Solution of a polynomial with degree 25

[jonjacson]

Homework Statement

You have a fixed payment loan, you know the quantity you need to pay every year, the years until maturity and (I suppose to) the loan value and you need to calculate which is the yield to maturity.

Homework Equations

Loan Value =Fp/(1+i) + Fp/(1+i)²+...+ Fp/(1+i)ⁿ

Fp= Fixed payment annually, in this case it is 85,81$

i = Yield to maturity

n= number of years, in this case it´s 25 years.

The Attempt at a Solution

I found that equation in an economics book, all previous problems use a loan value of 1000 so I´m assuming in this case is the same, unfortunately the writer of the book doesn´t say it explicitly.

The point is that he says that the solution to that equation is yield to maturity i=7%, and I wonder how did he arrived to that conclusion because I see a polynomial with degree 25 that I could solve using mathematica but obviously I get 25 solutions to that equation that satisfy the equality.

Any ideas about how has he arrived at the conclusion that i must be 7%?

 

[phyzguy]

Try looking up the sum of a geometric series.

 

[jonjacson]

Thanks phyzguy, wikipedia explains that formula as a example of geometric serie.

I assume that if we had a polynomial without the geometric progression property we would have the problem of solving the 25 degree polynomial, now as every term is related with the previous one we can use that property to do the sum.

Am I right?

 

[Ray Vickson]

Thanks phyzguy, wikipedia explains that formula as a example of geometric serie.

I assume that if we had a polynomial without the geometric progression property we would have the problem of solving the 25 degree polynomial, now as every term is related with the previous one we can use that property to do the sum.

Am I right?

I cannot understand your question. The sum is
$$F_p \left( \frac{1-(i+1)^{-n}}{i}\right)$$
so you need to solve the equation
$$\frac{85.81}{i} -\frac{85.81}{i(1+i)^{25}}=1000.$$
Doing this in Maple gives one real solution $i \doteq 0.06999933091 \doteq 0.07,$ and 24 complex solutions. The real solution is the one you want. Alternatively, in Maple (and, I guess, in Mathematica---but I don't know the commands) one can look for a numerical solution in a specified range, by writing fsolve(eq,i=0..1) to get i in the range 0--1.

 

[jonjacson]

You are right I got confused with the answers in mathematica because the letter for the imaginary part i, was the same as the unknown quantity i. Now I changed that and I saw what you say, 24 imaginary solutions and one real.

Do you know why is there only one real solution?

I thought that we had those 25 solutions because of the fundamental theorem of algebra, but I studied this some time ago and I don´t remember well the details.
I thought that if you have this for example x^3+x^1-x^2=9 due to that theorem you have three possible solutions that satisfy the equation.

I hope not being annoying you with my questions.

 

[Ray Vickson]

You are right I got confused with the answers in mathematica because the letter for the imaginary part i, was the same as the unknown quantity i. Now I changed that and I saw what you say, 24 imaginary solutions and one real.

Do you know why is there only one real solution?

I thought that we had those 25 solutions because of the fundamental theorem of algebra, but I studied this some time ago and I don´t remember well the details.
I thought that if you have this for example x^3+x^1-x^2=9 due to that theorem you have three possible solutions that satisfy the equation.

I hope not being annoying you with my questions.

Well, if we look for positive solutions x = 1/(1+i) > 0, then Descarte's Rule of signs says there is exactly one positive root, because the number of coefficient sign changes in
$$F x^{25} + F x^{24} + \cdots + F x - V = 0$$ is one (from + F to -V).

It is more complicated to count the negative roots (none, in this case). In fact, if we put x = -t (t>0), the polynomial p(t) has 24 coefficient sign-changes, so the number of roots t > 0 is 24 minus an even number between 0 and 24. In the current case that even number is 24, but this fact is not a direct consequence of the Rule of Signs, and I am not sure how one can know it ahead of time.

