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+b(n)n(n+1)=1nn+1
1, b=1
= ∞(

+ )
n 0
1n 1n 1
(1112)+(1213)+…+(1n1n+1)
111n+1
where n
→
∞
1
=0 fn(0)n!xn
∞
n=0 en0n!xn
∞
ex
for all n
1
n=0 xnn!
∞
he value of

e x2
urin for
= = ∞
!
ex n 0 xnn
urin for

= = ∞(
)
!
e x2 n 0
x2 nn
ne equation values to solve for t
Plug in t to find values
+ 
 +  + =
5 t 24 2t 3 2 3t 12
for point where the line meets
x=6.5
y=1
z=2.5
(6.5,1,2.5)
m parallel plane and containing a point
e parallel to plane

+
=
x 2y 3z 12
and
P(5,4,2)
plane H.
h P, and the normal of the given plane is
n of the new plane is
(x5)2(y4)+3(z+2)=0
2+x33+…+C= 0 xn+1n+1+C=1 xnn+C
∞
∞
…
x33
eries is conditionally
ent
2 2
∞
∞
e
nt
nes L
1
and L
2
:
r(t) = (3+t)i + (1t)j + (5+2t)k
d z values are equal
you have one, plug in to find intersection point.
he intersection is at (1,3,1)
en by s and t.
b3=abcos
θ
product values
ning point P
=
z 12
and
is <1,2,3>, the equation for the line is

=   = +
x 51 y 4 2 z 23
Increasing:
<
+
an an 1
for all values of n sufficiently large
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This note was uploaded on 01/26/2011 for the course MATH 2433 taught by Professor Guralnik during the Spring '08 term at The University of Oklahoma.
 Spring '08
 GURALNIK
 Math

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