Compound Interest Formula

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
How does (P + Pi) - P become P(1 + i)^n - 1? I know that factoring is involved, but I not sure what the steps are.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083
How does (P + Pi) - P become P(1 + i)^n - 1? I know that factoring is involved, but I not sure what the steps are.
Where did you see that "happening"? I think you are misquoting!
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
Where did you see that "happening"? I think you are misquoting!
I've seen this or something similar to it in finance books. Think of it in another way. Where does P(1 + i)^n - 1 come from? Isn't this the compound interest formula?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083
I've seen this or something similar to it in finance books. Think of it in another way. Where does P(1 + i)^n - 1 come from? Isn't this the compound interest formula?
However, you have NOT seen that

Most probably you have seen

Compound interest after n period = P[(1+i)^n -1]

Those [] are super-important. Now that you know the correct expression - google around a bit, you can find the derivation.
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
However, you have NOT seen that

Most probably you have seen

Compound interest after n period = P[(1+i)^n -1]

Those [] are super-important. Now that you know the correct expression - google around a bit, you can find the derivation.
Thanks for the reply! I'll search for an answer, but I thought that is what you "experts" are for -- to answer questions. Is there a simple reason why the brackets are important?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083
Thanks for the reply! I'll search for an answer, but I thought that is what you "experts" are for -- to answer questions. Is there a simple reason why the brackets are important?
"...I thought that is what you "experts" are for -- to answer questions"

Nope ... we are volunteers - we "choose" our actions - we GUIDE.

And we refuse to spoon-feed.....
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083

firemath

Full Member
Joined
Oct 29, 2019
Messages
587
Is there a simple reason why the brackets are important?
The very simple reason that you will get it wrong if you do not follow the order of operations.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,030
(P + Pi) - P = Pi as the P cancels out to 0.

P(1 + i)^n - 1 means to multiply P and(1 + i)^n and then subtract 1

P[(1 + i)^n - 1] means to multiply P and (1 + i)^n - 1

P[(1 + i)^n - 1] = P(1 + i)^n - P which is different from P(1 + i)^n - 1
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
"...I thought that is what you "experts" are for -- to answer questions"

Nope ... we are volunteers - we "choose" our actions - we GUIDE.

And we refuse to spoon-feed.....
I'm certainly glad that I do not have you as an instructor, if that is your occupation!
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,068
How does (P + Pi) - P become P(1 + i)^n - 1? I know that factoring is involved, but I not sure what the steps are.
It doesn't. You are mixing up different formulas.

\(\displaystyle (P + Pi) - P = Pi\).

That is the formula for calculating the interest paid on amount P for one period without compounding at an interest rate of 100 * i percent per period.

30000 dollars for one month at 1/4 % per month. What's i?

\(\displaystyle 100 * i = \dfrac{1}{4} = 0.25 \implies i = \dfrac{0.25}{100} = 0.0025.\)

So the interest paid is \(\displaystyle 30000 * 0.0025 = 75.\)

The other formula, which you got wrong, is

\(\displaystyle P\{(1 + i)^n - 1\}.\)

That is the formula for interest paid after n periods when interest at a rate of 100 * i percent per period is compounded rather than paid before maturity.

30,000 for 3 months at an interest rate of 1/4 % per month compounded monthly will involve payment of interest after 3 months of

\(\displaystyle 30000\{(1 + 0.0025)^3 - 1\} = 225.76 > 3 * 75.\)
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
It doesn't. You are mixing up different formulas.

\(\displaystyle (P + Pi) - P = Pi\).

That is the formula for calculating the interest paid on amount P for one period without compounding at an interest rate of 100 * i percent per period.

30000 dollars for one month at 1/4 % per month. What's i?

\(\displaystyle 100 * i = \dfrac{1}{4} = 0.25 \implies i = \dfrac{0.25}{100} = 0.0025.\)

So the interest paid is \(\displaystyle 30000 * 0.0025 = 75.\)

The other formula, which you got wrong, is

\(\displaystyle P\{(1 + i)^n - 1\}.\)

That is the formula for interest paid after n periods when interest at a rate of 100 * i percent per period is compounded rather than paid before maturity.

30,000 for 3 months at an interest rate of 1/4 % per month compounded monthly will involve payment of interest after 3 months of

\(\displaystyle 30000\{(1 + 0.0025)^3 - 1\} = 225.76 > 3 * 75.\)

My question is what is the original formula that becomes P{(1 + i)^n - 1} and how does it become P{(1 + i)^n - 1}?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,068
I can give you an intuitive answer or a proof by weak mathematical induction.
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
I can give you an intuitive answer or a proof by weak mathematical induction.
I have no idea what your reply indicates.

Does P + Pr = P(1 + r)? If yes, then can you show me the steps?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,068
I can show you steps if you understand weak mathematical induction. You do realize that n can be any of an infinite number of values. I am not going to show you detailed steps for when n = 365.

I can show you an answer for n = 2, or n = 3, but that is not a general answer. The way to give a general answer is through weak mathematical induction.
 

Explain this!

Junior Member
Joined
Feb 7, 2019
Messages
100
I can show you steps if you understand weak mathematical induction. You do realize that n can be any of an infinite number of values. I am not going to show you detailed steps for when n = 365.

I can show you an answer for n = 2, or n = 3, but that is not a general answer. The way to give a general answer is through weak mathematical induction.
I'm trying to keep this simple:
First of all P + Pr has a common factor of P, so it can be expressed as P(1 + r).

I want to know how the "P" gets in front of (1 +r) to become P(1 + r). If I divide P + Pr by P, I get (1 + r), so how does the P get in front of (1 + r) as in P(1 + r)?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,068
3 * 6 + 3 * 8 = 18 + 24 = 42

3 * (6 + 8) = 3 * 14 = 42.

Now you can try experimenting with other numbers until you have convinced yourself of the validity of this basic law of arithmetic:

\(\displaystyle a(b + c) \equiv ab + ac.\)

It's called the distributive law of multiplication over multiplication.

There is a proof using induction and the Peano Postulates, but I have long since forgotten it.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,030
P=P*1

P + Pr = P*1 + Pr = P(1+r). The P was factored out.
 
Top