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anonymous
 5 years ago
Hi Amistre...any luck
anonymous
 5 years ago
Hi Amistre...any luck

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3some... how about you? shall we dance :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0SURE!...I've been just reading over it all to try and make sense of it...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3difference equations go by a more common name of recursion formulas

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3go ahead and post your "problem" page and lets see what we can do with it now :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3I 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
 5 years ago
Best ResponseYou've already chosen the best response.3so we are on 2a right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.3We 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3These are also refered to as NEXT = NOW + C equations..... and they can be written as: X{n} = k X{n1} + C as well; they both mean the same thing right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3So 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
 5 years ago
Best ResponseYou've already chosen the best response.3that does 2a, now for 2b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I should start at 0 instead of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you get 1 when you plug in 0 for x sub n

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n+1} = (1/2) X{n} is the homo part;

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the first term is 0, its our starting term and is the first term :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok just kept going back and forth with the table

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3We need to come up with a solution that will make the homo part "fit" the data.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.31 = (1/2)(0) + C 1' 1/2 = (1/2)(1) + C is what I believe it is getting at

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0true...it is like trial and error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which looks like the limit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops missed negative so 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3yeah, thats what it was to begin with, which throws me off, why do this part if its just the same as the last ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0suppose to just drop constant to make it 0go figure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then have to combine general and particular for complete solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://media.pearsoncmg.com/aw/aw_gordon_functioning_2/Gordon_12_01.pdf

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3yeah :) Now to solve for any value of "n" we do this: X{n} = X{0} + (n1)d X{0} = 0 so we are left with (n1)d = X{n} right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3So from our table above we can input the values and find a solution for "d" (n1)d = X{n} keep in mind that this can be written as: (n)d = X{n+1}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is trial and error pretty much the method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using the patterns as a guide

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3(n)d = X{3} ; 2+1 = 3 (2)d = 1.75 d = 1.75/2 = .875 :: 7/8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3im 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
 5 years ago
Best ResponseYou've already chosen the best response.37/8 , i dropped the sign ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n} = (7/8)(n1) does this fit?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3all sequences start at n=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0checked it with 3...WORKS!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0of course it works for 1 was checking another ; )

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.34.375 is what you get right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3and it shouldnt be less than 2....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32 + (1/2)^n try that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0works for previous value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3If we want to know X{n}; we can use 2+(.5)^n X{0} = 0 2 +(1/2)^0 = 1..... 2 + (1/2)^(01) = 2 + 2 = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3yes ^(n1) looks good :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{5} = 1' 15/16 = 2 + (1/16) = 2 + (1/2)^(n1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3How we work this backwards tho..... to see how we got to it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32 +(1/2)^(n1) = X{n} = 2(X{n+1} 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0slowly but surely...can't say I'm comfortable with doing alone which is my goal

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32 +(1/2)^(n1) = X{n} = 2(X{n+1} +1) < signs again lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32 +(1/2)^(n1) = 2(X{n+1} +1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32 +(1/2)^(n1) = 2X{n+1} +2 (1/2)^(n1) = 2[X{n+1}] +4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{0} + 1/2^(n1) = X{n} right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{0} + X{n} * (1/2) (1/2)^(n1)......is what we need to be getiing at I think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n+1}  X{n} (1/2) [(1/2)^(n1)] = 0 Does that work out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got 1.9375 but lost track let me try again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0works for n=0 working on others

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3(1/2)^(n1) split into (1/2)^n  = 2[(1/2)^n] (1/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n+1}  X{n} 2 [(1/2)^n] = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm taking notes and trying to make sense of it and thinking about the final project I have to create...smh, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all for 1 credit, lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3lol.....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
 5 years ago
Best ResponseYou've already chosen the best response.3x{n+1} = (1/2)x{n} * (D)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i had originally, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just noticed it is blurry

