zeros and convergence or divergence in iterations The 2019 Stack Overflow Developer Survey Results Are InFunctional Iterations and order of convergenceImproving Newton's iteration where the derivative is near zero?Fix point iteration, rate of convergenceSome questions about variations of fixed point methodNewton-Raphson's methodPractice versus asymptotic statements about the speed of convergence.Rate of convergence of fixed-point iteration in higher dimensionsWhat is minimum number of iterations required in the bisection method to reach at the desired accuracy?Convergence of a variant of Newton's MethodAlmost sure convergence of Newton's method
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zeros and convergence or divergence in iterations
The 2019 Stack Overflow Developer Survey Results Are InFunctional Iterations and order of convergenceImproving Newton's iteration where the derivative is near zero?Fix point iteration, rate of convergenceSome questions about variations of fixed point methodNewton-Raphson's methodPractice versus asymptotic statements about the speed of convergence.Rate of convergence of fixed-point iteration in higher dimensionsWhat is minimum number of iterations required in the bisection method to reach at the desired accuracy?Convergence of a variant of Newton's MethodAlmost sure convergence of Newton's method
$begingroup$
I'm solving some numerical analysis exercises and I still can not find a way to solve the following
Consider the function $f (x) = x^3 - x - 1$. For the equation $f (x) = 0$ answer the following:
.
If we rearrange $f(x)=x(x^2-1)-1$, we arrive at the iteration:
$x_n+1 = frac1x_n^2-1$
When reorganizing $f (x) = (x-1)(x^2+x+1)-1$ we obtain:
$x_n+1 = 1 + fracxx^2+x+1$
demonstrate the equivalence of these two expressions.
How do these iterations behave? (Take start point $x_0 = 1.5$ and calculate some $x_n$ values).
Analyze the convergence or divergence of the previous interactions near zero $x$*.
How should I do it to get to the right way?
complex-analysis numerical-methods numerical-calculus
asked Mar 30 at 16:47
FmkitFmkit
52
$endgroup$
add a comment |
$begingroup$
I'm solving some numerical analysis exercises and I still can not find a way to solve the following
Consider the function $f (x) = x^3 - x - 1$. For the equation $f (x) = 0$ answer the following:
.
If we rearrange $f(x)=x(x^2-1)-1$, we arrive at the iteration:
$x_n+1 = frac1x_n^2-1$
When reorganizing $f (x) = (x-1)(x^2+x+1)-1$ we obtain:
$x_n+1 = 1 + fracxx^2+x+1$
demonstrate the equivalence of these two expressions.
How do these iterations behave? (Take start point $x_0 = 1.5$ and calculate some $x_n$ values).
Analyze the convergence or divergence of the previous interactions near zero $x$*.
How should I do it to get to the right way?
complex-analysis numerical-methods numerical-calculus
asked Mar 30 at 16:47
FmkitFmkit
52
$endgroup$
2
$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07
add a comment |
$begingroup$
I'm solving some numerical analysis exercises and I still can not find a way to solve the following
Consider the function $f (x) = x^3 - x - 1$. For the equation $f (x) = 0$ answer the following:
.
If we rearrange $f(x)=x(x^2-1)-1$, we arrive at the iteration:
$x_n+1 = frac1x_n^2-1$
When reorganizing $f (x) = (x-1)(x^2+x+1)-1$ we obtain:
$x_n+1 = 1 + fracxx^2+x+1$
demonstrate the equivalence of these two expressions.
How do these iterations behave? (Take start point $x_0 = 1.5$ and calculate some $x_n$ values).
Analyze the convergence or divergence of the previous interactions near zero $x$*.
How should I do it to get to the right way?
complex-analysis numerical-methods numerical-calculus
asked Mar 30 at 16:47
FmkitFmkit
52
$endgroup$
I'm solving some numerical analysis exercises and I still can not find a way to solve the following
Consider the function $f (x) = x^3 - x - 1$. For the equation $f (x) = 0$ answer the following:
.
If we rearrange $f(x)=x(x^2-1)-1$, we arrive at the iteration:
$x_n+1 = frac1x_n^2-1$
When reorganizing $f (x) = (x-1)(x^2+x+1)-1$ we obtain:
$x_n+1 = 1 + fracxx^2+x+1$
demonstrate the equivalence of these two expressions.
