Difference $\Delta P_t$ approaches 0, then its relative difference $\Delta P_t / P_(t-1) \approx ln(P_t / P_(t-1))$.

Question

When difference $\Delta P_t$ approaches 0, its relative difference $\dfrac{\Delta P_t}{P_{t-1}} \approx \ln(\dfrac{P_t}{P_{t-1}})$.

I know that it can be shown somehow with Taylor series: $\ln(1+x)=x-\frac12x^2+\frac13x^3...$ I just can't figure how.

Answer 1

Note that $\Delta P_t=P_t-P_{t-1}$, the right hand side $$\ln\frac{P_t}{P_{t-1}}=\ln\frac{P_{t-1}+\Delta P_t}{P_{t-1}}=\ln\left(1+\frac{\Delta P_t}{P_{t-1}}\right)\approx \frac{\Delta P_t}{P_{t-1}}$$