<p><strong>The answer: it is an assumption</strong> <br>
Solow-models usually assume that <span class="math-container">$f$</span> fulfills the <a href="https://en.wikipedia.org/wiki/Inada_conditions" rel="nofollow noreferrer">Inada conditions</a>, points 3. and 4. of which state
<span class="math-container">$$
\begin{equation*}
\lim_{k \to 0} f'(k) = \infty \\
\lim_{k \to \infty} f'(k) = 0.
\end{equation*}
$$</span></p>
<p><strong>A frequent special case</strong> <br>
If the production function <span class="math-container">$F(K,L)$</span> is of Cobb-Douglas type with constant returns to scale and has positive parameters, then the derived <span class="math-container">$f(k) = F(K,L)/L$</span> function will satisfy the Inada conditions.</p>
<p><strong>Caveat</strong> <br>
Based on the graphs they see, students of economics sometimes mistakenly assume that all strictly concave functions have these properties, but this is not the case. <span class="math-container">$f(k) = \sqrt{k} + k$</span> does not have the second one, while <span class="math-container">$f(k) = \sqrt{1+k}$</span> does not have the first one. Thus properties like <span class="math-container">$f$</span> being concave or non-linear are not sufficient, one has to assume the conditions.</p>

CC BY-SA 4.0
2023-09-15T11:05:12.563Z
2023-09-20T10:20:00.833Z
2023-09-20T10:20:00.833Z
Giskard
<p>Because <span class="math-container">$\delta k$</span> is linear function, so the slope of the function is always constant, whereas <span class="math-container">$sf(k)$</span> is typically nonlinear function exponential function like <span class="math-container">$f(k)=k^\alpha$</span> with <span class="math-container">$0&lt;\alpha&lt;1$</span>, so mathematically such function starts with steep slope, but this slope always gets smaller (this can be seen from <span class="math-container">$d/dk= \alpha k^{\alpha -1}$</span>.</p>
<p>However, note depending on value of parameter <span class="math-container">$\delta$</span>, <span class="math-container">$\delta k$</span> could have equal or  steeper slope but in such case capital accumulation is impossible because capital depreciates faster than it can be accumulated given <span class="math-container">$s$</span>.</p>

CC BY-SA 4.0
2023-09-15T11:08:16.467Z
Can you please give me a depreciation parameter $\delta$ at which the slope of $\delta k$ will always be steeper than the slope of, let's say, $0.5\sqrt{k}$? If not, can you please explain what you mean by your second paragraph.
Giskard
CC BY-SA 4.0
2023-09-15T12:52:25.030Z
@Giskard $\delta$ that satisfies $\delta > \max \frac{1}{\sqrt{k}}$ for k>0. Also savings rate can be just zero and in that case it’s trivial
1muflon1
CC BY-SA 4.0
2023-09-15T13:19:32.413Z
But you specify "always", when you write "always steeper slope", so I don't think a $\delta > \max_{k>0} \frac{1}{\sqrt{k}}$ exists? You are right about the edge case of $s = 0$, but that seems quite specific, not useful for the general case.
Giskard
CC BY-SA 4.0
2023-09-15T13:44:50.297Z
@Giskard I edited out the word always
1muflon1
CC BY-SA 4.0
2023-09-15T13:47:25.543Z
Interesting choice; I am still not sure what you are trying to say with the paragraph though!
Giskard
CC BY-SA 4.0
2023-09-16T20:41:01.130Z
@1muflon1  Adding to Giskard's point, surely $\lim_{k \to 0} \frac{1}{\sqrt{k}}=\infty$?
Adam Bailey
CC BY-SA 4.0
2023-09-20T11:05:08.950Z
@AdamBailey yes what should I have said is always greater or equal than, hence my edit
1muflon1
CC BY-SA 4.0
2023-09-15T10:12:53.387Z
2023-09-15T13:44:05.593Z
2023-09-15T13:44:05.593Z
1muflon1
<p>I wonder why at the beginning <span class="math-container">$sf(k(t))$</span> is steeper than <span class="math-container">$\delta k(t)$</span>, but at some point, it starts to become flatter than <span class="math-container">$\delta k(t)$</span>?</p>

