anonymous
  • anonymous
Need help with difference equation...
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • 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
  • amistre64
found it :)

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anonymous
  • anonymous
great!
1 Attachment
anonymous
  • anonymous
I attached so I will not have to type out entire problems
amistre64
  • amistre64
i see it. how much additive is remaining at the time of each fill up..
anonymous
  • anonymous
I truly appreciate your help...I love getting on here helping others so I share your passion!
amistre64
  • amistre64
pistachios are more of my passion i think :)
amistre64
  • amistre64
this is more of a hobby lol
anonymous
  • anonymous
lol...well you're AWESOME!
anonymous
  • anonymous
ok so do you understand the question
amistre64
  • amistre64
lets assume that we start with a clean tank..right? at halhway we fill it up and add the additive.
anonymous
  • anonymous
ok
amistre64
  • 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
  • anonymous
yes
amistre64
  • amistre64
the quart we put in becomes half a quart, and we add more gas and anouter quart... right?
anonymous
  • anonymous
ok
amistre64
  • amistre64
this loks to be one of those exponential or log setups... from first glance
anonymous
  • anonymous
don't think so
amistre64
  • amistre64
I could be wrong, I allow that :)
amistre64
  • 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
  • anonymous
yes!
amistre64
  • amistre64
lets plug in the numbers and see if theres a pattern to follow :)
anonymous
  • anonymous
would it be easier to stay with gallons since the tank is in gallons A1=.25+7?
amistre64
  • 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
  • anonymous
ok
amistre64
  • amistre64
1 = 1 2 = 1.5 3 = 1.75 4 = 1.875 5 = 1.9375 6 = 1.96875
amistre64
  • amistre64
7 = 1.984375
amistre64
  • amistre64
i see a pattern emerging, just gotta put my finger on it :)
anonymous
  • anonymous
I got it An=.5A0+1?
amistre64
  • amistre64
\[\frac{\frac{\frac{\frac{1}{2}+1}{2}+1}{2}+1}{2}+1...\]
amistre64
  • amistre64
Does this pattern remind you of anything?
anonymous
  • anonymous
Fibonacci?
amistre64
  • amistre64
it does kinda look fibonaci like :)
amistre64
  • amistre64
i cant place it yet....
amistre64
  • amistre64
but its growing to a limit of 1.99999999999999999999
anonymous
  • anonymous
Can you help with others/check
anonymous
  • anonymous
yes I have 1.998 for n=10
amistre64
  • amistre64
I can try; have you finished them?
anonymous
  • anonymous
Still have to answer rest of 2, almost done with 3, 4 complete-easy
anonymous
  • anonymous
the formula on p.368 is \[D _{n}=L-(L-D _{0})a ^{n}\]
amistre64
  • amistre64
and what do we do with this formula :)
anonymous
  • anonymous
c) Find the solution from a using equation
anonymous
  • anonymous
or maybe we use L =B/1-a as n approaches infinity
amistre64
  • amistre64
have we found our difference equation yet?
anonymous
  • anonymous
I am using Asub n = .5Asub0 + 1
anonymous
  • anonymous
typed it earlier...do you agree?
amistre64
  • amistre64
is that the same as .05(n) + 1?
amistre64
  • amistre64
.5(n) +1
anonymous
  • anonymous
yes!
amistre64
  • amistre64
at n = 3 we get: 2.5 with that one, so it doesnt match our data.
amistre64
  • amistre64
at any rate, we are adding 1 and a fraction, so anything over 2 is not correct :)
anonymous
  • anonymous
using previous values of Asubn to get others
anonymous
  • anonymous
\[A _{n}=.5A _{n+1}+1\]
amistre64
  • 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
  • amistre64
I got something :) needs to be ironed out tho
anonymous
  • anonymous
\[A _{n+1}=.5A _{n}+1\]
amistre64
  • amistre64
2n+1 An = 1 + ------ 2^n
anonymous
  • anonymous
I see it but can't explain in formula correctly
amistre64
  • amistre64
\[1,\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \frac{31}{32}, \frac{63}{64}\]
amistre64
  • amistre64
that fits it to a "T" :)
amistre64
  • amistre64
\[A _{n} = 1+\frac{2n+1}{2^n}\]
amistre64
  • amistre64
its off at n=0; do we start at 0?
