Divided Difference Examples

7/26/2019 Divided Difference Examples

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Chapter 05.03Newtons Divided Difference Interpolation More ExamplesElectrical Enineerin

Example 1

Thermistors are used to measure the temperature of bodies. Thermistors are based on materials

change in resistance with temperature. To measure temperature, manufacturers provide you with

a temperature vs. resistance calibration curve. If you measure resistance, you can find thetemperature. A manufacturer of thermistors makes several observations with a thermistor, which

are given in Table 1.

Table 1 Temperature as a function of resistance.

R( )ohm T ( )C

111.!11."

#"#.

$%1.1

&%.11"".1"1

$.1&

%.1&'

%.".1


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(ewtons )ivided )ifference Interpolation*+ore -amples lectrical ngineering %.".&

Figure 1 /esistance vs. temperature.

)etermine the temperature corresponding to 0%$.' ohms using (ewtons divided difference

method of interpolation and a first order polynomial.

Solution

or linear interpolation, the temperature is given by2323 1 RRbbRT +=

4ince we want to find the temperature at '.0%$=R and we are using a first order polynomial,

we need to choose the two data points that are closest to '.0%$=R that also bracket '.0%$=R

to evaluate it. The two points are ".!11=R and .#"#=R .Then

,".!11 =R 1"1."23 =RT

,.#"#1 =R 1&.$23 1 =RT

gives

23 RTb = 1"1."=

1

11

2323

RR

RTRTb

=

".!11.#"#

1"1."1&.$

=

"#&'$.=


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(ewtons )ivided )ifference Interpolation*+ore -amples lectrical ngineering %."."

5ence

2323 1 RRbbRT +=

2,".!113"#&'$.1"1." = R ".!11.#"# RAt '.0%$=R

2".!11'.0%$3"#&'$.1"1."2'.0%$3 =T

C'!."% =If we e-pand

2,".!113"#&'$.1"1."23 = RRT ".!11.#"# Rwe get

,"#&'$.1!0.#"23 RRT = ".!11.#"# RThis is the same e-pression that was obtained with the direct method.

Example 2

Thermistors are used to measure the temperature of bodies. Thermistors are based on materialschange in resistance with temperature. To measure temperature, manufacturers provide you with

a temperature vs. resistance calibration curve. If you measure resistance, you can find thetemperature. A manufacturer of thermistors makes several observations with a thermistor, which

are given in Table &.


R( )ohm T ( )C

111.

!11."#"#.

$%1.1

&%.11"

".1"1$.1&

%.1&'

)etermine the temperature corresponding to 0%$.' ohms using (ewtons divided difference

method of interpolation and a second order polynomial. ind the absolute relative appro-imateerror for the second order polynomial appro-imation.

Solution

or 6uadratic interpolation, the temperature is given by

22332323 1&1 RRRRbRRbbRT ++=

4ince we want to find the temperature at '.0%$=R and we are using a second order polynomial,

we need to choose the three data points that are closest to '.0%$=R that also bracket '.0%$=R

to evaluate it. The three points are,".!11 =R .#"#1 =R and 1.$%1& =R .

Then

,".!11 =R 1"1."23 =RT

,.#"#1 =R 1&.$23 1 =RT

,1.$%1& =R 1&'.%23 & =RT


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%.".$ Chapter %."

gives

23 RTb =

1"1."=

1

1

1

2323

RR

RTRTb

=

".!11.#"#

1"1."1&.$

=

"#&'$.=

&

1

1

1&

1&

&

23232323

RR

RR

RTRT

RR

RTRT

b

=

".!111.$%1

".!11.#"#

1"1."1&.$

.#"#1.$%1

1&.$1&'.%

=

&.$#"#&'$.%$1&0.

+=

%1'001." =

5ence

22332323 1&1 RRRRbRRbbRT ++=

2,.#"#23".!1131'001."2".!113"#&'$.1"1."%

+= RRR

".!111.$%1 R

At ,'.0%$=R

2.#"#'.0%$23".!11'.0%$31'001."

2".!11'.0%$3"#&'$.1"1."2'.0%$3

%+

=

T

C'!."% =

The absolute relative appro-imate error a

obtained between the results from the first and

second order polynomial is

1'!."%

'!."%'!."%

=a

7%$".&=

If we e-pand,

2,.#"#23".!1131'001."2".!113"#&'$.1"1."23%

+= RRRRT

".!111.$%1 R

we get

( ) ,1'001."!#&0%.##'.'% &%RRRT += ".!111.$%1 RThis is the same e-pression that was obtained with the direct method.

