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Consider the following $C$ function.

For large values of $y$, the return value of the function $f$ best approximates

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;
}

  1. $x^y$
  2. $e^x$
  3. $\text{ln} (1+x)$
  4. $x^x$
asked in Algorithms by Veteran (69k points)
edited by | 1.3k views
No need to remember series expansion just use Taylor series for getting expansion of f(x).

2 Answers

+14 votes
Best answer
The simplified version of above program can be rewritten as
float f(float x, int y) {
    float p=1, s=1; 
    int i;
    for (i=1; i<y; i++) {
        p = p * (x/i);
        s = s + p;
    }
    return s;
}

we can take p=1,s=1 intialization outside of for loop cz there is no condition checking in for loop involving p,s .

i p= p * (x/i) s= s + p
1 x 1+x
2 $\frac{x^{2}}{2}$ 1+x+$\frac{x^{2}}{2}$
3 $\frac{x^{3}}{6}$ 1+x+$\frac{x^{2}}{2}$+$\frac{x^{3}}{6}$
4 $\frac{x^{4}}{24}$ 1+x+$\frac{x^{2}}{2}$+$\frac{x^{3}}{6}$+$\frac{x^{4}}{24}$
n $\frac{x^{n}}{n!}$ ex

 

 esum (n=0..inf) xn / n!
    = 1+x+$\frac{x^{2}}{2}$+$\frac{x^{3}}{6}$ +$\frac{x^{4}}{24}$+ ...+$\frac{x^{n}}{n!}$

 

Hence Option B is Ans.

 

answered by Veteran (23.4k points)
selected by
+23 votes

$$\begin{array}{l}
i = 1 &\rm then& p = x &\&& s = 1 + x\\[1em]
i = 2 &\rm then& p = \frac{x^2}{2} &\&& s = 1+x+\frac{x^2}{2}
\end{array}$$

As $y$ tends to infinity, $s$ tends to  $e^x$.

Hence, the correct answer is option B.

answered by Veteran (34.3k points)
edited by
@Pooja Thanks for compact answers.I have been following your answers. :) just a minute correction needed in the above answer when i=1 p=x.

@Pooja Palod good answer but just small improvement, Please put factorial symbol in denominator.

if i do with example then e^2 is 7.38 but it gives 5 ... why is this difference ?
@sid1221 :- e^x is an infinite series,so for large value y,if you run,the output will approximate to you answer
Answer:

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