calculation problem with math.h predefined methods

[mattmao]

Okay, I was asked by a friend about the result of this limit:

Not n->zero but n-> + infinite

I really know nothing about advanced math, so I wrote a C program to help me:

(BTW, I guess the result would be ln2.)

#include <stdio.h>
# include <math.h>
 
int main()
{
  double result;
  double n;
 
  printf("input n, followed by the enter key.\n");
  scanf("%lf", &n);
 
  result = log(2+sqrt(atan(n)*sin(1/n)));
 
  printf("The result for number %f is: %f", n, result);
 
  result = log(2.00);
  printf("The result for ln2 is: %f", result);
 
  return 0;
}

I used the gcc to compile my code, it rejected, saying there were some problems:

bash-3.2$ gcc calculation.c
/tmp/ccyzzqks.o: In function `main':
calculation.c:(.text+0x37): undefined reference to `atan'
calculation.c:(.text+0x49): undefined reference to `sin'
calculation.c:(.text+0x70): undefined reference to `sqrt'
calculation.c:(.text+0x86): undefined reference to `log'
calculation.c:(.text+0xb1): undefined reference to `log'
collect2: ld returned 1 exit status

I googled for the error message and some expert said to use the -lm parameter to compile.

Yeah, this time it worked and I got the results.

But why?




PS. I tried to issue this commands:
bash-3.2$ gcc -lm calculation.c
bash-3.2$ ./a.out

input n, followed by the enter key.
0.0000
The result for number 0.000000 is: nanThe result for ln2 is: 0.693147

input n, followed by the enter key.
0.000001
The result for number 0.000001 is: nanThe result for ln2 is: 0.693147

input n, followed by the enter key.
9999999999999999
The result for number 10000000000000000.000000 is: 0.693147The result for ln2 is: 0.693147

nan = Not a Number ?!

Again, why? (If that is a math question, ignore it plz...)

[weaknessforcats]

Yoiur code compiles and links without error using Visual Studio.NET 2005.

[JosAH]

Again, why? (If that is a math question, ignore it plz...)

The lim(x->oo, atan(x)) == pi/2, lim(x->oo, sin(1/x)) == 0; so that entire root
has a limit of 0 (zero). Did you want to ask this? Or is it about C/C++ techincalities?

kind regards,

Jos

[mattmao]

The lim(x->oo, atan(x)) == pi/2, lim(x->oo, sin(1/x)) == 0; so that entire root
has a limit of 0 (zero). Did you want to ask this? Or is it about C/C++ techincalities?

kind regards,

Jos

Hello Jos.

It's not about math, my question is for the -lm parameter in gcc. Why would my code be compiled successfully if I use it? I am trying to make my code independent from any compiler restrictions, such as some specific parameters like the "-lm".

I reckon the problems would be with the sin(x) atan(x) methods in math.h.
What would be the "proper" ways for calling them when you want to use them?

Thanks for your help in math :)