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minus_twelve_._note [2017/07/02 22:57] nikolaj |
minus_twelve_._note [2017/11/25 23:02] nikolaj |
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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 | ||
$$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ |