Question

Question 7iii and iv pleasee http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w02_qp_3.pdf

Answer 1

7iii should be easy :
\[x_{n+1} = 50\sin x_n - 48.5x_n\]
Suppose the above recursive sequence converges to \(L\), then we have \(L = \lim \limits_{n\to \infty} x_{n+1} = \lim \limits_{n\to \infty} x_{n}\) :
\[\lim \limits_{n\to \infty} x_{n+1} = \lim \limits_{n\to \infty} 50\sin x_n - \lim \limits_{n\to \infty} 48.5x_n\]
\[L = 50\sin L - 48.5L\]
\[2L = 100\sin L - 97L\]
\[99L = 100\sin L \]
\[\dfrac{99}{100}L = \sin L\]

Answer 2

Oh wait. I didn't learn that deep yet.

Answer 3

what i did was f'(x) = 50cosx-48.5

Answer 4

this question is not about newton method

Answer 5

what's newton method? D:

Answer 6

It's iterative method.

Answer 7

am i supposed to substitute 0.1 to 0.5 inside the f'x equation?

Answer 8

derivatives are not needed here

Answer 9

then? this is just A Levels. I'm not doing a degree level of maths here btw.

Answer 10

just see that when a sequence converges, we have \(\lim\limits_{n\to \infty} x_{n+1} =\lim\limits_{n\to \infty} x_{n} \)

Answer 11

limits are simpler than derivatives

Answer 12

when we dont use limits in this syllabus.

Answer 13

how about substitutions?

Answer 14

have you read my first reply atleast once before giving up saying it is too advanced for you ?

Answer 15

it is not advanced actually, it uses one simple limit property and rest everyting is just algebra

Answer 16

Yeah the thing is i dont understand what you wrote. hence i said its too advanced. im not saying this ques is too advanced for me. it's not. im supposed to do this for exam also.

Answer 17

i just need a clarification. am i supposed to substitute 0.1 to 0.5 to show that it's converge or not.

Answer 18

im not sure how that allows us to conclude the given recursive sequence converges

Answer 19

how about iv?

Answer 20

@satellite73

Answer 21

is it 7iii?

Answer 22

and iv also

Answer 23

for iii do i just substitute 0.1 to 0.5 in the equation??

Answer 24

caz it says f'x <1 is converge

Answer 25

ok well what @ganeshie8 wrote is pretty much all i can write
the gimmick is this: when you want to find the limit of a recursion, the limit for \(x_n\) is the limit for \(x_{n+1}\) so lets call that limit \(L\) in other words replace \(x_n\) and \(x_{n+1}\) by \(L\)

Answer 26

\[L=50\sin(L)-4.8L\] in other words

Answer 27

damn another typo, must be late \[L=50\sin(L)-48.5L\]

Answer 28

dont we need to differentiate? im so confused, caz my textbook says must diffeetntiate.

Answer 29

got nothing to do with differentiation

Answer 30

are we on the right topic hahaha.

Answer 31

and you don't have to show that the limit exists, it just says "IF" it is exists show it is ...

Answer 32

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w02_ms_3.pdf take a look at the marking scheme. :)

Answer 33

got it
i used L, as did @ganeshie8 but they used \(x\) same thing

Answer 34

they replaced \(x_n\) and \(x_{n+1}\) by the limit, which they called \(x\) and wrote
\[x=50\sin(L)-48.5x\]

Answer 35

ohh kay. so x=50sinx- 48.5 x then?

Answer 36

yes, the variable is of course unimportant

Answer 37

what do i do next?

Answer 38

where you see an \(x_n\) and an \(x_{n+1}\) replace it by just a \(x\) which is the presumed limit
you get exactly that
\[x=50\sin(x)-48.5x\]

Answer 39

that's all? :O

Answer 40

that is all, then arrange it to look the the equation they want it to look like
using a small amount of algebra turn \[x=50\sin(x)-48.5x\] in to \[\frac{99}{10}x=\sin(x)\]

Answer 41

well actually \[\frac{99}{100}x=\sin(x)\]

Answer 42

pretty much nothing to do there right? add \(48.5x\) to both sides, then divide by \(50\)

Answer 43

But i aint showing it's converge means f'x <1. i wanna cry now.

Answer 44

i am confused by your confusion
lets read what it asks for you to do in that question
it does not as for you to SHOW that it converges

Answer 45

it says:
Show that IF it converges, then converges to the equation in part one.
that is all it asks for

Answer 46

ah, okay. :/ how about iv?

Answer 47

that is a calculator exercise

Answer 48

ok not really it is this
\[x_2=50\sin(.25)-48.5\times .25\]

Answer 49

then? :/ how to find alpha? :(

Answer 50

compute that number, and you get \[x_2=0.245198\]

Answer 51

then lather, rinse, repeat

Answer 52

\[x_3=50\sin(0.245198)-48.5\times 0.245198\]

Answer 53

it is clear what i am doing? that is using the iteration

Answer 54

it is a total calculator exercise, plug and chug

Answer 55

tryna digest. when do i stop all this?

Answer 56

that is an intelligent question, when to stop!

Answer 57

until the digits are almost the same right?

Answer 58

before we get to that however, it is clear what i am doing ? first i plugged in \(x_1=.25\) to find \(x_2\) etc

Answer 59

uh huh

Answer 60

then plug in \(x_2\) to get \(x_3\) and so on
clear or no?

Answer 61

yess

Answer 62

ok so how many decimal place accuracy does it ask for?

Answer 63

3

Answer 64

then when you get the same 3 decimals twice, stop

Answer 65

ohh x2 and x3 is the same

Answer 66

idk i didn't do it

Answer 67

okayy i got it! thanks so much! you finally clear my doubts. :) appreciate it

Answer 68

yeah you get \(.245\) twice in a row

Answer 69

yw
hope you see it was actually pretty easy, your problem was understanding what you had to do
really almost nothing to these parts
iii) replace \(x-n\) and \(x_{n+1}\) by \(x\) and do a small amount of algebra
iv) calculator exercise

Answer 70

yes. :)