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minus_twelve_._note [2016/11/11 21:48] nikolaj |
minus_twelve_._note [2017/07/02 21:59] nikolaj |
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| ===== Minus twelve . Note ===== | ===== Minus twelve . Note ===== | ||
| - | ==== Note ==== | + | ==== Note === |
| - | http://math.stackexchange.com/questions/1327812/limit-approach-to-finding-1234-ldots/1332159#1332159 | ||
| - | ----- | + | https://upload.wikimedia.org/wikipedia/commons/thumb/4/4b/Geometric_progression_convergence_diagram.svg/350px-Geometric_progression_convergence_diagram.svg.png |
| - | $1+1+1+1+\dots = -\tfrac{1}{2}$ | + | |
| - | $1+2+3+4+\dots = -\tfrac{1}{12}$ | + | 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 |
| - | <code> | + | $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots = 2$$ |
| - | Sum[(1 - d)^k, {k, 0, n - 1}] - Integrate[(1 - d)^k, {k, 0, m}] | + | |
| - | Series[1/d + 1/Log[1 - d], {d, 0, 2}] | + | which can be proven, in analysis. Here's another claim |
| - | </code> | + | |
| - | ----- | + | $$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots = \dfrac{\pi^2}{6} \approx 1.64493$$ |
| - | The geometric series for $z\in(0,1)$ is | + | This goes by the name //Basel problem// and was proven, in a context similar to what is analysis today, by Leonhard Euler in 1735. This one is interesting in various aspects. In particular, it shows how an infinite quantity of rational numbers can sum up to be an irrational number! |
| - | $\sum_{n=0}^\infty z^n = \dfrac{1}{1-z}$ | + | Well, then. You may or may not have heard of the claim |
| - | The smooth analogous is | + | $$"1+2+3+4+\dots = - \dfrac{1}{12}"$$ |
| + | |||
| + | An infinite sum of natural numbers is supposed to be a negative rational number? | ||
| + | |||
| + | Let's take a step back. | ||
| + | |||
| + | $1=1$ | ||
| + | |||
| + | $1+2=3$ | ||
| + | |||
| + | $1+2+3=6$ | ||
| + | |||
| + | $1+2+3+4=10$ | ||
| + | |||
| + | $1+2+3+4+5=\dots$ | ||
| + | |||
| + | Here we take a finite sum of a number and, in each step, add another positive number, resulting in a longer sum. In the roughly 150 year old theory of Peano arithmetic (to which I'll soon come in my series), we can prove that each such sum is bigger than the previous one. In particular, if we add only positive number, we can prove that the sum will always be positive. | ||
| + | |||
| + | However, we can't use Peano arithmetic to prove a statement about infinite sums. An infinite sum is a creature of, for example, the theory of analysis. And, indeed, in calculus the claim with $-\frac{1}{12}$ is false! In particular, it's easy to show that | ||
| + | |||
| + | $$\dfrac{1}{1+2+3+4+\dots} = 0$$ | ||
| + | |||
| + | If you add more and more numbers in the denominator, the resulting sequence of rational numbers will come closer and closer to zero, which is exactly what this says. In particular, this reciprocal of the sum is definitely not $-12$! | ||
| + | |||
| + | Why would it be $\dfrac{1}{-12}$, or all numbers? Well, the fact is that this number isn't random. | ||
| + | |||
| + | 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 | ||
| + | $$\int_a^b f(n)\,{\mathrm d}n = \sum_{n=a}^{b-1} f(n) + \left(\lim_{x\to b}-\lim_{x\to a}\right)\left(\dfrac{1}{2}-\dfrac{1}{12}\dfrac{d}{dx}+\dots\right)f(x)$$ | ||
| + | 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 | ||
| + | $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | ||
| + | equals $\sum_{n=a}^{b-1} n^{k-1}$ minus some lower order corrections. | ||
| + | |||
| + | So as an example, with $k=3,a=2,b=4$ you get the identity | ||
| + | $$\frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ | ||
| + | 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 $\tfrac{56}{3}$. | ||
| + | |||
| + | Okay, but why $-\frac{1}{12}$? I'd like to tell you something about the theory of combinatorics at this point, but we'll only get the much later in my series. | ||
| + | So besides just making you curious, the goal of this post is to show an innocent way in which you still can tie a sum that's very related to $-\frac{1}{12}$. Sadly, I've already written a whole lot and I don't want to make this boring either. So I'm gonna assume you're familiar with the logarithmic function and go from there. As a side note, tell me below which part of this are tough and which work will. | ||
| + | |||
| + | So the geometric series for a real number $z$ in the interval $(0,1)$ is | ||
| + | |||
| + | $$\sum_{n=0}^\infty z^n = \dfrac{1}{1-z}$$ | ||
| + | |||
