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Style exercise/solutions to make more distinct from surroundings
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‎Book/style.css‎

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@@ -126,6 +126,19 @@ blockquote {
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border-left: 3px solid #BABABA;
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}
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details {
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margin-bottom: 30px;
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}
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/* Exercise solutions / spoilers */
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details > div {
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margin-top: 10px;
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margin-left: 30px;
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padding: 10px 20px;
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border-radius: 10px;
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background-color: #fcfbf0;
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}
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/* VVVV GENERATED BY PANDOC BELOW VVV */
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code {

‎Physics/src/Calculus/DifferentialCalc.lhs‎

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@@ -306,8 +306,18 @@ And the limit
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$$\lim_{x \to 0} \frac{sin x}{x} = 1$$
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which can be proved using the unit circle and squeeze theorem, but we
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won't do that here.
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the proof of which is left as an exercise to the reader
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**Exercise.** Prove the limit $\lim_{x \to 0} \frac{sin x}{x} = 1$
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<details>
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<summary>**Hint**</summary>
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<div>
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This limit can be proven using the [unit circle](https://en.wikipedia.org/wiki/Unit_circle) and [squeeze theorem](https://en.wikipedia.org/wiki/Squeeze_theorem)
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</div>
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</details>
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Then, the differentiation
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> derive Sin = Cos
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I'll leave the proving of the rest of the implementations as an exercise to you, the reader.
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**Exercise.** Derive the rest of the cases using the definition of the derivative
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<details>
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<summary>**Solution**</summary>
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<div>
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> derive Exp = Exp
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> derive Log = Const 1 :/ Id
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> derive (f :- g) = derive f :- derive g
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> derive (f :* g) = derive f :* g :+ f :* derive g
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> derive (f :/ g) = (derive f :* g :- f :* derive g) :/ (g:^(Const 2))
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> derive (f :^ g) = f:^(g :- Const 1) :* (g :* derive f :+ f :* (Log :. f) :* derive g)
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> derive (f :^ g) = f:^(g :- Const 1)
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> :* (g :* derive f :+ f :* (Log :. f) :* derive g)
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> derive Id = Const 1
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> derive (Const _) = Const 0
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> derive (f :. g) = derive g :* (derive f :. g)
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> derive (I c f) = f
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</div>
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</details>
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Keep it simple
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----------------------------------------------------------------------
@@ -452,6 +468,21 @@ Exponentiation is not commutative, and further has no (two-sided) identity eleme
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> (f', Const 1) -> f'
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> (f', g') -> f' :^ g'
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**Exercises.** Look up (or prove by yourself) more identities (of
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expressions, not identity elements) for exponentiation and implement
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them.
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<details>
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<summary>**Solution**</summary>
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<div>
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For example, there is the identity of negative exponents. For any integer $n$ and nonzero $b$
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$$b^{-n} = \frac{1}{b^n}$$
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</div>
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</details>
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Intuitively, the identity function is the identity element for function composition
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> simplify (f :. g) = case (simplify f, simplify g) of
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< (cos + (2 * id))
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A sight for sore eyes!
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**Exercise.** Think of more ways an expression can be "simplified", and add your cases to the implementation.

‎Physics/src/Calculus/IntegralCalc.lhs‎

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@@ -396,6 +396,19 @@ which at least implies that we can use polynomials.
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> (Id, Const c) -> Id:^(Const (c+1)) :/ Const (c+1)
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> (_, _) -> error "Can't integrate integrals like that!"
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**Exercise.** Find more rules of integrating exponentials and add to
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the implementation.
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<details>
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<summary>**Solution**</summary>
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<div>
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Wikipedia has [a nice list of integrals of exponentials](https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
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</div>
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</details>
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Integration of function composition is, simply said, somewhat
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complicated. The technique to use is called "integration by
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substitution", and is something like a reverse of the chain-rule of

