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 Post subject: Solve for N
PostPosted: Fri, 9 May 2008 01:11:06 UTC 
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Joined: Fri, 9 May 2008 00:57:41 UTC
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I am a freshman in algebra II and I have a little question for all you in a higher math class than myself. About a month ago we talked about interest formulas (basic, compound, continuous). I get bored frequently in math class and do a little programming on my ti-83. I was wondering, in the compound interest equation

A=P(1+r/n)^nt

How would you isolate n? I proposed this question to my math teacher and he was stumped, so don't worry if you don't get it, but please post so I know people are reading this. Thank you!


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PostPosted: Fri, 9 May 2008 06:28:54 UTC 
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The formula in your equation needs grouping symbols, such as parentheses,
to make it correct:

A = P(1 + r/n)^(nt)

Otherwise, a quantity would have been raised to n, and THEN multiplied by t.

With the version I typed, this will make sure the quantity will be raised to the
product nt.

= . . . . . = . . . . . = . . . . . = . . . . . = . . . . . = . . . . . = . . . . . = . . . . . =

As far as my knowledge goes, it is not possible to isolate n, but you can
approximate isolating n.

A = P(1 + r/n)^(nt)

A/P = (1 + r/n)^(nt)

Take specifically natural logs of each side:

ln(A/P) = ln(1 + r/n)^(nt)

ln(A/P) = (nt)ln(1 + r/n)

Note: ln(1 + r/n) is approximately equal to
r/n - (1/2)(r/n)^2 + (1/3)(r/n)^3 - ...

Continuing,

ln(A/P) ~ (nt)[r/n - (1/2)(r/n)^2 + (1/3)(r/n)^3]

Use the corresponding equation to this approximation.
Distribute the n from inside the parentheses into the brackets:

ln(A/P) = (t)[r - r^2/(2n) + r^3/(3n^2)]

Distribute the t and clear the brackets:

ln(A/P) = tr - tr^2/(2n) + tr^3/(3n^2)
ln(A/P) - tr = -tr^2/(2n) + tr^3/(3n^2)

Multiply both sides by the L.C.D. of 6n^2:

[ln(A/P) - tr]6n^2 = -3tnr^2 + 2tr^3

Get all the terms to one side:

[ln(A/P) - tr](6)n^2 + 3tnr^2 - 2tr^3 = 0

n = {-3tr^2 +/- sqrt[9(t^2)(r^4) - 4(ln(A/P) - tr)(6)(-2tr^3)]}/{2[ln(A/P) - tr](6)}

EDIT: More simplifying in this quadratic formula could be done/shown,
with certain constants being multiplied together and certain constants
and/or variables having their order changed.

There should be one legitimate value here,
and the rounded value for n should be an
integer such as (but not limited to)
1, 2, 4, 12, 52, or 365 to correspond,
respectively to, annually, semiannually,
quarterly, monthly, weekly, or daily.


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 Post subject: Re: Solve for N
PostPosted: Tue, 6 Sep 2011 13:15:34 UTC 
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Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 4181
Location: Ottawa Ontario
Deerek wrote:
A=P(1+r/n)^nt
How would you isolate n?

It's a "hit and miss" procedure; known as numerical solving or iteration.
Strange your teacher did not know this.
Go here for an idea:
http://www.mathsrevision.net/gcse/pages.php?page=41

Look at something similar, but simpler: a = u^u ; solve for u :
u = log(a) / log(u) , right?

_________________
I'm just an imagination of your figment...


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