you implement this in C++, since I do absolutely not know where to start.

EDIT

Thanks for the comments (I cannot comment yet). I was indeed a bit too fast with the coding, but I do nevertheless want to write the function myself, because that is exactly the thing I want to learn.

Generally, the limits for `f(0)` will depend only on the smallest exponent of the normalized polynom.

The normalized polynom (as I call it) is the polynom, where all `a` values, that belong to a repeated `b` value are added up and only the non-zero `a` values are kept.

1. If the smallest exponent is greater than 0, then `f(0) = 0`
2. If the smallest exponent is 0, the corresponding `a` value is the result for `f(0)`
3. If the smallest exponent is smaller than 0, the limits are positive and/or negative infinity
• Even smallest exponent means the upper and lower limit goes same direction
• Odd smallest exponent means the upper and lower limit goes opposite direction

This approach only works for whole numbers `b`. At least I didn’t investigate all the details for other cases.

```#include <iostream>
#include <limits>
#include <map>

using namespace std;

double f0(int* a, int* b, int n);

int main()
{
int a[] = {2, 4, 6, -2, 5, -4};
int b[] = {2, 1, -1, -1, 0, -1};
// number of values in array a and b
int n = 6;

double result = f0(a, b, n);

cout << "f(0) = " << result << endl;

return 0;
}

double f0(int* a, int* b, int n)
{
map<int, int> exponents;

for (int i = 0; i < n; ++i)
{
exponents[b[i]] += a[i];
// debug printing intermediate sums per exponent
cout << b[i] << ": " << exponents[b[i]] << endl;
}

int minExp = 0;

for (auto it = exponents.begin(); it != exponents.end(); ++it)
{
if (it->second != 0 && it->first < minExp)
{
minExp = it->first;
}
}

// no negative exponent. f(0) is defined by 0 exponents
if (minExp == 0) return exponents[0];

// minimum exponent is even => positive or negative infinity limit
if (minExp % 2 == 0)
{
return exponents[minExp] > 0
? numeric_limits<double>::infinity()
: -numeric_limits<double>::infinity();
}

// minimum exponent is odd => f(0) limits approach both positive AND negative infinity
return numeric_limits<double>::quiet_NaN();
}
```