anonymous
  • anonymous
that is giving greater than 2
anonymous
  • anonymous
no n=1
amistre64
  • amistre64
its close, but needs to be tweeked
amistre64
  • amistre64
2(n-1)+1
amistre64
  • amistre64
\[A_{n} = 1 + \frac{2(n-1)+1}{2^n}\]
amistre64
  • amistre64
not yet eh lol
anonymous
  • anonymous
\[A _{n}=.5A _{n-1}+1\]
anonymous
  • anonymous
we are taking half of previous output and adding 1 right?
amistre64
  • amistre64
yes
amistre64
  • amistre64
yours looks good like that :)
amistre64
  • amistre64
half of what it was, then add 1....yep
amistre64
  • amistre64
What is A_0-1? or are we starting n=1 :)
anonymous
  • anonymous
geeshhh. I had that thinking too hard ; (
amistre64
  • amistre64
lol
anonymous
  • anonymous
n=1
amistre64
  • amistre64
good, we got an equation that we like...now we use it in C right? or B first
anonymous
  • anonymous
a
amistre64
  • amistre64
that was A; find the equation ;)
amistre64
  • amistre64
B is; use it to make a table....
anonymous
  • anonymous
yes...done
amistre64
  • amistre64
We just happened to make a table first lol
anonymous
  • anonymous
working backwards in a problem solving strategy ; )
anonymous
  • anonymous
is
amistre64
  • amistre64
Dn =L−(L−D0 )a^n So what does this formula mean? its greek to me :)
anonymous
  • anonymous
me too...uh oh
amistre64
  • amistre64
368 should tell us something about it...like what D is and L
amistre64
  • amistre64
pg 368 that is
anonymous
  • anonymous
B/1-a in place of L for solutions growing towards infinity
anonymous
  • anonymous
sorry keep forgetting to his post
amistre64
  • amistre64
whats the chapter heading for your material?
anonymous
  • anonymous
Modeling with Difference Equations
amistre64
  • amistre64
does the formula have a name for it? its probably online somewhere and I can get a better grip on it.
anonymous
  • anonymous
difference equation...
anonymous
  • anonymous
using Asubn+1=.5Asubn +1
amistre64
  • amistre64
we found a first order linear difference equation, if I see it right..
amistre64
  • amistre64
Tn+1 − Tn = k(Tn − S) looks useful Dn =L−(L−D0 )a^n but I dunno.... are there any other formulas on page 368?
anonymous
  • anonymous
\[x _{n=}L+(x _{0}-L)a ^{n}\]
anonymous
  • anonymous
used that in c got b working on e using L=B/1- to find limiting value
anonymous
  • anonymous
finished d
amistre64
  • amistre64
you got it?
anonymous
  • anonymous
need e and f
amistre64
  • amistre64
the limit is 2 I beleive.... is that right? as n increases without bound we approach 2
anonymous
  • anonymous
yes but trying to get that from equation...working backwards again
amistre64
  • amistre64
:) I plugged in alot of iteration in google and gets 2 :)
amistre64
  • amistre64
are we allowed to use the definition of a limt? f(n+h) - f(n)/h ?
amistre64
  • amistre64
thats the limit as it approaches zero....not infiinity tho....
amistre64
  • 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
  • amistre64
unless you got a formula for that too :)
anonymous
  • 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
  • anonymous
since n=0
anonymous
  • anonymous
e) 1/1-.5
amistre64
  • amistre64
1/.5 = 2
amistre64
  • amistre64
1/(1-.5) right?
anonymous
  • anonymous
YEP!!!!
