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