Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | Next revision Both sides next revision | ||
|
minus_twelve_._note [2017/07/02 21:59] nikolaj |
minus_twelve_._note [2017/07/02 22:57] nikolaj |
||
|---|---|---|---|
| Line 82: | Line 82: | ||
| We have the Taylor expansion for the logarithm | We have the Taylor expansion for the logarithm | ||
| - | $$\log(1+r) = \sum_{k=1}^{\infty}\dfrac{1}{k}(-r)^k = r - \dfrac{1}{2}r^2 + \dfrac{1}{3}r^3 + {\mathcal O}(r^4) = r \left(1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) \right) $$ | + | $$\log(1+r) = \sum_{k=1}^{\infty}\dfrac{1}{k}(-r)^k = r - \dfrac{1}{2}r^2 + \dfrac{1}{3}r^3 + {\mathcal O}(r^4)= $$ |
| + | $$= r \left(1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) \right) $$ | ||
| and using the geometric series expansion, we get | and using the geometric series expansion, we get | ||
| - | $$\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } = \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$$ | + | $$\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } =$$ |
| + | $$= \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$$ | ||
| With $r=z-1$ we see | With $r=z-1$ we see | ||
