Newton's Method

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I'm currently new to Mathematica and have been trying to solve this problem. I was digging around and found this code:

newtonmethod[error_, initial_, maxiteration_, errorpower_] := 
Module[,
g[x_] := D[f[x], x];
h[t_] := t - f[t]/g[t];
guess = initial;
tol = error;
errorset = ;
ratios = ;
Do[p = h[t] /. t -> guess;
tol = Abs[p - guess];
AppendTo[errorset, tol];
Print["n = ", n, ", x= ", N[ p], ", error =", N[ tol]];
guess = p; 
If[tol <= error [Or] Chop[g[t] /. t -> guess] == 0, 
Goto["errorcalculation"]], n, 1, maxiteration];
Label["errorcalculation"];
Do[AppendTo[ratios, errorset[[i + 1]]/errorset[[i]]^errorpower], i,
 1, Length[errorset] - 1];
Print["Here are the error ratios n"];
TableForm[N[ratios]]
]

I'm not really sure on how to use it and/or if it's enough to complete the problem. Any help would be greatly appreciated.

equation-solving function-construction numerics homework

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

asked 6 hours ago

JVang10

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
  – bbgodfrey
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

I'm currently new to Mathematica and have been trying to solve this problem. I was digging around and found this code:

newtonmethod[error_, initial_, maxiteration_, errorpower_] := 
Module[,
g[x_] := D[f[x], x];
h[t_] := t - f[t]/g[t];
guess = initial;
tol = error;
errorset = ;
ratios = ;
Do[p = h[t] /. t -> guess;
tol = Abs[p - guess];
AppendTo[errorset, tol];
Print["n = ", n, ", x= ", N[ p], ", error =", N[ tol]];
guess = p; 
If[tol <= error [Or] Chop[g[t] /. t -> guess] == 0, 
Goto["errorcalculation"]], n, 1, maxiteration];
Label["errorcalculation"];
Do[AppendTo[ratios, errorset[[i + 1]]/errorset[[i]]^errorpower], i,
 1, Length[errorset] - 1];
Print["Here are the error ratios n"];
TableForm[N[ratios]]
]

I'm not really sure on how to use it and/or if it's enough to complete the problem. Any help would be greatly appreciated.

equation-solving function-construction numerics homework

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

asked 6 hours ago

JVang10

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
  – bbgodfrey
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I'm currently new to Mathematica and have been trying to solve this problem. I was digging around and found this code:

newtonmethod[error_, initial_, maxiteration_, errorpower_] := 
Module[,
g[x_] := D[f[x], x];
h[t_] := t - f[t]/g[t];
guess = initial;
tol = error;
errorset = ;
ratios = ;
Do[p = h[t] /. t -> guess;
tol = Abs[p - guess];
AppendTo[errorset, tol];
Print["n = ", n, ", x= ", N[ p], ", error =", N[ tol]];
guess = p; 
If[tol <= error [Or] Chop[g[t] /. t -> guess] == 0, 
Goto["errorcalculation"]], n, 1, maxiteration];
Label["errorcalculation"];
Do[AppendTo[ratios, errorset[[i + 1]]/errorset[[i]]^errorpower], i,
 1, Length[errorset] - 1];
Print["Here are the error ratios n"];
TableForm[N[ratios]]
]

I'm not really sure on how to use it and/or if it's enough to complete the problem. Any help would be greatly appreciated.

equation-solving function-construction numerics homework

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

asked 6 hours ago

JVang10

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I'm currently new to Mathematica and have been trying to solve this problem. I was digging around and found this code:

newtonmethod[error_, initial_, maxiteration_, errorpower_] := 
Module[,
g[x_] := D[f[x], x];
h[t_] := t - f[t]/g[t];
guess = initial;
tol = error;
errorset = ;
ratios = ;
Do[p = h[t] /. t -> guess;
tol = Abs[p - guess];
AppendTo[errorset, tol];
Print["n = ", n, ", x= ", N[ p], ", error =", N[ tol]];
guess = p; 
If[tol <= error [Or] Chop[g[t] /. t -> guess] == 0, 
Goto["errorcalculation"]], n, 1, maxiteration];
Label["errorcalculation"];
Do[AppendTo[ratios, errorset[[i + 1]]/errorset[[i]]^errorpower], i,
 1, Length[errorset] - 1];
Print["Here are the error ratios n"];
TableForm[N[ratios]]
]

I'm not really sure on how to use it and/or if it's enough to complete the problem. Any help would be greatly appreciated.

equation-solving function-construction numerics homework

equation-solving function-construction numerics homework

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

asked 6 hours ago

JVang10

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

asked 6 hours ago

JVang10

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 34 secs ago

Michael E2

141k11191457

edited 34 secs ago

Michael E2

141k11191457

edited 34 secs ago

Michael E2

141k11191457

141k11191457

asked 6 hours ago

JVang10

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

JVang10

111

asked 6 hours ago

JVang10

111

111

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

JVang10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
  – bbgodfrey
  6 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
  – bbgodfrey
  6 hours ago
  

  
  
  
  

1

1

Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
– bbgodfrey
6 hours ago

Define the function for which a zero is desired, for instance, f[x_] := (x - 1)^2. Then, execute newtonmethod, for instance, newtonmethod[.0001, .1, 20, 2].
– bbgodfrey
6 hours ago

add a comment | 

1 Answer
1

active

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up vote
3
down vote

I offer an implementation more of Mathematica's style, I suppose,

Clear[findRootByNewton]
findRootByNewton[f : _Symbol | _Function, initPt_Real, η_:1.*^-6] := Module[df,
 df = Derivative[1][func];
 NestWhileList[# - f[#]/df[#] &, initPt, Abs[Subtract[##]] >= η &, 2]
 ]

One needs to provide the function (of pure function form or defined by SetDelayed (:=)) whose root to be found, the initial guess and an optional precision goal with a default value of $ 1.0times 10^-6 $.