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3trying to go that route just messes me up lol....do youhave to follow a certain step by step from the book?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.31/0 does not equal 0.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at this point...no just want to make sense of it however I can...more than one way to skin a cat, right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.31 + 1 = 0; whicn means (1/2)X{0} = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3.5 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^(n1)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3wait, 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
 5 years ago
Best ResponseYou've already chosen the best response.3(1/2)D needs to equal (1/2)^(n1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n} = (1/2)D  2; D = X{n1}

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the general solution gives us a "family" of curves to choose from right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one for each initial value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32[(1/2)^n] is our general soluton then right? maybe?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3we know we need it; and then we pinpoint it with 2 as a constant right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=plot+%282+*+%281%2F2%29%5Ex%29++2+from+x%3D0+to+20

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3this is the "family" of curves.... the general solution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3if we look at it head on its like this.... but how to get to the family lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3So: X{x+1} = (1/2)X{n} = 2[(1/2)^(n1)]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3(1/2)X{n} = 2[(1/2)^(n1)] + C .....should we bother to figure out the inbetweens :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3B should be 2 * (1/2)^(n1) + C as a general solution; but how to get to it the way they suggest is beyond me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know this works so no prob

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the general solution will be: 2*(1/2)^n

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.33c) 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
 5 years ago
Best ResponseYou've already chosen the best response.3http://www.math.kth.se/math/student/courses/5B1203/F/200304/linrek.pdf might be helpful

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3In 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
 5 years ago
Best ResponseYou've already chosen the best response.0are you going to be on later...have to fix lunch for daughter and work on final presentation for another course...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3I got 5 hours left online today....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I will study tonight and hope to see you tomorrow...thanks! I'll email you

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3After more reading; the general solution is: C * (1/2)^n that starting to look like the material your studying off of....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I had that...just didn't know how, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now I have a better understanding

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the rest of it there is still confusing me; I keep reading over the material, but it is quite frankly abismal..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0must get get 24 some how

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n+11}  (1/2) C(1/2)^(n1) = 1 X{n}  C (1/2)(1/2)^(n1)= 1 C(1/2)^n  C(1/2)(1/2)^(n1) = 1 C(1/2)^n [1  (1/2) (1/2)^1] = 1 C(1/2)^n = 1/(11) = 1/0 ..... thats just upsetting lol C(1/2)^n

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the answer for 2c is: 2, but how to get it .....ack!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3the 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
 5 years ago
Best ResponseYou've already chosen the best response.3the particular solution is simply a multiple of the nonhomo term: A(1) ; where A = 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0professor said to use trial solutions and figure out the constanthardest part

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{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
 5 years ago
Best ResponseYou've already chosen the best response.3n * X(n+1) = An *(1) (n1) * X{n} = A(n1) *(1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{n} = A *(1) then...... now what :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3or rather: (n+1) * X({n+1} = A(n+1)(1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{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
 5 years ago
Best ResponseYou've already chosen the best response.3X{n+1} = (1/2)^(n) + 2(1) X{n} = (1/2)^(n1) 2 is what we got to begin with without going thru this headache lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3took 3 months to get to the answer, an answer that we got the first time lol......

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3So 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
 5 years ago
Best ResponseYou've already chosen the best response.3the homo part is: X{n+1} = (1/2)*X{n} X{n} = C * (1/2)^n ....right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3teh solution we are getting at tho is: 2n + (1/2)^(n1) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this 3...my table is wrong

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3X{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
 5 years ago
Best ResponseYou've already chosen the best response.3yeah, working on 3 now..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3A(n+1)  (1/2) A(n) = n An + A An/2 = n 2An +2a An  = n 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.32An An +2A = 2n An + 2A = 2n A(n + 2) = 2n A = 2n/(n+2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3If 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
 5 years ago
Best ResponseYou've already chosen the best response.3or maybe it is :) let see..... (1/2)^(n1) + 2n/3 ..... nah, that aint it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.3when 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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