Example 3

Thermistors are used to measure the temperature of bodies. Thermistors are based on materials

change in resistance with temperature. To measure temperature, manufacturers provide you with

a temperature vs. resistance calibration curve. If you measure resistance, you can find the


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(ewtons )ivided )ifference Interpolation*+ore -amples lectrical ngineering %.".%

temperature. A manufacturer of thermistors makes several observations with a thermistor, which

are given in Table ".


R( )ohm T ( )C

111.

!11."

#"#.

$%1.1

&%.11"

".1"1

$.1&

%.1&'

a2 )etermine the temperature corresponding to 0%$.' ohms using (ewtons divided

difference method of interpolation and a third order polynomial. ind the absolute

relative appro-imate error for the third order polynomial appro-imation.b2 The actual calibration curve used by industry is given by

2ln23lnln23lnln3ln2ln23lnln3ln2ln3ln

1

&1"1&1 RRRRRRbRRRRbRRbbT+++=

substituting,

1

Ty=

and ,ln Rx= the calibration curve is given by

22323322332323 &1"1&1 xxxxxxbxxxxbxxbbxy +++=

Table 4 +anipulation for the given data.

R ( )ohm T ( )C x( )Rln y

T

1

111.!11."

#"#.

$%1.1

&%.11"".1"1

$.1&

%.1&'

0.$#.'1$!

#.$%%&

#.1110

."!'&.""1''

.&$!&%

.1!!$!

ind the calibration curve and find the temperature corresponding to 0%$.' ohms. 8hat is thedifference between the results from part 3a29 Is the difference larger using results from part 3a2 or

part 3b2, if the actual measured value at 0%$.' ohms is C&'%."% 9

Solution

a2 or cubic interpolation, the temperature is given by

22323322332323 &1"1&1 RRRRRRbRRRRbRRbbRT +++=

4ince we want to find the temperature at ,'.0%$=R we need to choose the four data points thatare closest to '.0%$=R that also bracket '.0%$=R to evaluate it. The four data points are

,.111 =R ,".!111 =R .#"#& =R and1.$%1" =R .

Then

,.111 =R 11".&%23 =RT


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%.".# Chapter %."

,".!111 =R 1"1."23 1 =RT

,.#"#& =R 1&.$23 & =RT

,1.$%1" =R 1&'.%23 " =RT

gives

:; RTb =

23 RT=

11".&%=

:,; 11 RRTb =

1

1 2323

RR

RTRT

=

.111".!11

11".&%1"1."

=

$%&.=

:,,; 1&& RRRTb =

&

11&:,;:,;

RR

RRTRRT

=

1&

1&1&

2323:,;

RR

RTRTRRT

=

".!11.#"#

1"1."1&.$

=

"#&'$.=

$%&.:,; 1 =RRT

&

11&

&

:,;:,;

RR

RRTRRT

b

=

.111.#"#

$%&."#&'$.

+

=

%111$$.& =

:,,,; 1&"" RRRRTb =

"

1&1&" :,,;:,,;

RR

RRRTRRRT

=

1"

1&&"

1&"

:,;:,;:,,;

RR

RRTRRTRRRT

=

&"

&"

&"

2323:,;

RR

RTRTRRT

=

.#"#1.$%1

1&.$1&'.%

=

%$1&0.=

"#&'$.:,;1& =RRT


7/10

(ewtons )ivided )ifference Interpolation*+ore -amples lectrical ngineering %.".0

1"

1&&"1&"

:,;:,;:,,;

RR

RRTRRTRRRT

=

".!111.$%1

"#&'$.%$1&0.

+

=

%

1'001."

