Eindeutigkeit Grenzwert konvergenter Folgen
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
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Sei Xd ein metrischer Raum. Jede konvergente Folge in X besitzt einen eindeutigen Grenzwert.
Solution:
Angenommen dass die Folge a_n sowohl gegen a als auch gegen b konvergiert. Zu zeigen ist: ab. Nach der Definition von Grenzwert ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n a iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-a| epsilon Andererseits ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n b iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-b| epsilon Man kombiniert nun diese zwei Bedingungen. Für ein fixes epsilon wählt man N : textmaxN_N_ damit beide Bedingungen gleichzeitig erfüllt sind. Es gilt dann |a-b| |a-a_n+a_n-b| leq |a-a_n|+|a_n-b| epsilon + epsilon epsilon Da epsilon beliebig klein wählbar ist bedeutet es dass |a-b| eine beliebig kleine Grösse ist also gleich Null ist. Somit gilt ab was zu zeigen war.
Sei Xd ein metrischer Raum. Jede konvergente Folge in X besitzt einen eindeutigen Grenzwert.
Solution:
Angenommen dass die Folge a_n sowohl gegen a als auch gegen b konvergiert. Zu zeigen ist: ab. Nach der Definition von Grenzwert ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n a iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-a| epsilon Andererseits ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n b iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-b| epsilon Man kombiniert nun diese zwei Bedingungen. Für ein fixes epsilon wählt man N : textmaxN_N_ damit beide Bedingungen gleichzeitig erfüllt sind. Es gilt dann |a-b| |a-a_n+a_n-b| leq |a-a_n|+|a_n-b| epsilon + epsilon epsilon Da epsilon beliebig klein wählbar ist bedeutet es dass |a-b| eine beliebig kleine Grösse ist also gleich Null ist. Somit gilt ab was zu zeigen war.
Meta Information
Exercise:
Sei Xd ein metrischer Raum. Jede konvergente Folge in X besitzt einen eindeutigen Grenzwert.
Solution:
Angenommen dass die Folge a_n sowohl gegen a als auch gegen b konvergiert. Zu zeigen ist: ab. Nach der Definition von Grenzwert ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n a iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-a| epsilon Andererseits ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n b iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-b| epsilon Man kombiniert nun diese zwei Bedingungen. Für ein fixes epsilon wählt man N : textmaxN_N_ damit beide Bedingungen gleichzeitig erfüllt sind. Es gilt dann |a-b| |a-a_n+a_n-b| leq |a-a_n|+|a_n-b| epsilon + epsilon epsilon Da epsilon beliebig klein wählbar ist bedeutet es dass |a-b| eine beliebig kleine Grösse ist also gleich Null ist. Somit gilt ab was zu zeigen war.
Sei Xd ein metrischer Raum. Jede konvergente Folge in X besitzt einen eindeutigen Grenzwert.
Solution:
Angenommen dass die Folge a_n sowohl gegen a als auch gegen b konvergiert. Zu zeigen ist: ab. Nach der Definition von Grenzwert ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n a iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-a| epsilon Andererseits ist lim limits_n to inftya_n a gleichbedeut mit lim limits_n to inftya_n b iff forall epsilon exists N_ in mathbbN: forall n geq N_ : |a_n-b| epsilon Man kombiniert nun diese zwei Bedingungen. Für ein fixes epsilon wählt man N : textmaxN_N_ damit beide Bedingungen gleichzeitig erfüllt sind. Es gilt dann |a-b| |a-a_n+a_n-b| leq |a-a_n|+|a_n-b| epsilon + epsilon epsilon Da epsilon beliebig klein wählbar ist bedeutet es dass |a-b| eine beliebig kleine Grösse ist also gleich Null ist. Somit gilt ab was zu zeigen war.
Contained in these collections: