- Cuchulainn
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[$]1\,=\,\frac{d}{dy} \, F ( F\,^{\,-\,1} (y)\, =\, \frac{d F ( x )}{dx} \frac{d x }{d y} [$]The inverse of F is actually what you are trying to prove, and you seem to be using it as an assumption. This is circular reasoning? I think it invalidates the rest.

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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- Cuchulainn
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QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.An counterexample; f(t) = t. If I did my calculus correctly I get x = + sqrt(2 y) or x = - sqrt(2 y), so the mapping is not bijective (1:1 onto). Furthermore, let y < 0 and you get pure imaginary complex numbers. There is (must be) a close connection between f(t) and y.OK, Define the (contraction) operator [$] T(x) = \, y \, - \, \int_0^x f ( t ) \;dt [$] If we can prove [$] T(x) = \, x \ [$]has a unique fixed point then we are finished. I reduce the scope by considering functions which are essentially like probability pdf (less than 1 in value);The steps are:1. We can prove that T is a contraction by proving that its derivative is less than 1 in absolute value (use the Mean Value Theorem for derivatives)2. Us the Banach-fixed point theorem to prove [$] T(x) = \, x \ [$] has a unique fixed point.3. And also we can define an iterative scheme to solve [$] x_{n+1} = \ T(x_n)[$] as I have posted on Numerics forum and here4. It can be computed using fixed point, Newton, ODE, erf_inv etc.QED 5. I leave it as an exercise to find the contraction mapping when the values of f are in the interval [a,b]. Then the mapping will be [$] T(x) = \, y/m \, - \, \int_0^x f ( t )/m \;dt [$]for some constant m to ensure that T remains a contraction.6. No closed, exact solution is possible. Looking back: we have used real and functional analysis to prove existence and uniqueness results and then numerical analysis to compute a solution.

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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1) And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? 2) And take f(t) = t.i.e. can you x without the following integral because the integral does not get us anywhere?[$]x ( y ) \, = \, \int_0^y f ^{\,-\,1}( u )\, du [$]3) BTW why is your f(t) monotonic increasing? You are excluding all the pdfs of statistics?(?)===============================================================3) it was made an assumption that f ( t ) [$]\ge[$] 0 and therefore f ( t ) can be equal to 1 + sin t , t [$]\ge[$] 0 or any density.2) an answer was given in Sun Dec 13, 15 06:40 PM. btw "can you x without ..." probably sound worse than my "multivariate ..." though it can be understand. 1) the answer is given in the message: Mon Dec 14, 15 01:29 AM. In particular inverse function is defined as [$]x\,=\, \sqrt{\,-\,\ln\,y}[$]. Here [$]0 \,< \,y\,< 1[$] and for any y from the domain of the inverse function we can find one value of x. For example if [$]y\, =\, e\,{-\,1}[$] then x = 1.

- Cuchulainn
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QuoteOriginally posted by: list11) And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? 2) And take f(t) = t.i.e. can you x without the following integral because the integral does not get us anywhere?[$]x ( y ) \, = \, \int_0^y f ^{\,-\,1}( u )\, du [$]3) BTW why is your f(t) monotonic increasing? You are excluding all the pdfs of statistics?(?)===============================================================3) it was made an assumption that f ( t ) [$]\ge[$] 0 and therefore f ( t ) can be equal to 1 + sin t , t [$]\ge[$] 0 or any density.2) an answer was given in Sun Dec 13, 15 06:40 PM. btw "can you x without ..." probably sound worse than my "multivariate ..." though it can be understand. 1) the answer is given in the message: Mon Dec 14, 15 01:29 AM. In particular inverse function is defined as [$]x\,=\, \sqrt{\,-\,\ln\,y}[$]. Here [$]0 \,< \,y\,< 1[$] and for any y from the domain of the inverse function we can find one value of x. For example if [$]y\, =\, e\,{-\,1}[$] then x = 1.I don't think so. You seem to have missed 1) again.Anyways, I am closing now.

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

QuoteOriginally posted by: Cuchulainn[$]1\,=\,\frac{d}{dy} \, F ( F\,^{\,-\,1} (y)\, =\, \frac{d F ( x )}{dx} \frac{d x }{d y} [$]The inverse of F is actually what you are trying to prove, and you seem to be using it as an assumption. This is circular reasoning? I think it invalidates the rest.A problem is to find inverse function. I a function is given by an integral its inverse can be found by the relationship between derivatives of function and its inverse. The derivatives of the inverse in more complex case can be say an a differential equation too, ie knowledge of derivative leads to integration. Though writing derivative one can state in particular that inverse function is the solution of the ode.

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.An counterexample; f(t) = t. If I did my calculus correctly I get x = + sqrt(2 y) or x = - sqrt(2 y), so the mapping is not bijective (1:1 onto). Furthermore, let y < 0 and you get pure imaginary complex numbers. There is (must be) a close connection between f(t) and y.OK, Define the (contraction) operator [$] T(x) = \, y \, - \, \int_0^x f ( t ) \;dt [$] If we can prove [$] T(x) = \, x \ [$]has a unique fixed point then we are finished. I reduce the scope by considering functions which are essentially like probability pdf (less than 1 in value);The steps are:1. We can prove that T is a contraction by proving that its derivative is less than 1 in absolute value (use the Mean Value Theorem for derivatives)2. Us the Banach-fixed point theorem to prove [$] T(x) = \, x \ [$] has a unique fixed point.3. And also we can define an iterative scheme to solve [$] x_{n+1} = \ T(x_n)[$] as I have posted on Numerics forum and here4. It can be computed using fixed point, Newton, ODE, erf_inv etc.QED 5. I leave it as an exercise to find the contraction mapping when the values of f are in the interval [a,b]. Then the mapping will be [$] T(x) = \, y/m \, - \, \int_0^x f ( t )/m \;dt [$]for some constant m to ensure that T remains a contraction.6. No closed, exact solution is possible. Looking back: we have used real and functional analysis to prove existence and uniqueness results and then numerical analysis to compute a solution.I do not see a question. Btw you talk about operator T and use functional analysis terminology but T ( x ) is a function of scalar or vector variable x.

Last edited by list1 on December 14th, 2015, 11:00 pm, edited 1 time in total.

- Cuchulainn
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The problem is solved.

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- Cuchulainn
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QuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.An counterexample; f(t) = t. If I did my calculus correctly I get x = + sqrt(2 y) or x = - sqrt(2 y), so the mapping is not bijective (1:1 onto). Furthermore, let y < 0 and you get pure imaginary complex numbers. There is (must be) a close connection between f(t) and y.OK, Define the (contraction) operator [$] T(x) = \, y \, - \, \int_0^x f ( t ) \;dt [$] If we can prove [$] T(x) = \, x \ [$]has a unique fixed point then we are finished. I reduce the scope by considering functions which are essentially like probability pdf (less than 1 in value);The steps are:1. We can prove that T is a contraction by proving that its derivative is less than 1 in absolute value (use the Mean Value Theorem for derivatives)2. Us the Banach-fixed point theorem to prove [$] T(x) = \, x \ [$] has a unique fixed point.3. And also we can define an iterative scheme to solve [$] x_{n+1} = \ T(x_n)[$] as I have posted on Numerics forum and here4. It can be computed using fixed point, Newton, ODE, erf_inv etc.QED 5. I leave it as an exercise to find the contraction mapping when the values of f are in the interval [a,b]. Then the mapping will be [$] T(x) = \, y/m \, - \, \int_0^x f ( t )/m \;dt [$]for some constant m to ensure that T remains a contraction.6. No closed, exact solution is possible. Looking back: we have used real and functional analysis to prove existence and uniqueness results and then numerical analysis to compute a solution.I do not see a question. Btw you talk about operator T and use functional analysis terminology but T ( x ) is a function of scalar or vector variable x.Read the statement of the theorem. It talks about metric spaces, not scalars but in your case the metric space is the real line with any distance metric you like.Clear?/

