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e
пятая буква английского алфавита; - прописная, - строчная
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1. Well, if that's true, then of course, S together with
E
2. BV & KT: (both fingerspell) K-A-T-
E
-L-Y-N
3. that it is an element in
E
, but not in S.
4. contains both S and
E
. So it contains the n
5. KT: K-
E
-T-
E
-L-Y-N
6.
e
^(pi i) = -1
7. where we assume that
E
is not in
E
star.
8. Well, there exists as edge in
E
minus S--
9. but
e
^x alone is the one that's actually equal to its own derivative
10. now you might ask "Why
e
? Why not some other base?"
11. So it's not spread like an "
e
" sound or an "ahh" sound.
12. for any pair of vertices, even after removal of
e
,
13. It has edges
E
star.
14. Looks like
e
is not listening to you.
15. alright, so
e
^x transforming the plane
16. So suppose this does not contain
e
.
17. after removing
e
from T.
18. does not contain this particular edge
e
that I remove here.
19. always a subset of
E
, right?
20. KT:(fingerspells) L-I-N-D-S-
E
-Y
21. is also a subset of
E
star.
22. part of
E
star.
23. contains both S and also
E
.
24. Because I have removed
e
from a cycle.
25. If this path does not contain the edge
e
,
26. "o" and "
e
", so there's many different spellings for this.
27. T that has the vertex set V and an edge set
E
.
28. BV: ok hey, K-A KT:K-A BV: see "A" yes BV/KT: fingerspell K-A-T-
E
-L-Y-N
29. that S is a subset of
E
.
30. the graph T minus this edge
e
is still connected.
31. I mean, that's the birthplace of
e
, and where it's even defined
32. The main thing is to remember that "a", "
e
", "i", "o", "u", and sometimes "y" are vowels.
33. BV: ok, I looked away K-A-T-
E
-L-Y-N
34.
E
has exactly n minus 1 edges, because it's a tree.
35. KT: K-A-T-
E
-L-Y-N
36. Well, I know that
E
is the edge set
37. connected in the new graph where I've removed this
e
.
38. what makes the number
e
special is that when the exponential
e
^x map vertical slides to rotations
39. It might be in the shape of an "o", it might be "o" and "
e
", it might be "o" and "a", "ow",
40. that the empty set is a subset of
E
. Well, the empty set is
41.
E
has successfully n minus 1 edges.
42. The "d", "r", and the "w", they're consonants, and the "
e
" is a vowel.
43. And we know that S is actually a subset of
E
star.
44. The first case could be that
E
is actually already
45. So we need to prove something about the union of S and
E
.
46. KT:(fingerspells) L-I-N-D-S-
E
-Y
47. So I have here "a", "
e
", "i", "o", "u", and sometimes "y".
48. the "
e
" is a vowel and so is the "o".
49. to finish things off here, I want to show a way you can think about this function
e
^x as a transformation of the complex plane
50.
e
-s-c-r-i-m-a
51. and more over,
e
^x does this in a very special way, that ensures that a vertical slide of pi units corresponds to rotation of exactly pi radians
52. And we know that S is actually a subset of
E
star.
53. "o" and "
e
", "o" and "u", and "
e
" and "w".
54. BV: Oh that, (fingerspells) K-A-T-
E
-L-Y-N. OK.
55. So let
E
denote the edge that I added in the m plus 1 step.
56. Girl: (fingerspells) K-A-T-
E
-L-Y-N
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