Wikipedia:Reference desk/Archives/Mathematics/2017 October 14

= 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)


 * 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)

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)


 * 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)

Dikipewia, you'd need to have the same exponent on both sides. For example:. Alternatively, it is also true that, 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)