Taylorpolynomial approximation
Let fi ( ai b ) -0 112 be n - times differentiable, for some
WEIN and let us denote by f' "
c × ) its km derivative
at x which exists for all he 3014 - - - in } .
( here f' "
cx ) = f Cx ) ) . Given Xo E La ,b)
,the Taylor
formula for f of order n with starting point to is
~
ftp.7f Cx ) = £
g.Cx - xD
"
t e
' I? C x ),
theCa , b).
k = 0
where e' I? C x ) is said to be the error at order W.
( n )
Taylor 's theorem states that line exo G 7-
= O
x → Xo ( x - AT
and moreover if f is endowed of a Ceti )th
derivative
in Ca , b) ,I Cx ) = f "x ) ex - xoj
"
for some
Tx E ( xo,
× ).
C htt ) !
Before proving the theorem,
we want to
convince ourselves that the approximationit gives makes sense in a couple of cases i
Fix Xo E Ca , b) and assume we want to find a
polynomial of degree 0 ( i.e.
a constant function )which
agrees with f Cx ) at the point Xo.
This is clearly the polynomial Xo,
whose graphis that of the line y = Xo
L
tix y-- f exo )
I
Xo X
Tf we ask for an approximation by a 1st order polynomialwe have to specify a parameter : every
1st order
polynomial whose graph passes through the point ( xo,f- Kd )
can be written as tcxo ) t in C x - Xo ) ,meth
.
Its graph is one of the straightlines y= fast in K - Xo )
We can now further ask which of those line
approximatefun better.
We need to define an approximationerror i e !! Cx ) = f HI - ftHolt mfx - Xo ))
,m
Note that fun l !? C xo) = o
in
Wesay
that the error is the best if m is sucha )
e Cx )
that lion Him=
o
x → Xo ( X - Xo )
this condition gives
• =
him f Cx ) - fcxo ) - m ( x - Xo )=
him txl- in
x. → xo C x - Xo ) x -2 XoX - Xo
a )
⇐ b in -
- f Go ) ( = f'
Go , )
andy
= fcxo ) tf'
Got C x - Xo ) is the tangent line
to the graph of f at the point ( xo, flexor ) -
For such m,
we cell the error eat Cx ).
x y = tcxdtftx.lk - Xo )
9 ex ) = error at x if we approximate text• i y = tho )
with the tangent line at Xo .
:
I D
Xo XX
We repeat the question assuming to use an
approximationby a second - order polynomial .
y = f Chol t m Cx -
xo ) + ccx- Xo )2
The error e!!m,ccx) = f Cx ) . f- Cbl - mcx - xo ) - ccx . %)2
a )e Cx )
Imposing now lim * mic=
o we get m = flew)x → xo X - Xo
(2)e Cx )
To optimize over C we now requirehim
bit'
case= O
X -7 Xo Cx - Xo )2a)
Note that lion e⇐ 7
xoyfkxo),
C
X -3×0 ¥=
=lim text - fcxo ) - think - a ) - CK - xo5
x -3 to Cx - Xo )2
deltas. him thx ) - ttxo ) - 2 CG - xo )
=
xoxo 2 C x - Xo )
= lim thx ) - f'
↳ ,
x -
not -c = o ⇐ is c= If
"
Go )
-"
{ F''
Go )
For the choice m=f' exo ) and c=f' we
cell the 2nd order error eatin,
och : = e !! Cx ).
we here proved that fcx ) = ¥of "L cx-xdhte.Y.kz
with e' ! CM→ o
-
( x - a)2 Xoxo
As x → x .we call Cx - edh infinitesimal of order he
.
The Taylor polynomial of order n of fcx ) with startingpoint Xo approximates fcx ) with an error LIK )
which is an infinitesimal of higher order with
respect to Cx - xoT .
Example text = et we want the
Taylor polynomial of order 4 with shorting point
Xo = O.
Remember thot f = et t k =p f' 403=2--1
The Taylor polynomial of order 4 is
4
I I ,
Cx - of= it x t I t +
Ik=0 24
( remember that k ! = k Ck - r ) ( k - 2) .
. - .
. 1)Ft is
very Ompontant to find explicit expressions
for the error , of course .
The following theorem uses Lagrange theorem
to write the error -
Theorem ( Taylor with Lagrange error )Let fi Ca ,
b ) -0112 be Cute ) - times differentiable
in Ca , b) .
let Xo E Ca , b ),
then Fxtxo there exists
§ in the internal of extreme x andx -such that
ex? 'cx)= F" s × - * 5th ( Lagrange error )@ ti ) !
Proof Fix for simplicity xsxo ( the case × > Xo
can be dealt with similarly ) .
We set
bunk ) : = Cut , ) ! extra )
( x - a)htt
'
2- the internal Ex , xo ] we consider the function
goys = In f "I ex . et t hit )k= o
the function g is continuous in Ex ,to ] and differentiable
in Cx , Xo ) j moreau
gcxi-E.tt?Yycx-xsht hncx ) = tix )
- -
" do
f-' 7×7.1a-
gcx.is-E.tk?gy3cx-xo5+4-xoI..hncx) = Fix )
( htt ) !
x )
Since gCx ) = g Ko ) we can apply Rolle 's theorem
to get that F EE Cx , Xo ) : g'
C 5×7=0
We now compute g' Cy ) :
guys ⇐IIE ,F'
"
In ex - y )h
+
InI ,
I Ty ) .mx . yin :c . i )
- ht r
ftp.K-y/hnCx7--EitIIIcx-si-i-E.III.cx-n"
he
- ÷,
Cx - y )"
- hncx )
( htt )= tnc Cx - yjn - HII.
Cx -
y5=n ?
= 4-2,5 E ft"
Ty ) - him ]
g'
CE ) = o ⇐ is f"
E) = huey = @th ! eii( X - Xo )
htt
⇐ b e !? =
f-' F) E) ¥5
"
( htt ) !⑤
In the previous example
e×= If I ,
xh +e' It ex )
By the previous theorem eY" ↳ =6×5%5 )I
⇒×
'= e
120
For xe
fi, i ) we have 1¥ Iset
.
120
This allows us to say that the error we make ifwe approximate et with it x txt Ig t x÷ for
x E C- I, e) is at most ↳ = 0.023
We now ask ourselves which order in the expansion
should we take if we want an error less than
to-
6
.
since et 'e I @5"n} .
" Is e÷'
,
it is enough to have ÷ s to- °
,
or
equivalently n ! 3 e. too which is satisfied byin 3 to . Eventually we have proved thot the
approximation of ex in C- in ) with a polynomialof order n= 10 produces an error less then to
- 6-
Example him cos x - I
x → o cos C x' 'a ) - ttxlz
cos x = i - Eg t ¥4 - sing xs
cos = i - Ez t XI,
- sin Eh
=D cos x - t =- XI + sink ×3
6
as- it In
,
- si Eh
him II*.
-- E:
⇒is¥x - so
¥4 - sin x%=
=
him
MC- It sink .
× )× - so
= = -k
Note that it is notnecessary
to have the error expressed
in the Lagrange form to complete the exercise i
Osx - is = - I +
ej"ex ,
ask "7 - it { = ¥,
+ e ?'c x
's)
⇒
him %*¥ -- Liao
Mfk+ )
= -iz since
⇐ o
HCE,
+
ei"g)
fins.
⇒, E's
.
= fi: ei"c=o( x' E) 4