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| 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 | ||
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| Consider this little gimmick: The difference between the integral and the sum of a smooth function is given by a very particular sum that involves $\dfrac{1}{-12}$ at the second place. It starts as out as | Consider this little gimmick: The difference between the integral and the sum of a smooth function is given by a very particular sum that involves $\dfrac{1}{-12}$ at the second place. It starts as out as | ||
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| $$\int_a^b f(n)\,{\mathrm d}n = \sum_{n=a}^{b-1} f(n) + \left(\lim_{x\to b}-\lim_{x\to a}\right)\left(\dfrac{1}{2}-\dfrac{1}{12}\dfrac{d}{dx}+\dots\right)f(x)$$ | $$\int_a^b f(n)\,{\mathrm d}n = \sum_{n=a}^{b-1} f(n) + \left(\lim_{x\to b}-\lim_{x\to a}\right)\left(\dfrac{1}{2}-\dfrac{1}{12}\dfrac{d}{dx}+\dots\right)f(x)$$ | ||
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| + | e.g. | ||
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| + | $$\int _m^n f(x)~{\rm d}x=\sum _{i=m}^n f(i)-\frac 1 2 \left( f(m)+f(n) \right) -\frac 1{12}\left( f'(n)-f'(m)\right) + \frac 1{720}\left( f'''(n)-f'''(m)\right) + \cdots.$$ | ||
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| and if you want to see a full version, check out the 300 year old //Euler–Maclaurin formula//. The building blocks of many functions are monomials $f(n)=n^{k-1}$ and for those the formula is particularly simple, because most all high derivatives vanish. The formula then tells us that | and if you want to see a full version, check out the 300 year old //Euler–Maclaurin formula//. The building blocks of many functions are monomials $f(n)=n^{k-1}$ and for those the formula is particularly simple, because most all high derivatives vanish. The formula then tells us that | ||
| $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | ||
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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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| 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 | ||
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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. | ||
