anonymous
  • anonymous
Hi Amistre...any luck
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
some... how about you? shall we dance :)
anonymous
  • anonymous
SURE!...I've been just reading over it all to try and make sense of it...
amistre64
  • amistre64
difference equations go by a more common name of recursion formulas

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anonymous
  • anonymous
yes!
amistre64
  • amistre64
go ahead and post your "problem" page and lets see what we can do with it now :)
anonymous
  • anonymous
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amistre64
  • 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
  • amistre64
so we are on 2a right?
anonymous
  • 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
  • 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
  • anonymous
ok
amistre64
  • 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
  • anonymous
yes
amistre64
  • 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
  • amistre64
that does 2a, now for 2b
anonymous
  • anonymous
so I should start at 0 instead of -1
anonymous
  • anonymous
I had 0 originally
anonymous
  • anonymous
but you get -1 when you plug in 0 for x sub n
amistre64
  • amistre64
X{n+1} = (1/2) X{n} is the homo part;
amistre64
  • amistre64
the first term is 0, its our starting term and is the first term :)
anonymous
  • anonymous
ok just kept going back and forth with the table
amistre64
  • amistre64
We need to come up with a solution that will make the homo part "fit" the data.
amistre64
  • amistre64
-1 = (1/2)(0) + C -1' 1/2 = (1/2)(-1) + C is what I believe it is getting at
anonymous
  • anonymous
true...it is like trial and error
anonymous
  • anonymous
-2
anonymous
  • anonymous
which looks like the limit?
anonymous
  • anonymous
oops missed negative so -1
amistre64
  • 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
  • anonymous
suppose to just drop constant to make it 0--go figure
anonymous
  • anonymous
then have to combine general and particular for complete solution
anonymous
  • anonymous
http://media.pearsoncmg.com/aw/aw_gordon_functioning_2/Gordon_12_01.pdf
amistre64
  • 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
  • anonymous
correct!
amistre64
  • 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
  • anonymous
so is trial and error pretty much the method
anonymous
  • anonymous
using the patterns as a guide
amistre64
  • amistre64
(n)d = X{3} ; 2+1 = 3 (2)d = -1.75 d = -1.75/2 = .875 :: 7/8
amistre64
  • 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
  • amistre64
-7/8 , i dropped the sign ;)
anonymous
  • anonymous
i caught it
amistre64
  • amistre64
X{n} = (-7/8)(n-1) does this fit?
anonymous
  • anonymous
checking on calc
amistre64
  • amistre64
for n=1
amistre64
  • amistre64
all sequences start at n=1
anonymous
  • anonymous
checked it with 3...WORKS!
anonymous
  • anonymous
of course it works for 1 was checking another ; )
amistre64
  • amistre64
check for X{6} :)
anonymous
  • anonymous
not getting answer
amistre64
  • amistre64
-4.375 is what you get right?
anonymous
  • anonymous
yes!
amistre64
  • amistre64
and it shouldnt be less than -2....
amistre64
  • amistre64
-2 + (1/2)^n try that :)
anonymous
  • anonymous
-1.984375
anonymous
  • anonymous
works for previous value
anonymous
  • anonymous
so raised to n-1?
amistre64
  • 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
  • amistre64
yes ^(n-1) looks good :)
amistre64
  • amistre64
X{5} = -1' 15/16 = -2 + (1/16) = -2 + (1/2)^(n-1)
amistre64
  • amistre64
How we work this backwards tho..... to see how we got to it :)
amistre64
  • amistre64
-2 +(1/2)^(n-1) = X{n} = 2(X{n+1} -1)
anonymous
  • anonymous
slowly but surely...can't say I'm comfortable with doing alone which is my goal
amistre64
  • amistre64
-2 +(1/2)^(n-1) = X{n} = 2(X{n+1} +1) <- signs again lol
amistre64
  • amistre64
-2 +(1/2)^(n-1) = 2(X{n+1} +1)
amistre64
  • amistre64
-2 +(1/2)^(n-1) = 2X{n+1} +2 (1/2)^(n-1) = 2[X{n+1}] +4
amistre64
  • amistre64
X{0} + 1/2^(n-1) = X{n} right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
X{0} + X{n} * (1/2) (1/2)^(n-1)......is what we need to be getiing at I think
amistre64
  • amistre64
X{n+1} - X{n} (1/2) [(1/2)^(n-1)] = 0 Does that work out?
