Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
minus_twelve_._note [2017/11/25 23:02] nikolaj |
minus_twelve_._note [2019/09/22 01:30] nikolaj removed |
||
---|---|---|---|
Line 7: | Line 7: | ||
There are theories in math that give meaning to infinite sums, and the standard one, analysis (or, to some reach, calculus) has a million applications for practical applications, in particular physics and engineering. The picture above demonstrates the claim | There are theories in math that give meaning to infinite sums, and the standard one, analysis (or, to some reach, calculus) has a million applications for practical applications, in particular physics and engineering. The picture above demonstrates the claim | ||
- | $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots = 2$$ | + | $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots = 2$$ |
which can be proven, in analysis. Here's another claim | which can be proven, in analysis. Here's another claim | ||
Line 56: | Line 56: | ||
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 $\tfrac{56}{3}$. | ||
Line 68: | Line 68: | ||
We've seen this above, actually, in the special case of $z=-\frac{1}{2}$. Indeed, $\frac{1}{1-1/2}=\frac{1}{1/2}=2$ and the first formula in this post was | We've seen this above, actually, in the special case of $z=-\frac{1}{2}$. Indeed, $\frac{1}{1-1/2}=\frac{1}{1/2}=2$ and the first formula in this post was | ||
- | $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots = 2$$ | + | $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots = 2$$ |
The smooth analogous to the sum with $z$ is | The smooth analogous to the sum with $z$ is | ||
Line 120: | Line 120: | ||
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? | ||
Line 138: | Line 138: | ||
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. |