Haskell Function that calculates e^x

Implement a function that calculates the value of e ^ x, x is a parameter of the function, an integer. To do this, use the Taylor series expansion to calculate the potency of e. The function will receives as a parameter, in addition to the exponent x, the number of terms of the series, which will operate as a maximum value of n. For the resolution of this function, recursion must be used.

I made this:

factorial 0 = 1
factorial n = n * factorial (n-1)

consigna3::Int->Int->Float
consigna3 _ 0 = 1
consigna3 x n = (fromIntegral(x^n) / fromIntegral(factorial n)) + consigna3 x (n-1)

But some results are wrong, this is what I expected:

Ejemplo 1: Main> funcion3 1 1
           2.0
Ejemplo 2: Main> funcion3 1 10
           2.718282
Ejemplo 3: Main> funcion3 2 10
           7.388997
Ejemplo 4: Main> funcion3 10 20
           21991.48
Ejemplo 5: Main> funcion3 10 30
           22026.46
Ejemplo 6: Main> funcion3 0 30
           1.0

The results (10 20) and (10 30) do not match what the function I did returns. What I am doing wrong? Thanks and sorry for my English.

Solution

You are using Int for calculations that will overflow an Int. Instead, convert to Float right away, and then using Float for everything. So:

consigna3 x n = ((fromIntegral x)^n / factorial (fromIntegral n)) + consigna3 x (n-1)

There are two critical changes from Int to Float here: first, you do x^n where x :: Int, but I do fromIntegral x^n where fromIntegral x :: Float; second, you do factorial n where n :: Int, but I do factorial (fromIntegral n) where fromIntegral n :: Float.