Question

Hi Amistre...any luck

Answer 1

some... how about you? shall we dance :)

Answer 2

SURE!...I've been just reading over it all to try and make sense of it...

Answer 3

difference equations go by a more common name of recursion formulas

Answer 4

yes!

Answer 5

go ahead and post your "problem" page and lets see what we can do with it now :)

Answer 6

I you tubed a few recurrsions and have a better idea of what they are trying to say..... that material of yours is simply atrocious to parse information out of :)

Answer 7

so we are on 2a right?

Answer 8

I've been googling and reading for the past couple of days but nothing seems to fully explain it or provide examples with solutions ; ) to guide me

Answer 9

We have 2 parts to every recurrsion formula; a start and a function

1) X{0} = 0

2) X{n+1}= 1/2 X{n} -1

Answer 10

ok

Answer 11

These are also refered to as NEXT = NOW + C equations.....

and they can be written as:

X{n} = k X{n-1} + C as well; they both mean the same thing right?

Answer 12

yes

Answer 13

So lets go ahead and list the first 6 terms;

X{0} = 0
X{1} = -1
X{2} = -1' 1/2
X{3} = -1' 3/4
X{4} = -1' 7/8
X{5} = -1' 15/16
X{6} = -1' 31/32

Answer 14

that does 2a, now for 2b

Answer 15

so I should start at 0 instead of -1

Answer 16

I had 0 originally

Answer 17

but you get -1 when you plug in 0 for x sub n

Answer 18

X{n+1} = (1/2) X{n} is the homo part;

Answer 19

the first term is 0, its our starting term and is the first term :)

Answer 20

ok just kept going back and forth with the table

Answer 21

We need to come up with a solution that will make the homo part "fit" the data.

Answer 22

-1 = (1/2)(0) + C
-1' 1/2 = (1/2)(-1) + C is what I believe it is getting at

Answer 23

true...it is like trial and error

Answer 24

-2

Answer 25

which looks like the limit?

Answer 26

oops missed negative so -1

Answer 27

yeah, thats what it was to begin with, which throws me off, why do this part if its just the same as the last ;)

Answer 28

suppose to just drop constant to make it 0--go figure

Answer 29

then have to combine general and particular for complete solution

Answer 30

http://media.pearsoncmg.com/aw/aw_gordon_functioning_2/Gordon_12_01.pdf

Answer 31

yeah :) Now to solve for any value of "n" we do this:

X{n} = X{0} + (n-1)d

X{0} = 0 so we are left with (n-1)d = X{n} right?

Answer 32

correct!

Answer 33

So from our table above we can input the values and find a solution for "d"

(n-1)d = X{n}

keep in mind that this can be written as: (n)d = X{n+1}

Answer 34

so is trial and error pretty much the method

Answer 35

using the patterns as a guide

Answer 36

(n)d = X{3} ; 2+1 = 3

(2)d = -1.75

d = -1.75/2 = .875 :: 7/8

Answer 37

im not sure about the trial and error method; if you are given an equationand asked to find the constant, then there are ways to determine its value. The trail and error might apply but it is reduced considerably

Answer 38

-7/8 , i dropped the sign ;)

Answer 39

i caught it

Answer 40

X{n} = (-7/8)(n-1) does this fit?

Answer 41

checking on calc

Answer 42

for n=1

Answer 43

all sequences start at n=1

Answer 44

checked it with 3...WORKS!

Answer 45

of course it works for 1 was checking another ; )

Answer 46

check for X{6} :)

Answer 47

not getting answer

Answer 48

-4.375 is what you get right?

Answer 49

yes!

Answer 50

and it shouldnt be less than -2....

Answer 51

-2 + (1/2)^n try that :)

Answer 52

-1.984375

Answer 53

works for previous value

Answer 54

so raised to n-1?

Answer 55

If we want to know X{n}; we can use -2+(.5)^n
X{0} = 0

-2 +(1/2)^0 = -1.....

-2 + (1/2)^(0-1) = -2 + 2 = 0

Answer 56

yes ^(n-1) looks good :)

Answer 57

X{5} = -1' 15/16 = -2 + (1/16) = -2 + (1/2)^(n-1)

Answer 58

How we work this backwards tho..... to see how we got to it :)

Answer 59

-2 +(1/2)^(n-1) = X{n} = 2(X{n+1} -1)

Answer 60

slowly but surely...can't say I'm comfortable with doing alone which is my goal

Answer 61

-2 +(1/2)^(n-1) = X{n} = 2(X{n+1} +1) <- signs again lol

Answer 62

-2 +(1/2)^(n-1) = 2(X{n+1} +1)

Answer 63

-2 +(1/2)^(n-1) = 2X{n+1} +2

(1/2)^(n-1) = 2[X{n+1}] +4

Answer 64

X{0} + 1/2^(n-1) = X{n} right?

Answer 65

yes

Answer 66

X{0} + X{n} * (1/2) (1/2)^(n-1)......is what we need to be getiing at I think

Answer 67

X{n+1} - X{n} (1/2) [(1/2)^(n-1)] = 0 Does that work out?

Answer 68

got -1.9375 but lost track let me try again

Answer 69

works for n=0 working on others

Answer 70

get -2 for n=1

Answer 71

(1/2)^(n-1) split into

(1/2)^n
------- = 2[(1/2)^n]
(1/2)

Answer 72

X{n+1} - X{n} 2 [(1/2)^n] = 0

Answer 73

I'm taking notes and trying to make sense of it and thinking about the final project I have to create...smh, lol

Answer 74

:)

Answer 75

all for 1 credit, lol

Answer 76

that was b correct

Answer 77

lol.....its so hard to tell these days; but yeah, i think that is what b is going to end up looking like; it is works out.

Answer 78

x{n+1} = (1/2)x{n} * (D)

Answer 79

what i had originally, lol

Answer 80

just noticed it is blurry

Answer 81

trying to go that route just messes me up lol....do youhave to follow a certain step by step from the book?

Answer 82

-1/0 does not equal 0.....

Answer 83

at this point...no just want to make sense of it however I can...more than one way to skin a cat, right?

Answer 84

that is undefined

Answer 85

-1 + 1 = 0; whicn means -(1/2)X{0} = 1

Answer 86

.5 X{n} = X{n+1} ..... but that is nowhere near the answers we get in the table; do we just punch thru thill we get a parttern for 1/2^(n-1)?

Answer 87

wait, I think I see it;

X{n+1} = (1/2) * X{n} only works if we do:

X{n+1} = (1/2) * X{n} - 1.....when we combine the 2 we get

X{n+1} = (1/2)D -2 ;X{n} becomes the variable.... and -1 -1 = -2

Answer 88

(1/2)D needs to equal (1/2)^(n-1)

Answer 89

d=2

Answer 90

X{n} = (1/2)D - 2; D = X{n-1}

Answer 91

the general solution gives us a "family" of curves to choose from right?

Answer 92

yes

Answer 93

one for each initial value

Answer 94

2[(1/2)^n] is our general soluton then right? maybe?

Answer 95

looks good

Answer 96

we know we need it; and then we pinpoint it with -2 as a constant right?