Question

Consider the following C function.
float f (float x, int y)
{
float p, s; int i;
for (s = 1, p = 1, i = 1; i < y; i ++)
{
p* = x/i;
s + = p:
}
return s;
}
For large values of y, the return value of the function f best approximates

• A. ln(1 + x)
• B. $x^x$
• C. $e^x$
• D. $x^y$

Answer 1

The correct option is C $e^x$

The given function f is not recursive, so consider the folowing iteration method.
i p s

p =p*$\frac{x}{i}$ s = s + p

initilize 1 1 1

1 p = x s = 1 + x

2 p = x. $\frac{x}{2}$ s = 1 + x + $\frac{x^2}{2}$

3 p = $\frac{x^2}{2}$.$\frac{x}{3}$ s = 1 + x + $\frac{x^2}{2}$ + $\frac{x^3}{6}$

4 p = $\frac{x^3}{6}$. $\frac{x}{4}$ s = 1 + x + $\frac{x^2}{2}$ + $\frac{x^3}{6}$ + $\frac{x^4}{24}$

5 p = $\frac{x^4}{24}$. $\frac{x}{5}$ s = 1 + x + $\frac{x^2}{2}$ + $\frac{x^3}{6}$ + $\frac{x^4}{24}$ + $\frac{x^5}{120}$

For large value of y assume y $\to$ $\infty$ then i also tends to infinite it means increment of for loop may tends to infinite. In the given function we choose y as a large integer but not infinite. The return value of the function f is s.

s = 1 + x + $\frac{x^2}{2}$ + $\frac{x^3}{6}$ + $\frac{x^4}{24}$ + $\frac{x^5}{120}$... + $\infty$

s = 1 + x + $\frac{x^2}{!2}$ + $\frac{x^3}{!3}$ + $\frac{x^4}{!4}$ + $\frac{x^5}{!5}$... + $\infty$

s = $e^x$