user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  lea    D57(%rip),%rdi
  call   <puts@plt>
  lea    D118(%rip),%rdi
  call   <puts@plt>
  pxor   %xmm0,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbx
  pxor   %xmm0,%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm3
  divsd  D50(%rip),%xmm3
  pxor   %xmm2,%xmm2
  movapd %xmm2,%xmm0
  mov    $0x0,%esi
  lea    D40(%rip),%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  mov    $0x1,%r12d
  mov    D117(%rip),%rbx
  movsd  D78(%rip),%xmm5
  movsd  %xmm5,0x8(%rsp)
  movsd  D62(%rip),%xmm5
  movsd  %xmm5,0x10(%rsp)
  lea    D17(%rip),%r13
L7:
  movsd  0x8(%rsp),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm2
  movsd  0x10(%rsp),%xmm7
  mulsd  %xmm7,%xmm2
  movq   %xmm2,%rbp
  movapd %xmm7,%xmm0
  call   <f@plt>
  movsd  0x8(%rsp),%xmm6
  mulsd  %xmm6,%xmm0
  movq   %rbp,%xmm4
  subsd  %xmm0,%xmm4
  movsd  %xmm4,0x18(%rsp)
  movapd %xmm6,%xmm0
  call   <f@plt>
  movq   %xmm0,%rbp
  movsd  0x10(%rsp),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  movsd  0x8(%rsp),%xmm6
  movsd  %xmm6,0x10(%rsp)
  movq   %rbp,%xmm0
  subsd  %xmm1,%xmm0
  movsd  0x18(%rsp),%xmm4
  divsd  %xmm0,%xmm4
  movapd %xmm4,%xmm0
  mov    %rbx,%r14
  movsd  %xmm4,0x8(%rsp)
  subsd  %xmm6,%xmm0
  movapd %xmm0,%xmm3
  andpd  D28(%rip),%xmm3
  movq   %xmm3,%rbx
  movapd %xmm3,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbp
  movq   %r14,%xmm0
  call   <log@plt>
  movq   %rbp,%xmm2
  divsd  %xmm0,%xmm2
  movq   %xmm2,%rbp
  movsd  0x8(%rsp),%xmm0
  call   <f@plt>
  movapd %xmm0,%xmm1
  mov    %r12d,%esi
  add    $0x1,%r12d
  movq   %rbp,%xmm3
  movq   %rbx,%xmm2
  movsd  0x8(%rsp),%xmm0
  mov    %r13,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  call   <f@plt>
  pxor   %xmm3,%xmm3
  ucomisd %xmm3,%xmm0
  jp     L7
  jne    L7
  movsd  0x8(%rsp),%xmm5
  movapd %xmm5,%xmm0
  subsd  %xmm5,%xmm0
  andpd  D28(%rip),%xmm0
  comisd D107(%rip),%xmm0
  ja     L7
  movapd %xmm5,%xmm0
  lea    D84(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D57
type_guessed: string
name: read_data
arguments: data_tag: D118
type_guessed: string
name: read_data
arguments: data_tag: D50
type_guessed: byte[8]
name: read_data
arguments: data_tag: D40
type_guessed: string
name: read_data
arguments: data_tag: D117
type_guessed: f64
name: read_data
arguments: data_tag: D78
type_guessed: f64
name: read_data
arguments: data_tag: D62
type_guessed: f64
name: read_data
arguments: data_tag: D17
type_guessed: string
name: read_data
arguments: data_tag: D28
type_guessed: qword[2]
name: read_data
arguments: data_tag: D107
type_guessed: f64[2]
name: read_data
arguments: data_tag: D84
type_guessed: string

tool: D57: Solution of e^x = 2x +1 using secants method

tool: D118: Initial guess: 1

tool: {"D50": [0xef,0x39,0xfa,0xfe,0x42,0x2e,0xe6,0x3f]}

tool: D40: i	Root		f(Root)		Error		order of convergence


tool: D117: 2

tool: D78: 3

tool: D62: 1

tool: D17: %d	%f	%f	%f	%f


tool: {"D28": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D107": [1e-6,null]}

tool: D84: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution of e^x = 2x +1 using secants method\n");
  printf("Initial guess: 1\n");
  double veryOldRoot = 1, oldRoot = 3, newRoot;
  double oldError = 2;
  double newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\torder of convergence\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
  do {
    newRoot = ((veryOldRoot * f(oldRoot) - oldRoot * f(veryOldRoot)) / (f(oldRoot) - f(veryOldRoot)));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
    veryOldRoot = oldRoot;
    oldRoot = newRoot;
    oldError = newError;
  } while ((f(newRoot) != 0) || (fabs(newRoot - oldRoot) > 0.000001));
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```