# [SOLVED]Taylor Series

#### dwsmith

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

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
$1+v_{t+1} = (1+v_t)\exp\left(-rv_{t-1}\right)\approx (1+v_t)(1-rv_{t-1})$

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
If $x$ is small, than $e^{x} \approx 1 + x$.

Edit - too slow

#### dwsmith

##### Well-known member
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
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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