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Applications
of differentiation - the graph of a function and its derivatives |
Taylor's
theorem (Taylor's
formula) - The extended mean value theorem |
The proof of Thaylor's theorem |
Maclaurin's
formula or Maclaurin's theorem |
Representing
polynomial using Maclaurin's and Taylor's formula |
Representing
polynomial using Maclaurin's and Taylor's formula examples |
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Taylor's
theorem (Taylor's
formula) - The extended mean value theorem |
Suppose
f
is continuous
on the closed interval [x0,
x0
+ h]
with continuous
derivatives to
(n
-
1)th
order on the
interval and its nth
derivative defined on
(x0,
x0
+ h)
then, |
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is called Taylor's theorem. |
To prove Taylor's theorem we
substitute x0
= a,
x0
+ h = b
and introduce the function |
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where Q
is still undetermined and p
> 0. |
Since
j
(b) = 0
we
can define
Q
by setting j
(a)
= 0
too, |
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then
we compute j
' (x)
to apply Rolle's
theorem. |
Note
that successive terms cancel (by the product rule) when differentiating j
(x), so we get |
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Thus,
by Rolle's theorem there is some intermediate point c
inside the interval [a,
b]
such that j'
(c)
= 0, |
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which
gives |
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By
plugging the Q
back into above equation j
(a)
= 0
while returning substitutions, |
a
= x0,
b
= x0
+ h,
b
-
a
= h
and where |
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we
obtain Taylor's theorem to be proved. |
The
last term in Taylor's formula |
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is
called the remainder and denoted Rn
since it follows after n
terms. |
By
plugging, a) p
= n into
Rn
we get |
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the
Lagrange
form of the remainder, |
while if b) p
= 1
we get |
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the
Cauchy
form of the remainder. |
Thus,
by substituting x0
+ h = x
obtained is |
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Taylor's
formula with Lagrange form of the remainder. |
Therefore, Taylor's
formula gives values of a function f
inside the interval [x0,
x0
+ h]
using its value and the values
of its derivatives to
(n
-
1)th
order at the point x0 in the form |
f (x) =
Pn -
1(x
-
x0) + R n |
where,
Pn
-
1(x
-
x0)
is
(n
-
1)th
order Taylor polynomial for f
given by the first n
terms in the above
formula and Rn
is one of the given remainders. |
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Maclaurin's
formula or Maclaurin's
theorem |
The
formula obtained from
Taylor's
formula by setting x0
= 0
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that
holds in an open neighborhood of the origin, is called Maclaurin's
formula or Maclaurin's
theorem. |
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Consider
the polynomial
fn
(x)
= anxn
+ an -
1xn -
-
1 + · · · + a3x3
+ a2x2 + a1x
+ a0, |
let
evaluate the polynomial and its successive derivatives at the origin, |
f
(0) =
a0, f '(0) = 1· a1,
f ''(0) = 1· 2a2,
f '''(0) = 1· 2· 3a3, . . .
, f (n)(0) =
n!an |
we
get the coefficients, |
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Therefore,
the Taylor polynomial of a function f
centered at x0
is the polynomial of degree n
which has the same
derivatives as f
at x0,
up to order n. |
If
a
function f
is infinitely differentiable on an interval about a point x0
or the origin, as are for example ex and
sin
x,
then |
P0 (x)
= f (x0), |
P1 (x) =
f (x0)
+ (x
-
x0)
f ' (x0), |
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P0,
P1,
P2,
. . . is a sequence of increasingly approximating polynomials for
f. |
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Representing
polynomial using Maclaurin's and Taylor's formula |
If
given is an n-th
degree polynomial |
Pn(x) =
anxn + an -
1xn -
1 + ·
· · + a2x2
+ a1x + a0 then, |
Pn(0) =
a0,
Pn' (0) = a1,
Pn''(0) = 2!
a2,
Pn'''(0) = 3!
a3,
. . . , Pn(n)(0) = n!
an
and
Pn(n
+ 1) = 0 |
so,
the coefficients of the polynomial |
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therefore,
applying Maclaurin's formula, every polynomial can be written as |
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since
Pn(n
+ 1) = 0,
the remainder vanishes. |
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Representing
polynomial using Maclaurin's and Taylor's formula, examples |
Example: Represent
the quintic y
= 2x5 + 3x4 -
5x3 + 8x2 -
9x + 1 using Maclaurin's
formula. |
Solution: Let write
all successive derivatives of the given quintic and evaluate them at the
origin, |
y' (x)
= 10x4 + 12x3 -
15x2 + 16x -
9,
y' (0)
= -
9 |
y'' (x)
= 40x3 + 36x2 -
30x + 16,
y'' (0)
= 16 |
y''' (x)
= 120x2 + 72x -
30,
y''' (0)
= -
30 |
yIV
(x)
= 240x + 72,
yIV
(0)
= 72 |
yV
(x)
= 240,
yV
(0)
= 240 |
yVI
= 0 and the last term of the polynomial a0
= y(0)
= P5(0) =
1, |
then
substitute obtained values into Maclaurin's
formula |
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Example: Represent
the quartic y
= x4 -
4x3 + 4x2 + x -
4 at x0
= 2 using
Taylor's
formula. |
Solution: Let write
all successive derivatives of the given quartic and evaluate them at x0
= 2, |
y' (x)
= 4x3 -
12x2 + 8x + 1,
y' (2)
= 1 |
y'' (x)
= 12x2 -
24x + 8,
y'' (2)
= 8 |
y''' (x)
= 24x -
24,
y''' (2)
= 24 |
yIV
(x)
= 24,
yIV
= 24 |
yV
(x)
= 0
and the last term, a0
=
y(2)
= P4(2) = -
2, |
then
substitute obtained values into Taylor's
formula |
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