Compound Interest Formula

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Explain this!

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I am curious to know how the compound interest formula is expressed as P(1 + r)t.

The simple interest formula, P(1 + rt), is derived from P + Prt.

Is the compound interest formula, P(1 + r)t, derived from (P + Pr)t?

I searched online and in a few textbooks, but I could not determine how the compound interest formula becomes P(1 + r)t.
 
Explain this! said:
I am curious to know how the compound interest formula is expressed as P(1 + r)t.

The simple interest formula, P(1 + rt), is derived from P + Prt.

Is the compound interest formula, P(1 + r)t, derived from (P + Pr)t?...........................No

I searched online and in a few textbooks, but I could not determine how the compound interest formula becomes P(1 + r)t.
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It is derived from (P + Prt) and sum of geometric series.

Have you studied geometric progression/series in algebra class yet?
 
Subhotosh Khan said:
It is derived from (P + Prt) and sum of geometric series.

Have you studied geometric progression/series in algebra class yet?
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I am somewhat familiar with geometric series, but I do not fully understand it enough to obtain the compound formula.
 
I think Khan may be talking about a different formula; there's no need to sum anything for basic compound interest.

Compound interest just means that you add in interest each period. In your version of the formula (there are several), you are taking r as the interest rate per period, and t as the number of periods. (Or it could be assuming annual compounding.)

So after one year (t = 1), you have P + Pr = P(1 + r). That's the same idea as simple interest.

The next year is as if you took that money and made a new loan for another year. Your principal is P(1 + r), and you are earning another year's interest, so that is multiplied by (1 + r). You now have P(1 + r)*(1 + r) = P(1 + r)^2.

After the third year, you have P(1 + r)^2*(1 + r) = P(1 + r)^3.

Each year, your money is multiplied by (1 + r); so over t years, it is multiplied by (1 + r), t times, which is the same as multiplying by (1 + r)^t.

Therefore, the amount at the end of t years is P(1 + r)^t.

I'm sure there are lots of sites that explain this, so you must have been unlucky in your search. Here's the first site I find when I search for "derive compound interest formula": https://www.mathsisfun.com/money/compound-interest-derivation.html
 
Explain this! said:
I am curious to know how the compound interest formula is expressed as P(1 + r)t.

The simple interest formula, P(1 + rt), is derived from P + Prt.

Is the compound interest formula, P(1 + r)t, derived from (P + Pr)t?

I searched online and in a few textbooks, but I could not determine how the compound interest formula becomes P(1 + r)t.
Click to expand...
No, it is not. First, your formula does not calculate interest at all. It calculates interest plus principal at the end of t periods, where r is the interest rate per period.

Think about this. Let's say the periods are 1 year in duration, the annual interest rate is 2% per year, and your initial principal is 100 dollars. So at the end of 1 year, you earn 2 dollars in interest, which is added to principal, giving a total of 102. In the second year, you earn interest of 0.02 times 102, which equals 2.04. When you add this to 102, you have 104.04 after 2 years. Now we could have done this another way. We could say that the principal plus interest at the end of year 1 was

[MATH]100 + 0.02 * 100 = 100(1 + 0.02). [/MATH]
Then the principal plus interest at the end of the second year is

[MATH]100(1 + 0.02) + 0.02\{100(1 + 0.02)\} = \{100(1 + 0.02)\} (1 + 0.02) = 100(1 + 0.02)^2.[/MATH]
What does that come out to?

[MATH]100(1 + 0.02)^2 = 100 * 1.02^2 = 100 * 1.0404 = 104.04.[/[/MATH]
 
Dr.Peterson said:
I think Khan may be talking about a different formula; there's no need to sum anything for basic compound interest.

Compound interest just means that you add in interest each period. In your version of the formula (there are several), you are taking r as the interest rate per period, and t as the number of periods. (Or it could be assuming annual compounding.)

So after one year (t = 1), you have P + Pr = P(1 + r). That's the same idea as simple interest.

The next year is as if you took that money and made a new loan for another year. Your principal is P(1 + r), and you are earning another year's interest, so that is multiplied by (1 + r). You now have P(1 + r)*(1 + r) = P(1 + r)^2.

After the third year, you have P(1 + r)^2*(1 + r) = P(1 + r)^3.

Each year, your money is multiplied by (1 + r); so over t years, it is multiplied by (1 + r), t times, which is the same as multiplying by (1 + r)^t.

