# Ruby math.pow equivalent

Hi,

I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?

@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))

Ive searched everywhere and cant seem to find anything.

thanks

Johnny B

On Fri, Mar 30, 2007 at 05:43:38AM +0900, John B. wrote:

I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?

@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))

Ive searched everywhere and cant seem to find anything.

http://www.rubycentral.com/book/ref_c_float.html#Float.Arithmeticoperations

Look for ‘exponentiation’

On 3/30/07, John B. [email protected] wrote:

Does
anyone know what the equivalent for math.pow in Java is for Ruby?

@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))

Is this what you are looking for?

p 4 ** 3

Harry

http://www.kakueki.com/ruby/list.html
Japanese Ruby List Subjects in English

John B. wrote:

Hi,

I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?

@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))

Ive searched everywhere and cant seem to find anything.

thanks

Johnny B

I think this is what you’re looking for:

irb(main):001:0> 2 ** 1
=> 2
irb(main):002:0> 2 ** 2
=> 4
irb(main):003:0> 2 ** 3
=> 8
irb(main):004:0> 2 ** 4
=> 16

Drew O. wrote:

John B. wrote:

Hi,

I am trying to calculate a monthly payment for a loan in Ruby. Does
anyone know what the equivalent for math.pow in Java is for Ruby?

@monthlypayment = @principal*@rate/(1-Math.pow(1/(1+@rate),@payments))

Ive searched everywhere and cant seem to find anything.

thanks

Johnny B

I think this is what you’re looking for:

irb(main):001:0> 2 ** 1
=> 2
irb(main):002:0> 2 ** 2
=> 4
irb(main):003:0> 2 ** 3
=> 8
irb(main):004:0> 2 ** 4
=> 16

Yes this is what i am looking for. How do i then use it to calculate
amonthly payment, i have tried quite a few things but cant get the
syntax right: @monthlypayment =
@principal*@rate/(1-**(1/(1+@rate),@payments))

So for the example below:

Suppose you finance your car with a loan of \$12000 at a yearly interest
rate of 11% for four years, and make equal payments monthly. How much
will your payments have to be? Here the parameters are principal P =
\$12000, interest rate i = 0.11, number of years n = 4, and number of
periods per year q = 12. Then the monthly car payment M is given by

M = Pi/[q(1-[1+(i/q)]-nq)],
= (\$12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))],
= \$110/(1-1.009166666...-48),
= \$310.15.

Anyone done this before?

Yes, thanks.

Alle venerdÃ¬ 30 marzo 2007, John B. ha scritto:

irb(main):003:0> 2 ** 3
So for the example below:
= \$310.15.

Anyone done this before?

** is an operator, just like + or ; it’s used this way: base**exponent.
In