smart math guys, compound interest question

SixFive

bonswa
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Mar 12, 2001
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Can somebody please post for me a formula for calculating compound interest?

Specifically, I want to be able to input a number, say 2,000 for x amount of years at x rate of interest return to show as an example to some young people to help them understand the importance of saving while young. Thanks!
 

SixFive

bonswa
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Mar 12, 2001
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Red & Black said:
Take 1 plus the interest rate to the power of the number of years. Then multiply by your original investment. Hope this helps.

huh? Can u put than in a formula that I can punch in on my calculator? Thanks. Also, I want to include the same investment amount each year.
 

saint

Go Heels
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Jan 10, 2002
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FV=PV(1+i)^N

FV-future value
PV-present value
i-interest rate in percent per period
n-number of periods

Nevermind...just saw your last post this isn't it.
 

KMA

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May 25, 2003
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There's really not much difference in interest compounded yearly and daily, it's just a question of how much interest is accrued in a certain length of time. They both amount to a finite geometric sequence, and to deal with those:


The interest rate, p, is the ratio of successive terms in the sequence (where p is a number like 1.05 or thereabouts). Then if you start with a certain amount, a, in your account, after n pay periods you have a*p^n in your account.

To find out how interest compounded continuously works, you use limits:
saywe've got "p" interest compounded continuously, where p is a number like .05 or thereabouts. Then if we subdivide the year into m sections, each section will compound p/m interest, and we'll compound it m times. So after the first section of the year we'll have:

a + a*p/m = a(m+p)/m, after the second section we'll have

a(m+p)/m + a(m+p)/m *p/m = a(m+p)^2/m^2, and then

a(m+p)^2/m^2 + a(m+p)^2/m^2 *p/m = a(m+p)^3/m^3, and then

a(m+p)^2/m^3 + a(m+p)^3*m^3 *p/m = a(m+p)^4/m^4, .......

So at the end of the year (i.e. after m sections) we'll have

a(m+p)^m/m^m = a*((m+p)/m)^m in our account.

If we take the limit as m goes to infinity, this will go to a*e^p, where e is the base of the natural logarithm. So it's similar to the non-continuous case.
 

KMA

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Or depending on how young they are you ought to teach them how inflation erodes saving, and the only way to grow money is to earn an interest at a rate ABOVE inflation:

Suppose you open an account that pays a guaranteed interest rate, compounded annually. You make no further contributions; you just leave your money alone and let compound interest work its magic.

The balance your account has grown to at some point in the future is known as the future value of your starting principal.


Starting principal = $1,000
Future Value (=investment value at maturity)= $2, 653

Growth Rate 5%-------------------------------$2,000

Years=20----------------------------------------$1,000



To find a formula for future value, write P for your starting principal, and r for the rate of return expressed as a decimal. (So if the interest rate is 5%, r equals .05).


Your balance will grow according to the following schedule:

Year Balance
Now P
1 P + rP
2 (P + rP) + r(P + rP)



This starts to get messy in a hurry. But you can simplify it by noticing that you can keep pulling out factors of (1 + r) from each line. If you do that, the balances collapse to a simple pattern:

Year Balance
Now P
1 P(1 + r)
2 P(1 + r)2



If you follow this pattern out for n years, you get the general formula for future value:

FV = P(1 + r)n
 
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