Satz Bolzano-Weierstrass
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
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\(\LaTeX\)
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Exercise:
Für eine reelle beschränkte Folge a_n_n gilt lim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n genau dann wenn a_n Jede beschränkte reelle Folge hat eine konvergente Teilfolge.
Solution:
Beweis. Angenommen Alim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n und epsilon . Dann existiert nach Satz . ein N s.d. a_n leq A+fracepsilon für alle n geq N und a_n geq A-fracepsilon für alle n geq N. Gemeinsam ergibt sich |a_n-A| epsilon für alle n geq N was zu zeigen war. Angenommen dass Alim limes_n rightarrow infty a_n existiert. Sei epsilon . Dann existiert ein N s.d. A-epsilon a_n A+epsilon für alle n geq N. Aus I_n textinfa_k|k geq n und S_n textsupa_k|k geq n folgt nun A-epsilon leq I_N leq I_n leq S_n leq S_N leq A+epsilon für alle n geq N und daher A-epsilon leq lim textinf_n rightarrow infty a_n leq lim textsup_n rightarrow infty a_n leq A+epsilon Da dies für alle epsilon gilt folgt daraus Alim textinf_n rightarrow infty a_nlim textsup_n rightarrow infty a_n.
Für eine reelle beschränkte Folge a_n_n gilt lim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n genau dann wenn a_n Jede beschränkte reelle Folge hat eine konvergente Teilfolge.
Solution:
Beweis. Angenommen Alim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n und epsilon . Dann existiert nach Satz . ein N s.d. a_n leq A+fracepsilon für alle n geq N und a_n geq A-fracepsilon für alle n geq N. Gemeinsam ergibt sich |a_n-A| epsilon für alle n geq N was zu zeigen war. Angenommen dass Alim limes_n rightarrow infty a_n existiert. Sei epsilon . Dann existiert ein N s.d. A-epsilon a_n A+epsilon für alle n geq N. Aus I_n textinfa_k|k geq n und S_n textsupa_k|k geq n folgt nun A-epsilon leq I_N leq I_n leq S_n leq S_N leq A+epsilon für alle n geq N und daher A-epsilon leq lim textinf_n rightarrow infty a_n leq lim textsup_n rightarrow infty a_n leq A+epsilon Da dies für alle epsilon gilt folgt daraus Alim textinf_n rightarrow infty a_nlim textsup_n rightarrow infty a_n.
Meta Information
Exercise:
Für eine reelle beschränkte Folge a_n_n gilt lim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n genau dann wenn a_n Jede beschränkte reelle Folge hat eine konvergente Teilfolge.
Solution:
Beweis. Angenommen Alim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n und epsilon . Dann existiert nach Satz . ein N s.d. a_n leq A+fracepsilon für alle n geq N und a_n geq A-fracepsilon für alle n geq N. Gemeinsam ergibt sich |a_n-A| epsilon für alle n geq N was zu zeigen war. Angenommen dass Alim limes_n rightarrow infty a_n existiert. Sei epsilon . Dann existiert ein N s.d. A-epsilon a_n A+epsilon für alle n geq N. Aus I_n textinfa_k|k geq n und S_n textsupa_k|k geq n folgt nun A-epsilon leq I_N leq I_n leq S_n leq S_N leq A+epsilon für alle n geq N und daher A-epsilon leq lim textinf_n rightarrow infty a_n leq lim textsup_n rightarrow infty a_n leq A+epsilon Da dies für alle epsilon gilt folgt daraus Alim textinf_n rightarrow infty a_nlim textsup_n rightarrow infty a_n.
Für eine reelle beschränkte Folge a_n_n gilt lim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n genau dann wenn a_n Jede beschränkte reelle Folge hat eine konvergente Teilfolge.
Solution:
Beweis. Angenommen Alim textinf_n rightarrow infty a_n lim textsup_n rightarrow infty a_n und epsilon . Dann existiert nach Satz . ein N s.d. a_n leq A+fracepsilon für alle n geq N und a_n geq A-fracepsilon für alle n geq N. Gemeinsam ergibt sich |a_n-A| epsilon für alle n geq N was zu zeigen war. Angenommen dass Alim limes_n rightarrow infty a_n existiert. Sei epsilon . Dann existiert ein N s.d. A-epsilon a_n A+epsilon für alle n geq N. Aus I_n textinfa_k|k geq n und S_n textsupa_k|k geq n folgt nun A-epsilon leq I_N leq I_n leq S_n leq S_N leq A+epsilon für alle n geq N und daher A-epsilon leq lim textinf_n rightarrow infty a_n leq lim textsup_n rightarrow infty a_n leq A+epsilon Da dies für alle epsilon gilt folgt daraus Alim textinf_n rightarrow infty a_nlim textsup_n rightarrow infty a_n.
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