Teorema di Lagrange 11
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
Teorema del punto fisso di Banach Sia f una funzione differenziabile su mathbbR e tale che f'x. Allora f ha almeno un punto fisso ossia un valore di x tale che fxx.
Solution:
Sia x_ un qualsiasi punto nel dominio e costruiamo la sequenza x_n:f^nxffffs fxs ossia dove f viene applicata n volte. Se il limite esiste allora notiamo subito che per x^starlim_nto inftyf^nx vale fx^starflim_nto inftyf^nxlim_nto inftyf^nxx^star dunque si tratterebbe di un punto fisso. Dobbiamo dunque mostrare che la sequenza converge e per questo dimostriamo che è Cauchy. Sia q:sup_xin mathbbR|f'x|. Notiamo che per il Teorema del valor medio tra x_n+ e x_n vale * |x_n+-x_n||fx_n-fx_n-| &hspac.cmoversetTVM|x_n-x_n-| |f'c| &|x_n-x_n-| q |fx_n--fx_n-| q &|x_n--x_n-| q^ & s & q^n|x_-x_|. * e quindi per mgeq ngeq N * |x_m-x_n||x_m-x_m-+x_m--x_m-+x_m--x_m-s+x_n+-x_n| &leq |x_m-x_m-|+|x_m--x_m-|+s+|x_n+-x_n| &q^m+q^m-+q^n|fx_-x_| &fracq^n-q |fx_-x_| &leq fracq^N-q |fx_-x_| * e il limite va a zero per Nto infty. La sequenza è dunque Cauchy e quindi converge.
Teorema del punto fisso di Banach Sia f una funzione differenziabile su mathbbR e tale che f'x. Allora f ha almeno un punto fisso ossia un valore di x tale che fxx.
Solution:
Sia x_ un qualsiasi punto nel dominio e costruiamo la sequenza x_n:f^nxffffs fxs ossia dove f viene applicata n volte. Se il limite esiste allora notiamo subito che per x^starlim_nto inftyf^nx vale fx^starflim_nto inftyf^nxlim_nto inftyf^nxx^star dunque si tratterebbe di un punto fisso. Dobbiamo dunque mostrare che la sequenza converge e per questo dimostriamo che è Cauchy. Sia q:sup_xin mathbbR|f'x|. Notiamo che per il Teorema del valor medio tra x_n+ e x_n vale * |x_n+-x_n||fx_n-fx_n-| &hspac.cmoversetTVM|x_n-x_n-| |f'c| &|x_n-x_n-| q |fx_n--fx_n-| q &|x_n--x_n-| q^ & s & q^n|x_-x_|. * e quindi per mgeq ngeq N * |x_m-x_n||x_m-x_m-+x_m--x_m-+x_m--x_m-s+x_n+-x_n| &leq |x_m-x_m-|+|x_m--x_m-|+s+|x_n+-x_n| &q^m+q^m-+q^n|fx_-x_| &fracq^n-q |fx_-x_| &leq fracq^N-q |fx_-x_| * e il limite va a zero per Nto infty. La sequenza è dunque Cauchy e quindi converge.
Meta Information
Exercise:
Teorema del punto fisso di Banach Sia f una funzione differenziabile su mathbbR e tale che f'x. Allora f ha almeno un punto fisso ossia un valore di x tale che fxx.
Solution:
Sia x_ un qualsiasi punto nel dominio e costruiamo la sequenza x_n:f^nxffffs fxs ossia dove f viene applicata n volte. Se il limite esiste allora notiamo subito che per x^starlim_nto inftyf^nx vale fx^starflim_nto inftyf^nxlim_nto inftyf^nxx^star dunque si tratterebbe di un punto fisso. Dobbiamo dunque mostrare che la sequenza converge e per questo dimostriamo che è Cauchy. Sia q:sup_xin mathbbR|f'x|. Notiamo che per il Teorema del valor medio tra x_n+ e x_n vale * |x_n+-x_n||fx_n-fx_n-| &hspac.cmoversetTVM|x_n-x_n-| |f'c| &|x_n-x_n-| q |fx_n--fx_n-| q &|x_n--x_n-| q^ & s & q^n|x_-x_|. * e quindi per mgeq ngeq N * |x_m-x_n||x_m-x_m-+x_m--x_m-+x_m--x_m-s+x_n+-x_n| &leq |x_m-x_m-|+|x_m--x_m-|+s+|x_n+-x_n| &q^m+q^m-+q^n|fx_-x_| &fracq^n-q |fx_-x_| &leq fracq^N-q |fx_-x_| * e il limite va a zero per Nto infty. La sequenza è dunque Cauchy e quindi converge.
Teorema del punto fisso di Banach Sia f una funzione differenziabile su mathbbR e tale che f'x. Allora f ha almeno un punto fisso ossia un valore di x tale che fxx.
Solution:
Sia x_ un qualsiasi punto nel dominio e costruiamo la sequenza x_n:f^nxffffs fxs ossia dove f viene applicata n volte. Se il limite esiste allora notiamo subito che per x^starlim_nto inftyf^nx vale fx^starflim_nto inftyf^nxlim_nto inftyf^nxx^star dunque si tratterebbe di un punto fisso. Dobbiamo dunque mostrare che la sequenza converge e per questo dimostriamo che è Cauchy. Sia q:sup_xin mathbbR|f'x|. Notiamo che per il Teorema del valor medio tra x_n+ e x_n vale * |x_n+-x_n||fx_n-fx_n-| &hspac.cmoversetTVM|x_n-x_n-| |f'c| &|x_n-x_n-| q |fx_n--fx_n-| q &|x_n--x_n-| q^ & s & q^n|x_-x_|. * e quindi per mgeq ngeq N * |x_m-x_n||x_m-x_m-+x_m--x_m-+x_m--x_m-s+x_n+-x_n| &leq |x_m-x_m-|+|x_m--x_m-|+s+|x_n+-x_n| &q^m+q^m-+q^n|fx_-x_| &fracq^n-q |fx_-x_| &leq fracq^N-q |fx_-x_| * e il limite va a zero per Nto infty. La sequenza è dunque Cauchy e quindi converge.
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| Title | Matched on |
|---|---|
| Teorema di Lagrange 7 | tagstitle |
| Teorema di Lagrange 8 | tagstitle |
| Teorema di Lagrange 9 | tagstitle |
| Teorema di Lagrange 10 | tagstitle |
| Teorema di Lagrange 12 | tagstitle |
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