See, eg., http://en.wikipedia.org/wiki/Descartes'_rule_of_signs for more information.

 

[Bill Simpson]

Do you know why is there only one real solution?

For some problems you can think for a few minutes and realize there should only be a single solution.

Suppose for this problem you discover you can pay off your loan with a $6 payment every month OR you can pay off the exact same loan in the exact same number of months with a $11523647 payment every month. (or even with up to 23 other different payments)

If you think about that for a moment I suspect you would think something is terribly broken somewhere if there is more than one specific amount that pays off the loan in the given time. Even 21st century economics hasn't found a way that this would make sense.

 

[jonjacson]

Well, if we look for positive solutions x = 1/(1+i) > 0, then Descarte's Rule of signs says there is exactly one positive root, because the number of coefficient sign changes in
$$F x^{25} + F x^{24} + \cdots + F x - V = 0$$ is one (from + F to -V).
It is more complicated to count the negative roots (none, in this case). In fact, if we put x = -t (t>0), the polynomial p(t) has 24 coefficient sign-changes, so the number of roots t > 0 is 24 minus an even number between 0 and 24. In the current case that even number is 24, but this fact is not a direct consequence of the Rule of Signs, and I am not sure how one can know it ahead of time.
See, eg., http://en.wikipedia.org/wiki/Descartes'_rule_of_signs for more information.

Thanks Ray I had never heard about this rule, it´s very interesting and gives information about polynomials that I didn´t know.

For some problems you can think for a few minutes and realize there should only be a single solution.

Suppose for this problem you discover you can pay off your loan with a $6 payment every month OR you can pay off the exact same loan in the exact same number of months with a $11523647 payment every month. (or even with up to 23 other different payments)

If you think about that for a moment I suspect you would think something is terribly broken somewhere if there is more than one specific amount that pays off the loan in the given time. Even 21st century economics hasn't found a way that this would make sense.

Bill you are right in that point, if we think only about the meaning of the equation it must have only one solution... but if I forget the meaning I just see, only from a mathematical point of view, any nth degree polynomial has exactly n roots and I was searching for an explanation of the fact that there is only one real solution.

The book says:

"For example , in the case of a 25 -year loan with yearly payments
of $85.8 1, the yield to maturity that solves Equation 2 is 7%. Real estate brokers always
have a pocket calculator that can solve such equations so that they can immediately tell
the prospective house buyer exactly what the yearly (or momhly) paymems will be if
the house purchase is financed by taking out a mortgage."

The fact that they have a calculator didn´t seem to me a good explanation, it didn´t even seem an explanation.

 

[SteamKing]

The calculator takes the place of published tables which used to be found in any financial institution or loan office.

These tables would tell a person how much interest an amount of $1 would accrue given the interest rate, how often it was compounded, and the term of the loan. Once you looked it up for a dollar, a simple multiplication for the true principal would give you the loan payment. People used to learn this stuff in school, and then Congress jumped in with APRs, and the calculations were not as simple. As a result, except for some easy problems with simple interest, most of this isn't taught anymore, except to hard core financial types.

 

[Ray Vickson]

The calculator takes the place of published tables which used to be found in any financial institution or loan office.

These tables would tell a person how much interest an amount of $1 would accrue given the interest rate, how often it was compounded, and the term of the loan. Once you looked it up for a dollar, a simple multiplication for the true principal would give you the loan payment. People used to learn this stuff in school, and then Congress jumped in with APRs, and the calculations were not as simple. As a result, except for some easy problems with simple interest, most of this isn't taught anymore, except to hard core financial types.

1. What is an APR? Not being a U.S. resident I am unfamiliar with the term.
2. Nowadays, spreadsheet solvers and the like can easily deal with these issues, at least to a reasonable degree of accuracy.

 

[Mark44]

1. What is an APR? Not being a U.S. resident I am unfamiliar with the term.

Annual percentage rate. It's the interest rate that would be applicable if you made one payment per year.