Therefore, the amount at the end of t years is P(1 + r)^t.

I'm sure there are lots of sites that explain this, so you must have been unlucky in your search. Here's the first site I find when I search for "derive compound interest formula": https://www.mathsisfun.com/money/compound-interest-derivation.html
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Yes... I got mixed up with the formula for Mortgage payment (or annuity). Right before answering this post, I was toying with the idea of increasing my monthly payment and possible effects there of.
 
Subhotosh Khan said:
Yes... I got mixed up with the formula for Mortgage payment (or annuity). Right before answering this post, I was toying with the idea of increasing my monthly payment and possible effects there of.
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If P + Pr = P(1 + r), what would the calculation look like if the P was not divided out from P + Pr? Would P(1 + r)3 look something like P + Pr + Pr + Pr ?
 
Explain this! said:
If P + Pr = P(1 + r), what would the calculation look like if the P was not divided out from P + Pr? Would P(1 + r)3 look something like P + Pr + Pr + Pr ?
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Your question doesn't quite make sense. If you didn't factor out P from P + Pr, then you wouldn't get P(1 + r)3 in the first place! The reason we do factor out the P is so that we have a process we can apply repeatedly by using an exponent: each year, we multiply the amount by (1 + r).

But if you expand P(1 + r)3 = P(1 + r)(1 + r)(1 + r), you get P(1 + 3r + 3r2 + r3) = P + 3Pr + 3Pr2 + Pr3. That's the other part of why the factored form is better: It's much more compact.
 
Dr.Peterson said:
Your question doesn't quite make sense. If you didn't factor out P from P + Pr, then you wouldn't get P(1 + r)3 in the first place! The reason we do factor out the P is so that we have a process we can apply repeatedly by using an exponent: each year, we multiply the amount by (1 + r).

But if you expand P(1 + r)3 = P(1 + r)(1 + r)(1 + r), you get P(1 + 3r + 3r2 + r3) = P + 3Pr + 3Pr2 + Pr3. That's the other part of why the factored form is better: It's much more compact.
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This is what I was trying to determine: P + 3Pr + 3Pr2 + Pr3 from P(1 + r) 3, but I do not understand why the 3 is used in 3Pr. (?)
 
Try it for yourself. What do you get when you expand (1 + r)(1 + r)(1 + r) - that is, "foil" or distribute?
 
Dr.Peterson said:
Try it for yourself. What do you get when you expand (1 + r)(1 + r)(1 + r) - that is, "foil" or distribute?
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I tried this instead P(1 +r)2 equals P + P (r)2 equals P + P (r)(r). If r is 5% and P is 100, it should be the same as 100(1 + 0.05)2. 100.25 is the result, I think. (?)

It's been a number of years since I had algebra!
 
You'd better review your algebra!

No, [MATH](1 + r)^2 = (1 + r)(1 + r) = 1 + 2r + r^2[/MATH], not [MATH]1 + r^2[/MATH].

Of course, as I suggested, we don't do this in working with compound interest; we just evaluate the expression as it is. The reason you need to know your algebra is that if you don't, you'll think you can do things that you can't.

Check it out: You claim that [MATH]100(1 + 0.05)^2 = 100.25[/MATH], but you can see for yourself that it is not. As written, it is [MATH]100(1.05)^2 = 100(1.05)(1.05) = 110.25[/MATH]. Just do it on your calculator! What you did, apparently, is [MATH]100(1 + 0.05^2) = 100(1.0025) = 100.25[/MATH]. That is not the same thing.

If you don't know that you can't do that, then you will make mistakes all over the place. But at the very least, you must check everything you do, and never just say "it should be the same as" if you don't know.
 
Explain this! said:
If P + Pr = P(1 + r), what would the calculation look like if the P was not divided out from P + Pr? Would P(1 + r)3 look something like P + Pr + Pr + Pr ?
Click to expand...
No!!

(1+r)3 = 1 + 3 * r + 3 * r2 + r3
Explain this! said:
I tried this instead P(1 +r)2 equals P + P (r)2 equals P + P (r)(r). If r is 5% and P is 100, it should be the same as 100(1 + 0.05)2. 100.25 is the result, I think. (?)

It's been a number of years since I had algebra!
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P(1 +r)2 equals P + P (r)2 equals P + P (r)(r) .... No it does not

P(1+r)2 = P * [(1+r) * (1+r)] = P * [ 1 * (1 + r) + r * (1 + r)] ..................... Do you understand this step?
 
