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Mathematics 14 Online
OpenStudy (loser66):

If you invest $1000 with 5% interest annually but you immediately withdraw 100 after the interest is paid. a) find the recursive formula for investment b) find the investment after the third year c) find formula for a_n d) when the investment =0 Please, help

OpenStudy (loser66):

for a) \(I_0 = 1000\\I_1 = 1.05 I_0 -100\\I_2= 1.05 I_1 -100\\I_3= 1.05 I_2 -100\\I_4 = 1.05 I_3-100\\-------\\I_n = 1.05 I_{n-1}-100\)

OpenStudy (loser66):

for b) I got 894.... but it is easy to find out I stuck at c and d for c) I worked back ward and got the weird formula in which I can't calculate d :(

ganeshie8 (ganeshie8):

for c, are you getting \[a(n)=2000-1000*1.05^n\] ?

ganeshie8 (ganeshie8):

If so, setting that equal to 0 and solving \(n\) is easy right ?

OpenStudy (loser66):

how? a(n) = what you showed?

ganeshie8 (ganeshie8):

your answer for part b should be 842.375 http://www.wolframalpha.com/input/?i=Table%5B2000-1000*1.05%5En%2C%7Bn%2C0%2C5%7D%5D

OpenStudy (loser66):

yes, I need c and d. :)

ganeshie8 (ganeshie8):

we want to solve \[a_n-1.05a_{n-1}=-100\]

ganeshie8 (ganeshie8):

Notice that it is a non homogeneous recurrence relation

ganeshie8 (ganeshie8):

so the complete solution will be of form : `particular` + `null`

ganeshie8 (ganeshie8):

lets find the null solution first

ganeshie8 (ganeshie8):

for `null solution`, we solve \[a_n-1.05a_{n-1}=0\]

ganeshie8 (ganeshie8):

for `null solution`, we solve \[a_n-1.05a_{n-1}=0\tag{1}\] suppose the solution is of form \(a_n = c*r^n\) plugging this in \((1)\) gives \(c*r^n - 1.05c*r^{n-1}=0\) \(r-1.05 = 0\) \(r=1.05\) so the `null solution` is \(c*1.05^n\)

ganeshie8 (ganeshie8):

see if that looks okay so far

ganeshie8 (ganeshie8):

next we need to find a particular solution

OpenStudy (loser66):

oh, I didn't think that way. Ok, let me try. Thanks a lot

ganeshie8 (ganeshie8):

this `particular` + `null` thingy shows up everywhere you see a `linear` equation

ganeshie8 (ganeshie8):

you must have seen this pattern while solving linear diophantine eqations using reverse euclid method

OpenStudy (loser66):

oh, no, it is not my diophantine equation, this is the problem from a student I tutored. I have to solve this problem base on his background, not my background. If I tell something about diophantine, he won't get.

ganeshie8 (ganeshie8):

you don't need to tell him anything about diophantine

OpenStudy (loser66):

got you :)

ganeshie8 (ganeshie8):

how are you going to find a particular solution ?

OpenStudy (loser66):

I need find and A such that A - 1.05 A =100 hence A = 100/ -0.5

ganeshie8 (ganeshie8):

Looks good except for a typo

ganeshie8 (ganeshie8):

I need find and A such that A - 1.05 A =-100 hence A = 100/ 0.5

OpenStudy (loser66):

oh

ganeshie8 (ganeshie8):

the typo is in first line

OpenStudy (loser66):

yup, -100 , not 100

OpenStudy (loser66):

ok, hence A = 200 and the whole thing is a(n) = C*(1.05)^n +200 and for a(0) =1000, we solve for C, right?

ganeshie8 (ganeshie8):

I need find and A such that A - 1.05 A =-100 hence A = 100/ 0.05 = 2000

OpenStudy (loser66):

:) my bad

ganeshie8 (ganeshie8):

ok, hence A = 2000 and the whole thing is a(n) = C*(1.05)^n +2000 and for a(0) =1000, we solve for C, right?

OpenStudy (loser66):

Yes, I got what you got. and d is easy now. :)

OpenStudy (loser66):

Thank you so so so much.

ganeshie8 (ganeshie8):

np:)

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