‎Physics/src/Dimensions/Quantity.lhs‎

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@@ -106,11 +106,16 @@ Note that the `Quantity` data type has both value-level and type-level dimension
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TODO: What is the purpose (semantics, intended meaning or use) of Quantity'?
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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< data Quantity' (d1 :: T.Dim) (d2 :: T.Dim) where
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< Quantity' :: Rational -> Rational -> Quantity' d1 d2
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</div>
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</details>
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A taste of types
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----------------
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![Are you a cowboy?](Cowboy.png){.float-img-right}
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**Solution.** Hold on cowboy! Not so fast. We'll come back later to subtraction and division, so check your solution then.
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<details>
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<summary>**Solution**</summary>
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<div>
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Hold on cowboy! Not so fast. We'll come back later to subtraction and division, so check your solution then.
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</div>
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</details>
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Now on to some example values and types.
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**Exercise.** What is the volume of a cuboid with sides $1.2\ m$, $0.4\ m$ and $1.9\ m$? Create values for the involved quantities. Use `Float` instead of `Double`.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> side1 :: Quantity Length Float
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> side1 = Quantity V.length 1.2
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> volume :: Quantity Volume Float
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> volume = quantityMul side1 (quantityMul side2 side3)
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</div>
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</details>
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Pretty-printer
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--------------
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**Exercise.** Make `Quantity` an instance of `Ord`.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> quantityCompare :: (Ord v) => Quantity d v ->
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> Quantity d v -> Ordering
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> instance (Ord v) => Ord (Quantity d v) where
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> compare = quantityCompare
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</div>
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</details>
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We often use `Double` as the value holding type. Doing exact comparsions isn't always possible to due rounding errors. Therefore, we'll create a `~=` function for testing if two quantities are almost equal.
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> infixl 4 ~=
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**Exercise.** `Quantity One` has compatible types. Make it an instance of `Num`, `Fractional`, `Floating` and `Functor`.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> instance (Num v) => Num (Quantity One v) where
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> (+) = (+#)
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> instance Functor (Quantity One) where
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> fmap = qmap
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</div>
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</details>
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Syntactic sugar
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---------------
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**Exercise.** Do the rest.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> current :: (Num v) => Quantity Current v
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> current = Quantity V.current 0
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> one :: (Num v) => Quantity One v
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> one = Quantity V.one 0
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</div>
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</details>
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And now the sugar.
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TODO: Please multiply the values instead of throwing them away. I'm pretty sure (#) should behave as "scale" of a vector space. So that (x#(y#z)) == ((x*y)#z).
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**Exericse.** Define pre-made values for velocity, acceleration, force, momentum and energy.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> velocity = length /# time
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> acceleration = velocity /# time
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> force = mass *# acceleration
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> momentum = force *# time
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> energy = force *# length
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</div>
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</details>
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---
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Alternative sugar with support for units
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< ghci> distance
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< 8046.72 m
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> metre = 1 # length
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< ghci> velocity1 +# velocity2
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< 67.77 m/s
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</div>
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</details>
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A physics example
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-----------------
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‎Physics/src/Dimensions/TypeLevel.lhs‎

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**Exercise.** Create types for velocity, acceleration and the scalar.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> type Velocity = 'Dim Pos1 Zero Neg1 Zero Zero Zero Zero
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> type Acceleration = 'Dim Pos1 Zero Neg2 Zero Zero Zero Zero
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> type One = 'Dim Zero Zero Zero Zero Zero Zero Zero
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</div>
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</details>
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Multiplication and division
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---------------------------
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**Exercise.** As you would suspect, division is very similar, so why don't you try 'n implement it yourself?
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> type family Div (d1 :: Dim) (d2 :: Dim) where
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> Div ('Dim le1 ma1 ti1 cu1 te1 su1 lu1)
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> 'Dim (le1-le2) (ma1-ma2) (ti1-ti2) (cu1-cu2)
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> (te1-te2) (su1-su2) (lu1-lu2)
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</div>
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</details>
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**Exercise.** Implement a type-level function for raising a dimension to the power of some integer.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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< type family Power (d :: Dim) (n :: TypeInt) where
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< Power ('Dim le ma ti cu te su lu) n =
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< 'Dim (le*n) (ma*n) (ti*n) (cu*n) (te*n) (su*n) (lu*n)
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</div>
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Now types for dimensions can be created by combining exisiting types, much like we did for values in the previous chapter.
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**Exercise.** Create types for velocity, area, force and impulse.
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**Solution.**
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<details>
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<summary>**Solution**</summary>
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<div>
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> type Velocity' = Length `Div` Time
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> type Area = Length `Mul` Length
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</div>
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</details>
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Perhaps not very exiting so far. But just wait 'til we create a data type for quantities. Then the strenghts of type-level dimensions will be clearer.

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