=

%

1& 111$$.&:,,; =RRRT

"

1&1&""

:,,;:,,;

RR

RRRTRRRTb

=

.1111.$%1

111$$.&1'001." %%

=

'101&$.& =

5ence


2".!1123.1113111$$.&2.1113$%&.11".&%

%+=

RRR .1111.$%12,.#"#23".!1123.1113101&$.&

'

RRRR

At ,'.0%$=R

2".!11'.0%$23.111'.0%$3111$$.&2.111'.0%$3$%&.11".&%2'.0%$3 % += T

2.#'.0%$23".!11'.0%$23.111'.0%$3101&$.&'

C&$&."% =

The absolute relative appro-imate error a

obtained between the results from the second and

third order polynomial is

1&$&."%

'!."%&$&."%

=a

7$"$%'.=

If we e-pand

( ) 2".!1123.1113111$$.&2.1113$%&.11".&%%

+= RRRRT

.1111.$%12,.#"#23".!1123.1113101&$.&'

RRRR

we get

( ) .1111.$%1,101&$.&1&!0%.!1"!".0%!.!& "'&% += RRRRRTThis is the same e-pression that was obtained with the direct method.

b2 inding the cubic interpolant using (ewtons divided difference for


re6uires that we first calculate the new values of x and y .


8/10

%.".' Chapter %."

x( )Rlny

T

1

0.$

#.'1$!

#.$%%.1110

."!'&

.""1''

.&$!&%.1!!$!

Then

( ) "!'&.,$.0 == xyx

( ) ""1''.,'1$!.# 11 == xyx

( ) &$!&%.,$%%&.# && == xyx( ) 1!!$!.,1110.# "" == xyx

gives

:; xyb =

23 xy=

"!'&.=:,; 11 xxyb =

1

1 2323

xx

xyxy

=

$.0'1$!.#

"!'&.""1''.

=

"%#!.=

:,,; 1&& xxxyb =

&

11& :,;:,;

xx

xxyxxy

=

1&

1&

1&

2323:,;

xx

xyxyxxy

=

'1$!.#$%%&.#

""1''.&$!&%.

=

&&!0$.=

"%#!.:,; 1 =xxy

&

11&&

:,;:,;

xx

xxyxxyb

=

$.0$%%&.#

"%#!.&&!0$.

=

&&$.=

:,,,; 1&"" xxxxyb =


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(ewtons )ivided )ifference Interpolation*+ore -amples lectrical ngineering %.".!

"

1&1&" :,,;:,,;

xx

xxxyxxxy

=

1"

1&&"

1&"

:,;:,;:,,;

xx

xxyxxyxxxy

=

&"

&"&"

2323:,;xx

xyxyxxy

=

$%%&.#1110.#

&$!&%.1!!$!.

=

1$$'0.=

&&!0$.:,; 1& =xxy

1"

1&&"1&"

:,;:,;:,,;

xx

xxyxxyxxxy

=

'1$!.#1110.#

&&!0$.1$$'0.

=

1&0.=

&&$.:,,; 1& =xxxy

"

1&1&"

"

:,,;:,,;

xx

xxxyxxxyb

=

$.01110.#

&&$.1&0.

=

1110".=5ence


$.01110.#2,$%%&.#23'1$!.#23$.031110".

2'1$!.#23$.03&&!0$.2$.03"%#!."!'&.

+++=xxxx

xxx

4ince were looking for the temperature at '.0%$=R , we will be using2'.0%$ln3=x

#%.#=

At ,#%.#=x

2$%%&.##%.#23'1$!.##%.#23$.0#%.#3111'&.

2'1$!.##%.#23$.0#%.#3&&!0&.

2$.0#%.#3"%01."!'&.2#%.#3

++

+=y

&'&'%.=

inally, since,

1

Ty=

yT

1=

&'&'%.

1=


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%.".1 Chapter %."

C"%%."% =

4ince the actual measured value at 0%$.' ohms is C,&'%."% the absolute relative true error forthe value found in part 3a2 is

1&'%."%

&$&."%&'%."%

=t

71&&%".=

and for part 3b2 is

1&'%."%

"%%."%&'%."%

=

t

71!'&%.=

Therefore, the cubic polynomial interpolant given by (ewtons divided difference method, that

is,


obtained more accurate results than the calibration curve of

2ln23lnln23lnln3ln2ln23lnln3ln2ln3ln

1&1"1&1 RRRRRRbRRRRbRRbbT +++=

I(T/ATI=(

Topic (ewtons )ivided )ifference Interpolation4ummary -amples of (ewtons divided difference interpolation.

+a?or lectrical ngineering

Authors Autar @aw)ate une 1, &1#

8eb 4ite httpBBnumericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu/http://numericalmethods.eng.usf.edu/

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