Solutions
Block
2 :
Ordinary D i f f e r e n t i a l Equations
Unit
8:
The
U s e
of Power Series
a .
Since two of the t h r e e summations involve xn and only one involves
n-1
x
,
it
s e e m s
l o g i c a l t o "jack up" the index i n xn'l
by
1.
That
is,
w e
replace n by n
+
1
i n s i d e the 1-sign and lower the
s t a r t i n g point of t h e summation by
1.
W e
obtain
s o t h a t
W e
then " s p l i t o f f " t h e f i r s t
term
i n each of the summation
i n (2) which begin with n
=
0
(in t h i s way, each summation
begins with n
=
1)
.
W e
therefore rewrite (2) as:
Note:
Especially f o r those who may
still
be "edgy" about extensive
use of the
C
-notation,
it
may be worthwhile t o show the equi-
valence of
(1)
and (3) by long hand techniques.
In t h i s way,
one does not review what we've done without recourse t o the
G
-notation, and one may a l s o learn t o appreciate b e t t e r the
compactness of the
C
-notation.
To begin with,
(1)
may be expanded as:
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Solutions
Block 2: Ordinary Differential Equations
Unit 8: The
U s e
of Power Series
2.8.1(L) continued
and t h i s agrees with (3) since the expansion of (3) yields
equals
It
should also be noted that i n e i t h e r approach, we used the
idea of absolute convergence t o justify our taking the liberty
of re-arranging and combining terms i n the way we did.
b.
Replacing
(1)
by (31, we see t h a t
implies t h a t
OD
Now, since the only way that two convergent power series can be
identical
is
for t h e i r coefficients t o be equal term-by-term
and
since
03
w e
see from (5) t h a t
and
S.2.8.2
Solutions
Block 2: Ordinary D i f f e r e n t i a l Equations
Unit 8: The
U s e
of Power Series
2.8.1
(L)
continued
f o r n
-
>
1;
(n
+
l)an+l
-
(n
-
l ) a n
=
0.
That
is,
n
-
>
1
implies
t h a t
From
(6)
we
see t h a t a.
may be chosen a t random, whereupon
and we may now use (7) t o compute a2 from al, a3 from a2,
a4
from a3, e t c .
For example with n
=
1,
(7) becomes
b u t once a 2
=
0, (7) t e l l s us t h a t f o r n
>
2, an
=
0.
That
is,
from (7) an+l
is
a multiple of an s o t h a t an+l must be 0 once
an
=
0.
Thus,
w e
have t h a t a.
is
a r b i t r a r y , al
=
-a
'
and an
=
0 f o r
0
n
-
>
2.
Since
%.re
have that
rlas a solution of the form
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Solutions
Block
2:
Ordinary Differential Equations
Unit
8:
The
U s e
of Power Series
2.8.1
(L)
continued
then
so t h a t (10) becomes
Notice t h a t equation
(11)
is
precisely equation
( 4 )
of part (b)
[where
-
the equality i n
(11)
is
understood t o be our identity.
We
used the identity symbol i n p a r t (b) f o r emphasis, but other
than that, we revert t o the usual notation that L(y)
=
0 means
L(y)
01.
Hence, from part (b) we deduce from
(11)
t h a t y
=
a.
(1
-
X)
.
Note
#1:
A s
usual we have elected t o s t a r t with an example t h a t could be
solved more easily by a more familiar technique.
In equation
(10)
we
could separate variables t o obtain that f o r x
#
1
and
y
#
0
,
dy/y
=
dx/x
-
1;
o r lnlyl= lnlx
-
11
+
cl; o r
y
=
+
-
ecl(x
-
1)
=
c ( x
-
l),
where c
= +
-
e
#
0.
- Winter '18
