Mixing
Write a C program to find a root of $x^3 - 3*x + 1 = 0$ by fixed point iteration method (including...
1
$begingroup$
The code works fine. But I want to include the convergence criterion which is as follows:
if the equation is written in the form $x=g(x)$, then condition of convergence is: $g'(x)<1$.
Note that: $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
Please modify my code for convergence.
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include<stdio.h>
#include<conio.h>
#include<math.h>
float g(float x)
{
return(pow((3*x-1), (.3333 )));
}
void main()
{
int n, i ;
float x0 ,xl ;
clrscr() ;
printf ("Enter xO , n");
scanf ("%f%d" ,&x0,&n);
for (i= 1;i<=n;i++ )
{
xl=g(x0);
printf("tt i=%d tt x1=%fnn", i,xl);
if(fabs (xl-x0)<.0001)
break ;
else
x0=xl;
}
printf ("The root is %£ ", xl);
getch();
}
numerical-methods fixed-point-theorems
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
$endgroup$
|
show 3 more comments
1
$begingroup$
The code works fine. But I want to include the convergence criterion which is as follows:
if the equation is written in the form $x=g(x)$, then condition of convergence is: $g'(x)<1$.
Note that: $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
Please modify my code for convergence.
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include<stdio.h>
#include<conio.h>
#include<math.h>
float g(float x)
{
return(pow((3*x-1), (.3333 )));
}
void main()
{
int n, i ;
float x0 ,xl ;
clrscr() ;
printf ("Enter xO , n");
scanf ("%f%d" ,&x0,&n);
for (i= 1;i<=n;i++ )
{
xl=g(x0);
printf("tt i=%d tt x1=%fnn", i,xl);
if(fabs (xl-x0)<.0001)
break ;
else
x0=xl;
}
printf ("The root is %£ ", xl);
getch();
}
numerical-methods fixed-point-theorems
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
$endgroup$
3
$begingroup$
What's your question then?
$endgroup$
– caverac
Jan 21 at 11:36
$begingroup$
@caverac code for convergence
$endgroup$
– user1942348
Jan 21 at 11:38
$begingroup$
1) Have you found $g$? 2) Can you differentiate $g$?
$endgroup$
– mathreadler
Jan 21 at 11:47
$begingroup$
@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
$endgroup$
– user1942348
Jan 21 at 11:53
$begingroup$
yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
$endgroup$
– mathreadler
Jan 21 at 11:55
|
show 3 more comments
1
1
1
1
$begingroup$
The code works fine. But I want to include the convergence criterion which is as follows:
if the equation is written in the form $x=g(x)$, then condition of convergence is: $g'(x)<1$.
Note that: $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
Please modify my code for convergence.
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include<stdio.h>
#include<conio.h>
#include<math.h>
float g(float x)
{
return(pow((3*x-1), (.3333 )));
}
void main()
{
int n, i ;
float x0 ,xl ;
clrscr() ;
printf ("Enter xO , n");
scanf ("%f%d" ,&x0,&n);
for (i= 1;i<=n;i++ )
{
xl=g(x0);
printf("tt i=%d tt x1=%fnn", i,xl);
if(fabs (xl-x0)<.0001)
break ;
else
x0=xl;
}
printf ("The root is %£ ", xl);
getch();
}
numerical-methods fixed-point-theorems
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
$endgroup$
The code works fine. But I want to include the convergence criterion which is as follows:
if the equation is written in the form $x=g(x)$, then condition of convergence is: $g'(x)<1$.
Note that: $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
Please modify my code for convergence.
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include<stdio.h>
#include<conio.h>
#include<math.h>
float g(float x)
{
return(pow((3*x-1), (.3333 )));
}
void main()
{
int n, i ;
float x0 ,xl ;
clrscr() ;
printf ("Enter xO , n");
scanf ("%f%d" ,&x0,&n);
for (i= 1;i<=n;i++ )
{
xl=g(x0);
printf("tt i=%d tt x1=%fnn", i,xl);
if(fabs (xl-x0)<.0001)
break ;
else
x0=xl;
}
printf ("The root is %£ ", xl);
getch();
}
numerical-methods fixed-point-theorems
numerical-methods fixed-point-theorems
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
edited Jan 21 at 11:54
user1942348
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
asked Jan 21 at 11:34
user1942348user1942348
1,3851934
1,3851934
3
$begingroup$
What's your question then?
