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minus_twelve_._note [2017/07/02 21:59] 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 | ||
| $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | $$\int_a^b n^{k-1}\,{\mathrm d}n=\frac{b^k}{k}-\frac{a^k}{k}$$ | ||
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| We have the Taylor expansion for the logarithm | We have the Taylor expansion for the logarithm | ||
| - | $$\log(1+r) = \sum_{k=1}^{\infty}\dfrac{1}{k}(-r)^k = r - \dfrac{1}{2}r^2 + \dfrac{1}{3}r^3 + {\mathcal O}(r^4) = r \left(1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) \right) $$ | + | $$\log(1+r) = \sum_{k=1}^{\infty}\dfrac{1}{k}(-r)^k = r - \dfrac{1}{2}r^2 + \dfrac{1}{3}r^3 + {\mathcal O}(r^4)= $$ |
| + | $$= r \left(1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) \right) $$ | ||
| and using the geometric series expansion, we get | and using the geometric series expansion, we get | ||
| - | $$\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } = \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$$ | + | $$\dfrac {1} { \log(1+r)} = \dfrac {1} {r} \dfrac {1} {1 - \dfrac{1}{2}r + \dfrac{1}{3}r^2 + {\mathcal O}(r^3) } =$$ |
| + | $$= \dfrac {1} {r} \left( 1 + \dfrac{1}{2} r - \dfrac{1}{2\cdot 2\cdot 3} r^2 + {\mathcal O}(r^3) \right) = \dfrac {1} {r} + \dfrac{1}{2} - \dfrac{1}{12} r + {\mathcal O}(r^2)$$ | ||
| With $r=z-1$ we see | With $r=z-1$ we see | ||
