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-rw-r--r--sem3/BfCI/husk.txt14
-rw-r--r--sem3/BfCI/mm1.md177
-rw-r--r--sem3/BfCI/mm3.md19
3 files changed, 210 insertions, 0 deletions
diff --git a/sem3/BfCI/husk.txt b/sem3/BfCI/husk.txt
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+
+
+log(log(x^4)) = log(log(x))
+
+Husk at 4 flytter sig uden for log(x).
+
+
+x^2-1 = (x-1)(x+1)
+
+lim(to inf) f(x) over g(x) = {
+ 0 -> f is O(g)
+ const -> f(x) = THETA(g(x))
+ INF -> f is OMEGA(g)
+}
diff --git a/sem3/BfCI/mm1.md b/sem3/BfCI/mm1.md
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+# Logic
+
+TODO Manger at lave en summary.
+
+## Opgaver
+
+### 1.
+
+A) Yes, T
+B) Yes, T
+C) Yes, T
+D) Yes, T
+E) No
+F) No
+
+### UNKNOWN ORIGIN (12.)
+
+A) if you have the flu, you miss the final examination
+B) you pass the course if and only if you do not miss the final examination
+C) if you miss the final examination, you do not pass the course
+D) you have the flu, miss the final examination and pass the couse
+
+### 15. (13.)
+
+A) ¬p
+B) p ^ ¬q
+C) p -> q
+D) ¬p -> ¬q
+E) p -> q
+F) q ^ ¬p
+G) q -> p
+
+### 20. (18.)
+
+A) T
+B) T
+C) F
+D) T
+
+### UNKNOWN ORIGIN(31.)
+
+A)
+
+| p | ¬p | p ^ ¬p |
+| -- | -- | -- |
+| T | F | F |
+| F | T | F |
+
+B)
+
+| p | ¬p | p v ¬p |
+| -- | -- | -- |
+| T | F | T |
+| F | T | T |
+
+C)
+
+| p | q | p v ¬q | ( p v ¬q) -> q |
+| -- | -- | -- | -- |
+| T | T | T | T |
+| T | F | T | F |
+| F | T | F | T |
+| F | F | T | F |
+
+D)
+
+| p | q | ( p v q) -> ( p ^ q ) |
+| -- | -- | -- |
+| T | T | T -> T = T |
+| T | F | T -> F = F |
+| F | T | T -> F = F |
+| F | F | F -> F = T |
+
+E)
+
+| p | q | (p -> q) | (¬q -> ¬p) | (p -> q) <-> (¬q -> ¬p ) |
+| -- | -- | -- | -- | -- |
+| T | T | T | T | T |
+| T | F | F | F | T |
+| F | T | T | T | T |
+| F | F | T | T | T |
+
+F)
+
+| p | q | (p -> q) | ( q -> p ) | (p -> q) -> ( q -> p) |
+| -- | -- | -- | -- | -- |
+| T | T | T | T | T |
+| T | F | F | T | T |
+| F | T | T | F | F |
+| F | F | T | T | T |
+
+
+### (40. )
+
+All the paranterees must be true, they can be split up.
+
+r is in two parantesees ( q v ¬r ) and ( r v ¬p ).
+
+If r is true q must also be true for ( q v ¬p ).
+
+Therefore in ( p v ¬p ) p must also be true.
+
+Therefore it creates a kind of circle, that also works then one is false.
+
+### (44. )
+
+A)
+
+ 01011
+ 11011
+v-----
+ 11011
+ 11000
+^-----
+ 11000
+
+B)
+
+ 01111
+ 10101
+^-----
+ 00101
+ 01000
+v-----
+ 01101
+
+C)
+
+ 01010
+ 11011
+x-----
+ 10001
+ 01000
+x-----
+ 11001
+
+D)
+
+ 11011
+ 01010
+v-----
+ 11011
+
+ 10001
+ 11011
+v-----
+ 11011
+
+ 11011
+ 11011
+^-----
+ 11011
+
+
+### (7. )
+
+Already done this
+
+### (9. )
+
+The rest is just stupid.
+
+### (20. )
+
+| p | q | p x q | p <-> q |
+| -- | -- | -- | -- |
+| T | T | F | T |
+| T | F | T | F |
+| F | T | T | F |
+| F | F | F | T |
+
+The two last columns are the negatives of each other thus
+
+p x q = p <-> q
+
+
+
diff --git a/sem3/BfCI/mm3.md b/sem3/BfCI/mm3.md
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+# Lektier
+
+## Steps til strong induction bevis
+
+1. *[Basis step]* start med at bevis at P(1) er sandt. Altså hvad resten vil bygge på.
+2. *[Inductive step]* bevis at hvis de første P(1 til k) er sandt er P(k+1) også sandt.
+
+## Recursive function
+
+En funktion der regner factorial.
+Altså f(x) = 1 * 2 * 3 * .. * x
+
+Her vil f(1) = 1.
+Og resten vil være f(x) = x * f(x-1).
+
+f(1) = 1
+f(2) = 2 * f(1) = 2 * 1 = 2
+f(3) = 3 * f(2) = 3 * 2 = 6
+f(4) = 4 * f(3) = 4 * 6 = 16