Solve for i?

Question

Solve for i.

(1+i)^360 - 1 = 1000*i

I can't seem to solve it.. wondering if its possible?

show steps please

Answer 1

You could definitely solve that using Newton's method.
f(i) = (1+i)^360 - 1000*i - 1
f'(i) = 360 *(1+i)^359 - 1000
Starting with i = 0.01
the solution converges to 4.98 x 10^-3

NOTE apart from i = 4.98 x 10^-3, there are no other positive solutions. Using the binomial expansion,
f(i) = -640i + a2*i^2 + a3*i^3 + ... + a360i^360
= i * (-640 + a2*i + a3*i^2 + ... + a360i^359)
All the coefficients an are positive, hence applying Descartes rules of signs, since there is only one sign change in the coefficients, there can be at most 1 positive root.
However, considering f(-i), there can be as many as 358 real negative solutions.

Answer 2

(1+i)^360 - 1 = 1000*i
(1+0)^360 - 1 = 1000*0
(1)^360 - 1 = 0
1 - 1 = 0

i = 0
is at least one root but i doubt you can solve for the other 359 roots lol. To solve only a cubic ( ax³ + bx² + cx + d = 0) it takes an extremely long an elaborate formula to get the roots, much more complicated than the quadratic formula. So it's hard to fathom solving for the roots of a 360th degree polynomial.
As far as I know, the quartic is the highest degree polynomial that can be solved for.

Check these out:

http://planetmath.org/encyclopedia/CubicFormula.html
http://planetmath.org/encyclopedia/QuarticFormula.html

So now try to imagine a 360th degree polynomial formula.

Answer 3

This is an algebraic manipulation of the formula for the Future Value Interest Factor of an Annuity (FVIFA), where i is the interest factor, 1,000 would be the actual future value of the annuity, and there would be 360 payments.

The formula for the Future Value of an annuity is

FVA=Pmt*((1+i)^n-1)/i

so you could look at it as if the FVA is $1000, the Payment is $1, and it will be compounded 360 times, but I still can't make the math work.

Generally, you would have to resort to the FVIFA charts to find this, but I'm not aware of any that go to 360 payments. Sorry! Maybe my thoughts will spur someone else's answer.

Answer 4

There is no general method to solve an arbitrary 360th degree polynomial, but there may be a method to solve this particular one. I suspect that if you rearrange the terms and then take logs on both sides you may be able to find a solution that way. However, my quick attempts using that approach didn't give me a solution, so that my not be the right approach after all.

Alternatively, you could use numerical methods, such as the Newton-Rhapson Method to give a solution. There is a solution around i = 0.00498, which at least tells us there is a solution other than i = 0.

Answer 5

I dont believe anyone can solve this.

I can tell you that i=0 is one solution, I can easily see that. But that is only one of 360 solutions, some or all of which may have the same value.

Expanding and solving for a polynomial in the 360th degree is not humanly possibly. There is no easy method like the quadratic formula.

Answer 6

if we're looking for nonzero solutions, we can divide by i to get
((1+i)^360 - 1)/i = 1000,
which you can interpret geometrically as saying that, for the curve y = x^360, the slope of the secant line from x=1 to x=1+i is 1000.

since the curve is concave up everywhere, the slope of the secant line strictly monotonic with respect to i. the derivative at x=1 is 360, so negative values of i will always correspond to slopes that are less than 360, whereas positive values will correspond to slopes that are greater than 360. other answerers have already found the solution at i = 0.00498, and the monotonic property that I described tells us that there are no other real solutions, either positive or negative.

so i=0 and i = 0.00498 are the only real solutions.

Answer 7

You can solve it by iteration by using EXCEL Goal Seek. Enter in A1 the expression
=((1+A2)^365-1)/A2
Then, /Tools > Goal Seek > Set cell A1 to value 1000 by changing cell A2 > OK
This gives i = 0.00485.