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

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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. $(ab+ac{)}^2=(a(b+c){)}^2=a^2(b+c{)}^2\ne a(b+c{)}^2.$ 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: 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)