Then I use $ f(x)=x^2-2 $ as an example. Either

findRootByNewton[#^2 - 2 &, 1.]

or

f[x_] := x^2 - 2
findRootByNewton[f, 1.]

returns

1., 1.5, 1.41667, 1.41422, 1.41421, 1.41421

The result shows root approximations found at each iteration, until the preset precision goal is reached.

---

P.S.

If one just wants the final result, use NestWhile instead of NestWhileList.

share|improve this answer

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

I offer an implementation more of Mathematica's style, I suppose,

Clear[findRootByNewton]
findRootByNewton[f : _Symbol | _Function, initPt_Real, η_:1.*^-6] := Module[df,
 df = Derivative[1][func];
 NestWhileList[# - f[#]/df[#] &, initPt, Abs[Subtract[##]] >= η &, 2]
 ]

One needs to provide the function (of pure function form or defined by SetDelayed (:=)) whose root to be found, the initial guess and an optional precision goal with a default value of $ 1.0times 10^-6 $.

Then I use $ f(x)=x^2-2 $ as an example. Either

findRootByNewton[#^2 - 2 &, 1.]

or

f[x_] := x^2 - 2
findRootByNewton[f, 1.]

returns

1., 1.5, 1.41667, 1.41422, 1.41421, 1.41421

The result shows root approximations found at each iteration, until the preset precision goal is reached.

---

P.S.

If one just wants the final result, use NestWhile instead of NestWhileList.

share|improve this answer

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

add a comment | 

up vote
3
down vote

I offer an implementation more of Mathematica's style, I suppose,

Clear[findRootByNewton]
findRootByNewton[f : _Symbol | _Function, initPt_Real, η_:1.*^-6] := Module[df,
 df = Derivative[1][func];
 NestWhileList[# - f[#]/df[#] &, initPt, Abs[Subtract[##]] >= η &, 2]
 ]

One needs to provide the function (of pure function form or defined by SetDelayed (:=)) whose root to be found, the initial guess and an optional precision goal with a default value of $ 1.0times 10^-6 $.

Then I use $ f(x)=x^2-2 $ as an example. Either

findRootByNewton[#^2 - 2 &, 1.]

or

f[x_] := x^2 - 2
findRootByNewton[f, 1.]

returns

1., 1.5, 1.41667, 1.41422, 1.41421, 1.41421

The result shows root approximations found at each iteration, until the preset precision goal is reached.

---

P.S.

If one just wants the final result, use NestWhile instead of NestWhileList.

share|improve this answer

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

add a comment | 

up vote
3
down vote

up vote
3
down vote

I offer an implementation more of Mathematica's style, I suppose,

Clear[findRootByNewton]
findRootByNewton[f : _Symbol | _Function, initPt_Real, η_:1.*^-6] := Module[df,
 df = Derivative[1][func];
 NestWhileList[# - f[#]/df[#] &, initPt, Abs[Subtract[##]] >= η &, 2]
 ]

One needs to provide the function (of pure function form or defined by SetDelayed (:=)) whose root to be found, the initial guess and an optional precision goal with a default value of $ 1.0times 10^-6 $.

Then I use $ f(x)=x^2-2 $ as an example. Either

findRootByNewton[#^2 - 2 &, 1.]

or

f[x_] := x^2 - 2
findRootByNewton[f, 1.]

returns

1., 1.5, 1.41667, 1.41422, 1.41421, 1.41421

The result shows root approximations found at each iteration, until the preset precision goal is reached.

---

P.S.

If one just wants the final result, use NestWhile instead of NestWhileList.

share|improve this answer

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

I offer an implementation more of Mathematica's style, I suppose,

Clear[findRootByNewton]
findRootByNewton[f : _Symbol | _Function, initPt_Real, η_:1.*^-6] := Module[df,
 df = Derivative[1][func];
 NestWhileList[# - f[#]/df[#] &, initPt, Abs[Subtract[##]] >= η &, 2]
 ]

One needs to provide the function (of pure function form or defined by SetDelayed (:=)) whose root to be found, the initial guess and an optional precision goal with a default value of $ 1.0times 10^-6 $.

Then I use $ f(x)=x^2-2 $ as an example. Either

findRootByNewton[#^2 - 2 &, 1.]

or

f[x_] := x^2 - 2
findRootByNewton[f, 1.]

returns

1., 1.5, 1.41667, 1.41422, 1.41421, 1.41421

The result shows root approximations found at each iteration, until the preset precision goal is reached.

---

P.S.

If one just wants the final result, use NestWhile instead of NestWhileList.

share|improve this answer

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

share|improve this answer

share|improve this answer

edited 36 mins ago

edited 36 mins ago

edited 36 mins ago

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

answered 49 mins ago

Αλέξανδρος Ζεγγ

2,304725

2,304725

add a comment | 

add a comment | 

JVang10 is a new contributor. Be nice, and check out our Code of Conduct.

 

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