Need help with difference equation...

- anonymous

Need help with difference equation...

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- schrodinger

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- anonymous

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.

- anonymous

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?

- amistre64

found it :)

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## More answers

- anonymous

great!

##### 1 Attachment

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

pistachios are more of my passion i think :)

- amistre64

this is more of a hobby lol

- anonymous

lol...well you're AWESOME!

- anonymous

ok so do you understand the question

- amistre64

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

- anonymous

ok

- amistre64

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?

- anonymous

yes

- amistre64

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

- anonymous

ok

- amistre64

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

- anonymous

don't think so

- amistre64

I could be wrong, I allow that :)

- amistre64

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?

- anonymous

yes!

- amistre64

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

- anonymous

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

- amistre64

"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

- anonymous

ok

- amistre64

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

- amistre64

7 = 1.984375

- amistre64

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

- anonymous

I got it An=.5A0+1?

- amistre64

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

- amistre64

Does this pattern remind you of anything?

- anonymous

Fibonacci?

- amistre64

it does kinda look fibonaci like :)

- amistre64

i cant place it yet....

- amistre64

but its growing to a limit of 1.99999999999999999999

- anonymous

Can you help with others/check

- anonymous

yes I have 1.998 for n=10

- amistre64

I can try; have you finished them?

- anonymous

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

- anonymous

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

- amistre64

and what do we do with this formula :)

- anonymous

c) Find the solution from a using equation

- anonymous

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

- amistre64

have we found our difference equation yet?

- anonymous

I am using Asub n = .5Asub0 + 1

- anonymous

typed it earlier...do you agree?

- amistre64

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

- amistre64

.5(n) +1

- anonymous

yes!

- amistre64

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

- amistre64

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

- anonymous

using previous values of Asubn to get others

- anonymous

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

- amistre64

lets test that equation and see if it gets us what we know.
.5(3+1) + 1 = 2+1 = 3. its not working yet :)

- amistre64

I got something :) needs to be ironed out tho

- anonymous

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

- amistre64

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

- anonymous

I see it but can't explain in formula correctly

- amistre64

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

- amistre64

that fits it to a "T" :)

- amistre64

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

- amistre64

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

- anonymous

that is giving greater than 2

- anonymous

no n=1

- amistre64

its close, but needs to be tweeked

- amistre64

2(n-1)+1

- amistre64

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

- amistre64

not yet eh lol

- anonymous

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

- anonymous

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

- amistre64

yes

- amistre64

yours looks good like that :)

- amistre64

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

- amistre64

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

- anonymous

geeshhh. I had that thinking too hard ; (

- amistre64

lol

- anonymous

n=1

- amistre64

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

- anonymous

a

- amistre64

that was A; find the equation ;)

- amistre64

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

- anonymous

yes...done

- amistre64

We just happened to make a table first lol

- anonymous

working backwards in a problem solving strategy ; )

- anonymous

is

- amistre64

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

- anonymous

me too...uh oh

- amistre64

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

- amistre64

pg 368 that is

- anonymous

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

- anonymous

sorry keep forgetting to his post

- amistre64

whats the chapter heading for your material?

- anonymous

Modeling with Difference Equations

- amistre64

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

- anonymous

difference equation...

- anonymous

using Asubn+1=.5Asubn +1

- amistre64

we found a first order linear difference equation, if I see it right..

- amistre64

Tn+1 âˆ’ Tn = k(Tn âˆ’ S) looks useful
Dn =Lâˆ’(Lâˆ’D0 )a^n but I dunno....
are there any other formulas on page 368?

- anonymous

\[x _{n=}L+(x _{0}-L)a ^{n}\]

- anonymous

used that in c got b working on e using L=B/1- to find limiting value

- anonymous

finished d

- amistre64

you got it?

- anonymous

need e and f

- amistre64

the limit is 2 I beleive.... is that right? as n increases without bound we approach 2

- anonymous

yes but trying to get that from equation...working backwards again

- amistre64

:) I plugged in alot of iteration in google and gets 2 :)

- amistre64

are we allowed to use the definition of a limt? f(n+h) - f(n)/h ?