Subhotosh Khan said:
No!!

(1+r)3 = 1 + 3 * r + 3 * r2 + r3

P(1 +r)2 equals P + P (r)2 equals P + P (r)(r) .... No it does not

P(1+r)2 = P * [(1+r) * (1+r)] = P * [ 1 * (1 + r) + r * (1 + r)] ..................... Do you understand this step?
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Thanks for the reply! No, I do not understand the next step. :confused:

I think that in order to get P + P(r) to produce the correct amount is to omit the exponent on the P(r) and just use repeated multiplication and addition. It would look something like P + P(r) + P + P(r) + ... The value of P would change so that, for example 100 + 100(0.05) + 105 + 105(0.05) + 110.25 + 105.25(0.05) = 115.76. The value for P is different throughout the solution. I would not suggest doing this to solve for 100 at 5% for 3 year. I just wanted to see how this could be solved without using the P(1 + r)n method. [P + P(r)]3 might work with some tweaking. I tried it, and I got 1157625, but it is missing the decimal.

Thanks again for your efforts to explain this to me! I do especially like the Voltaire quote!
 
Explain this! said:
Thanks for the reply! No, I do not understand the next step. :confused:

I think that in order to get P + P(r) to produce the correct amount is to omit the exponent on the P(r) and just use repeated multiplication and addition. It would look something like P + P(r) + P + P(r) + ... The value of P would change so that, for example 100 + 100(0.05) + 105 + 105(0.05) + 110.25 + 105.25(0.05) = 115.76. The value for P is different throughout the solution. I would not suggest doing this to solve for 100 at 5% for 3 year. I just wanted to see how this could be solved without using the P(1 + r)n method. [P + P(r)]3 might work with some tweaking. I tried it, and I got 1157625, but it is missing the decimal.

Thanks again for your efforts to explain this to me! I do especially like the Voltaire quote!
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NO.

The P does not get added each period. If you think that 100 invested at 5% compounded annually will give you

[MATH]100 + 5 + 105 + 105(0.05) = 210 + 5.25 = 215.25[/MATH]
at the end of 2 years, you will be sadly disappointed.

What you will have is

[MATH]100 + 5 + 105 * 0.05 = 105 + 5.25 = 110.25 = 100(1.1025) = 100(1.05)^2.[/MATH]
Your point about principal changing is correct in the legal sense, but in the formula P stands for the initial principal, which of course does not change.

It is actually more comprehensible to present the formula as:

[MATH]P_n = P_0(1 + r)^n, \text { where}[/MATH]
[MATH]P_n = \text {principal after } n \text { compounding periods,}[/MATH]
[MATH]P_0 = \text { initial principal before any interest accrues. and}[/MATH]
[MATH]r = \text {interest rate in each compounding period,}[/MATH]
Note that the formula only works if r is the same for each period.
 
Explain this! said:
The following is what I am trying to represent. See http://www.moneychimp.com/articles/finworks/fmfutval.htm
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Year......Balance

Now P

Year 1: P + rP = P *(1 +r)

Year 2: (P + rP) + r(P + rP) = P * ( 1 + 2r + r2) = P * (1 + r)2 = P * (1 + r) * (1 + r)

Year 3: What would this year be? = P * (1 + r)2 * (1 + r) = P * (1 + r)3

Year 4: What would this year be? = P * (1 + r)3 * (1 + r) = P * (1 + r)4

Year 5: etc...... and so on.....
 
Subhotosh Khan said:
Year......Balance

Now P

Year 1: P + rP = P *(1 +r)

Year 2: (P + rP) + r(P + rP) = P * ( 1 + 2r + r2) = P * (1 + r)2 = P * (1 + r) * (1 + r)

Year 3: What would this year be? = P * (1 + r)2 * (1 + r) = P * (1 + r)3

Year 4: What would this year be? = P * (1 + r)3 * (1 + r) = P * (1 + r)4

Year 5: etc...... and so on.....
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Thanks for the reply, but I did not want the factored out version!

What would come after (P + rP) + r(P + rP) for the third year? Would it be another r(P +rP)?
 
Explain this! said:
Thanks for the reply, but I did not want the factored out version!

What would come after (P + rP) + r(P + rP) for the third year? Would it be another r(P +rP)?
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No -

That would be:

[(P + r*P) + r*(P + r*P)] * (1 + r)

Now multiply it out......
 
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