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本帖最后由 hbghlyj 于 2023-8-17 13:54 编辑 Leibniz, Bernoulli and The Logarithms of Negative Numbers
...In 1702, Bernoulli reached the result that
$$\arctan (x)=\frac{1}{2 i} \log \left(\frac{1+i x}{1-i x}\right)$$
Letting $x = 1$, we have that
$$\log (i)=i \frac{\pi}{2}$$
...Bernoulli responded that since $\frac{d x}{x}=\frac{-d x}{-x}$, by integration we have that $\log (x)=\log (-x)$. So he claims that the logarithmic curve has two symmetric branches, similar to the hyperbola. Leibniz responds to this by saying that the differentiation rule for $\log (x)$ does not apply to $\log (-x)$.
In his response he also mentions that $\log (-2)$ does not exist because then $\frac{1}{2} \log (-2)=\log (\sqrt{-2})$ which he again considers absurd.
Bernoulli responds to this claim by denying that $\frac{1}{2} \log (-2)=\log (\sqrt{-2})$ even though $\frac{1}{2} \log (2)=\log (\sqrt{2})$. He says that $\sqrt{2}$ is a mean proportional between 1 and 2 , but $\sqrt{-2}$ is not a mean proportional between $-1$ and $-2$. So he says that just as $\log (\sqrt{1 \times 2})=\frac{1}{2} \log (2)$, we also have that $\log (\sqrt{-1 \times-2})=\frac{1}{2} \log (-2)$
...Bernoulli then proceeds to give a geometric argument, as he would often do. He draws the two branches of the hyperbola and then constructs two branches of the logarithmic curve by using proportional areas under the hyperbola. Bernoulli’s argument involves passing across the asymptotes of the hyperbola, and uses the assumption that $∞-∞ = 0$.
\[\log(z)=\int_1^z\frac1zdz\]\[⇒\log(-1)\color{red}=\int_1^{-1}\frac1zdz=\int_{-1}^0-\frac1zdz-\int_0^1\frac1zdz=\infty-\infty=0\] |
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