Mathematics desk
< October 13	<< Sep \| October \| Nov >>	October 15 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 14

Basic math, compound interest, multiply out

In the common compound interest formula: Capital C, interest rate i and periods p:

C * (1 + i)^p

multiplying out would lead to:

(C + iC)^p

which is a completely different result, so the parenthesis has to be solved first. But isn't algebraically valid transforming:

(ab+ac)^2

into

a(b+c)^2

or back? Shouldn't the rule rather be expressed as:

C * ((1 + i)^p)?

--Dikipewia (talk) 12:46, 14 October 2017 (UTC)[reply]

No.

(ab+ac)^{2}=(a(b+c))^{2}=a^{2}(b+c)^{2}\neq a(b+c)^{2}.

And the extra set of parentheses aren't necessary (but aren't wrong either) because exponentiation is generally understood to have higher precedence than multiplication. --Deacon Vorbis (talk) 12:57, 14 October 2017 (UTC)[reply]

You cannot get from

C * (1 + i)^p

to

(C + iC)^p

because power (Exponentiation) comes first then multiplication. 110.22.20.252 (talk) 15:04, 14 October 2017 (UTC)[reply]

Another way to prove that (ab+ac)^2 ≠ a(b+c)^2, is to show one counterexample, so let's use a=2,b=3,c=4:

(ab+ac)^2         ≠ a(b+c)^2
((2)(3)+(2)(4))^2 ≠ 2(3+4)^2
(6 + 8)^2         ≠ 2(7)^2
(14)^2            ≠ (2)(49)
 196              ≠ 98

StuRat (talk) 18:06, 15 October 2017 (UTC)[reply]

Dikipewia, you'd need to have the same exponent on both sides. For example: C^p * (1 + i)^p = (C + iC)^p. Alternatively, it is also true that C * (1 + i)^p = [C^(1/p) + i * C^(1/p)]^p, but as C is not a dimensionless quantity, the practical value of this equation is pretty much nil. 78.0.216.92 (talk) 03:05, 16 October 2017 (UTC)[reply]