 154 Chapter 5 Let’s Learn About Taylor Expansions!
When a general function f(x) (provided it is differentiable infinitely many
times) can be expressed as
f x a a x a x a x a x
n
n
( )
= + + + + + +
0 1 2
2
3
3
... ...
the right side is called the Taylor expansion of f(x) (about x = 0).
Left side =
=
1
1 2
1
Right side
= + + + + +
1 2 4 8 16 ...
It turns out that expression u is correct for all
x satisfying −1 < x < 1, which is the allowed interval
of a Taylor expansion. In technical terms, the inter-
val −1 < x < 1 is called the circle of convergence.
S? The two
sides are not
equal.
For example, substituting x = 2
in both sides of expreion u,
This means that f(x) perfectly coincides with an
infinite-degr polynomial in a definite interval
including
x = 0. It should be noted, however,
that the right side may become meaningle
because it may not have a single defined value
outside the interval. How to Obtain a Taylor Expansion 155
How to Obtain a Taylor Expansion
When we have
v
f x a a x a x a x a x
n
n
( )
= + + + + + +
0 1 2
2
3
3
... ...
let’s find the coefficient a
n
.
Substituting x = 0 in the above equation and noting f(0) = a
0
, we find that
the 0th-degree coefficient a
0
is f(0).
We then differentiate v.
w
( )
= + + + + +
f x a a x a x na x
n
n
1 2 3
2 1
2 3 . .. ...
Substituting x = 0 in w and noting f(0) = a
1
, we find that the 1st-degree
coefficient a
1
is f(0).
We differentiate w to get
( )
= + + +
( )
+
f x a a x n n a x
n
n
2 6 1
2 3
2
... ...
Substituting x = 0 in , we find that the 2nd-degree coefficient a
2
is
1
2
0
( )
f
.
Differentiating
, we get
( )
= + +
( )
( )
+
f x a n n n a x
n
n
6 1 2
3
3
... ...
From this, we find that the 3rd-degree coefficient a
3
is
1
6
0
( )
f
.
Repeating this differentiation operation n times, we get
f x n n a
n
n
( )
( )
=
( )
× × +1 2 1... ...
where f
(n)
(x) is the expression obtained after differentiating f(x) n times.
From this result, we find
nth-degree coefficient
a
n
f
n
n
=
( )
( )
1
0
!
n! is read “n factorial” and means
n n n×
( )
×
( )
× × ×1 2 2 1... . 156 Chapter 5 Let’s Learn About Taylor Expansions!
We, that
introduction
was a lile
t long...
So, why is our
company’s
predicament
the Taylor
expansion?
I mean that if
f(x) is a
function that describes
Burnham Chemicals
their suort of
our paper could be
considered the third
term of a Taylor
expansion. f(x) = The Japan
Times + The Kyodo News +
TheAsagake Times
Thats right.
Actuay for
BurnhamChemical,
the amount of money
they spend for
us is only a very
sma amountthe
3rd-degr term,
obtained after
dierentiating
thrtimes.
Since it’s
insignificant for
them anyway,
they’ likely
suort us like
they did before
even if they
change their
executives.
Mr. Seki, where
did you go out
for drinks when
you worked at
the main oice?
What?
A the
way at
the end? How to Obtain a Taylor Expansion 157
You know,
drinking with
after work,
sue stories...
Oh.
We’re done
with our work.
So, sha we
go out for a
drink?
Okay,
let’s go.
Yes!
(Open 24Hours) 158 Chapter 5 Let’s Learn About Taylor Expansions!
Nice
atmosphere,
isn’t it?
Ah, yes. Are a
these people
journalists?
Lk, that’s Ishizuka,
the photographer
who is the youngest
wier of the Japan
Photographic Prize.
And that’s Mr. Nakata,
a heavyweight in the
designers’ circle.
The guys over
there are from the
Sanda City Post.
Mr.
Calculus!?
Hey, Calculus,
long time no
What a
nickname! But,
it certainly
fits.
Let’s s. I
hope I can
listen to their
profeional
discuion.
Bierman
got diabetes
recently.
Mr. Stack
now ss
a doctor
because of
high bld
preure...
Maybe I
should think