Checking to see if is correctHigh iteration Newton's MethodGlobally Convergent Methods for Nonlinear Systems of EquationsHow to correctly apply Newton-Raphson method to Backward Euler method?Optimizing trigonometric and nonlinear functionsbisection method, Numerical AnalysisCan someone explain Newton's method to me?Relating convergence theorem for Newton-Raphson method to Newton fractalHow to solve for the interval of convergence in Newton's Method?Newton's Method for nonlinear systemsNewton Raphson method initial guessRecommendations for Numerical Analysis book covering specific requirements?
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Checking to see if is correct
High iteration Newton's MethodGlobally Convergent Methods for Nonlinear Systems of EquationsHow to correctly apply Newton-Raphson method to Backward Euler method?Optimizing trigonometric and nonlinear functionsbisection method, Numerical AnalysisCan someone explain Newton's method to me?Relating convergence theorem for Newton-Raphson method to Newton fractalHow to solve for the interval of convergence in Newton's Method?Newton's Method for nonlinear systemsNewton Raphson method initial guessRecommendations for Numerical Analysis book covering specific requirements?
$begingroup$
Suppose Newton’s method is applied to the function f(x) = 1/x. If the initial guess is
x0 = 1, find x50.
Solution
F(x) =1/x x0=1 where n=1,2,3,...
F'(x) =-1/(x^2)
Xn+1=xn-(f(xn))/(f'(xn))
And I had x50=1125899907*10^6
numerical-methods
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose Newton’s method is applied to the function f(x) = 1/x. If the initial guess is
x0 = 1, find x50.
Solution
F(x) =1/x x0=1 where n=1,2,3,...
F'(x) =-1/(x^2)
Xn+1=xn-(f(xn))/(f'(xn))
And I had x50=1125899907*10^6
numerical-methods
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
1
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21
add a comment |
$begingroup$
Suppose Newton’s method is applied to the function f(x) = 1/x. If the initial guess is
x0 = 1, find x50.
Solution
F(x) =1/x x0=1 where n=1,2,3,...
F'(x) =-1/(x^2)
Xn+1=xn-(f(xn))/(f'(xn))
And I had x50=1125899907*10^6
numerical-methods
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose Newton’s method is applied to the function f(x) = 1/x. If the initial guess is
x0 = 1, find x50.
Solution
F(x) =1/x x0=1 where n=1,2,3,...
F'(x) =-1/(x^2)
Xn+1=xn-(f(xn))/(f'(xn))
And I had x50=1125899907*10^6
numerical-methods
numerical-methods
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 29 at 2:38
Andrew Odonkor
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
asked Mar 29 at 2:32
Andrew OdonkorAndrew Odonkor
61
61
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andrew Odonkor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
1
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21
add a comment |
$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
1
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21
$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
1
1
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I get that
$$x_n+1 = x_n - fracf(x_n)f'(x_n) = x_n - fracfrac1x_n-frac1x_n^2 = x_n + x_n = 2x_n,$$
thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have
$$x_50 = 2^50 approx 1.1258999 cdot 10^15,$$
which appears roughly the same as what you've got.
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
I get that
$$x_n+1 = x_n - fracf(x_n)f'(x_n) = x_n - fracfrac1x_n-frac1x_n^2 = x_n + x_n = 2x_n,$$
thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have
$$x_50 = 2^50 approx 1.1258999 cdot 10^15,$$
which appears roughly the same as what you've got.
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
$endgroup$
add a comment |
$begingroup$
I get that
$$x_n+1 = x_n - fracf(x_n)f'(x_n) = x_n - fracfrac1x_n-frac1x_n^2 = x_n + x_n = 2x_n,$$
thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have
$$x_50 = 2^50 approx 1.1258999 cdot 10^15,$$
which appears roughly the same as what you've got.
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
$endgroup$
add a comment |
$begingroup$
I get that
$$x_n+1 = x_n - fracf(x_n)f'(x_n) = x_n - fracfrac1x_n-frac1x_n^2 = x_n + x_n = 2x_n,$$
thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have
$$x_50 = 2^50 approx 1.1258999 cdot 10^15,$$
which appears roughly the same as what you've got.
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
$endgroup$
I get that
$$x_n+1 = x_n - fracf(x_n)f'(x_n) = x_n - fracfrac1x_n-frac1x_n^2 = x_n + x_n = 2x_n,$$
thus $x_n = 2^n$, given $x_0 = 1$. According to Google, we have
$$x_50 = 2^50 approx 1.1258999 cdot 10^15,$$
which appears roughly the same as what you've got.
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
answered Mar 29 at 2:42
Theo BenditTheo Bendit
20.2k12353
20.2k12353
add a comment |
add a comment |
Andrew Odonkor is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Odonkor is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Odonkor is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Odonkor is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Your answer makes sense, considering that $f(x) = 1/x$ has no zeros.
$endgroup$
– D.B.
Mar 29 at 2:38
$begingroup$
Boy... this is the worst posting title I have ever seen.
$endgroup$
– David G. Stork
Mar 29 at 5:17
1
$begingroup$
Possible duplicate of High iteration Newton's Method
$endgroup$
– Carl Christian
Mar 29 at 18:53
$begingroup$
You may find this answer to your question instructive.
$endgroup$
– Carl Christian
Mar 29 at 22:21