LATEX Practice
Aleph Continuum
February 10, 2017
1 Beginner
1.1 18 April 2016
n
n(n + 1)
1. k=
2
k=1
n
sin k
2.
k
k=1
3. x2 + y 2 = z 2
4.
5. C, f (x) R[x] such that f () = 0
6.
7.
8.
9. , 1 2 , 1 2
10. ABC DEF, ABC
= DEF
11. ~, u
12.
13.
[[ ((
[[ ((
14. [ ( |
1
f
15. ,
x
16. f : (x, y, z) 7 (x + 2y, x + z, 3y z)
v
17. x
y
18. (f g)(x) = f (g(x))
19.
20. Z 1
21. [[x]]
22.
23. H
4Q 1Q
24. Q(3 + i) = Q(4 (1 i)) = Q(4 + (1 + i)) = Q(1 + i) = Q((1 i)) = Q(1 i).
= Q[x]/x2 2 Q[x]/x2 3
25. Q( 2) = Q( 3)?
1 0 0
26. I3 = 0 1 0
0 0 1
{
x, if x 0
27. |x| =
x, if x < 0
0, if x = 0 = y
28. u(x, y) = x3 3xy 2
, otherwise
x2 + y 2
u u(x, 0) u(0, 0)
29. = lim
x (0,0) x0 x
sin t
30. Evaluate dt.
0 ln(1 + x + t)
31. F2 F2 = {(a, b) | a, b F2 }
32. lim xn = x
n
3
9x2
3
33. Use polar coordinates to evaluate (x + xy 2 ) dy dx.
0 9x2
2
34. Let X be a Banach space and let f : B R be a bounded linear transformation on X. The norm of
f , denoted by f , is defined by
f = inf{K [0, +) : |f (x)| Kx for all x X}.
35. In non-relativistic wave mechanics, the wave function (r, t) of a particle satisfies the Schr
odinger
Wave Equation ( )
~2 2 2 2
i~ = + + + V .
t 2m x2 y 2 z 2
It is customary to normalize the wave equation by demanding that
2
|(r, 0)| dx dy dz = 1.
R3
A simple calculation using the Schr
odinger wave equation shows that
d 2
|(r, t)| dx dy dz = 0,
dt R3
and hence
2
|(r, t)| dx dy dz = 1
R3
for all times t. If we normalize the wave function in this way then, for any (measurable) subset V of
R3 and time t,
2
|(r, t)| dx dy dz
V
represents the probability that the particle is to be found within the region V at time t
36. ( )
7x + 5
f (x, y, z) = 3y 2 z 3 + .
1 + y2
37. ( )
3
4x + x + 42 .
1+x
4
ln(x) du
38. Given u(x) = , find .
sin(e2x ) + 1 dx x=0
39.
cos(2) = cos2 () sin2 ()
= 2 cos2 () 1.
3
40. If h 12 | z| then
1
| z h| | z|
2
and hence
1 1 ( z) ( z h)
=
z h z ( z h)( z)
h
=
( z h)( z)
2|h|
.
| z|2
41.
1
n sin( n ) [ 1
n sin( n ) ]
4 1
lim dx = lim 4 dx
n+ 0 1 + x2 n+ 0 1 + x2
[ n sin( n1 ) ]
= lim 4 tan1 (x)
n+
[ 0
]
= lim 4 tan [n sin( n1 )] tan1 (0)
1
n+
[ ]
= lim 4 tan1 [n sin( n1 )] 0
n+
= lim 4 tan1 [n sin( n1 )]
n+
= 4 lim tan1 [n sin( n1 )]
n+
1
Since tan (x) is continuous on R,
[ ]
= 4 tan1 lim n sin( n1 )
n+
[ sin( n1 ) ]
= 4 tan1 lim 1
n+
n
1 sin( n1 )
Since 0 as n +, lim 1 = 1,
n n+
n
()
= 4 tan1 (1) = 4 = (voila!)
4
42.
f (x1 , x2 , . . . , xn ) = x21 + x22 + + x2n
43.
1 xn+1
= 1 + x + x2 + + xn
1x
44.
M = {f V : f (m) = 0 for all m M }.
4
I
1 1 U
45. Isnt this beautiful: ds ?
4 r n
46. The definite integral of f on [a, b] is
b
n
f (x) dx = lim f (xk )xk .
a P 0
k=1
47. (i )
(ii )
(iii )
48. a n , s n
49. In writing text: Schr
odinger. In math mode: a
n , sn
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
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3 Master
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