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. 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
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)