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[SOLVED] Taylor Series

dwsmith

Well-known member
Feb 1, 2012
1,673
$1+v_{t+1} = (1+v_t)\exp\left(-rv_{t-1}\right)\approx (1+v_t)(1-rv_{t-1})$

The book is linearizing the model where we generally use a Taylor Series.

How was the expression expanded in the Taylor Series to get the approximate answer?

Thanks.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
$1+v_{t+1} = (1+v_t)\exp\left(-rv_{t-1}\right)\approx (1+v_t)(1-rv_{t-1})$

The book is linearizing the model where we generally use a Taylor Series.

How was the expression expanded in the Taylor Series to get the approximate answer?

Thanks.
The second set of brackets contains the first two terms of the series expansion of

\[ \exp(-rv_{t-1}) =\sum_{k=0}^{\infty}\frac{(-rv_{t-1})^k}{k!}= 1-rv_{t-1}+{\text{O}}((rv_{t-1})^2)\]

CB
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
If $x$ is small, than $e^{x} \approx 1 + x$.

Edit - too slow :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The second set of brackets contains the first two terms of the series expansion of

\[ \exp(-rv_{t-1}) =\sum_{k=0}^{\infty}\frac{(-rv_{t-1})^k}{k!}= 1-rv_{t-1}+{\text{O}}((rv_{t-1})^2)\]

CB
Now if i was dealing $K+v_{t+1}=(K+v_t)\left[1 + r \left(1- \frac{K+v_t}{K}\right)\right]$, would I want to put it in the form $(K+v_t)g$ and then expand g in the Taylor series to the linear terms?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Now if i was dealing $K+v_{t+1}=(K+v_t)\left[1 + r \left(1- \frac{K+v_t}{K}\right)\right]$, would I want to put it in the form $(K+v_t)g$ and then expand g in the Taylor series to the linear terms?
Or do we expand the Taylor Series of

$$
\left[1 + r \left(1- \frac{K+v_t}{K}\right)\right]
$$


with the \(K+v_t\) inside?
 
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