Hi Amistre...any luck

- anonymous

Hi Amistre...any luck

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

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

- anonymous

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

- amistre64

difference equations go by a more common name of recursion formulas

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

yes!

- amistre64

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

- anonymous

##### 1 Attachment

- amistre64

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 :)

- amistre64

so we are on 2a right?

- anonymous

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

- amistre64

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

- anonymous

ok

- amistre64

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?

- anonymous

yes

- amistre64

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

- amistre64

that does 2a, now for 2b

- anonymous

so I should start at 0 instead of -1

- anonymous

I had 0 originally

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

ok just kept going back and forth with the table

- amistre64

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

- amistre64

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

- anonymous

true...it is like trial and error

- anonymous

-2

- anonymous

which looks like the limit?

- anonymous

oops missed negative so -1

- amistre64

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

- anonymous

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

- anonymous

then have to combine general and particular for complete solution

- anonymous

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

- amistre64

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?

- anonymous

correct!

- amistre64

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}

- anonymous

so is trial and error pretty much the method

- anonymous

using the patterns as a guide

- amistre64

(n)d = X{3} ; 2+1 = 3
(2)d = -1.75
d = -1.75/2 = .875 :: 7/8

- amistre64

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

- amistre64

-7/8 , i dropped the sign ;)

- anonymous

i caught it

- amistre64

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

- anonymous

checking on calc

- amistre64

for n=1

- amistre64

all sequences start at n=1

- anonymous

checked it with 3...WORKS!

- anonymous

of course it works for 1 was checking another ; )

- amistre64

check for X{6} :)

- anonymous

not getting answer

- amistre64

-4.375 is what you get right?

- anonymous

yes!

- amistre64

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

- amistre64

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

- anonymous

-1.984375

- anonymous

works for previous value

- anonymous

so raised to n-1?

- amistre64

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

- amistre64

yes ^(n-1) looks good :)

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

yes

- amistre64

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

- amistre64

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

- anonymous

got -1.9375 but lost track let me try again

- anonymous

works for n=0 working on others

- anonymous

get -2 for n=1

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

:)

- anonymous

all for 1 credit, lol

- anonymous

that was b correct

- amistre64

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.

- amistre64

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

- anonymous

what i had originally, lol

##### 1 Attachment

- anonymous

just noticed it is blurry

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

that is undefined

- amistre64

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

- amistre64

.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)?

- amistre64

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

- amistre64

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

- anonymous

d=2

- amistre64

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

- amistre64

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

- anonymous

yes

- anonymous

one for each initial value

- amistre64

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

- anonymous

looks good

- amistre64

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

- anonymous

yeah

- amistre64

http://www.wolframalpha.com/input/?i=plot+%282+*+%281%2F2%29%5Ex%29+-+2+from+x%3D0+to+20

- amistre64

this is the "family" of curves.... the general solution

##### 1 Attachment

- anonymous

nice!

- amistre64

if we look at it head on its like this....
but how to get to the family lol

##### 1 Attachment

- amistre64

So:
X{x+1} = (1/2)X{n} = 2[(1/2)^(n-1)]

- amistre64

+ a constant...

- amistre64

(1/2)X{n} = 2[(1/2)^(n-1)] + C .....should we bother to figure out the inbetweens :)

- anonymous

no is that b

- amistre64

B should be 2 * (1/2)^(n-1) + C as a general solution; but how to get to it the way they suggest is beyond me :)

- anonymous

I know this works so no prob

- amistre64

the general solution will be: 2*(1/2)^n

- amistre64

3c) Find a particular solution P{n} of the difference equation
X{n+1}=(1/2) X{n} -1 ; X0=0
from Part a. Show your work....

- amistre64

http://www.math.kth.se/math/student/courses/5B1203/F/200304/linrek.pdf might be helpful

- amistre64

In order to solve such a recursion, we need only solve the corresponding
homogeneous recursion and then find one particular solution, say apart
n , to the inhomogeneous recursion. Then any solution can be written as:
an = a{n}hom + a{n}part ;

- anonymous

are you going to be on later...have to fix lunch for daughter and work on final presentation for another course...

- amistre64

I got 5 hours left online today....

- anonymous

ok I will study tonight and hope to see you tomorrow...thanks! I'll email you

- amistre64

After more reading; the general solution is:
C * (1/2)^n that starting to look like the material your studying off of....

