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- Thread starter dwsmith
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- #1

- Jan 26, 2012

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The second set of brackets contains the first two terms of the series expansion of$1+v_{t+1} = (1+v_t)\exp\left(-rv_{t-1}\right)\approx (1+v_t)(1-rv_{t-1})$

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.

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

CB

- Jan 26, 2012

- 183

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

Edit - too slow

Edit - too slow

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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?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

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- #5

Or do we expand the Taylor Series ofNow 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?

$$

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

$$

with the \(K+v_t\) inside?

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