<h2>Intuition</h2>

<p>Well, when <span class="math-container">$r$</span> is zero, you are just getting value <span class="math-container">$A$</span> per time unit, so total present value will be <span class="math-container">$A \cdot t$</span>.</p>

<p><br></p>

<h2>The approximation</h2>

<p>You are asking about the approximation
<span class="math-container">$$
\frac A {-r}(e^{-rt}-1) \approx A \cdot t.
$$</span>
This is equivalent to
<span class="math-container">$$
\frac 1{rt}(1-e^{-rt}) \approx 1,
$$</span>
which is true for small <span class="math-container">$rt$</span> values. You can see this by taking the limit of the left hand side and using L'Hospital's rule.
<span class="math-container">$$
\lim_{rt \to 0} \frac{\frac{\text{d} (1 - e^{-rt})}{\text{d} rt}}{\frac{\text{d} rt}{\text{d} rt}}
= \lim_{rt \to 0} \frac{e^{-rt}}{1} = 1.
$$</span></p>

<p><br></p>

<h2>The approximation in another form</h2>

<p>The approximation
<span class="math-container">$$
1-e^{-h} \approx h
$$</span>
seems to be a variant of the more well known approximation
<span class="math-container">$$
\ln(1+h) \approx h
$$</span>
for small values of <span class="math-container">$h$</span>.</p>

<p><br></p>

<h2>How good is this approximation</h2>

<p>To get a sense of the error term for large values using your approximation, you can use <a href="http://www.math.ubc.ca/~CLP/CLP1/clp_1_dc/ssec_taylor_error.html" rel="nofollow noreferrer">Taylor-approximation</a> for the function
<span class="math-container">$$
f(x) = \frac{1-e^{-x}}{x}
$$</span>
starting from some very small <span class="math-container">$x$</span> value.</p>

CC BY-SA 4.0
2019-11-24T09:05:50.650Z
2019-11-25T06:54:13.630Z
2019-11-25T06:54:13.630Z
Giskard
<p>No idea if this is the right stackexchange for this,  feel free to point me elsewhere!</p>

<p>I'm teaching business calculus and one of the problems the students have is to figure out the present value of a continuous income stream <span class="math-container">$\$$$A$</span> per year earning <span class="math-container">$r$</span> interest rate.  The integral comes to <span class="math-container">$$P.V=A\int_0^te^{-rz}dz=\frac A {-r}(e^{-rt}-1)$$</span>
I've noticed as a heuristic that for relatively short values of t (<span class="math-container">$t&lt;10$</span>) and small values of r (<span class="math-container">$r&lt;.04$</span>) (the parameters the learning software uses),   the present value comes to be just a bit less than <span class="math-container">$A\cdot t$</span>. (usually within about 10 percent)  Does anyone know an economics rational why this is the case,  and would it hold for larger values of <span class="math-container">$r$</span> or <span class="math-container">$t$</span>?</p>

<mathematical-economics>
Present value heuristics problem
Alan
# Present value heuristics problem

(Asked by: username_0 on 2019-11-24)

No idea if this is the right stackexchange for this, feel free to point me elsewhere!

I'm teaching business calculus and one of the problems the students have is to figure out the present value of a continuous income stream $\$$$A$ per year earning $r$ interest rate. The integral comes to $$P.V=A\int\_0^te^{-rz}dz=\frac A {-r}(e^{-rt}-1)$$
I've noticed as a heuristic that for relatively short values of t ($t<10$) and small values of r ($r<.04$) (the parameters the learning software uses), the present value comes to be just a bit less than $A\cdot t$. (usually within about 10 percent) Does anyone know an economics rational why this is the case, and would it hold for larger values of $r$ or $t$?

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2019-11-24. Score: 6 | Accepted answer

## Intuition

Well, when $r$ is zero, you are just getting value $A$ per time unit, so total present value will be $A \cdot t$.

## The approximation

You are asking about the approximation
$$
\frac A {-r}(e^{-rt}-1) \approx A \cdot t.
$$
This is equivalent to
$$
\frac 1{rt}(1-e^{-rt}) \approx 1,
$$
which is true for small $rt$ values. You can see this by taking the limit of the left hand side and using L'Hospital's rule.
$$
\lim\_{rt \to 0} \frac{\frac{\text{d} (1 - e^{-rt})}{\text{d} rt}}{\frac{\text{d} rt}{\text{d} rt}}
= \lim\_{rt \to 0} \frac{e^{-rt}}{1} = 1.
$$

## The approximation in another form

The approximation
$$
1-e^{-h} \approx h
$$
seems to be a variant of the more well known approximation
$$
\ln(1+h) \approx h
$$
for small values of $h$.

## How good is this approximation

To get a sense of the error term for large values using your approximation, you can use [Taylor-approximation](http://www.math.ubc.ca/~CLP/CLP1/clp_1_dc/ssec_taylor_error.html) for the function
$$
f(x) = \frac{1-e^{-x}}{x}
$$
starting from some very small $x$ value.


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