Ruby math.pow equivalent

[John Butler]

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

 

[Brian Candler]

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'

 

[Harry]

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

 

[Drew Olson]

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

 

[John Butler]

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?

 

[Stefano Crocco]

Alle venerd=C3=AC 30 marzo 2007, John Butler ha scritto:

I think this is what you're looking for:

irb(main):001:0> 2 ** 1
=3D> 2
irb(main):002:0> 2 ** 2
=3D> 4
irb(main):003:0> 2 ** 3
=3D> 8
irb(main):004:0> 2 ** 4
=3D> 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 =3D
@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 =3D
$12000, interest rate i =3D 0.11, number of years n =3D 4, and number of
periods per year q =3D 12. Then the monthly car payment M is given by

M =3D Pi/[q(1-[1+(i/q)]-nq)],
=3D ($12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))],
=3D $110/(1-1.009166666...-48),
=3D $310.15.


Anyone done this before?

** is an operator, just like + or *; it's used this way: base**exponent. In=
=20
your case, you should write:
@principal*@rate/(1-1/(1+@rate)**@payments)

I hope this helps

Stefano

 

[John Butler]

Yes, thanks.