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minus_twelve_._note [2017/11/25 23:02] nikolaj |
minus_twelve_._note [2019/09/09 22:45] 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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So as an example, with $k=3,a=2,b=4$ you get the identity | So as an example, with $k=3,a=2,b=4$ you get the identity | ||
- | $$\frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ | + | $$\frac{1}{3}(4^3-2^3)=(2^2+(4-1)^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ |
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 $\tfrac{56}{3}$. | 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 $\tfrac{56}{3}$. | ||
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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 |