$endgroup$
– caverac
Jan 21 at 11:36
$begingroup$
@caverac code for convergence
$endgroup$
– user1942348
Jan 21 at 11:38
$begingroup$
1) Have you found $g$? 2) Can you differentiate $g$?
$endgroup$
– mathreadler
Jan 21 at 11:47
$begingroup$
@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
$endgroup$
– user1942348
Jan 21 at 11:53
$begingroup$
yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
$endgroup$
– mathreadler
Jan 21 at 11:55
|
show 3 more comments
3
$begingroup$
What's your question then?
$endgroup$
– caverac
Jan 21 at 11:36
$begingroup$
@caverac code for convergence
$endgroup$
– user1942348
Jan 21 at 11:38
$begingroup$
1) Have you found $g$? 2) Can you differentiate $g$?
$endgroup$
– mathreadler
Jan 21 at 11:47
$begingroup$
@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
$endgroup$
– user1942348
Jan 21 at 11:53
$begingroup$
yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
$endgroup$
– mathreadler
Jan 21 at 11:55
3
3
$begingroup$
What's your question then?
$endgroup$
– caverac
Jan 21 at 11:36
$begingroup$
What's your question then?
$endgroup$
– caverac
Jan 21 at 11:36
$begingroup$
@caverac code for convergence
$endgroup$
– user1942348
Jan 21 at 11:38
$begingroup$
@caverac code for convergence
$endgroup$
– user1942348
Jan 21 at 11:38
$begingroup$
1) Have you found $g$? 2) Can you differentiate $g$?
$endgroup$
– mathreadler
Jan 21 at 11:47
$begingroup$
1) Have you found $g$? 2) Can you differentiate $g$?
$endgroup$
– mathreadler
Jan 21 at 11:47
$begingroup$
@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
$endgroup$
– user1942348
Jan 21 at 11:53
$begingroup$
@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
$endgroup$
– user1942348
Jan 21 at 11:53
$begingroup$
yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
$endgroup$
– mathreadler
Jan 21 at 11:55
$begingroup$
yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
$endgroup$
– mathreadler
Jan 21 at 11:55
|
show 3 more comments
1 Answer
1
active
oldest
votes
1
$begingroup$
Cleaned the code a bit, and used another function $g(x) = (x^3 + 1)/3$ and $g'(x) = x^2$. So as long as you start in $|x|<1$ convergence will work
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include <stdio.h>
#include <math.h>
float g(float x)
{
return (x * x * x + 1) / 3.;
}
float dg(float x)
{
return x * x;
}
int main()
{
int n, i;
float x0, xl, d;
char convergence;
printf ("Enter x0, n: ");
scanf ("%f %d", &x0, &n);
for (i = 1; i <= n; i ++)
{
xl = g(x0);
d = dg(x0);
convergence = fabs(d) < 1 ? 't' : 'f';
printf("tt i = %d tt x1 = %f tt g'(x1) = %f tt convergence = %cn", i, xl, d, convergence);
if(fabs (xl - x0) < .0001)
break;
else
x0 = xl;
}
printf ("The root is %fn", xl);
}
This is a test
Enter x0, n: 0.1 10
i = 1 x1 = 0.333667 g'(x1) = 0.010000 convergence = t
i = 2 x1 = 0.345716 g'(x1) = 0.111333 convergence = t
i = 3 x1 = 0.347107 g'(x1) = 0.119520 convergence = t
i = 4 x1 = 0.347273 g'(x1) = 0.120483 convergence = t
i = 5 x1 = 0.347294 g'(x1) = 0.120599 convergence = t
The root is 0.347294
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
$begingroup$
@user1942348 That is short forif (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience
$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
1
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
Cleaned the code a bit, and used another function $g(x) = (x^3 + 1)/3$ and $g'(x) = x^2$. So as long as you start in $|x|<1$ convergence will work
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include <stdio.h>
#include <math.h>
float g(float x)
{
return (x * x * x + 1) / 3.;
}
float dg(float x)
{
return x * x;
}
int main()
{
int n, i;
float x0, xl, d;
char convergence;
printf ("Enter x0, n: ");
scanf ("%f %d", &x0, &n);
for (i = 1; i <= n; i ++)
{
xl = g(x0);
d = dg(x0);
convergence = fabs(d) < 1 ? 't' : 'f';
