# Euler equation guess and verify

So I have the Euler condition written as follows: $$\frac{1}{k_t^a-k_{t+1}}=\frac{a\beta k_{t+1}^{a-1}}{k_{t+1}^a-k_{t+2}}$$ and it says that $$k_{t+1}$$ takes the form $$gk_t^a$$, where g is an unknown to be determined. I know the result is $$k_{t+1}=a\beta k_t^a$$, but don't know how the derivation goes to reach this result.

From the hint, I got that $$๐^๐_{๐ก+1}โ๐_{๐ก+2}=๐๐๐^{๐โ1}_{๐ก+1}๐^๐_๐กโ๐๐๐^๐_{๐ก+1}$$, we divide by ๐^๐_{๐ก+1} and what I get is $$1โ๐=๐๐๐โ๐๐$$. Now based on the results, $$g$$ should somehow be $$๐๐$$, but I'm stuck here

• Are you sure it's not that $k_{t+1}=gk_t^a$? Commented Feb 4, 2021 at 19:31
• @HerrK. yes, my mistake Commented Feb 4, 2021 at 19:34

Note that $$$$k_{t+1}=gk_t^a \quad\Rightarrow\quad g=\frac{k_{t+1}}{k_t^a}=\frac{k_{t+2}}{k_{t+1}^a}.$$$$
Rearrange your Euler condition into a form such that you can use $$g$$ to sub-out the above ratios. Then it should be straightforward to solve for $$g$$.
• @MaybelineLee: Try cross-multiplying the denominators in the original equation, and then divide both sides by $k_{t+1}^a$. Simplify using the hint to get an equation that only involves $g,a,\beta$. Lastly solve for $g$. Commented Feb 4, 2021 at 22:32
• So I'm going through the derivations again on this one and it's still confusing me. We have $k_{t+1}^a-k_{t+2}=abk_{t+1}^{a-1}k_t^a-abk_{t+1}^a$, we divide by $k_{t+1}^a$ and what I get is $1-g=\frac{ab}{g}-ab$. Now based on the results, g should somehow be $ab$ but I'm just not getting it. Commented Sep 16, 2021 at 10:19
• @MaybelineLee: Solving $1-g=\frac{ab}{g}-ab$ for $g$ yields $g=ab$ or $g=1$. Commented Sep 16, 2021 at 12:59