- amistre64

thats the limit as it approaches zero....not infiinity tho....

- amistre64

I think we would need a more rigid equation for that than just An=.5A(n-1) + 1 tho, becasue there is no way to determine the value of A(n-1) at any given point.

- amistre64

unless you got a formula for that too :)

- anonymous

limit is to infinity ...it also states that n can be replaced by n-1 to convert a forward difference equation to an equivalent backward de

- anonymous

since n=0

- anonymous

e) 1/1-.5

- amistre64

1/.5 = 2

- amistre64

1/(1-.5) right?

- anonymous

YEP!!!!

- amistre64

thats 2 :)

- amistre64

thats 2 :)

- anonymous

f)

- amistre64

.5 would change to a different number

- amistre64

.6 I believe

- amistre64

40%full...... .4

- anonymous

I concur! g)

- amistre64

.4(An-1) + 1

- anonymous

forgot post button again

- amistre64

and for the last one; the size of the tank doesnt matter, its the ratio.

- amistre64

16 gal filled every half is the same result

- anonymous

what about limiting value for f

- amistre64

id venture to guess that is it 2; but let me check my sources :)

- amistre64

1.666666666666666666666666666......

- amistre64

if I did it right :)

- anonymous

just got taht on cal!!!!

- anonymous

GREAT!

- anonymous

I have 3a and b can you check c for me Bsub15 = 80000(1.06)to the n - 10,000 so

- amistre64

\[80000(1.06)n -10000\] looks good

- amistre64

or is it n-1?

- amistre64

\[B_n = 80000(1.06)(n-1) - 10000\]

- amistre64

at 1 that would be a balance of -10000; so prolly not

- anonymous

\[B _{n+1}=1.06B _{n}-10000\]

- anonymous

noooo...you were about to make me scream!

- amistre64

lol

- anonymous

so c I keep going back and forth on

- anonymous

got $181,724.66

- amistre64

well, if he keeps taking 10000 out, he runs outta money before then I think.

- amistre64

80000(1.06) = 84800 - 10000 = 74800 right?

- anonymous

yes for n=1

- amistre64

79288 - 10000 = 69288

- amistre64

73445.28 - 10000 = 63445.28

- amistre64

67251.9968 - 10000 = 57251.9968

- amistre64

60687.116608 - 10000= 50687.116608
53728.34360448 - 10000 = 43728.34
46352.0404 - 10000 = 36352.0404

- amistre64

were running outta money :)

- amistre64

8th year:
38533.17 - 10000 = 28533.17
9th year:
30245.16 - 10000 = 20245.16
10th year:
21459.87 - 10000 = 11459.87
11th year:
12147.47 -10000 = 2147.47
12th year:
2276.32 - 10000 ...... cant do it :)

- anonymous

lol...my 2 year was off by 1,000 wrote wrong digit from cal....geesh

- amistre64

year 12 shes broke... at 15 theres nothing left;

- amistre64

if anything she owes 37723.68 by year 15

- amistre64

so what amount can she withdraw each year that will result in a zero balance at year 15 is the question lol

- anonymous

yeah i was just reading it again, too lmbo

- amistre64

its between 8000 and 9000 :)

- amistre64

between 8200 and 8300 :)

- amistre64

between 8230 and 8245

- anonymous

what would be my explanation...you working backwards

- amistre64

im just reiterating it 15 times on google and narrowing it down :)

- amistre64

once we get a number, we can plug it in and see if it works :)

- amistre64

same equation, just a different withdrawal amount

- anonymous

ok

- anonymous

would we have to log this

- amistre64

-8237 = 0.491505273 :)

- amistre64

- 8 237.03 = -0.206773824
cant get any less
- 8 237.02 = 0.0259858751

- amistre64

explain how you got your answer: trial and error :)

- amistre64

the way all good people have done it for thousands of years; we narrowed it down til something stuck

- anonymous

I might add that, lol!

- amistre64

:)

- amistre64

plug it into your calculator and see if it gets you a good deal...

- anonymous

I will finish e bc I know you want to move on...THANKS SOOOO MUCH!

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