| + | 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 | ||
| + | |||
| + | $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots = 2$$ | ||
| + | |||
| + | The smooth analogous to the sum with $z$ is | ||
| + | |||
| + | $$\int_0^\infty z^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{t\log(z)}\,{\mathrm d}t=\frac{1}{-\log(z)}$$ | ||
| - | $\int_0^\infty z^t\,{\mathrm d}t=\int_0^\infty {\mathrm e}^{t\log(z)}\,{\mathrm d}t=\dfrac{1}{-\log(z)}$ | + | where I used $\int_0^y {\mathrm e}^{-a\cdot{}x}\,{\mathrm d}x = \frac{1}{-a}({\mathrm e}^{-y}-1)$. Note that the logarithm is negative for $z$ in the interval $(0,1)$. |
| - | Applying $z\dfrac{d}{dz}$ to the first yields | + | Applying the derivative $z\frac{d}{dz}$ to the first yields |
| - | $z\dfrac{d}{dz}\dfrac{1}{1-z} = \sum_{n=0}^\infty n \, z^n$ | + | $$z\dfrac{d}{dz}\dfrac{1}{1-z} = \sum_{n=0}^\infty n \, z^n$$ |
| and applying it to the second yields | and applying it to the second yields | ||
| - | $z\dfrac{z}{dz}\dfrac{1}{-\log(z)} = \dfrac{1}{\log(z)^2}$ | + | $$z\dfrac{z}{dz}\dfrac{1}{-\log(z)} = \dfrac{1}{\log(z)^2}$$ |
| How do those two differ? | How do those two differ? | ||
| - | We have the Taylor expansion | + | 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) $ | + | $$\log(1+r) = \sum_{k=1}^{\infty}\dfrac{1}{k}(-r)^k = r - \dfrac{1}{2}r^2 + \dfrac{1}{3}r^3 + {\mathcal O}(r^4) = r \left(1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) \right) $$ |
| - | Using the geometric series expansion, we get | + | and using the geometric series expansion, we get |
| - | $\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } = \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$ | + | $$\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } = \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$$ |
| - | + | ||
| - | <code> | + | |
| - | Plot[{ | + | |
| - | Log[1 + r] | + | |
| - | , 1/Log[1 + r] - 1/r | + | |
| - | , 1/Log[1 + r] | + | |
| - | , 1/2 + (-1/12) r | + | |
| - | }, {r, -0.5, 1.5}, PlotRange -> {-0.5, 1.5}] | + | |
| - | </code> | + | |
| With $r=z-1$ we see | With $r=z-1$ we see | ||
| - | $\dfrac {1} { \log(z)} = - \dfrac {1} {1-z} + \dfrac{1}{2} - \dfrac{1}{12} (z-1) + {\mathcal O}((z-1)^2)$ | + | $$\dfrac {1} { \log(z)} = - \dfrac {1} {1-z} + \dfrac{1}{2} - \dfrac{1}{12} (z-1) + {\mathcal O}((z-1)^2)$$ |
| and taking the derivative, we get | and taking the derivative, we get | ||
| - | $\sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1))$ | + | $$\sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1))$$ |
| - | For z to 1 this says 1+2+3+4+... minus a 1/log^2 divergence is -1/12. That's a number symptomatic between discrete and smooth expressions. | + | For $z$ approaching $1$, which is the upper bound of the interval $(0,1)$ that we considered, this says $1+2+3+4+\cdot $ minus a \dfrac {1} { \log(z)^2} is exactly $- \dfrac{1}{12}$. That's a number symptomatic between discrete and smooth expressions. We can visualize this: |
| + | |||
| + | Here the blue and the red functions are $\sum_{n=0}^\infty n \, z^n$ and $\frac {1} { \log(z)^2}$ and both diverge at $z=1$. However, the diverge in a way that their difference converges exactly to $-\frac{1}{12}$ | ||
| + | |||
| + | You can use the computer or a size like WolframAlpha to verify this result. | ||
| + | |||
| + | Tell me if that was interesting and if you'd like to hear more about this. | ||
| ----- | ----- | ||
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| At $z=1$ they both vanish, of course, but the second function also diverges at $z=0$. This is also the behaviour of $\log(z)$ or, for that matter, for any $\log(z)^k$ for any nonzero $k$. | At $z=1$ they both vanish, of course, but the second function also diverges at $z=0$. This is also the behaviour of $\log(z)$ or, for that matter, for any $\log(z)^k$ for any nonzero $k$. | ||
| The next paragraph has the purpose of motivating why to look at the divergent behaviour of the $\log$ function. | The next paragraph has the purpose of motivating why to look at the divergent behaviour of the $\log$ function. | ||
| + | ----- | ||
| + | |||
| + | |||
| + | http://math.stackexchange.com/questions/1327812/limit-approach-to-finding-1234-ldots/1332159#1332159 | ||
| + | |||
| + | ----- | ||
| + | $1+1+1+1+\dots = -\tfrac{1}{2}$ | ||
| + | |||
| + | $1+2+3+4+\dots = -\tfrac{1}{12}$ | ||
| + | |||
| + | <code> | ||
| + | Sum[(1 - d)^k, {k, 0, n - 1}] - Integrate[(1 - d)^k, {k, 0, m}] | ||
| + | |||
| + | Series[1/d + 1/Log[1 - d], {d, 0, 2}] | ||
| + | </code> | ||
| + | |||
| ----- | ----- | ||
