/sci/ - Science & Math

[Anonymous No.15080106]

What's the difference between compounded, compounded continuously, and exponential growth? This is so stupid...

[Anonymous No.15080199]

>>15080106
>compounded vs compounded continuously
This originates in basically a scam from hundreds of years ago. Look at some examples, say it’s a loan with a $1000 principle
>30% compounded annually
works as expected, after 1 year it’s $1300, after 2 years it’s $1690 etc.
>30% compounded monthly
actually means it’s 2.5% per month, so that after 1 year it’s 1000(1.025)^12=$1345 and after 2 years it’s $1808
>30% compounded daily
about $1350 and $1822

it doesn’t make sense but then it doesn’t have to, because again, it was a scam designed to rip people off who didn’t read the fine print. But it got people wondering: if you keep making the interval shorter and shorter, does it approach a limit? It turns out there is a limit which we call “compounded continuously”, and it turns out to be connected to this famous number [math]e[/math]. In fact, as far as I know, this question about the theoretical limits of a financial scam is how [math]e[/math] was first discovered

[Anonymous No.15080209]

>>15080199
>a scam from hundreds of years ago
I should add that this is illegal now. In most places the bank is required to advertise the “effective annual interest rate” which in these examples is just 30%, 34.5%, and 35% respectively

[Anonymous No.15080322]

>>15080199
Thanks for the explanation... Why are radioactive half lives expressed in continuously compounded instead of compounded intervals?

[Anonymous No.15080342]

[math] \displaystyle

\\ \text{Continuous compounding}
\\ \displaystyle P(t)=P_{0} \, e^{rt}
\\ \text{ } P_0 \; \, \text{initial value}
\\ \text{ } r \quad \text{rate of growth}
\\ \text{ } t \quad \text{time}
\\ ~~~~----- \\
P(t_2) = 2P_0 \Rightarrow 2P_0 = P_{0} \, e^{rt_2} \\
2 = e^{rt_2} \\
e^{ln(2)} = e^{rt_2} \\
ln(2) = rt_2 \\
t_2 = \frac{ln(2)}{r} \approx \frac{70\%}{100r\%} \\
\\
P(t_{10}) = 10P_0 \Rightarrow 10P_0 = P_{0} \, e^{rt_{10}} \\
10 = e^{rt_{10}} \\
e^{ln(10)} = e^{rt_{10}} \\
ln(10) = rt_{10} \\
t_{10} = \frac{ln(10)}{r} \approx \frac{230\%}{100r\%}

[/math]

[Anonymous No.15080461]

>>15080322
Because atoms decay continuously