amistre64
  • amistre64
thats 2 :)
amistre64
  • amistre64
thats 2 :)
anonymous
  • anonymous
f)
amistre64
  • amistre64
.5 would change to a different number
amistre64
  • amistre64
.6 I believe
amistre64
  • amistre64
40%full...... .4
anonymous
  • anonymous
I concur! g)
amistre64
  • amistre64
.4(An-1) + 1
anonymous
  • anonymous
forgot post button again
amistre64
  • amistre64
and for the last one; the size of the tank doesnt matter, its the ratio.
amistre64
  • amistre64
16 gal filled every half is the same result
anonymous
  • anonymous
what about limiting value for f
amistre64
  • amistre64
id venture to guess that is it 2; but let me check my sources :)
amistre64
  • amistre64
1.666666666666666666666666666......
amistre64
  • amistre64
if I did it right :)
anonymous
  • anonymous
just got taht on cal!!!!
anonymous
  • anonymous
GREAT!
anonymous
  • anonymous
I have 3a and b can you check c for me Bsub15 = 80000(1.06)to the n - 10,000 so
amistre64
  • amistre64
\[80000(1.06)n -10000\] looks good
amistre64
  • amistre64
or is it n-1?
amistre64
  • amistre64
\[B_n = 80000(1.06)(n-1) - 10000\]
amistre64
  • amistre64
at 1 that would be a balance of -10000; so prolly not
anonymous
  • anonymous
\[B _{n+1}=1.06B _{n}-10000\]
anonymous
  • anonymous
noooo...you were about to make me scream!
amistre64
  • amistre64
lol
anonymous
  • anonymous
so c I keep going back and forth on
anonymous
  • anonymous
got $181,724.66
amistre64
  • amistre64
well, if he keeps taking 10000 out, he runs outta money before then I think.
amistre64
  • amistre64
80000(1.06) = 84800 - 10000 = 74800 right?
anonymous
  • anonymous
yes for n=1
amistre64
  • amistre64
79288 - 10000 = 69288
amistre64
  • amistre64
73445.28 - 10000 = 63445.28
amistre64
  • amistre64
67251.9968 - 10000 = 57251.9968
amistre64
  • amistre64
60687.116608 - 10000= 50687.116608 53728.34360448 - 10000 = 43728.34 46352.0404 - 10000 = 36352.0404
amistre64
  • amistre64
were running outta money :)
amistre64
  • 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
  • anonymous
lol...my 2 year was off by 1,000 wrote wrong digit from cal....geesh
amistre64
  • amistre64
year 12 shes broke... at 15 theres nothing left;
amistre64
  • amistre64
if anything she owes 37723.68 by year 15
amistre64
  • amistre64
so what amount can she withdraw each year that will result in a zero balance at year 15 is the question lol
anonymous
  • anonymous
yeah i was just reading it again, too lmbo
amistre64
  • amistre64
its between 8000 and 9000 :)
amistre64
  • amistre64
between 8200 and 8300 :)
amistre64
  • amistre64
between 8230 and 8245
anonymous
  • anonymous
what would be my explanation...you working backwards
amistre64
  • amistre64
im just reiterating it 15 times on google and narrowing it down :)
amistre64
  • amistre64
once we get a number, we can plug it in and see if it works :)
amistre64
  • amistre64
same equation, just a different withdrawal amount
anonymous
  • anonymous
ok
anonymous
  • anonymous
would we have to log this
amistre64
  • amistre64
-8237 = 0.491505273 :)
amistre64
  • amistre64
- 8 237.03 = -0.206773824 cant get any less - 8 237.02 = 0.0259858751
amistre64
  • amistre64
explain how you got your answer: trial and error :)
amistre64
  • amistre64
the way all good people have done it for thousands of years; we narrowed it down til something stuck
anonymous
  • anonymous
I might add that, lol!
amistre64
  • amistre64
:)
amistre64
  • amistre64
plug it into your calculator and see if it gets you a good deal...
anonymous
  • anonymous
I will finish e bc I know you want to move on...THANKS SOOOO MUCH!

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