Orthogonal projection characteristics
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:
P_U has the following properties: abcliste abc P_U is a linear map. abc textIm P_U U textKer P_U U^perp. abc forall vin V v-P_Uvin U^perp. abc forall uin U P_Uuu. abc If e_...e_r is any orthonormal basis for U then P_Uv _i^r langle ve_irangle e_i. tildeP_Uv from the previous proof. abcliste
Solution:
Proof. abcliste a-d follow from the general fact that if W_ W_ are vector subspaces of W and W_oplus W_W then the map p_W_:Wrightarrow W_ defined by P_W_ww_ where we write ww_+w_ is a unique way with w_in W_ w_in W_ is a linear map textIm P_W_W_ textKer P_W_W_ and forall win W we have w-P_W_win W_. So wP_W_w+w-P_W_w is the decoposition of w as ww_+w_. We also have P_W_wwquad forall win W_. For e write vleft _i^r langle v e_irangle e_iright + v-_i^r langle v e_irangle e_i *. Now left _i^r langle v e_irangle e_irighttildeP_Uv from the previous proof and we have also seen in that proof that v-_i^r langle v e_irangle e_iin U^perp. Longrightarrow * is the unique decomposition of v with respect to VUoplus U^perpLongrightarrow P_Uvleft _i^r langle v e_irangle e_irighttildeP_Uv. abcliste
P_U has the following properties: abcliste abc P_U is a linear map. abc textIm P_U U textKer P_U U^perp. abc forall vin V v-P_Uvin U^perp. abc forall uin U P_Uuu. abc If e_...e_r is any orthonormal basis for U then P_Uv _i^r langle ve_irangle e_i. tildeP_Uv from the previous proof. abcliste
Solution:
Proof. abcliste a-d follow from the general fact that if W_ W_ are vector subspaces of W and W_oplus W_W then the map p_W_:Wrightarrow W_ defined by P_W_ww_ where we write ww_+w_ is a unique way with w_in W_ w_in W_ is a linear map textIm P_W_W_ textKer P_W_W_ and forall win W we have w-P_W_win W_. So wP_W_w+w-P_W_w is the decoposition of w as ww_+w_. We also have P_W_wwquad forall win W_. For e write vleft _i^r langle v e_irangle e_iright + v-_i^r langle v e_irangle e_i *. Now left _i^r langle v e_irangle e_irighttildeP_Uv from the previous proof and we have also seen in that proof that v-_i^r langle v e_irangle e_iin U^perp. Longrightarrow * is the unique decomposition of v with respect to VUoplus U^perpLongrightarrow P_Uvleft _i^r langle v e_irangle e_irighttildeP_Uv. abcliste
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Exercise:
P_U has the following properties: abcliste abc P_U is a linear map. abc textIm P_U U textKer P_U U^perp. abc forall vin V v-P_Uvin U^perp. abc forall uin U P_Uuu. abc If e_...e_r is any orthonormal basis for U then P_Uv _i^r langle ve_irangle e_i. tildeP_Uv from the previous proof. abcliste
Solution:
Proof. abcliste a-d follow from the general fact that if W_ W_ are vector subspaces of W and W_oplus W_W then the map p_W_:Wrightarrow W_ defined by P_W_ww_ where we write ww_+w_ is a unique way with w_in W_ w_in W_ is a linear map textIm P_W_W_ textKer P_W_W_ and forall win W we have w-P_W_win W_. So wP_W_w+w-P_W_w is the decoposition of w as ww_+w_. We also have P_W_wwquad forall win W_. For e write vleft _i^r langle v e_irangle e_iright + v-_i^r langle v e_irangle e_i *. Now left _i^r langle v e_irangle e_irighttildeP_Uv from the previous proof and we have also seen in that proof that v-_i^r langle v e_irangle e_iin U^perp. Longrightarrow * is the unique decomposition of v with respect to VUoplus U^perpLongrightarrow P_Uvleft _i^r langle v e_irangle e_irighttildeP_Uv. abcliste
P_U has the following properties: abcliste abc P_U is a linear map. abc textIm P_U U textKer P_U U^perp. abc forall vin V v-P_Uvin U^perp. abc forall uin U P_Uuu. abc If e_...e_r is any orthonormal basis for U then P_Uv _i^r langle ve_irangle e_i. tildeP_Uv from the previous proof. abcliste
Solution:
Proof. abcliste a-d follow from the general fact that if W_ W_ are vector subspaces of W and W_oplus W_W then the map p_W_:Wrightarrow W_ defined by P_W_ww_ where we write ww_+w_ is a unique way with w_in W_ w_in W_ is a linear map textIm P_W_W_ textKer P_W_W_ and forall win W we have w-P_W_win W_. So wP_W_w+w-P_W_w is the decoposition of w as ww_+w_. We also have P_W_wwquad forall win W_. For e write vleft _i^r langle v e_irangle e_iright + v-_i^r langle v e_irangle e_i *. Now left _i^r langle v e_irangle e_irighttildeP_Uv from the previous proof and we have also seen in that proof that v-_i^r langle v e_irangle e_iin U^perp. Longrightarrow * is the unique decomposition of v with respect to VUoplus U^perpLongrightarrow P_Uvleft _i^r langle v e_irangle e_irighttildeP_Uv. abcliste
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