1. ## Re: Perpetuity

Originally Posted by romsek
what on Earth are you talking about?
To solve for "i" in the equation is akin to answering the following question

What came first? the chicken or the egg
And the only one who knows the answer to this is the one who created the "chicken or the egg"

And that would have to be the "scienceboy"

There are three perpetuities for the given problem to solve for "i"

Code:
1     1     1     1     1     1     1     1     1     1    .....
0.66  0.66  0.66  0.66  0.66  0.66  0.66  0.66  0.66  0.66 .....
0.34  0.34  0.34  0.34  0.34  0.34  0.34  0.34  0.34  0.34 .....
present value of 60 payments in amount of 1 dollar discounted at "i" is the same as present value of infinite payments in amount of 34 cents (amount rounded to 2 decimal places)

The IRR formula finds "i" for the perpetuity in amount of 66 cents (amount rounded to 2 decimal places) which is one dollar discounted at "i" for n periods

To find 34 cents, one has to at first find 66 cents and to find 66 cents one must know "i" but you know if we knew "i" we wouldn't be finding "i" in the first place

There are two infinite series whose values oscillate between each other

Code:
0.66  0.34  0.66  0.34  0.66  0.34  0.66  0.34  0.66  0.34 .....
0.34  0.66  0.34  0.66  0.34  0.66  0.34  0.66  0.34  0.66 .....
The present value of 60 payments in amount of 1 dollar is the same as sum of present value of first 60 terms of the two infinite series

2. ## Re: Perpetuity

Originally Posted by Scienceboy

It is indeed, as you say, an ordinary annuity.

I continued my research and it appears to be very tedious to find "i" algebraically.
The formula I have derived thus far gives me the following relationship between the rate and its solution

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

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### shrinking annuity formula

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