Can decay percent be less than $-100\%$?

Question

I have this problem:

$$A=A_0\left(1+\frac{r}{n}\right)^{nt}$$

where $A=100$, $A_0=25$, $n=1$, and $t=2$.

This leads to

\begin{align*} (1+r)^2=4 &\quad\Rightarrow\quad |1+r|=2\\ &\quad\Rightarrow\quad 1+r = +2 \text{ or } -2 \\ &\quad\Rightarrow\quad r= 1 \text{ or } -3. \end{align*}

Can I have $r$ which is rate of decay $= -3$, which is $-300\%$?

Although mathematics itself does not prevent you from getting $r=-3$ under this particular setting, the problem seems to have arisen either in physics or finance context. In such case, the factor $\left(1+\frac{r}{n}\right)^{nt}$ as a function of $t$ is often assumed to be defined for real $t$, and so, negative bases are depreciated. As such, $r=-3$ is naturally excluded as a feasible answer. — Sangchul Lee, Jun 03 '20 at 08:27

score 1 · Answer 1 · answered Jun 03 '20 at 08:21

1

Exponential functions $b^x$ imply $b\in(0,\infty)\setminus\{1\}$. Therefore, you do not have $-(1+r)$ is part of your solution.

answered Jun 03 '20 at 08:21

Andrew Chin

7,389

is there text book documenting this implication.if yes I can accept your answer – ahmed allam Jun 05 '20 at 21:48
Any book on introductory real analysis should have this point. – Andrew Chin Jun 05 '20 at 22:23

score 1 · Answer 2 · 2020-06-03T08:32:14.113

1

Assuming that $-3$ is a legit value, the sequence over time would be

$$25,-50,100,-200,400,\cdots$$ which I would not call a decay.

You should question the a priori values of $r$.

edited Jun 03 '20 at 08:32

answered Jun 03 '20 at 08:26

is -3 legit solution??! – ahmed allam Jun 05 '20 at 21:48
I mean i got the problem as i wrote it.i solved by the way I wrote it.i found 2 solutions ,one is negative??does this negative solution has meaning? – ahmed allam Jun 05 '20 at 21:51
@ahmedallam: only you know that. – Jun 05 '20 at 21:56

tommik · Answer 3 · 2020-06-03T08:52:35.297

There is a problem in your forumulas. Substituting your data in the formulas you get

$$25(1+r)^2=100$$

So is clearly meaning that $r$ is not a dacay rate but an increasing rate:

$r=\sqrt{4}-1=100$%

The result is correct and you can easily check it

Start: 25

1° period: 25+100%$\times$25=50

2° period: 50+100%$\times$50=100

If you want to have the decay rate you can modify your forumula in the following way:

$$100(1-r)^2=25$$

$r=1-\sqrt{\frac{1}{4}}=1-\frac{1}{2}=50$%

The result is correct and you can easily check it

Start: 100

1° period: 100-50%$\times$100=50

2° period: 50-50%$\times$50=25

Notes:

$n=1$ means that the result of $r$ is expressed as ratio per 1 period (one year, one month a.s.o.)

$r$ as a ratio, is always meant non negative

Can decay percent be less than $-100\%$?

3 Answers3