2.2k views

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. $x^x$
edited | 2.2k views
+1
No need to remember series expansion just use Taylor series for getting expansion of f(x).

A simplified version of the given program can be written 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 $p=1,s=1$ initialization outside of for loop because there is no condition checking in for loop involving $p,s$ .

 $i$ $p= p$*$(x/i)$ $s= s + p$ $1$ $x$ $1+x$ $2$ $\frac{x^{2}}{2}$ $1+x+\frac{x^{2}}{2}$ $3$ $\frac{x^{3}}{6}$ $1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$ $4$ $\frac{x^{4}}{24}$ $1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}$ $n$ $\frac{x^{n}}{n!}$ $e^x$
 $e^x = \displaystyle{\sum_{n=0}^{\infty} \frac{x^n}{n!}} = 1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} +\frac{x^{4}}{24}+ \ldots+\frac{x^{n}}{n!}$

edited

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

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

Hence, the correct answer is option B.

edited
0
@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.
0

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

0
if i do with example then e^2 is 7.38 but it gives 5 ... why is this difference ?
+1
@sid1221 :- e^x is an infinite series,so for large value y,if you run,the output will approximate to you answer

I am using substitution method to get answer

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