Stetigkeit Anwendung
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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Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Sei I ein offenes Intervall und f:I rightarrow mathbbR eine stetige Abbildung. Sei x_ in I so dass fx_ neq gilt. Beweisen Sie dass es ein delta gibt s.d. fy neq für alle y in x_-deltax_+delta gilt.
Solution:
Beweis. Da f bei x_ stetig ist gilt dass für alle epsilon ein delta existiert s.d. für alle x in I mit |x-x_| delta |fx-fx_| epsilon gilt. Weiter gelten: abcliste abc |x-x_| delta ist äquivalent zu x in x_-delta x_+delta abc |fx-fx_| epsilon ist äquivalent zu fx in fx_-epsilon fx_+epsilon abcliste Man setzt epsilon : frac|fx_|. Bemerke: epsilon und notin fx_-epsilonfx_+epsilon. Da f bei x_ stetig ist und da epsilon ist existiert nach Definition der Stetigkeit ein delta s.d. für alle x in x_-deltax_+delta fxin fx_-epsilonfx_+epsilon folgt. Da notin fx_-epsilonfx_+epsilon ist die Behauptung bewiesen.
Sei I ein offenes Intervall und f:I rightarrow mathbbR eine stetige Abbildung. Sei x_ in I so dass fx_ neq gilt. Beweisen Sie dass es ein delta gibt s.d. fy neq für alle y in x_-deltax_+delta gilt.
Solution:
Beweis. Da f bei x_ stetig ist gilt dass für alle epsilon ein delta existiert s.d. für alle x in I mit |x-x_| delta |fx-fx_| epsilon gilt. Weiter gelten: abcliste abc |x-x_| delta ist äquivalent zu x in x_-delta x_+delta abc |fx-fx_| epsilon ist äquivalent zu fx in fx_-epsilon fx_+epsilon abcliste Man setzt epsilon : frac|fx_|. Bemerke: epsilon und notin fx_-epsilonfx_+epsilon. Da f bei x_ stetig ist und da epsilon ist existiert nach Definition der Stetigkeit ein delta s.d. für alle x in x_-deltax_+delta fxin fx_-epsilonfx_+epsilon folgt. Da notin fx_-epsilonfx_+epsilon ist die Behauptung bewiesen.
Meta Information
Exercise:
Sei I ein offenes Intervall und f:I rightarrow mathbbR eine stetige Abbildung. Sei x_ in I so dass fx_ neq gilt. Beweisen Sie dass es ein delta gibt s.d. fy neq für alle y in x_-deltax_+delta gilt.
Solution:
Beweis. Da f bei x_ stetig ist gilt dass für alle epsilon ein delta existiert s.d. für alle x in I mit |x-x_| delta |fx-fx_| epsilon gilt. Weiter gelten: abcliste abc |x-x_| delta ist äquivalent zu x in x_-delta x_+delta abc |fx-fx_| epsilon ist äquivalent zu fx in fx_-epsilon fx_+epsilon abcliste Man setzt epsilon : frac|fx_|. Bemerke: epsilon und notin fx_-epsilonfx_+epsilon. Da f bei x_ stetig ist und da epsilon ist existiert nach Definition der Stetigkeit ein delta s.d. für alle x in x_-deltax_+delta fxin fx_-epsilonfx_+epsilon folgt. Da notin fx_-epsilonfx_+epsilon ist die Behauptung bewiesen.
Sei I ein offenes Intervall und f:I rightarrow mathbbR eine stetige Abbildung. Sei x_ in I so dass fx_ neq gilt. Beweisen Sie dass es ein delta gibt s.d. fy neq für alle y in x_-deltax_+delta gilt.
Solution:
Beweis. Da f bei x_ stetig ist gilt dass für alle epsilon ein delta existiert s.d. für alle x in I mit |x-x_| delta |fx-fx_| epsilon gilt. Weiter gelten: abcliste abc |x-x_| delta ist äquivalent zu x in x_-delta x_+delta abc |fx-fx_| epsilon ist äquivalent zu fx in fx_-epsilon fx_+epsilon abcliste Man setzt epsilon : frac|fx_|. Bemerke: epsilon und notin fx_-epsilonfx_+epsilon. Da f bei x_ stetig ist und da epsilon ist existiert nach Definition der Stetigkeit ein delta s.d. für alle x in x_-deltax_+delta fxin fx_-epsilonfx_+epsilon folgt. Da notin fx_-epsilonfx_+epsilon ist die Behauptung bewiesen.
Contained in these collections
| Title | Matched on |
|---|---|
| Beweis Stetigkeit unter Verknüpfung | tags |
| Stetigkeit und Division | tags |
| Stetigkeit, Injektivität, Monotonie | tags |
| Über den Konvergenzradius | tags |
| Gleichmässige Stetigkeit (Heine) | tags |