anonymous
  • anonymous
got -1.9375 but lost track let me try again
anonymous
  • anonymous
works for n=0 working on others
anonymous
  • anonymous
get -2 for n=1
amistre64
  • amistre64
(1/2)^(n-1) split into (1/2)^n ------- = 2[(1/2)^n] (1/2)
amistre64
  • amistre64
X{n+1} - X{n} 2 [(1/2)^n] = 0
anonymous
  • 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
  • amistre64
:)
anonymous
  • anonymous
all for 1 credit, lol
anonymous
  • anonymous
that was b correct
amistre64
  • 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
  • amistre64
x{n+1} = (1/2)x{n} * (D)
anonymous
  • anonymous
what i had originally, lol
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anonymous
  • anonymous
just noticed it is blurry
amistre64
  • amistre64
trying to go that route just messes me up lol....do youhave to follow a certain step by step from the book?
amistre64
  • amistre64
-1/0 does not equal 0.....
anonymous
  • 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
  • anonymous
that is undefined
amistre64
  • amistre64
-1 + 1 = 0; whicn means -(1/2)X{0} = 1
amistre64
  • 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
  • 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
  • amistre64
(1/2)D needs to equal (1/2)^(n-1)
anonymous
  • anonymous
d=2
amistre64
  • amistre64
X{n} = (1/2)D - 2; D = X{n-1}
amistre64
  • amistre64
the general solution gives us a "family" of curves to choose from right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
one for each initial value
amistre64
  • amistre64
2[(1/2)^n] is our general soluton then right? maybe?
anonymous
  • anonymous
looks good
amistre64
  • amistre64
we know we need it; and then we pinpoint it with -2 as a constant right?
anonymous
  • anonymous
yeah
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=plot+%282+*+%281%2F2%29%5Ex%29+-+2+from+x%3D0+to+20
amistre64
  • amistre64
this is the "family" of curves.... the general solution
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anonymous
  • anonymous
nice!
amistre64
  • amistre64
if we look at it head on its like this.... but how to get to the family lol
1 Attachment
amistre64
  • amistre64
So: X{x+1} = (1/2)X{n} = 2[(1/2)^(n-1)]
amistre64
  • amistre64
+ a constant...
amistre64
  • amistre64
(1/2)X{n} = 2[(1/2)^(n-1)] + C .....should we bother to figure out the inbetweens :)
anonymous
  • anonymous
no is that b
amistre64
  • 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
  • anonymous
I know this works so no prob
amistre64
  • amistre64
the general solution will be: 2*(1/2)^n
amistre64
  • 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
  • amistre64
http://www.math.kth.se/math/student/courses/5B1203/F/200304/linrek.pdf might be helpful
amistre64
  • 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
  • anonymous
are you going to be on later...have to fix lunch for daughter and work on final presentation for another course...
amistre64
  • amistre64
I got 5 hours left online today....
anonymous
  • anonymous
ok I will study tonight and hope to see you tomorrow...thanks! I'll email you
amistre64
  • amistre64
After more reading; the general solution is: C * (1/2)^n that starting to look like the material your studying off of....
anonymous
  • anonymous
I had that...just didn't know how, lol
anonymous
  • anonymous
now I have a better understanding
amistre64
  • amistre64
the rest of it there is still confusing me; I keep reading over the material, but it is quite frankly abismal..
anonymous
  • anonymous
I know thanks ...
anonymous
  • anonymous
must get get 2-4 some how
amistre64
  • 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
  • amistre64
the answer for 2c is: -2, but how to get it .....ack!!
amistre64
  • 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
  • amistre64
the particular solution is simply a multiple of the nonhomo term: A(-1) ; where A = 2
anonymous
  • anonymous
professor said to use trial solutions and figure out the constant---hardest part
amistre64
  • 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
  • amistre64
n * X(n+1) = An *(-1) (n-1) * X{n} = A(n-1) *(-1)
amistre64
  • amistre64
X{n} = A *(-1) then...... now what :)
amistre64
  • amistre64
or rather: (n+1) * X({n+1} = A(n+1)(-1)
amistre64
  • 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
  • amistre64
that did it ;)
amistre64
  • 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
  • anonymous
WOw, yep
amistre64
  • amistre64
took 3 months to get to the answer, an answer that we got the first time lol......
anonymous
  • anonymous
Scary
amistre64
  • 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
  • amistre64
the homo part is: X{n+1} = (1/2)*X{n} X{n} = C * (1/2)^n ....right?
amistre64
  • amistre64
teh solution we are getting at tho is: 2n + (1/2)^(n-1) right?
anonymous
  • anonymous
is this 3...my table is wrong
amistre64
  • 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
  • amistre64
yeah, working on 3 now..
amistre64
  • amistre64
A(n+1) - (1/2) A(n) = n An + A -An/2 = n 2An +2a -An ----------- = n 2
amistre64
  • amistre64
2An -An +2A = 2n An + 2A = 2n A(n + 2) = 2n A = 2n/(n+2)
amistre64
  • 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
  • amistre64
or maybe it is :) let see..... (1/2)^(n-1) + 2n/3 ..... nah, that aint it
amistre64
  • 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.....

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