Coordinate map
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:
Phi_mathcalB is an isomorphism. Phi_mathcalB:Vlongrightarrow textK^n Phi_mathcalBv:v_mathcalBin textK^n
Solution:
Proof. We first show that Phi_mathcalB is a linear map. Let abin K vv'in V. Write vv' as linear combination of the elements of mathcalB: va_v_+...+a_nv_n v'b_v_+...+b_nv_n with a_ia_i'in K. By definition v_mathcalBleftarrayc a_ vdots a_n arrayrightquad v'_mathcalBleftarrayc b_ vdots b_n arrayright. Now av+bv'aa_+bb_v_+...+aa_n+bb_nv_n hence Phi_mathcalBav+bv'av+bv'_mathcalB leftarrayc aa_+bb_ vdots aa_n+bb_n arrayright aleftarrayc a_ vdots a_n arrayright+bleftarrayc b_ vdots b_n arrayright a Phi_mathcalBv+b Phi_mathcalBv'. This proves Phi_mathcalB is linear. We now claim that Phi_mathcalB is an isomorphism. By previous results it's enough to show Phi_mathcalB is bijective. Indeed: If vin textKerPhi_mathcalB then v v_+...+ v_n Longrightarrow textKerPhi_mathcalB. This shows that Phi_mathcalB is injective. Surjectivity of Phi_mathcalB: Let leftarrayc a_ vdots a_n arrayright in K^n Put va_v_+...+a_nv_n. Then Phi_mathcalBvleftarrayc a_ vdots a_n arrayright. Longrightarrow textImPhi_mathcalBK^n.
Phi_mathcalB is an isomorphism. Phi_mathcalB:Vlongrightarrow textK^n Phi_mathcalBv:v_mathcalBin textK^n
Solution:
Proof. We first show that Phi_mathcalB is a linear map. Let abin K vv'in V. Write vv' as linear combination of the elements of mathcalB: va_v_+...+a_nv_n v'b_v_+...+b_nv_n with a_ia_i'in K. By definition v_mathcalBleftarrayc a_ vdots a_n arrayrightquad v'_mathcalBleftarrayc b_ vdots b_n arrayright. Now av+bv'aa_+bb_v_+...+aa_n+bb_nv_n hence Phi_mathcalBav+bv'av+bv'_mathcalB leftarrayc aa_+bb_ vdots aa_n+bb_n arrayright aleftarrayc a_ vdots a_n arrayright+bleftarrayc b_ vdots b_n arrayright a Phi_mathcalBv+b Phi_mathcalBv'. This proves Phi_mathcalB is linear. We now claim that Phi_mathcalB is an isomorphism. By previous results it's enough to show Phi_mathcalB is bijective. Indeed: If vin textKerPhi_mathcalB then v v_+...+ v_n Longrightarrow textKerPhi_mathcalB. This shows that Phi_mathcalB is injective. Surjectivity of Phi_mathcalB: Let leftarrayc a_ vdots a_n arrayright in K^n Put va_v_+...+a_nv_n. Then Phi_mathcalBvleftarrayc a_ vdots a_n arrayright. Longrightarrow textImPhi_mathcalBK^n.
Meta Information
Exercise:
Phi_mathcalB is an isomorphism. Phi_mathcalB:Vlongrightarrow textK^n Phi_mathcalBv:v_mathcalBin textK^n
Solution:
Proof. We first show that Phi_mathcalB is a linear map. Let abin K vv'in V. Write vv' as linear combination of the elements of mathcalB: va_v_+...+a_nv_n v'b_v_+...+b_nv_n with a_ia_i'in K. By definition v_mathcalBleftarrayc a_ vdots a_n arrayrightquad v'_mathcalBleftarrayc b_ vdots b_n arrayright. Now av+bv'aa_+bb_v_+...+aa_n+bb_nv_n hence Phi_mathcalBav+bv'av+bv'_mathcalB leftarrayc aa_+bb_ vdots aa_n+bb_n arrayright aleftarrayc a_ vdots a_n arrayright+bleftarrayc b_ vdots b_n arrayright a Phi_mathcalBv+b Phi_mathcalBv'. This proves Phi_mathcalB is linear. We now claim that Phi_mathcalB is an isomorphism. By previous results it's enough to show Phi_mathcalB is bijective. Indeed: If vin textKerPhi_mathcalB then v v_+...+ v_n Longrightarrow textKerPhi_mathcalB. This shows that Phi_mathcalB is injective. Surjectivity of Phi_mathcalB: Let leftarrayc a_ vdots a_n arrayright in K^n Put va_v_+...+a_nv_n. Then Phi_mathcalBvleftarrayc a_ vdots a_n arrayright. Longrightarrow textImPhi_mathcalBK^n.
Phi_mathcalB is an isomorphism. Phi_mathcalB:Vlongrightarrow textK^n Phi_mathcalBv:v_mathcalBin textK^n
Solution:
Proof. We first show that Phi_mathcalB is a linear map. Let abin K vv'in V. Write vv' as linear combination of the elements of mathcalB: va_v_+...+a_nv_n v'b_v_+...+b_nv_n with a_ia_i'in K. By definition v_mathcalBleftarrayc a_ vdots a_n arrayrightquad v'_mathcalBleftarrayc b_ vdots b_n arrayright. Now av+bv'aa_+bb_v_+...+aa_n+bb_nv_n hence Phi_mathcalBav+bv'av+bv'_mathcalB leftarrayc aa_+bb_ vdots aa_n+bb_n arrayright aleftarrayc a_ vdots a_n arrayright+bleftarrayc b_ vdots b_n arrayright a Phi_mathcalBv+b Phi_mathcalBv'. This proves Phi_mathcalB is linear. We now claim that Phi_mathcalB is an isomorphism. By previous results it's enough to show Phi_mathcalB is bijective. Indeed: If vin textKerPhi_mathcalB then v v_+...+ v_n Longrightarrow textKerPhi_mathcalB. This shows that Phi_mathcalB is injective. Surjectivity of Phi_mathcalB: Let leftarrayc a_ vdots a_n arrayright in K^n Put va_v_+...+a_nv_n. Then Phi_mathcalBvleftarrayc a_ vdots a_n arrayright. Longrightarrow textImPhi_mathcalBK^n.
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