Differenzierbarkeit via Komponenten
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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Exercise:
Sei Usubseteq mathbbR^n offen und f:Urightarrow mathbbR^m eine Funktion. Dann ist f genau dann bei x_in U differenzierbar wenn die Komponenten f_kpi_kcirc f für jedes kin ...m bei x_ differenzierbar sind. In diesem Fall gilt textD_x_f pmatrix textD_x_f_ vdots textD_x_f_m pmatrix.
Solution:
Beweis. Es wird nur eine Implikation bewiesen andere ist Übung. Angenommen f_k ist für jedes kin ...m bei x_ differenzierbar. Dann gilt für kin ...m f_kx_+hf_kx_+textD_x_f_kh+alpha_kx_h für gewisse Funktionen alpha_k mit alpha_kx_ho||h|| für hrightarrow . Daraus folgt aber fx_+h pmatrix f_x_+h vdots f_mx_+h pmatrix pmatrix f_x_ vdots f_mx_ pmatrix + pmatrix textD_x_f_ vdots textD_x_f_m pmatrixh + alphax_h wobei alphax_h pmatrix alpha_x_+h vdots alpha_mx_+h pmatrix o||h|| für hrightarrow gilt. Also ist f differenzierbar und es gilt die behauptete Formel für textD_x_f.
Sei Usubseteq mathbbR^n offen und f:Urightarrow mathbbR^m eine Funktion. Dann ist f genau dann bei x_in U differenzierbar wenn die Komponenten f_kpi_kcirc f für jedes kin ...m bei x_ differenzierbar sind. In diesem Fall gilt textD_x_f pmatrix textD_x_f_ vdots textD_x_f_m pmatrix.
Solution:
Beweis. Es wird nur eine Implikation bewiesen andere ist Übung. Angenommen f_k ist für jedes kin ...m bei x_ differenzierbar. Dann gilt für kin ...m f_kx_+hf_kx_+textD_x_f_kh+alpha_kx_h für gewisse Funktionen alpha_k mit alpha_kx_ho||h|| für hrightarrow . Daraus folgt aber fx_+h pmatrix f_x_+h vdots f_mx_+h pmatrix pmatrix f_x_ vdots f_mx_ pmatrix + pmatrix textD_x_f_ vdots textD_x_f_m pmatrixh + alphax_h wobei alphax_h pmatrix alpha_x_+h vdots alpha_mx_+h pmatrix o||h|| für hrightarrow gilt. Also ist f differenzierbar und es gilt die behauptete Formel für textD_x_f.
Meta Information
Exercise:
Sei Usubseteq mathbbR^n offen und f:Urightarrow mathbbR^m eine Funktion. Dann ist f genau dann bei x_in U differenzierbar wenn die Komponenten f_kpi_kcirc f für jedes kin ...m bei x_ differenzierbar sind. In diesem Fall gilt textD_x_f pmatrix textD_x_f_ vdots textD_x_f_m pmatrix.
Solution:
Beweis. Es wird nur eine Implikation bewiesen andere ist Übung. Angenommen f_k ist für jedes kin ...m bei x_ differenzierbar. Dann gilt für kin ...m f_kx_+hf_kx_+textD_x_f_kh+alpha_kx_h für gewisse Funktionen alpha_k mit alpha_kx_ho||h|| für hrightarrow . Daraus folgt aber fx_+h pmatrix f_x_+h vdots f_mx_+h pmatrix pmatrix f_x_ vdots f_mx_ pmatrix + pmatrix textD_x_f_ vdots textD_x_f_m pmatrixh + alphax_h wobei alphax_h pmatrix alpha_x_+h vdots alpha_mx_+h pmatrix o||h|| für hrightarrow gilt. Also ist f differenzierbar und es gilt die behauptete Formel für textD_x_f.
Sei Usubseteq mathbbR^n offen und f:Urightarrow mathbbR^m eine Funktion. Dann ist f genau dann bei x_in U differenzierbar wenn die Komponenten f_kpi_kcirc f für jedes kin ...m bei x_ differenzierbar sind. In diesem Fall gilt textD_x_f pmatrix textD_x_f_ vdots textD_x_f_m pmatrix.
Solution:
Beweis. Es wird nur eine Implikation bewiesen andere ist Übung. Angenommen f_k ist für jedes kin ...m bei x_ differenzierbar. Dann gilt für kin ...m f_kx_+hf_kx_+textD_x_f_kh+alpha_kx_h für gewisse Funktionen alpha_k mit alpha_kx_ho||h|| für hrightarrow . Daraus folgt aber fx_+h pmatrix f_x_+h vdots f_mx_+h pmatrix pmatrix f_x_ vdots f_mx_ pmatrix + pmatrix textD_x_f_ vdots textD_x_f_m pmatrixh + alphax_h wobei alphax_h pmatrix alpha_x_+h vdots alpha_mx_+h pmatrix o||h|| für hrightarrow gilt. Also ist f differenzierbar und es gilt die behauptete Formel für textD_x_f.
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