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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OK, Define the (contraction) operator [$] T(x) = \, y \, - \, \int_0^x f ( t ) \;dt [$] If we can prove [$] T(x) = \, x \ [$]has a unique fixed point then we are finished. I reduce the scope by considering functions which are essentially like probability pdf (less than 1 in value);-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------The setting of the problem does not clear. If 'operator' T ( x ) is a function of x then it is not clear why we can call it contractor. To say that we should have | T ( x ) | < 1. For example if y = 2 and f= 0 then operator does not a contractor. You might wish to say that operator T depends on parameter y T ( y ) and transform functions f ( x ) , x [$]\, \in\, [$] [0 , 1 ] f [$]\, \in\, [$] M to M. To state that it should be defined a norm and a complete normed space. I also do not know what is a question or it just a comment

- Cuchulainn
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QuoteOriginally posted by: list1OK, Define the (contraction) operator [$] T(x) = \, y \, - \, \int_0^x f ( t ) \;dt [$] If we can prove [$] T(x) = \, x \ [$]has a unique fixed point then we are finished. I reduce the scope by considering functions which are essentially like probability pdf (less than 1 in value);-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------The setting of the problem does not clear. If 'operator' T ( x ) is a function of x then it is not clear why we can call it contractor. To say that we should have | T ( x ) | < 1. For example if y = 2 and f= 0 then operator does not a contractor. You might wish to say that operator T depends on parameter y T ( y ) and transform functions f ( x ) , x [$]\, \in\, [$] [0 , 1 ] f [$]\, \in\, [$] M to M. To state that it should be defined a norm and a complete normed space. I also do not know what is a question or it just a commentWell, no.Wrong definition of contraction and wrong example. As mentioned, first do some reading and then post.Do you understand? Your posts are confused and incorrect. How come you misintrepret/misunderstand what people write?

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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- Cuchulainn
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Here is a quiz:Prove the following are contractions1. x = cos(x)2. x = (x + 2/x)/2Stick to metric spaces, don't worry about Banach spaces for the moment. Once contraction is clear, some meaningful discussion becomes possible. ==edit I get the impression list is talking about inverse function theorem

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

QuoteOriginally posted by: CuchulainnHere is a quiz:Prove the following are contractions1. x = cos(x)2. x = (x + 2/x)/2Stick to metric spaces, don't worry about Banach spaces for the moment. Once contraction is clear, some meaningful discussion becomes possible. ==edit I get the impression list is talking about inverse function theorem*) If one looks at the contractions reference the we see that it relates to a map which is realized by f ( x) and there is a metric d ( x , y ). In 1.-2. we have two equations and it is not clear following the reference what is f and what is d. Also one should define closed area when map is contraction with respect to chosen metric d. Natural domain of f could be broader than area where f is a contractor.**) Of course I talked about inverse function and after my second question I asked how to change 'inverse number' in title on 'inverse function' and I got there an advice.

- Cuchulainn
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For 1, see thisFor my example 2, it is the fixed point form for sqrt(2) known to the ancient Babylonians.The ideal of trying to find an exact formula for the inverse (quantile) is a waste of time IMO. I have shown that the solution exists, is unique and 4 ways to construct it. But I suppose that's the never-ending discussion between pure and constructive mathematics.

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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QuoteOriginally posted by: CuchulainnFor 1, see thisFor my example 2, it is the fixed point form for sqrt(2) known to the ancient Babyloninansin this thei no words similar to contractor. two curves intersected in one point and how does this graph is related to contractor notion?My question was how 1-2 are related to contractors?

- Cuchulainn
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QuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnFor 1, see thisFor my example 2, it is the fixed point form for sqrt(2) known to the ancient Babyloninansin this thei no words similar to contractor. two curves intersected in one point and how does this graph is related to contractor notion?My question was how 1-2 are related to contractors?First, it is called contraction. Second, you need to find it, it is easy. It is high-school piddling calculus.hintfor x = g(x) we need to have |g'(x)| < 1.now let g(x) = cos(x). BTW this is a contractor

Last edited by Cuchulainn on December 14th, 2015, 11:00 pm, edited 1 time in total.

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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

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