Why does this iterative way of solving of equation work? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving a Recurrence Relation/Equation, is there more than 1 way to solve this?Iterative method for matrix differential equationHow does this simplification work?Solution of $x^2(y')^2-2(xy-4)y'+y^2=0$For which values does this series converge?Why does changing variables work?Why does this method for solving recurrence relations work in some cases and not in others?What's wrong in this method of solving a difference equation?Solving a non-linear recurrence equationWhy variation of constant work to solve first order ODE?
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Why does this iterative way of solving of equation work?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving a Recurrence Relation/Equation, is there more than 1 way to solve this?Iterative method for matrix differential equationHow does this simplification work?Solution of $x^2(y')^2-2(xy-4)y'+y^2=0$For which values does this series converge?Why does changing variables work?Why does this method for solving recurrence relations work in some cases and not in others?What's wrong in this method of solving a difference equation?Solving a non-linear recurrence equationWhy variation of constant work to solve first order ODE?
$begingroup$
I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:
$$ T = dfrac7020dfrac32ln(T)+12.$$
I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.
Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.
Questions.
1) What is intuitive way to see why this method works?
2) How can I show rigoursly that this method actualy converges to solution of equation?
3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?
4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?
Thank you for any help.
real-analysis sequences-and-series recurrence-relations irrational-numbers
$endgroup$
add a comment |
$begingroup$
I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:
$$ T = dfrac7020dfrac32ln(T)+12.$$
I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.
Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.
Questions.
1) What is intuitive way to see why this method works?
2) How can I show rigoursly that this method actualy converges to solution of equation?
3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?
4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?
Thank you for any help.
real-analysis sequences-and-series recurrence-relations irrational-numbers
$endgroup$
2
$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
3 hours ago
4
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
3 hours ago
1
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
3 hours ago
add a comment |
$begingroup$
I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:
$$ T = dfrac7020dfrac32ln(T)+12.$$
I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.
Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.
Questions.
1) What is intuitive way to see why this method works?
2) How can I show rigoursly that this method actualy converges to solution of equation?
3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?
4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?
Thank you for any help.
real-analysis sequences-and-series recurrence-relations irrational-numbers
$endgroup$
I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:
$$ T = dfrac7020dfrac32ln(T)+12.$$
I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.
Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.
Questions.
1) What is intuitive way to see why this method works?
2) How can I show rigoursly that this method actualy converges to solution of equation?
3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?
4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?
Thank you for any help.
real-analysis sequences-and-series recurrence-relations irrational-numbers
real-analysis sequences-and-series recurrence-relations irrational-numbers
asked 3 hours ago
ThomThom
381111
381111
2
$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
3 hours ago
4
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
3 hours ago
1
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
3 hours ago
add a comment |
2
$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
3 hours ago
4
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
3 hours ago
1
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
3 hours ago
2
2
$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
3 hours ago
$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
3 hours ago
4
4
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
3 hours ago
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
3 hours ago
1
1
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
3 hours ago
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This method works because you are looking at a discrete dynamical of the form
$$x_n+1 = f(x_n)$$
where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.
$endgroup$
add a comment |
$begingroup$
The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.
The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$
In the linked page, you will see the different steps.
Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$
$endgroup$
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
add a comment |
$begingroup$
In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).
You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:
Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This method works because you are looking at a discrete dynamical of the form
$$x_n+1 = f(x_n)$$
where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.
$endgroup$
add a comment |
$begingroup$
This method works because you are looking at a discrete dynamical of the form
$$x_n+1 = f(x_n)$$
where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.
$endgroup$
add a comment |
$begingroup$
This method works because you are looking at a discrete dynamical of the form
$$x_n+1 = f(x_n)$$
where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.
$endgroup$
This method works because you are looking at a discrete dynamical of the form
$$x_n+1 = f(x_n)$$
where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.
answered 3 hours ago
Tony S.F.Tony S.F.
3,51421030
3,51421030
add a comment |
add a comment |
$begingroup$
The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.
The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$
In the linked page, you will see the different steps.
Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$
$endgroup$
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
add a comment |
$begingroup$
The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.
The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$
In the linked page, you will see the different steps.
Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$
$endgroup$
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
add a comment |
$begingroup$
The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.
The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$
In the linked page, you will see the different steps.
Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$
$endgroup$
The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.
The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$
In the linked page, you will see the different steps.
Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$
answered 2 hours ago
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
add a comment |
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
$endgroup$
– marty cohen
45 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
$begingroup$
@martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
$endgroup$
– Claude Leibovici
28 mins ago
add a comment |
$begingroup$
In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).
You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:
Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.
$endgroup$
add a comment |
$begingroup$
In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).
You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:
Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.
$endgroup$
add a comment |
$begingroup$
In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).
You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:
Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.
$endgroup$
In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).
You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:
Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.
answered 1 hour ago
Robert IsraelRobert Israel
331k23221477
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2
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Looks like the solution happens to be an attractive fixed point.
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– Brevan Ellefsen
3 hours ago
4
$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
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– Ian
3 hours ago
1
$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
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– hardmath
3 hours ago