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minus_twelve_._note [2017/07/02 22:57] nikolaj |
minus_twelve_._note [2019/09/09 22:33] nikolaj |
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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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| 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 | ||