CC BY-SA 4.0
2023-09-15T10:48:36.737Z
The standard chart used to explain the Solow model plots output per worker (vertical) against capital per worker (horizontal).  It does not have a time axis.  So when you say "at the beginning", a more accurate description would be "if capital per worker is very low".
Adam Bailey
CC BY-SA 4.0
2023-09-15T11:10:38.283Z
Hi! Did you know you can accept answers to [your questions](https://economics.stackexchange.com/users/44577/huy-l%C3%AA-thanh?tab=questions)?
Giskard
<macroeconomics><mathematical-economics><economic-growth><solow><endogenous-growth>
The Solow growth model
Huy Lê Thanh
# The Solow growth model

(Asked by: username_0 on 2023-09-15)

I wonder why at the beginning $sf(k(t))$ is steeper than $\delta k(t)$, but at some point, it starts to become flatter than $\delta k(t)$?

**Tags:** macroeconomics, mathematical-economics, economic-growth, solow, endogenous-growth
> **Comments:**
> **username_1**: The standard chart used to explain the Solow model plots output per worker (vertical) against capital per worker (horizontal).  It does not have a time axis.  So when you say "at the beginning", a more accurate description would be "if capital per worker is very low".  
> **username_2**: Hi! Did you know you can accept answers to [your questions](https://economics.stackexchange.com/users/44577/huy-l%C3%AA-thanh?tab=questions)?  

---

## 2 answers

### Answer by username_2 on 2023-09-15. Score: 4

**The answer: it is an assumption**   
Solow-models usually assume that $f$ fulfills the [Inada conditions](https://en.wikipedia.org/wiki/Inada_conditions), points 3. and 4. of which state
$$
\begin{equation\*}
\lim\_{k \to 0} f'(k) = \infty \\
\lim\_{k \to \infty} f'(k) = 0.
\end{equation\*}
$$

**A frequent special case**   
If the production function $F(K,L)$ is of Cobb-Douglas type with constant returns to scale and has positive parameters, then the derived $f(k) = F(K,L)/L$ function will satisfy the Inada conditions.

**Caveat**   
Based on the graphs they see, students of economics sometimes mistakenly assume that all strictly concave functions have these properties, but this is not the case. $f(k) = \sqrt{k} + k$ does not have the second one, while $f(k) = \sqrt{1+k}$ does not have the first one. Thus properties like $f$ being concave or non-linear are not sufficient, one has to assume the conditions.


---

### Answer by username_3 on 2023-09-15. Score: 2

Because $\delta k$ is linear function, so the slope of the function is always constant, whereas $sf(k)$ is typically nonlinear function exponential function like $f(k)=k^\alpha$ with $0<\alpha<1$, so mathematically such function starts with steep slope, but this slope always gets smaller (this can be seen from $d/dk= \alpha k^{\alpha -1}$.

However, note depending on value of parameter $\delta$, $\delta k$ could have equal or steeper slope but in such case capital accumulation is impossible because capital depreciates faster than it can be accumulated given $s$.

> **Comments:**
> **username_2**: Can you please give me a depreciation parameter $\delta$ at which the slope of $\delta k$ will always be steeper than the slope of, let's say, $0.5\sqrt{k}$? If not, can you please explain what you mean by your second paragraph.  
> **username_3**: @Giskard $\delta$ that satisfies $\delta > \max \frac{1}{\sqrt{k}}$ for k>0. Also savings rate can be just zero and in that case it’s trivial  
> **username_2**: But you specify "always", when you write "always steeper slope", so I don't think a $\delta > \max_{k>0} \frac{1}{\sqrt{k}}$ exists? You are right about the edge case of $s = 0$, but that seems quite specific, not useful for the general case.  
> **username_3**: @Giskard I edited out the word always  
> **username_2**: Interesting choice; I am still not sure what you are trying to say with the paragraph though!  
> **username_1**: @1muflon1  Adding to Giskard's point, surely $\lim_{k \to 0} \frac{1}{\sqrt{k}}=\infty$?  
> **username_3**: @AdamBailey yes what should I have said is always greater or equal than, hence my edit  

---

