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Thread: Simple isolation question

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    Default Simple isolation question

    I'm sure there's an extremely simple method to solving this that I just can't remember. Anyone know how to calculate the discount rate "r" in a present value formula by hand (step by step)?

    So for example, for present value of 2, find (isolate) r:

    2 = 1/(1+r) + 1/(1+r)^2 + 1/(1+r)^3
    Last edited by Ashihtaka; 06-27-2012 at 10:09 AM.
    118/120 Dungeoneering

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    I dont understand what do u mean by "discount rate" (isn't it to do with $?)
    And in the example u are trying to find the value of r?
    if its 2=1 + 1/2 + 1/4 + 1/8 + 1/16 +.... then it make sense.

    Otherwise it just look like a geometric progression with ratio 1/(1+r). (oh, is this ratio what u are talking about?)

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    Sorry, the terminology just means it's a finance question (discount rate is what "r" is).

    The main question here is how do you find the value of "r" in that example.
    118/120 Dungeoneering

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    Oh i get what u mean now, I've no knowledge of financial accounting, but it can be solve rather easily from a pure mathematical perspective:

    If u multiply the whole equation by the largest degree of r, (1+r)^3 in this case, u will be able to linearize the equation:
    2(1+r)^3=(1+r)^2+(1+r)+1

    Expand everything out, bring everything to one side and u will get a standard polynomial with highest degree at r^3, something like
    a(r^3)+b(r^2)+cr+d=0

    Then work out the roots of the equation (through long division) and u will get 3 answers (unless the roots are identical or involves complex number)

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    Okay, I thought that was it, but it was actually finding the roots that confused me.
    I got up to:
    2(1+r)^3=(1+r)^2+(1+r)+1
    2r^3+6r^2+6r+2=1+2r+r^2+1+r+1
    2r^3+5r^2+3r-1=0,

    But I didn't know how to isolate the variable in a cubic function so thought I was doing something wrong, got it now.
    Thanks
    118/120 Dungeoneering

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    Har HAr Har HAr, Maple seems to think poorly of this.


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    i may be going out on a limb by saying this, but just last semester i had to solve for this r shit. it was... well... easy my way, but slightly harder using the teacher's method of what she called 'interpolation,' though i wouldnt have called it that.
    First... My method. Replace 1+r with x. It doesnt seem like much, but at times can make things seem much simpler. From there solve and stuff and get 0 = 2x^3 - x^2 - x - 1 and then use synthetic division or just graph it/solve using a calculator. This would result in x = 1.23 , so to get r, just subtract 1.
    Teacher's method. Now this to me was a little sketchy. I was in a class called 'Engineering economy' which was a required class for all engineering majors. The university assumes you take this as like a sophomore when you are barely in to calc2, so everything uses some round about ass way to do shit. I at this time was already done with all my math classes and had taken a few extra, so i knew a few more tricks... But anyway onto the method. I liked to call this method similar triangles or something. In our books, we were given thousands of tabulated values for interest rates, so formulas were rarely used. (i hated using tables for just this reason)
    We would start out by making a guess as to what the rate was... lets say we guessed it was 20%. we looked at the tables and found... well i dont actually remember. then we did one that we knew was higher than the actual rate (25%).
    With this we had two different results (in this case assuming the 2[meaning doubled] ww might end up with 1.8 and 2.3) so then we went ahead and made some sort of similar triangles sort of deal with (2.3-1.8)/(25%-20%) = 2/r (or something) and then solved for r. this only works if you have the tabulated values and could make any sense out of what i just said. so. best of luck!

    TL;DR: make x=1+r and solve.
    use similar triangles with tabulated values.

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    If that was the whole equation, then the above solutions are all right.

    however, if it's a sequence, then just multiply the equation by 1+r, so then you get 2 + 2r = 1 + 1/r+1 + 1/(r+1)^2...
    from there, you get 1 + 2r = 2 (because the 1/r+1 sequence remains the same), and 2r = 1, r = 1/2.

    I may have made a stupid mistake, I'm incredibly tired.

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    Quote Originally Posted by Hazzah View Post
    Har HAr Har HAr, Maple seems to think poorly of this.

    What program is that? 0.o

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