numerical solutions of equations (6)

8/9/2019 Numerical Solutions of Equations (6)

1/15

Kosturi Ash 12/3

A2 Mathematics Coursework C3

Year 12

Numerical solutions of equations

Solving 0 = ! "#! using the $Change %f Sign& Metho'

The method I will use to solve 0 = !"#! is the Change of Sign Metho' involving

the (ecimal Search method. I have drawn this graph using the Autograph Software,

and the print screen of this is elow!

1

This is the root of

this e"uation that I

need to find. See

)igure 1


2/15

Kosturi Ash 12/3

#rom m$ graph aove, I can see that the root of this e"uation is etween =1 and =

1*!. The tale of values and f+, values is shown elow. I can wor% out the f+,

values $ sustituting the #values into the e"uation.

1 1*1 1*2 1*3 1*- 1*!

f+, #3 #2*2./-/ #1*311. 0*012/3 1*.2- -*0/3!

#rom m$ tale of values aove, it is clear that the change of sign from negative to

positive occurs etween = 1*2 and = 1*3. So, I can narrow these values down

further to find another change of sign.

f+,

1*21 #1*1/2

1*22 #1*02/

1*23 #0*/!-/

1*2- #0*.2.3

1*2! #0*/.2-

1*2 #0*!-20

1*2 #0*-21

1*2. #0*2.-03

1*2/ #0*13/

1*30 0*012/3

I can see that the change of sign is etween = 1*2/ and = 1*30*

f+,

1*2/1 #0*122.3

1*2/2 #0*10/2

1*2/3 #0*0/2/

1*2/- #0*0/

1*2/! #0*02/3

1*2/ #0*0-.-

1*2/ #0*0321

1*2/. #0*01!-

1*2// #0*002331*300 0*012/3

The change of sign is in the interval 1*2// 1*3004

f+,

1*2//1 #0*000.0!

1*2//2 0*00020

1*2//3 0*0022--

The root of this e"uation lies in the interval 1*2//1 1*2//24. This means that I can

ta%e the root to e = 1*2//1! +! 'ecimal 5laces,*

2


3/15

Kosturi Ash 12/3

The ma&imum error of the root would e 0*0000!

'owever, there are some functions for which the (ecimal Search method would not

wor%. An e&ample is +1*!!"3*-,-=0* This would not wor% ecause the curve is

touching the &(a&is, so there would not e a change of sign. The graph of this function

and the calculations are elow!

I can see that the root lies etween = #2 and = #3.

f+,

#2 0*00.10000

#2*1 0*000--20!

#2*2 0*00000001

#2*3 0*000-120

#2*- 0*010-.!

#2*! 0*0!0/0-

#2* 0*1!!2/1

#2* 0*3/332!#2*. 0*.0-./

#2*/ 1*-30/!

#3*0 2*--1-02!

As I have rightl$ guessed, there is no change of sign according to m$ calculations

aove. This is ecause the e"uation has an even power in it. Therefore, the (ecimal

Search method has failed to find the root of +1*!!"3*-,-=0*

Solving 0=- "#1 using the $Newton#6a5hson& metho'

3

I would not e ale

to find this root

ecause there is no

change of sign. See

)igure 2


4/15

Kosturi Ash 12/3

As stated aove, the e"uation I will use for the $Newton#6a5hson& method is

-"#1=0* I can see from the graph that two roots need to e found using this method.

The graph of this is elow )or see )igure 3*.

The iterative formula for the $Newton#6a5hson& method is!

n"1=n# f+n,

f7+n,

#or m$ e"uation, f7+n, = -3"1 when m$ function is differentiated to find the

gradient function. The iterative formula I will use to find the roots of this function is!

n"1 = n# n- "n#1

-n3"1

I will start $ finding out the negative root first. +$ loo%ing at the diagram )see

)igure 3*, I can see that m$ starting value )1* is 81*!.

1 = #1*!

2 = #1*2/! )I sustituted 1 = #1*! into the iterative formula to get the value for 2*

3 = #1*22/0 )I sustituted 2 = #1*2/! into the iterative formula to get 3*

- = #1*220.12

! = #1*220--

= #1*220--0.!

= #1*220--0.!

I can see some convergence from !. There has een no change in the &(values

etween and for this numer of decimal places. I %now that this method for

finding the root has wor%ed.

I would need to find

these two roots of this

e"uation


5/15

Kosturi Ash 12/3

-$ negative root is = #1*220-- + 'ecimal 5laces,

ow I will find the positive root of this function. The starting value +1, is

appro&imatel$ 1*! when f+,=0*

1 = 1*!2 = 1*113/

3 = 0*.20.

