Question

Need help with difference equation...

Answer 1

Suppose your car has a 14 gallon gas tank that you fill as soon as the level drops below half a tank. Every time you fill up, you add 1 quart (1/4 gallon) of an additive that mixes completely with the gas and is used along with the gas.

Answer 2

What is the difference equation and initial value that models the amount A sub n of additive in the tank at the time of the nth fill-up?

Answer 3

found it :)

Answer 4

great!

Answer 5

I attached so I will not have to type out entire problems

Answer 6

i see it. how much additive is remaining at the time of each fill up..

Answer 7

I truly appreciate your help...I love getting on here helping others so I share your passion!

Answer 8

pistachios are more of my passion i think :)

Answer 9

this is more of a hobby lol

Answer 10

lol...well you're AWESOME!

Answer 11

ok so do you understand the question

Answer 12

lets assume that we start with a clean tank..right? at halhway we fill it up and add the additive.

Answer 13

ok

Answer 14

at half a tank we have only half of what we put still in..... and the ration is the same, 14: .25 or some such right?

Answer 15

yes

Answer 16

the quart we put in becomes half a quart, and we add more gas and anouter quart... right?

Answer 17

ok

Answer 18

this loks to be one of those exponential or log setups... from first glance

Answer 19

don't think so

Answer 20

I could be wrong, I allow that :)

Answer 21

So A0 = 1 quart of additive.

A1 = 1 quart + 1/2 thats left = 1.5 quarts

a2 = 1.5/2 +1 and iterates that way...makes sense?

Answer 22

yes!

Answer 23

lets plug in the numbers and see if theres a pattern to follow :)

Answer 24

would it be easier to stay with gallons since the tank is in gallons A1=.25+7?

Answer 25

"that models the amount An of additive in the tank at the time" leads me to believe that the gas itself is of no consquence, only to tell us when to fill up again

Answer 26

ok

Answer 27

1 = 1
2 = 1.5
3 = 1.75
4 = 1.875
5 = 1.9375
6 = 1.96875

Answer 28

7 = 1.984375

Answer 29

i see a pattern emerging, just gotta put my finger on it :)

Answer 30

I got it An=.5A0+1?

Answer 31

\[\frac{\frac{\frac{\frac{1}{2}+1}{2}+1}{2}+1}{2}+1...\]

Answer 32

Does this pattern remind you of anything?

Answer 33

Fibonacci?

Answer 34

it does kinda look fibonaci like :)

Answer 35

i cant place it yet....

Answer 36

but its growing to a limit of 1.99999999999999999999

Answer 37

Can you help with others/check

Answer 38

yes I have 1.998 for n=10

Answer 39

I can try; have you finished them?

Answer 40

Still have to answer rest of 2, almost done with 3, 4 complete-easy

Answer 41

the formula on p.368 is \[D _{n}=L-(L-D _{0})a ^{n}\]

Answer 42

and what do we do with this formula :)

Answer 43

c) Find the solution from a using equation

Answer 44

or maybe we use L =B/1-a as n approaches infinity

Answer 45

have we found our difference equation yet?

Answer 46

I am using Asub n = .5Asub0 + 1

Answer 47

typed it earlier...do you agree?

Answer 48

is that the same as .05(n) + 1?

Answer 49

.5(n) +1

Answer 50

yes!

Answer 51

at n = 3 we get: 2.5 with that one, so it doesnt match our data.

Answer 52

at any rate, we are adding 1 and a fraction, so anything over 2 is not correct :)

Answer 53

using previous values of Asubn to get others

Answer 54

\[A _{n}=.5A _{n+1}+1\]

Answer 55

lets test that equation and see if it gets us what we know.

.5(3+1) + 1 = 2+1 = 3. its not working yet :)

Answer 56

I got something :) needs to be ironed out tho

Answer 57

\[A _{n+1}=.5A _{n}+1\]

Answer 58

2n+1
An = 1 + ------
2^n

Answer 59

I see it but can't explain in formula correctly

Answer 60

\[1,\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \frac{31}{32}, \frac{63}{64}\]

Answer 61

that fits it to a "T" :)

Answer 62

\[A _{n} = 1+\frac{2n+1}{2^n}\]

Answer 63

its off at n=0; do we start at 0?

Answer 64

that is giving greater than 2

Answer 65

no n=1

Answer 66

its close, but needs to be tweeked

Answer 67

2(n-1)+1

Answer 68

\[A_{n} = 1 + \frac{2(n-1)+1}{2^n}\]

Answer 69

not yet eh lol

Answer 70

\[A _{n}=.5A _{n-1}+1\]

Answer 71

we are taking half of previous output and adding 1 right?

Answer 72

yes

Answer 73

yours looks good like that :)

Answer 74

half of what it was, then add 1....yep

Answer 75

What is A_0-1? or are we starting n=1 :)

Answer 76

geeshhh. I had that thinking too hard ; (

Answer 77

lol

Answer 78

n=1

Answer 79

good, we got an equation that we like...now we use it in C right? or B first

Answer 80

a

Answer 81

that was A; find the equation ;)

Answer 82

B is; use it to make a table....

Answer 83

yes...done

Answer 84

We just happened to make a table first lol

Answer 85

working backwards in a problem solving strategy ; )

Answer 86

is

Answer 87

Dn =L−(L−D0 )a^n So what does this formula mean? its greek to me :)

Answer 88

me too...uh oh

Answer 89

368 should tell us something about it...like what D is and L

Answer 90

pg 368 that is

Answer 91

B/1-a in place of L for solutions growing towards infinity

Answer 92

sorry keep forgetting to his post

Answer 93

whats the chapter heading for your material?

Answer 94

Modeling with Difference Equations

Answer 95

does the formula have a name for it? its probably online somewhere and I can get a better grip on it.

Answer 96

difference equation...

Answer 97

using Asubn+1=.5Asubn +1