How do these iterations behave? (Take start point $x_0 = 1.5$ and calculate some $x_n$ values).
Analyze the convergence or divergence of the previous interactions near zero $x$*.
How should I do it to get to the right way?
complex-analysis numerical-methods numerical-calculus
complex-analysis numerical-methods numerical-calculus
asked Mar 30 at 16:47
FmkitFmkit
52
asked Mar 30 at 16:47
FmkitFmkit
52
asked Mar 30 at 16:47
FmkitFmkit
52
asked Mar 30 at 16:47
FmkitFmkit
52
asked Mar 30 at 16:47
FmkitFmkit
52
52
2
$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07
add a comment |
2
$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07
2
2
$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case,
$$
|g'(x^*)| = left|dfrac-2x^*((x^*)^2-1)^2right|>1
$$
and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0in [1,2]$. Moreover, considering that $|g'|leq 3/49$, we can say that
$$
|x_n-x^*| leq (3/49)^n |x_0-x^*| leq (3/49)^n.
$$
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
$endgroup$
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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active
oldest
votes
$begingroup$
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case,
$$
|g'(x^*)| = left|dfrac-2x^*((x^*)^2-1)^2right|>1
$$
and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0in [1,2]$. Moreover, considering that $|g'|leq 3/49$, we can say that
$$
|x_n-x^*| leq (3/49)^n |x_0-x^*| leq (3/49)^n.
$$
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
$endgroup$
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
add a comment |
$begingroup$
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case,
$$
|g'(x^*)| = left|dfrac-2x^*((x^*)^2-1)^2right|>1
$$
and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0in [1,2]$. Moreover, considering that $|g'|leq 3/49$, we can say that
$$
|x_n-x^*| leq (3/49)^n |x_0-x^*| leq (3/49)^n.
$$
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
$endgroup$
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
add a comment |
$begingroup$
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case,
$$
|g'(x^*)| = left|dfrac-2x^*((x^*)^2-1)^2right|>1
$$
and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0in [1,2]$. Moreover, considering that $|g'|leq 3/49$, we can say that
$$
|x_n-x^*| leq (3/49)^n |x_0-x^*| leq (3/49)^n.
$$
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
$endgroup$
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case,
$$
|g'(x^*)| = left|dfrac-2x^*((x^*)^2-1)^2right|>1
$$
and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0in [1,2]$. Moreover, considering that $|g'|leq 3/49$, we can say that
$$
|x_n-x^*| leq (3/49)^n |x_0-x^*| leq (3/49)^n.
$$
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
edited Apr 1 at 16:56
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
answered Apr 1 at 8:59
PierreCarrePierreCarre
2,103214
2,103214
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
add a comment |
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
2
2
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
How do you get from $g'(x)=frac-2x(x^2-1)^2$ to your formula for $|g'(x^*)|$? I get $g'(x^*)=-2(x^*)^3=-2(x^*+1)$, which still proves the point but is a different formula.
$endgroup$
– LutzL
Apr 1 at 10:13
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
$begingroup$
@LutzL you are correct, I'll update the post.
$endgroup$
– PierreCarre
Apr 1 at 16:44
add a comment |
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$begingroup$
You can with any method find some approximation for $x^*$. The execution of task 2 with 10-20 iterations should also not be too difficult, you have a computer, thus you have some scripting language or a spreadsheet calculator available. Please do that and report your results. For task 1 you need to do exactly what is demanded, show that $f(x_n)=0$ if and only if $x_n+1=x_n$, which is the equivalency.
$endgroup$
– LutzL
Mar 30 at 20:10
$begingroup$
Hi, thanks for your response. Regarding task 1 I have doubts. What do you mean by showing that $ f (x_n) = 0 $ and that $ x_ n + 1 = x_n $ the task is done ?, with what method could you demonstrate this? Thank you very much again
$endgroup$
– Fmkit
Apr 1 at 0:36
$begingroup$
You have to show that the fixed points of the iterations are exactly the roots of the polynomial. There is no method involved, just an interpretation of the factorizations that are given.
$endgroup$
– LutzL
Apr 1 at 10:07