- = 0*-!1

! = 0*2!0-/

= 0*2--/2

= 0*2--/1/!/

. = 0*2--/1/!/

I can see some convergence from . There has een no change in &(values etween and . for this numer of decimal places. It is clear that another root has een

successfull$ found using this method.

-$ positive root is = 0*2--/1/!/ +/ 'ecimal 5laces,

The error ound is! 0*000000000!

'owever, the Newton#6a5hson method would not wor% with the graph elow!

If I start from the turning point of the curve, I do not thin% it will converge )see

)igure -*. The e"uation for this function is 0 = -"23"3#- f7+, = -3"2"3

The iterative formula is! n"1 = n# f+n,

f7+n,

So the iterative formula for this e"uation is! n"1 = n# n- "2n3 "3n#- -n3"n2"3

I wish to find this root


6/15

Kosturi Ash 12/3

0n m$ diagram of this function, I have ta%en m$ 1 value to e aout 81*

1 = #1*

2 = 3*.2!/1

3 = 2*!./- = 1*.//.!

! = 1*3!3/

= 1*0033!

= 0*.-11

. = 0*.1/12

/ = 0*.1.3/0

10 = 0*.1.3./

11 = 0*.1.3./

#rom that, I can "uite clearl$ see that there is a convergence eginning to show

towards the positive root. 'owever, m$ aim was to find the negative root, and I couldnot find it. Therefore, the Newton#6a5hson method has failed to find that particular

root, even though the starting value was close to it.

Another e&ample where the ewton(aphson would not wor% is the function

+"2192,192 ln +"2192, = 0. The graph and calculations of this is elow!

The function is! +"2192,192 ln +"2192, = 0

f7+, = ++"2192,192 X 1 , " +ln +"2192, X : +"2192, #192,

" 2 192

+"2192 ,192 " ln +"2192 , +"2192 , #192

+"2

192

,

1

2

I wish to find this

root


7/15

Kosturi Ash 12/3

1 " ln+"2192 ,

+"2192,192 2+"2192, 192

= 2" ln++"2192 ,

2+"2192, 192

The iterative formula for the Newton#6a5hson method is!

n"1 = n = f7+n,

f7+n,

n"1 = n # 2+"2192 ,ln+"2192 ,

2 " ln+"2192,

I choose m$ starting value of +1, to e aout 81*3.

1 = #1*3

2 = #-*220/

3 = not 'efine'

- =

Another e&ample where the Newton#6a5hson method would fail is with the function

3#!"0*1=0. The graph of this, along with the calculations is shown elow!

I wish to find this

root


8/15

Kosturi Ash 12/3

f+, = 3#!"0*1 f7+, = 32#!

The iterative formula for the Newton#6a5hson method is! n"1 = n# f+n,

f7+n,

-$ iterative formula is! n"1 = n# 3 #!"0*1

32#!

I will ta%e m$ starting value of )1* to e 1*1.

1 = 1*1

2 = #1*.003

3 = #2*-000!3

- = #2*2!/1.

I can see that there is no convergence to the positive root I am loo%ing for. Therefore,

the Newton#6a5hson method has failed to find that particular root despite ta%ing a

starting value close to it. This is shown graphicall$ in )igure ;.

Solving 0= ! "- 2 #2 using the $6earranging metho'&

The graph of this is shown elow )or see )igure !*!

4

)This is the same graph as

the one on previous page*.

I wish to find this root


9/15

Kosturi Ash 12/3

I will rearrange f+,=0 in the form of = g+,* As m$ e"uation is 0= !"-2#2 there

are possile rearrangements of this. I will onl$ pic% out two possile rearrangements.

earrangement 1! 0= !"-2#2

#!

= -2

#2!= #-2"2

= +#-2"2,19!

0n the Autograph software, I will draw the e"uation


10/15

Kosturi Ash 12/3


#-2= !#2

-2= #!"2

2= +#!"2,9-

= ++#!"2,9-,,192

0n Autograph software, I will draw the e"uation < = g+, = ++#!"2,9-,,192 and the line

< = . This is what it loo%s li%e elow )or see )igure *!

I can see from the diagram directl$ aove for earrangement 2 that g7+, is etween 1and 81, so there should e a convergence.