printf("tt i = %d tt x1 = %f tt g'(x1) = %f tt convergence = %cn", i, xl, d, convergence);
if(fabs (xl - x0) < .0001)
break;
else
x0 = xl;
}
printf ("The root is %fn", xl);
}
This is a test
Enter x0, n: 0.1 10
i = 1 x1 = 0.333667 g'(x1) = 0.010000 convergence = t
i = 2 x1 = 0.345716 g'(x1) = 0.111333 convergence = t
i = 3 x1 = 0.347107 g'(x1) = 0.119520 convergence = t
i = 4 x1 = 0.347273 g'(x1) = 0.120483 convergence = t
i = 5 x1 = 0.347294 g'(x1) = 0.120599 convergence = t
The root is 0.347294
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
$begingroup$
@user1942348 That is short forif (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience
$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
1
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
add a comment |
1
$begingroup$
Cleaned the code a bit, and used another function $g(x) = (x^3 + 1)/3$ and $g'(x) = x^2$. So as long as you start in $|x|<1$ convergence will work
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include <stdio.h>
#include <math.h>
float g(float x)
{
return (x * x * x + 1) / 3.;
}
float dg(float x)
{
return x * x;
}
int main()
{
int n, i;
float x0, xl, d;
char convergence;
printf ("Enter x0, n: ");
scanf ("%f %d", &x0, &n);
for (i = 1; i <= n; i ++)
{
xl = g(x0);
d = dg(x0);
convergence = fabs(d) < 1 ? 't' : 'f';
printf("tt i = %d tt x1 = %f tt g'(x1) = %f tt convergence = %cn", i, xl, d, convergence);
if(fabs (xl - x0) < .0001)
break;
else
x0 = xl;
}
printf ("The root is %fn", xl);
}
This is a test
Enter x0, n: 0.1 10
i = 1 x1 = 0.333667 g'(x1) = 0.010000 convergence = t
i = 2 x1 = 0.345716 g'(x1) = 0.111333 convergence = t
i = 3 x1 = 0.347107 g'(x1) = 0.119520 convergence = t
i = 4 x1 = 0.347273 g'(x1) = 0.120483 convergence = t
i = 5 x1 = 0.347294 g'(x1) = 0.120599 convergence = t
The root is 0.347294
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
$begingroup$
@user1942348 That is short forif (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience
$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
1
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
add a comment |
1
1
1
$begingroup$
Cleaned the code a bit, and used another function $g(x) = (x^3 + 1)/3$ and $g'(x) = x^2$. So as long as you start in $|x|<1$ convergence will work
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include <stdio.h>
#include <math.h>
float g(float x)
{
return (x * x * x + 1) / 3.;
}
float dg(float x)
{
return x * x;
}
int main()
{
int n, i;
float x0, xl, d;
char convergence;
printf ("Enter x0, n: ");
scanf ("%f %d", &x0, &n);
for (i = 1; i <= n; i ++)
{
xl = g(x0);
d = dg(x0);
convergence = fabs(d) < 1 ? 't' : 'f';
printf("tt i = %d tt x1 = %f tt g'(x1) = %f tt convergence = %cn", i, xl, d, convergence);
if(fabs (xl - x0) < .0001)
break;
else
x0 = xl;
}
printf ("The root is %fn", xl);
}
This is a test
Enter x0, n: 0.1 10
i = 1 x1 = 0.333667 g'(x1) = 0.010000 convergence = t
i = 2 x1 = 0.345716 g'(x1) = 0.111333 convergence = t
i = 3 x1 = 0.347107 g'(x1) = 0.119520 convergence = t
i = 4 x1 = 0.347273 g'(x1) = 0.120483 convergence = t
i = 5 x1 = 0.347294 g'(x1) = 0.120599 convergence = t
The root is 0.347294
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
$endgroup$
Cleaned the code a bit, and used another function $g(x) = (x^3 + 1)/3$ and $g'(x) = x^2$. So as long as you start in $|x|<1$ convergence will work
/* g (x) = (3x - l)^(l/3 ), x0=5 */
/* Fixed po int of Iteration Method */
#include <stdio.h>
#include <math.h>
float g(float x)
{
return (x * x * x + 1) / 3.;
}
float dg(float x)
{
return x * x;
}
int main()
{
int n, i;
float x0, xl, d;
char convergence;
printf ("Enter x0, n: ");
scanf ("%f %d", &x0, &n);
for (i = 1; i <= n; i ++)
{
xl = g(x0);
d = dg(x0);
convergence = fabs(d) < 1 ? 't' : 'f';
printf("tt i = %d tt x1 = %f tt g'(x1) = %f tt convergence = %cn", i, xl, d, convergence);
if(fabs (xl - x0) < .0001)
break;
