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()
(
)
1
2
0
1
1
−
=
−
−Π=
ε
s
n
s
n
xxD
D sub n minus one prime of x is equal to the product from s equal zero to n of,
parenthesis, one minus x sub s squared, close parentheses, to the power epsilon minus
one.
()
(
)
()
dw
kww
ztK
Пi
xt
w
−
=Κ
∫
=−
,
2
1
,
2
1
ρ
K of t and x is equal to one over two
π
i times the integral of K of t and z, over w
minus w of x, with respect to w along curve of the modulus of w minus one half, is
equal to rho.
()
0;0
4
2
2
>=∆∆+ aua
dt
ud
the second partial (derivative) of u with respect to t plus a to the fourth power, times
the Laplacian of the Laplacian of u, is equal to zero, where a is positive
() () ()
1;
2
1
>=
∫
∞+
∞−
cdw
w
x
w
Пi
xD
n
ic
ic
k
k
ζ
D sub k of x is equal to one over two
πι
times integral from c minus
ι
infinity to c
plus i infinity of dzeta to the k of w, x to the w over (or: divided by) w, with
respect to w, where c is greater than one.
2. Practice reading the following expressions by yourself, check your answer
using the keys
a.
()
xx
4
3
3
2
523
2
1
715
7
2
+=
−++
b.
()
Lxf
x
=
→
lim
1
c.
()
()
(
)
Sx
xfxSxf
xf
s
−+
=
→
lim
0
'
d.
dt
ds
S =
29
( )
n
Dn−1 ( x ) = Π 1 − x s2
ε −1
s =0
D sub n minus one prime of x is equal to the product from s equal zero to n of,
parenthesis, one minus x sub s squared, close parentheses, to the power epsilon minus
one.
1 K (t , z )
Κ (t , x ) = ∫ dw
2 Пi 1 w − w(k )
w− = ρ
2
K of t and x is equal to one over two πi times the integral of K of t and z, over w
minus w of x, with respect to w along curve of the modulus of w minus one half, is
equal to rho.
d 2u
+ a 4 ∆∆u = 0; (a > 0)
dt 2
the second partial (derivative) of u with respect to t plus a to the fourth power, times
the Laplacian of the Laplacian of u, is equal to zero, where a is positive
c + i∞
1 k xn
Dk ( x ) = (w) dw; (c > 1)
2 Пi c −∫i∞
ζ
w
D sub k of x is equal to one over two πι times integral from c minus ι infinity to c
plus i infinity of dzeta to the k of w, x to the w over (or: divided by) w, with
respect to w, where c is greater than one.
2. Practice reading the following expressions by yourself, check your answer
using the keys
2 1 2 3
a. 15 + 7 + 3(x − 2) = 5 + x
7 2 3 4
b. lim f (x ) = L
x →1
f ( x + S )x − f ( x )
c. f ' (x ) = lim
s →0 Sx
ds
d. S =
dt
29
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