## anonymous 5 years ago Need help with difference equation...

1. 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.

2. 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?

3. amistre64

found it :)

4. anonymous

great!

5. anonymous

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

6. amistre64

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

7. anonymous

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

8. amistre64

pistachios are more of my passion i think :)

9. amistre64

this is more of a hobby lol

10. anonymous

lol...well you're AWESOME!

11. anonymous

ok so do you understand the question

12. amistre64

13. anonymous

ok

14. 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?

15. anonymous

yes

16. amistre64

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

17. anonymous

ok

18. amistre64

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

19. anonymous

don't think so

20. amistre64

I could be wrong, I allow that :)

21. 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?

22. anonymous

yes!

23. amistre64

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

24. anonymous

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

25. 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

26. anonymous

ok

27. amistre64

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

28. amistre64

7 = 1.984375

29. amistre64

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

30. anonymous

I got it An=.5A0+1?

31. amistre64

$\frac{\frac{\frac{\frac{1}{2}+1}{2}+1}{2}+1}{2}+1...$

32. amistre64

Does this pattern remind you of anything?

33. anonymous

Fibonacci?

34. amistre64

it does kinda look fibonaci like :)

35. amistre64

i cant place it yet....

36. amistre64

but its growing to a limit of 1.99999999999999999999

37. anonymous

Can you help with others/check

38. anonymous

yes I have 1.998 for n=10

39. amistre64

I can try; have you finished them?

40. anonymous

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

41. anonymous

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

42. amistre64

and what do we do with this formula :)

43. anonymous

c) Find the solution from a using equation

44. anonymous

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

45. amistre64

have we found our difference equation yet?

46. anonymous

I am using Asub n = .5Asub0 + 1

47. anonymous

typed it earlier...do you agree?

48. amistre64

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

49. amistre64

.5(n) +1

50. anonymous

yes!

51. amistre64

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

52. amistre64

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

53. anonymous

using previous values of Asubn to get others

54. anonymous

$A _{n}=.5A _{n+1}+1$

55. 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 :)

56. amistre64

I got something :) needs to be ironed out tho

57. anonymous

$A _{n+1}=.5A _{n}+1$

58. amistre64

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

59. anonymous

I see it but can't explain in formula correctly

60. amistre64

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

61. amistre64

that fits it to a "T" :)

62. amistre64

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

63. amistre64

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

64. anonymous

that is giving greater than 2

65. anonymous

no n=1

66. amistre64

its close, but needs to be tweeked

67. amistre64

2(n-1)+1

68. amistre64

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

69. amistre64

not yet eh lol

70. anonymous

$A _{n}=.5A _{n-1}+1$

71. anonymous

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

72. amistre64

yes

73. amistre64

yours looks good like that :)

74. amistre64

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

75. amistre64

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

76. anonymous

geeshhh. I had that thinking too hard ; (

77. amistre64

lol

78. anonymous

n=1

79. amistre64

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

80. anonymous

a

81. amistre64

that was A; find the equation ;)

82. amistre64

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

83. anonymous

yes...done

84. amistre64

We just happened to make a table first lol

85. anonymous

working backwards in a problem solving strategy ; )

86. anonymous

is

87. amistre64

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

88. anonymous

me too...uh oh

89. amistre64

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

90. amistre64

pg 368 that is

91. anonymous

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

92. anonymous

sorry keep forgetting to his post

93. amistre64

94. anonymous

Modeling with Difference Equations

95. amistre64

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

96. anonymous

difference equation...

97. anonymous

using Asubn+1=.5Asubn +1

98. amistre64

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

99. 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?

100. anonymous

$x _{n=}L+(x _{0}-L)a ^{n}$

101. anonymous

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

102. anonymous

finished d

103. amistre64

you got it?

104. anonymous

need e and f

105. amistre64

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

106. anonymous

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

107. amistre64

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

108. amistre64

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

109. amistre64

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

110. 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.

111. amistre64

unless you got a formula for that too :)

112. 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

113. anonymous

since n=0

114. anonymous

e) 1/1-.5

115. amistre64

1/.5 = 2

116. amistre64

1/(1-.5) right?

117. anonymous

YEP!!!!

118. amistre64

thats 2 :)

119. amistre64

thats 2 :)

120. anonymous

f)

121. amistre64

.5 would change to a different number

122. amistre64

.6 I believe

123. amistre64

40%full...... .4

124. anonymous

I concur! g)

125. amistre64

.4(An-1) + 1

126. anonymous

forgot post button again

127. amistre64

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

128. amistre64

16 gal filled every half is the same result

129. anonymous

what about limiting value for f

130. amistre64

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

131. amistre64

1.666666666666666666666666666......

132. amistre64

if I did it right :)

133. anonymous

just got taht on cal!!!!

134. anonymous

GREAT!

135. anonymous

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

136. amistre64

$80000(1.06)n -10000$ looks good

137. amistre64

or is it n-1?

138. amistre64

$B_n = 80000(1.06)(n-1) - 10000$

139. amistre64

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

140. anonymous

$B _{n+1}=1.06B _{n}-10000$

141. anonymous

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

142. amistre64

lol

143. anonymous

so c I keep going back and forth on

144. anonymous

got \$181,724.66

145. amistre64

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

146. amistre64

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

147. anonymous

yes for n=1

148. amistre64

79288 - 10000 = 69288

149. amistre64

73445.28 - 10000 = 63445.28

150. amistre64

67251.9968 - 10000 = 57251.9968

151. amistre64

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

152. amistre64

were running outta money :)

153. 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 :)

154. anonymous

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

155. amistre64

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

156. amistre64

if anything she owes 37723.68 by year 15

157. amistre64

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

158. anonymous

yeah i was just reading it again, too lmbo

159. amistre64

its between 8000 and 9000 :)

160. amistre64

between 8200 and 8300 :)

161. amistre64

between 8230 and 8245

162. anonymous

what would be my explanation...you working backwards

163. amistre64

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

164. amistre64

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

165. amistre64

same equation, just a different withdrawal amount

166. anonymous

ok

167. anonymous

would we have to log this

168. amistre64

-8237 = 0.491505273 :)

169. amistre64

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

170. amistre64

171. amistre64

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

172. anonymous

173. amistre64

:)

174. amistre64

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

175. anonymous

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