else
x0 = xl;
}
printf ("The root is %fn", xl);
}
This is a test
Enter x0, n: 0.1 10
i = 1 x1 = 0.333667 g'(x1) = 0.010000 convergence = t
i = 2 x1 = 0.345716 g'(x1) = 0.111333 convergence = t
i = 3 x1 = 0.347107 g'(x1) = 0.119520 convergence = t
i = 4 x1 = 0.347273 g'(x1) = 0.120483 convergence = t
i = 5 x1 = 0.347294 g'(x1) = 0.120599 convergence = t
The root is 0.347294
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
answered Jan 21 at 12:27
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
$begingroup$
@user1942348 That is short forif (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience
$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
1
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
add a comment |
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
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@user1942348 That is short forif (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience
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– caverac
Jan 21 at 12:41
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This code works good for |x|<1 only
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– user1942348
Jan 21 at 12:48
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There is a root 1.5319 also. How to catch that root
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– user1942348
Jan 21 at 13:00
1
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@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
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– caverac
Jan 21 at 13:05
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Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
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– user1942348
Jan 21 at 12:30
$begingroup$
Thank you. Please explain the line "convergence = fabs(d) < 1 ? 't' : 'f';" Also, if g'(x1) >=1, please, I want to get a message of non-convergence
$endgroup$
– user1942348
Jan 21 at 12:30
1
1
$begingroup$
@user1942348 That is short for
if (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
@user1942348 That is short for
if (fabs(d)) < 1 then convergence = true, else convergence = false. You can modify that bit at your convenience$endgroup$
– caverac
Jan 21 at 12:41
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
This code works good for |x|<1 only
$endgroup$
– user1942348
Jan 21 at 12:48
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
$begingroup$
There is a root 1.5319 also. How to catch that root
$endgroup$
– user1942348
Jan 21 at 13:00
1
1
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
$begingroup$
@user1942348 You can manipulate the function $g$ to ensure convergence. For example $g(x) = (3x - 1)^{1/3}$ has $g'(1.53) < 1$, so you could try that version of $g$ instead
$endgroup$
– caverac
Jan 21 at 13:05
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3
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What's your question then?
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– caverac
Jan 21 at 11:36
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@caverac code for convergence
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– user1942348
Jan 21 at 11:38
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1) Have you found $g$? 2) Can you differentiate $g$?
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– mathreadler
Jan 21 at 11:47
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@mathreadler $g(x)=sqrt [3]{3,x-1}$ and $g'(x)=left( 3,x-1 right) ^{-2/3}$
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– user1942348
Jan 21 at 11:53
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yep, and you see how $g$ is implemented in the code, I presume. how can you copy the code for $g$ and modify it to do $g'$?
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– mathreadler
Jan 21 at 11:55