16

I wish to find this

intersection


11/15

Kosturi Ash 12/3

Therefore, m$ chosen rearrangement is earrangement 2, which is! = ++#!"2,9-,,192

So m$ iterative formula for this rearrangement is! n"1= ++#n!"2,9-,,192

I can see from m$ graph aove that the point of intersections of the two lines is

etween = 0 and = 1. So m$ starting value for +1, would e 0*!.

1 = 0*! )I sustitute 1 = 0*! into iterative formula to get 2*

2 = 0*01!1 )I sustitute 2 value into iterative formula to get 3*

3 = 0*3/

- = 0*.11/

! = 0*.000

= 0*.0.02

= 0*.02

. = 0*.0/.

/ = 0*.0.3

10 = 0*.0.11 = 0*.0.!

12 = 0*.0.!

I can see a convergence from . and there is no change etween 11 and 12 for this

numer of decimal places. I %now that this method has wor%ed successfull$ to find a

root. This is shown graphicall$ in )igure .

-$ root is! = 0*.0 +! 'ecimal 5laces,*

I %now that using earrangement 2 helped me find a root of 0=!"-2 82. 'owever,

earrangement 1 of this function would not have helped me find the same root. This

is e&plained with a diagram of earrangement 1 and its iterative formula elow. This

failure is shown graphicall$ in )igure !


11

I would not e ale to

find this root using

this rearrangement

ecause the gradient

of


12/15

Kosturi Ash 12/3

#!= -2#2

!= #-2"2

= +#-2"2,19!

The iterative formula is! n"1= +#-n2"2,19!

#rom the graph aove and from )igures I can see that the point of intersection

etween < = and < = g+, = +#-2"2,19! is etween = 0 and = 1. 7i%e efore, I

will ta%e m$ starting value of +1, to e 0*!. The same process of sustituting values

of & into the iterative formula, as in the last earrangement is carried out here.

1 = 0*!

2 = 1

3 = #1*1-./.

- = #1*2.010

! = #1*3-.1

I can immediatel$ see that there is no convergence in this rearrangement towards the

particular root I am loo%ing for, that I found in the previous rearrangement

)earrangement 2** Therefore, I have found a rearrangement of the same e"uation

)earrangement 1* where the iteration has failed to converge to the re"uired root.

Com5arison of metho's

I will now use the e"uation 0=!"-2#2 that I solved using the 6earrangement

metho' and see if I can find the same root using the other two methods( the (ecimal

Search method and the Newton#6a5hson method. I will start off with the (ecimal

Search method.

#rom the graph 0=!"-2#2, I can see that the root lies etween = 0*! an' = 1

f+,

0*! #0*/.!

0* #0*-.22-

0* 0*12.0

I can see that the change of sign is etween = 0* and = 0**

f+,

0*1 #0*-21-0

0*2 #0*30.

0*3 #0*3131!

0*- #0*2!-22

0*! #0*1/3/1

0* #0*1323

0* #0*0/3.

0*. #0*00!00

0*/ 0*00.03

12


13/15

Kosturi Ash 12/3

The change of sign is etween = 0*. and = 0*/

f+,

0*.0 #0*00!00

0*.1 0*001!10

The change of sign is etween = 0*.0 and = 0*.1

f+,

0*.01 #-*3!!!10#3

0*.02 #3*0-!--10#3

0*.03 #3*0!32/10#3

0*.0- #2*-01.210#3

0*.0! #1*!032210#3

0*.0 #1*0/.2/10#3

0*.0 #-*-/210

#-

0*.0. 2*0!1.10#-

The change of sign is etween = 0*.0 and = 0*.0.

f+,

0*.01 #3*.101210#-

0*.02 #3*1-0.310#-

0*.03 #2*!121-010#-

0*.0- #1*.01.210#-

0*.0! #1*20.21010#-

0*.0 #!*!223210#!

0*.0 /*!!10#

I can see that the change of sign is etween = 0*.0 and = 0*.0

As there is a change of sign, I can finall$ sa$ that the root of this function is in the

interval 0*.0 0*.04. Therefore, using this method, I can ta%e the root of this

e"uation to e = 0*.0!, which is similar to the answer I got when finding the

same root using the $6earrangement metho'&.

I will now use the Newton#6a5hson method to find the same root for this function.

0=!"-2#2 f7+, = !-".

The iterative formula for the Newton#6a5hson method is! n"1=n# f+n,

f7+n,

So m$ iterative formula is! n"1=n# n! "-n2 #2

!n-".nI thin% that a good starting value for 1 would e 1*2

1 = 1*2

2 = 0*..0.33 = 0*20!/

13


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15/15

numerical solutions of equations (6)

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