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minus_twelve_._note [2019/09/09 22:33] nikolaj |
minus_twelve_._note [2019/09/21 00:34] nikolaj |
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| So as an example, with $k=3,a=2,b=4$ you get the identity | So as an example, with $k=3,a=2,b=4$ you get the identity | ||
| - | $$\frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ | + | $$\frac{1}{3}(4^3-2^3)=(2^2+(4-1)^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ |
| and there are literally infinitely many identities involving $-\frac{1}{12}$ because of this formula. In case you're wondering, both sides of the equation above simplify to $\tfrac{56}{3}$. | and there are literally infinitely many identities involving $-\frac{1}{12}$ because of this formula. In case you're wondering, both sides of the equation above simplify to $\tfrac{56}{3}$. | ||
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| The geometric series for z in (0,1) is | The geometric series for z in (0,1) is | ||
| - | [math] \sum_{n=0}^\infty z^n = \dfrac {1} {1-z} [/math] | + | $ \sum_{n=0}^\infty z^n = \dfrac {1} {1-z} $ |
| The smooth analogous is | The smooth analogous is | ||
| - | [math] \int_0^\infty z^n \, d n = \int_0^\infty e^{ n \, \log (z) } \ d n = \dfrac {1} {-\log(z)} [/math] | + | $ \int_0^\infty z^n \, d n = \int_0^\infty e^{ n \, \log (z) } \ d n = \dfrac {1} {-\log(z)} $ |
| - | Applying [math] z \dfrac {d} {dz} [/math] to the first yields | + | Applying $ z \dfrac {d} {dz} $ to the first yields |
| - | [math] z \dfrac {z} {dz} \dfrac {1} {1-z} = \sum_{n=0}^\infty n \, z^n [/math] | + | $ z \dfrac {z} {dz} \dfrac {1} {1-z} = \sum_{n=0}^\infty n \, z^n $ |
| and applying it to the second yields | and applying it to the second yields | ||
| - | [math] z \dfrac {z} {dz} \dfrac{1} {-\log(z)} = \dfrac {1} { \log(z)^2 } [/math] | + | $ z \dfrac {z} {dz} \dfrac{1} {-\log(z)} = \dfrac {1} { \log(z)^2 } $ |
| How do those two differ? | How do those two differ? | ||
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| We have the Taylor expansion | We have the Taylor expansion | ||
| - | [math] \log(1+r) = \sum_{k=1}^{\infty} \dfrac {1} {k} (-r)^k = r - \dfrac {1} {2} r^2 + \dfrac {1} {3} r^3 + O (r^4) = r \left( 1 - \dfrac {1} {2} r + \dfrac {1} {3} r^2 + O (r^3) \right) [/math] | + | $ \log(1+r) = \sum_{k=1}^{\infty} \dfrac {1} {k} (-r)^k = r - \dfrac {1} {2} r^2 + \dfrac {1} {3} r^3 + O (r^4) = r \left( 1 - \dfrac {1} {2} r + \dfrac {1} {3} r^2 + O (r^3) \right) $ |
| Using the geometric series expansion, we get | Using the geometric series expansion, we get | ||
| - | [math] \dfrac {1} { \log (1+r) } = \dfrac {1} {r} \dfrac {1} {1 - \dfrac {1} {2} r + \dfrac {1} {3} r^2 + O (r^3) } = \dfrac {1} {r} \left( 1 + \dfrac {1} {2} r - \dfrac {1} { 2 \cdot 2 \cdot 3 } r^2 + O (r^3) \right) = \dfrac {1} {r} + \dfrac {1} {2} - \dfrac {1} {12} r + O (r^2) [/math] | + | $ \dfrac {1} { \log (1+r) } = \dfrac {1} {r} \dfrac {1} {1 - \dfrac {1} {2} r + \dfrac {1} {3} r^2 + O (r^3) } = \dfrac {1} {r} \left( 1 + \dfrac {1} {2} r - \dfrac {1} { 2 \cdot 2 \cdot 3 } r^2 + O (r^3) \right) = \dfrac {1} {r} + \dfrac {1} {2} - \dfrac {1} {12} r + O (r^2) $ |
| With r=z-1 we see | With r=z-1 we see | ||
| - | [math] \dfrac {1} { \log(z) } = - \dfrac {1} {1-z} + \dfrac {1} {2} - \dfrac {1} {12} (z-1) + O ((z-1)^2) [/math] | + | $ \dfrac {1} { \log(z) } = - \dfrac {1} {1-z} + \dfrac {1} {2} - \dfrac {1} {12} (z-1) + O ((z-1)^2) $ |
| and taking the derivative, we get | and taking the derivative, we get | ||
| - | [math] \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2 } = - \dfrac {1} {12} + O ((z-1)) [/math] | + | $ \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2 } = - \dfrac {1} {12} + O ((z-1)) $ |
| - | For z to 1 this says 1+2+3+4+... minus a 1/log^2 divergence is [math]- \dfrac {1} {12} [/math]. | + | For z to 1 this says 1+2+3+4+... minus a 1/log^2 divergence is $- \dfrac {1} {12} $. |
| - | For gathering the data bits [math]z^n[/math], that limit doesn’t exist for neither the operation [math]\sum_{n=0}^\infty[/math] nor [math]\int_0^\infty dn[/math], but it does for [math]\sum_{n=0}^\infty - \int_0^\infty dn[/math]. | + | For gathering the data bits $z^n$, that limit doesn’t exist for neither the operation $\sum_{n=0}^\infty$ nor $\int_0^\infty dn$, but it does for $\sum_{n=0}^\infty - \int_0^\infty dn$. |
| The space(time) your physical field theories are defined on generally fuck with you, but there are often such systematic renormalizations of your physical expressions, and for the addition of integers it relates to that relational. | The space(time) your physical field theories are defined on generally fuck with you, but there are often such systematic renormalizations of your physical expressions, and for the addition of integers it relates to that relational. | ||