- anonymous

I had that...just didn't know how, lol

- anonymous

now I have a better understanding

- amistre64

the rest of it there is still confusing me; I keep reading over the material, but it is quite frankly abismal..

- anonymous

I know thanks ...

- anonymous

must get get 2-4 some how

- amistre64

X{n+1-1} - (1/2) C(1/2)^(n-1) = -1
X{n} - C (1/2)(1/2)^(n-1)= -1
C(1/2)^n - C(1/2)(1/2)^(n-1) = -1
C(1/2)^n [1 - (1/2) (1/2)^-1] = -1
C(1/2)^n = -1/(1-1) = -1/0 ..... thats just upsetting lol
C(1/2)^n

- amistre64

the answer for 2c is: -2, but how to get it .....ack!!

- amistre64

the answer for 2d is:
X{n+1} = general homo + particular
X{n+1} = (1/2)^(n) + (-2)
but the instructions are not clear on how to get the particular solution to the nonhomo equation.... cause I end up with -1/0 which is absurd ;)

- amistre64

the particular solution is simply a multiple of the nonhomo term:
A(-1) ; where A = 2

- anonymous

professor said to use trial solutions and figure out the constant---hardest part

- amistre64

X{n+1} - (1/2) X{n} = A*(-1) is what the book say to do wo far...
it then say s to do this:
X{n} = A*(-1) but we have no "n" nalue over there; so it suggests elsewhere to multiply both sides by "n" first... i think lol

- amistre64

n * X(n+1) = An *(-1)
(n-1) * X{n} = A(n-1) *(-1)

- amistre64

X{n} = A *(-1) then...... now what :)

- amistre64

or rather:
(n+1) * X({n+1} = A(n+1)(-1)

- amistre64

X{n+1} = A * (-1)...... this feels redundant lol
X{n+1} - (1/2)A(-1) = -1...
A(-1) - -A/2 = -1
A(-1 +1/2) = -1
A(-1/2) = -1
A = 2....

- amistre64

that did it ;)

- amistre64

X{n+1} = (1/2)^(n) + 2(-1)
X{n} = (1/2)^(n-1) -2 is what we got to begin with without going thru this headache lol

- anonymous

WOw, yep

- amistre64

took 3 months to get to the answer, an answer that we got the first time lol......

- anonymous

Scary

- amistre64

So for the next one; we can see if this setup works the same..
X{n+1} = (1/2) X{n} + n ; X{0} = 0
1/2(0) + 0 ; X{0}=0
1/2(0)+1 ; X{1}=1
1/2(1)+2 ; X{2}=2' 1/2
1/2(2.5)+3 ; X{3}= 4' 1/4
1/2(4.25)+4 ; X{4}= 6' 1/8
1/2(6.125)+5 ; X{5}= 8' 1/16

- amistre64

the homo part is:
X{n+1} = (1/2)*X{n}
X{n} = C * (1/2)^n ....right?

- amistre64

teh solution we are getting at tho is:
2n + (1/2)^(n-1) right?

- anonymous

is this 3...my table is wrong

- amistre64

X{n+1} = (1/2)^(n) + An....
X{n} = A *n
X{n+1} = A (n+1); use this in our equation:
A(n+1) - (1/2)(1/2)^(n) = n

- amistre64

yeah, working on 3 now..

- amistre64

A(n+1) - (1/2) A(n) = n
An + A -An/2 = n
2An +2a -An
----------- = n
2

- amistre64

2An -An +2A = 2n
An + 2A = 2n
A(n + 2) = 2n
A = 2n/(n+2)

- amistre64

If we stop at:
An + 2A = 2n ; when=1 we get:
A +2A = 2
3A = 2
A = 2/3 ...... which is not what I like lol

- amistre64

or maybe it is :) let see.....
(1/2)^(n-1) + 2n/3 ..... nah, that aint it

- amistre64

when n = 0 tho we need n = -2 if we go this route...
X{0} = (1/2)^-1 + n = 0
2 + n = 0; n=-2; so A(0) = -2?
X{3} = (1/2)^2 + n = 4' 1/4
1/4 + n = 4' 1/4; A(3)= 4 A=4/3....
ugh.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.